Tag: maths class 12th NCERT Solutions

Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT

Question 1 to 12:

Q1. Prove that the function

$f (x) = 5 x - 3$

is continuous at $x = 0$ , $x = - 3$ and $x = 5$ .

Solution

The function $f (x) = 5 x - 3$ is a polynomial function.
We know from theory that every polynomial function is continuous at every real number.

To verify at specific points:

At $x = 0$

$f (0) = 5 (0) - 3 = - 3$ $\lim_{x \to 0} f (x) = \lim_{x \to 0} (5 x - 3) = 5 (0) - 3 = - 3$

Since

$\lim_{x \to 0} f (x) = f (0) \Rightarrow f (x) is continuous at x = 0$

At $x = - 3$

$f (- 3) = 5 (- 3) - 3 = - 18$ $\lim_{x \to - 3} (5 x - 3) = 5 (- 3) - 3 = - 18$

Thus

$\lim_{x \to - 3} f (x) = f (- 3)$

At $x = 5$

$f (5) = 5 (5) - 3 = 22$ $\lim_{x \to 5} (5 x - 3) = 22$

Conclusion

$f (x) is continuous at x = 0, - 3, 5$

Q2. Examine continuity of

$f (x) = 2 x^{2} - 1 at x = 3$

Solution

The function is defined at $x = 3$ :

$f (3) = 2 (3)^{2} - 1 = 2 (9) - 1 = 17$

Now find the limit:

$\lim_{x \to 3} f (x) = \lim_{x \to 3} (2 x^{2} - 1) = 2 (3)^{2} - 1 = 17$

Since

$\lim_{x \to 3} f (x) = f (3)$

Therefore, the function is continuous at $x = 3$ .

Q3. Examine the following functions for continuity

(a) $f (x) = x - 5$

Polynomial ⇒ continuous everywhere.

$\lim_{x \to c} (x - 5) = c - 5 = f (c)$

Continuous for all real x

(b) $f (x) = \frac{1}{x - 5}, x \neq 5$

The function is undefined at $x = 5$ , so its domain excludes 5.
Every rational function is continuous in its domain (Example 16)

Continuous for all $x \neq 5$
Discontinuous at $x = 5$

(c)

$f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5$

Factor numerator:

$x^{2} - 25 = (x - 5) (x + 5)$

Thus,

$f (x) = x - 5, x \neq - 5$

This becomes a linear function, but $x = - 5$ is excluded from the domain.

Continuous for $x \neq - 5$
Discontinuous at $x = - 5$

(d) $f (x) = ∣ x - 5 ∣$

Modulus function is continuous everywhere (Example 7)

Continuous for all real x

Q4. Prove that

$f (x) = x^{n} is continuous at x = n$

(where $n$ is a positive integer)

Solution

$f (n) = n^{n}$ $\lim_{x \to n} x^{n} = n^{n}$

Thus

$\lim_{x \to n} f (x) = f (n)$

So the function is continuous at $x = n$

Q5. Function

$f (x) = {\begin{cases} x, & x \leq 1 \\ 5, & x > 1 \end{cases}$

Check continuity at $x = 0$ , $x = 1$ , $x = 2$

At $x = 0$

$f (0) = 0$

Nearby values belong to $x \leq 1$ :

$\lim_{x \to 0} f (x) = \lim_{x \to 0} x = 0 = f (0)$

Continuous at 0

At $x = 1$

Left-hand limit:

$\lim_{x \to 1^{-}} f (x) = 1$

Right-hand limit:

$\lim_{x \to 1^{+}} f (x) = 5$

$f (1) = 1$

Since

$L H L = 1, R H L = 5, L H L \neq R H L$

Discontinuous at $x = 1$

At $x = 2$

$f (2) = 5, \lim_{x \to 2} f (x) = 5$

Continuous at $x = 2$

Point of discontinuity

$x = 1$

Find all points of discontinuity of f, where f is defined by

Q6. The function is:

$f (x) = {\begin{cases} 2 x + 3, & x \leq 2 \\ 2 x - 3, & x > 2 \end{cases}$

We must examine continuity at $x = 2$ .

Step 1: Check if the function is defined at $x = 2$

Since $x = 2$ satisfies $x \leq 2$ , we use the first expression:

$f (2) = 2 (2) + 3 = 4 + 3 = 7$

So, $f (2) = 7$

Thus, the function is defined at $x = 2$ .

Step 2: Left-Hand Limit (LHL) at $x = 2$

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (2 x + 3)$ $= 2 (2) + 3 = 4 + 3 = 7$

So,

$L H L = 7$

Step 3: Right-Hand Limit (RHL) at $x = 2$

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (2 x - 3)$ $= 2 (2) - 3 = 4 - 3 = 1$

So,

$R H L = 1$

Step 4: Compare limits and value at the point

Quantity Value

Left-hand limit (LHL) 7

Right-hand limit (RHL) 1

Actual value $f (2)$ 7

Condition for continuity

$f (x) is continuous at x = a if \lim_{x \to a} f (x) = f (a)$

Since

$L H L \neq R H L \Rightarrow Limit does not exist at x = 2$

So, the function cannot be continuous at that point.

FINAL ANSWER

$f (x) is discontinuous at x = 2.$

Question 7

The function is defined as:

$f (x) = {\begin{cases} ∣ x ∣ + 3, & x \leq - 3 \\ - 2 x, & - 3 < x < 3 \\ 6 x + 2, & x \geq 3 \end{cases}$

We must check continuity at the points where the definition changes, i.e.,

$x = - 3 and x = 3$

Check Continuity at $x = - 3$

Step 1: Find $f (- 3)$

Since $- 3 \leq - 3$ , use $∣ x ∣ + 3$ :

$f (- 3) = ∣ - 3 ∣ + 3 = 3 + 3 = 6$

Step 2: Left-hand limit (LHL) at $x = - 3$

For $x < - 3$ , expression is $∣ x ∣ + 3$

$\lim_{x \to - 3^{-}} (∣ x ∣ + 3) = ∣ - 3 ∣ + 3 = 3 + 3 = 6$

Step 3: Right-hand limit (RHL) at $x = - 3$

For $- 3 < x < 3$ , expression is $- 2 x$

$\lim_{x \to - 3^{+}} (- 2 x) = - 2 (- 3) = 6$

Compare values

Quantity Value

LHL 6

RHL 6

$f (- 3)$ 6

$Since L H L = R H L = f (- 3), f(x) is continuous at x = - 3$

Check Continuity at $x = 3$

Step 1: Find $f (3)$

Since $x \geq 3$ , use $6 x + 2$ :

$f (3) = 6 (3) + 2 = 18 + 2 = 20$

Step 2: Left-hand limit (LHL)

For $x < 3$ and $x > - 3$ , use $- 2 x$

$\lim_{x \to 3^{-}} (- 2 x) = - 2 (3) = - 6$

Step 3: Right-hand limit (RHL)

$\lim_{x \to 3^{+}} (6 x + 2) = 6 (3) + 2 = 20$

Compare values

Quantity Value

LHL -6

RHL 20

$f (3)$ 20

Since

$L H L \neq R H L$

Question 8

The function is defined as:

$f (x) = {\begin{cases} \frac{∣ x ∣}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

We need to check continuity at $x = 0$ because that is the only point where the definition changes.

Step 1: Find $f (0)$

Given directly:

$f (0) = 0$

Step 2: Find Left-Hand Limit (LHL) as $x \to 0^{-}$

When $x < 0$ , $∣ x ∣ = - x$

So,

$f (x) = \frac{∣ x ∣}{x} = \frac{- x}{x} = - 1$

Thus,

$\lim_{x \to 0^{-}} \frac{∣ x ∣}{x} = - 1$

Step 3: Find Right-Hand Limit (RHL) as $x \to 0^{+}$

When $x > 0$ , $∣ x ∣ = x$

So,

$f (x) = \frac{∣ x ∣}{x} = \frac{x}{x} = 1$

Thus,

$\lim_{x \to 0^{+}} \frac{∣ x ∣}{x} = 1$

Step 4: Compare values

Quantity Value

Left-hand limit −1

Right-hand limit 1

$f (0)$ 0

Condition for Continuity:

$Function is continuous at x = a if L H L = R H L = f (a)$

But here:

$L H L = - 1 \neq R H L = 1$

Therefore:

$\lim_{x \to 0} f (x) does not exist$ So,

$f(x) is NOT continuous at x = 0$

Question 9

The given function is:

$f (x) = {\begin{cases} \frac{x}{∣ x ∣}, & x < 0 \\ - 1, & x \geq 0 \end{cases}$

We must examine continuity at $x = 0$ , because that is the point where the rule changes.

Step 1: Find $f (0)$

Since $0 \geq 0$ , use the second part of the definition:

$f (0) = - 1$

Step 2: Find Left-Hand Limit (LHL) as $x \to 0^{-}$

For $x < 0$ , the expression is:

$f (x) = \frac{x}{∣ x ∣}$

When $x < 0$ , $∣ x ∣ = - x$ , so:

$\frac{x}{∣ x ∣} = \frac{x}{- x} = - 1$

Thus:

$\lim_{x \to 0^{-}} f (x) = - 1$

Step 3: Find Right-Hand Limit (RHL) as $x \to 0^{+}$

For $x > 0$ , use second part of definition (since all $x \geq 0$ map to $f (x) = - 1$ :

$f (x) = - 1$

Thus:

$\lim_{x \to 0^{+}} f (x) = - 1$

Step 4: Compare values

Quantity Value

LHL −1

RHL −1

Actual value $f (0)$ −1

So,

$L H L = R H L = f (0)$

Final Conclusion

$The function is continuous at x = 0$

FINAL ANSWER

$f (x) is continuous everywhere, including at x = 0.$

Question 10

The function is: $f (x) = {\begin{cases} x + 1, & x \geq 1 \\ x^{2} + 1, & x < 1 \end{cases}$

We need to examine continuity at $x = 1$ , because that is the point where the definition switches.

Step 1: Check whether $f (x)$ is defined at $x = 1$

Since $1 \geq 1$ , use first expression:

$f (1) = 1 + 1 = 2$

So, the function is defined at $x = 1$ , and $f (1) = 2$

Step 2: Find the Left-Hand Limit (LHL) as $x \to 1^{-}$

For values less than 1, use $x^{2} + 1$ :

$\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2$

So,

$L H L = 2$

Step 3: Find the Right-Hand Limit (RHL) as $x \to 1^{+}$

For values greater than or equal to 1, use $x + 1$ :

$\lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2$

So,

$R H L = 2$

Step 4: Compare LHL, RHL and $f (1)$

Quantity Value

LHL 2

RHL 2

$f (1)$ 2

Since

$L H L = R H L = f (1)$

FINAL CONCLUSION

$The function is continuous at x = 1.$

Final Answer

$f (x) is continuous at x = 1$

Question 11

The given function is: $f (x) = {\begin{cases} x^{3} - 3, & x \leq 2 \\ x^{2} + 1, & x > 2 \end{cases}$

We must check continuity at $x = 2$ , because that is where the definition changes.

Step 1: Check if the function is defined at $x = 2$

Since $2 \leq 2$ , we use the first expression:

$f (2) = 2^{3} - 3 = 8 - 3 = 5$

So,

$f (2) = 5$

The function is defined at $x = 2$ .

Step 2: Left-Hand Limit (LHL) as $x \to 2^{-}$

For $x < 2$ , use $x^{3} - 3$ :

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (x^{3} - 3) = 2^{3} - 3 = 8 - 3 = 5$

So,

$L H L = 5$

Step 3: Right-Hand Limit (RHL) as $x \to 2^{+}$

For $x > 2$ , use $x^{2} + 1$ :

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (x^{2} + 1) = 2^{2} + 1 = 4 + 1 = 5$

So, $R H L = 5$

Step 4: Compare Limits and Function Value

Quantity Value

LHL 5

RHL 5

$f (2)$ 5

Condition for continuity:

$f (x) is continuous at x = a if L H L = R H L = f (a)$

Here:

$L H L = R H L = f (2)$

FINAL CONCLUSION

$f (x) is continuous at x = 2$

FINAL ANSWER

$The function f (x) is continuous at x = 2.$

Question 12

The function is:

$f (x) = {\begin{cases} x^{10} - 1, & x \leq 1 \\ x^{2}, & x > 1 \end{cases}$

We need to examine continuity at $x = 1$ (the point where the definition changes).

Step 1: Check whether the function is defined at $x = 1$

Since $1 \leq 1$ , we use the first expression $x^{10} - 1$ :

$f (1) = 1^{10} - 1 = 1 - 1 = 0$

So, $f (1) = 0$

Step 2: Left-Hand Limit (LHL) at $x = 1$

For values $x < 1$ , use $x^{10} - 1$

$\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{10} - 1) = 1^{10} - 1 = 0$

So:

$L H L = 0$

Step 3: Right-Hand Limit (RHL) at $x = 1$

For values $x > 1$ , use $x^{2}$

$\lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x^{2}) = 1^{2} = 1$

So:

$R H L = 1$

Step 4: Compare values

Quantity Value

LHL 0

RHL 1

$f (1)$ 0

Condition for continuity

$Function is continuous at x = a if L H L = R H L = f (a)$

But here:

$L H L = 0 \neq 1 = R H L$

So the limit does not exist at $x = 1$

FINAL CONCLUSION

$f(x) is NOT continuous at x = 1$ $The function is discontinuous at x = 1$

November 22, 2025
Exercise-4.5-Part-2, Class 12th, Maths, NCERT

Question 15.
If

$A = (\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}),$

find $A^{- 1}$ . Using $A^{- 1}$ , solve the following system of equations:

${\begin{cases} 2 x - 3 y + 5 z = 11, \\ 3 x + 2 y - 4 z = - 5, \\ x + y - 2 z = - 3. \end{cases}$

Solution:

We have

$A = (\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array})$

Let $b = (\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array})$

We know that

$A x = b \Rightarrow x = A^{- 1} b$

Step 1: Find $A^{- 1}$

Using the adjoint and determinant method,

$\det (A) = - 1$

The adjugate (adjoint) of $A$ is:

$adj (A) = (\begin{array}{ccc} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array})$

Hence,

$A^{- 1} = \frac{1}{\det (A)} \cdot adj (A) = - 1 \times (\begin{array}{ccc} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}) = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}) .$

Step 2: Use $A^{- 1}$ to find $x, y, z$

$(\begin{array}{c} x \\ y \\ z \end{array}) = A^{- 1} b = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}) (\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array})$

Now multiply:

$\begin{aligned} x & = 0 (11) + 1 (- 5) + (- 2) (- 3) = - 5 + 6 = 1, \\ y & = - 2 (11) + 9 (- 5) + (- 23) (- 3) = - 22 - 45 + 69 = 2, \\ z & = - 1 (11) + 5 (- 5) + (- 13) (- 3) = - 11 - 25 + 39 = 3. \end{aligned}$

Final Answers:

$A^{- 1} = (\begin{array}{ccc} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}), x = 1, y = 2, z = 3$

Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60.
The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90.
The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70.
Find the cost of each item per kilogram using the matrix method.

Solution:

Let the cost per kilogram of onion, wheat and rice be ₹ $o$ , ₹ $w$ and ₹ $r$ respectively.

Then, according to the question, we have:

$\begin{aligned} 4 o + 3 w + 2 r & = 60, \\ 2 o + 4 w + 6 r & = 90, \\ 6 o + 2 w + 3 r & = 70 \end{aligned}$

Step 1: Write in matrix form

$(\begin{array}{ccc} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}) (\begin{array}{c} o \\ w \\ r \end{array}) = (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

That is,

$A x = b,$

where

$A = (\begin{array}{ccc} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}), x = (\begin{array}{c} o \\ w \\ r \end{array}), b = (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

Step 2: Find $A^{- 1}$

We have

$\det (A) = 50,$

and the adjugate of $A$ is

$adj (A) = (\begin{array}{ccc} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array})$

Hence,

$A^{- 1} = \frac{1}{50} (\begin{array}{ccc} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array}) = (\begin{array}{ccc} 0 & - \frac{1}{10} & \frac{1}{5} \\ \frac{3}{5} & 0 & - \frac{2}{5} \\ - \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \end{array})$

Step 3: Find $x = A^{- 1}$

$(\begin{array}{c} o \\ w \\ r \end{array}) = (\begin{array}{ccc} 0 & - \frac{1}{10} & \frac{1}{5} \\ \frac{3}{5} & 0 & - \frac{2}{5} \\ - \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \end{array}) (\begin{array}{c} 60 \\ 90 \\ 70 \end{array})$

Now multiplying:

$\begin{aligned} o & = 0 (60) + (- \frac{1}{10}) (90) + (\frac{1}{5}) (70) = - 9 + 14 = 5, \\ w & = (\frac{3}{5}) (60) + 0 (90) + (- \frac{2}{5}) (70) = 36 - 28 = 8, \\ r & = (- \frac{2}{5}) (60) + (\frac{1}{5}) (90) + (\frac{1}{5}) (70) = - 24 + 18 + 14 = 8. \end{aligned}$

Final Answers:

$\begin{aligned} Cost of onion per kg & = ₹ 5, \\ Cost of wheat per kg & = ₹ 8, \\ Cost of rice per kg & = ₹ 8 \end{aligned}$

October 29, 2025
Exercise4.4, Class 12th, Maths, Chapter 4, NCERT
Exercise 4.4

Question 1:
Find $adj A$ for

$A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$ Solution:
For a $2 \times 2$ matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array})$ $adj A = (\begin{array}{cc} d & - b \\ - c & a \end{array})$

So for

$A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$

$adj A = (\begin{array}{cc} 4 & - 2 \\ - 3 & 1 \end{array})$

(You can check $A (adj A) = (\det A) I$ :

$\det A = 1 \cdot 4 - 2 \cdot 3 = - 2$ , and indeed $A adj A = (- 2) I$ )

Question 2:
Find $adj A$

for $A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 2 & 3 & 5 \\ - 2 & 0 & 1 \end{array})$

Below I show the cofactors, the adjoint, and a verification

$A adj A = (\det A) I$

Minors and cofactors

Compute the $2 \times 2$ minors and cofactors

$A_{i j} = (- 1)^{i + j} M_{i j}$

$Cofactor matrix [A_{i j}] = (\begin{array}{ccc} 3 & - 12 & 6 \\ 1 & 5 & 2 \\ - 11 & - 1 & 5 \end{array})$

(Each entry is the cofactor of the corresponding element of

$A$ .)

$adj A$

(adjugate = transpose of cofactor matrix)

$adj A = (\begin{array}{ccc} 3 & 1 & - 11 \\ - 12 & 5 & - 1 \\ 6 & 2 & 5 \end{array})$

Determinant and verification

$\det A = 27$

Now verify

$A adj A = (\det A) I = 27 I$

Multiplying gives

$A adj A = (\begin{array}{ccc} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{array}) = 27 I,$

so the adjoint is correct.

___________________

Verify the identity for both matrices

$Show that A (adj A) = (adj A) A = ∣ A ∣ I$

Question 3:

$A = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array})$

Solution – Compute the determinant:

$∣ A ∣ = 2 (- 6) - 3 (- 4) = - 12 + 12 = 0.$

Compute the Adj (classical adjoint). For a $2 \times 2$ matrix

$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$adj A = (\begin{array}{cc} d & - b \\ - c & a \end{array})$

Thus,

$adj A = (\begin{array}{cc} - 6 & - 3 \\ 4 & 2 \end{array})$

Multiply:

$A (adj A) = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array}) (\begin{array}{cc} - 6 & - 3 \\ 4 & 2 \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Similarly

$(adj A) A = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Since

$∣ A ∣ = 0$ , we have

$∣ A ∣ I = 0 \cdot I = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Hence,

$A (adj A) = (adj A) A = ∣ A ∣ I,$

as required.

Question 4:

$A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array})$

Solution – Compute the determinant –

$∣ A ∣ = 11$

Compute the adjugate (transpose of cofactor matrix). The cofactors lead to

$adj A = (\begin{array}{ccc} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{array}) .$

Multiply:

$A (adj A) = (\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}) = 11 I,$

and likewise

$(adj A) A = 11 I$

Thus for this matrix also

$A (adj A) = (adj A) A = ∣ A ∣ I$

Conclusion

Both verifications hold:
- For
  $A = (\begin{array}{cc} 2 & 3 \\ - 4 & - 6 \end{array})$ $A (adj A) = (adj A) A = 0 = ∣ A ∣ I$
- For
  $A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array})$
$A (adj A) = (adj A) A = 11 I .$

Question 5:

$Find A^{- 1} if A = (\begin{array}{cc} 2 & - 2 \\ 4 & 3 \end{array}) .$

Formula for inverse of a 2×2 matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array})$

its inverse (if

$∣ A ∣ \neq 0$

is

$A^{- 1} = \frac{1}{a d - b c} (\begin{array}{cc} d & - b \\ - c & a \end{array}$

Compute determinant

$∣ A ∣ = (2) (3) - (- 2) (4) = 6 + 8 = 14$

Since

$∣ A ∣ \neq 0$

, the inverse exists.

Apply the formula

$A^{- 1} = \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array})$

Final Answer:

$A^{- 1} = \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{cc} 2 & - 2 \\ 4 & 3 \end{array}) \frac{1}{14} (\begin{array}{cc} 3 & 2 \\ - 4 & 2 \end{array}) = \frac{1}{14} (\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}) = I$

✔ Verified.

Question 6:

$Find A^{- 1} if A = (\begin{array}{cc} - 1 & 5 \\ - 3 & 2 \end{array})$

Formula for inverse

For any

$2 \times 2$

matrix

$A = (\begin{array}{cc} a & b \\ c & d \end{array}),$

if

$∣ A ∣ = a d - b c \neq 0$

, then

$A^{- 1} = \frac{1}{∣ A ∣} (\begin{array}{cc} d & - b \\ - c & a \end{array})$

Compute determinant

$∣ A ∣ = (- 1) (2) - (5) (- 3) = - 2 + 15 = 13$

$∣ A ∣ = 13 \neq 0$

, so the inverse exists.

Apply the formula

$A^{- 1} = \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array})$

Final Answer:

$A^{- 1} = \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{cc} - 1 & 5 \\ - 3 & 2 \end{array}) \frac{1}{13} (\begin{array}{cc} 2 & - 5 \\ 3 & - 1 \end{array}) = \frac{1}{13} (\begin{array}{cc} 13 & 0 \\ 0 & 13 \end{array}) = I$

✔ Verified.

Question 7:

$Find A^{- 1} if A = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array})$

Observation

$A$

is an upper triangular matrix, and for triangular matrices,

$∣ A ∣ = product of diagonal elements.$

Determinant

$∣ A ∣ = 1 \times 2 \times 5 = 10$

Since

$∣ A ∣ \neq 0$ , $A^{- 1}$ exists.

Use properties of triangular matrices

For an upper triangular matrix, its inverse is also upper triangular.
We find

$A^{- 1} = [a_{i j}]$

such that

$A A^{- 1} = I$

Let

$A^{- 1} = (\begin{array}{ccc} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{array})$

Then

$A A^{- 1} = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}) (\begin{array}{ccc} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{array}) = (\begin{array}{ccc} a & b + 2 d & c + 2 e + 3 f \\ 0 & 2 d & 2 e + 4 f \\ 0 & 0 & 5 f \end{array})$

We want this to equal the identity

$I = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})$

Compare entries

From the bottom:

$f = 1 ⟹ f = \frac{1}{5}$

$2 d = 1 ⟹ d = \frac{1}{2}$

$a = 1$

Now use off-diagonal equations:

$e + 4 f = 0 ⟹ e = - 2 f = - \frac{2}{5}$

$b + 2 d = 0 ⟹ b = - 2 d = - 1$

$c + 2 e + 3 f = 0 ⟹ c = - 2 e - 3 f$

Substitute

$e = - \frac{2}{5}$ and $f = \frac{1}{5}$ :

$c = - 2 (- \frac{2}{5}) - 3 (\frac{1}{5}) = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}$

Write the inverse

$A^{- 1} = (\begin{array}{ccc} 1 & - 1 & \frac{1}{5} \\ 0 & \frac{1}{2} & - \frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{array})$

Verification (optional):

$A A^{- 1} = (\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}) (\begin{array}{ccc} 1 & - 1 & \frac{1}{5} \\ 0 & \frac{1}{2} & - \frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{array}) = I$

✔ Verified.

Question 8:

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{array})$

Solution.

$A$ is lower triangular, so $A^{- 1}$ is also lower triangular.
Let

$A^{- 1} = (\begin{array}{ccc} a & 0 & 0 \\ b & c & 0 \\ d & e & f \end{array})$

and solve

$A A^{- 1} = I$

.Multiply $A$ and $A^{- 1}$ : From the first row: $a = 1$

.From the second row:

$3 a + 3 b = 0, 3 c = 1 \Rightarrow b = - 1, c = \frac{1}{3}$

From the third row:

$5 a + 2 b - d = 0, 2 c - e = 0, - f = 1$

Substituting

$a = 1, b = - 1, c = \frac{1}{3}$

$,$ gives

$5 (1) + 2 (- 1) - d = 0 \Rightarrow d = 3, 2 \cdot \frac{1}{3} - e = 0 \Rightarrow e = \frac{2}{3}, f = - 1.$

Thus

$A^{- 1} = (\begin{array}{ccc} 1 & 0 & 0 \\ - 1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & - 1 \end{array})$

(You can verify

$A A^{- 1} = I$

by multiplication.)

Question 9:

find $A^{- 1}$

for $A = (\begin{array}{ccc} 2 & 1 & 3 \\ 4 & - 1 & 0 \\ - 7 & 2 & 1 \end{array})$

Determinant

Compute $∣ A ∣$

(expand along the first row):

$∣ A ∣ = 2 ∣ \begin{array}{cc} - 1 & 0 \\ 2 & 1 \end{array} ∣ - 1 ∣ \begin{array}{cc} 4 & 0 \\ - 7 & 1 \end{array} ∣ + 3 ∣ \begin{array}{cc} 4 & - 1 \\ - 7 & 2 \end{array} ∣$

$= 2 (- 1) - 1 (4) + 3 (1) = - 2 - 4 + 3 = - 3$

So

$∣ A ∣ = - 3$

(non zero) — inverse exists.

Cofactor / Adjugate

Compute the matrix of cofactors (I show the cofactors

$C_{i j}$ ):

$[C_{i j}] = (\begin{array}{ccc} - 1 & - 4 & 1 \\ 5 & 23 & - 11 \\ 3 & 12 & - 6 \end{array})$

The adjugate (adj $A$ ) is the transpose of this:

$adj A = (\begin{array}{ccc} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{array})$

Inverse

$A^{- 1} = \frac{1}{∣ A ∣} adj A = - \frac{1}{3} (\begin{array}{ccc} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{array}) = (\begin{array}{ccc} \frac{1}{3} & - \frac{5}{3} & - 1 \\ \frac{4}{3} & - \frac{23}{3} & - 4 \\ - \frac{1}{3} & \frac{11}{3} & 2 \end{array})$

Equivalently-

$A^{- 1} = \frac{1}{3} (\begin{array}{ccc} 1 & - 5 & - 3 \\ 4 & - 23 & - 12 \\ - 1 & 11 & 6 \end{array})$

Quick check (first-row × first-column)

$2 \cdot \frac{1}{3} + 1 \cdot \frac{4}{3} + 3 \cdot (- \frac{1}{3}) = \frac{2 + 4 - 3}{3} = 1,$

so the product gives the identity as expected.

Question 10.

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{array})$

Solution:

Compute

$∣ A ∣$

. Expanding along the first row,

$\begin{aligned} ∣ A ∣ & = 1 \cdot ∣ \begin{array}{cc} 2 & - 3 \\ - 2 & 4 \end{array} ∣ - (- 1) \cdot ∣ \begin{array}{cc} 0 & - 3 \\ 3 & 4 \end{array} ∣ + 2 \cdot ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 2 \end{array} ∣ \\ = 1 (2 \cdot 4 - (- 3) (- 2)) + 1 (0 \cdot 4 - (- 3) \cdot 3) + 2 (0 \cdot (- 2) - 2 \cdot 3) \\ = 1 (8 - 6) + 1 (9) + 2 (- 6) = 2 + 9 - 12 = - 1 \end{aligned}$

So

$∣ A ∣ = - 1$

(nonzero) and the inverse exists.

Compute cofactors (minors

$M_{i j}$

and cofactors

$C_{i j} = (- 1)^{i + j} M_{i j}$

$\begin{array}{lll} M_{11} = ∣ \begin{array}{cc} 2 & - 3 \\ - 2 & 4 \end{array} ∣ = 2, & C_{11} = + 2, \\ M_{12} = ∣ \begin{array}{cc} 0 & - 3 \\ 3 & 4 \end{array} ∣ = 9, & C_{12} = - 9, \\ M_{13} = ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 2 \end{array} ∣ = - 6, & C_{13} = - 6, \\ M_{21} = ∣ \begin{array}{cc} - 1 & 2 \\ - 2 & 4 \end{array} ∣ = 0, & C_{21} = 0, \\ M_{22} = ∣ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} ∣ = - 2, & C_{22} = - 2, \\ M_{23} = ∣ \begin{array}{cc} 1 & - 1 \\ 3 & - 2 \end{array} ∣ = 1, & C_{23} = - 1, \\ M_{31} = ∣ \begin{array}{cc} - 1 & 2 \\ 2 & - 3 \end{array} ∣ = - 1, & C_{31} = - 1, \\ M_{32} = ∣ \begin{array}{cc} 1 & 2 \\ 0 & - 3 \end{array} ∣ = - 3, & C_{32} = + 3, \\ M_{33} = ∣ \begin{array}{cc} 1 & - 1 \\ 0 & 2 \end{array} ∣ = 2, & C_{33} = + 2, \end{array}$

Cofactor matrix

$C = [C_{i j}]$

is

$C = (\begin{array}{ccc} 2 & - 9 & - 6 \\ 0 & - 2 & - 1 \\ - 1 & 3 & 2 \end{array})$

Adjugate is transpose of cofactor matrix:

$adj A = C^{T} = (\begin{array}{ccc} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{array})$

Inverse:

$A^{- 1} = \frac{1}{∣ A ∣} adj A = - 1 \cdot adj A = (\begin{array}{ccc} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{array})$

Question 11.

$Find A^{- 1} for A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{array})$

Solution:

Write

$A = diag (1, M)$

where

$M = (\begin{array}{cc} \cos α & \sin α \\ \sin α & - \cos α \end{array})$

Compute

$\det M$

:

$\det M = (\cos α) (- \cos α) - (\sin α) (\sin α) = - (\cos^{2} α + \sin^{2} α) = - 1$

So

$∣ A ∣ = 1 \cdot \det M = - 1$ , hence

$A$ is invertible.

Now compute $M^{- 1}$ using the $2 \times 2$

formula:

$M^{- 1} = \frac{1}{\det M} (\begin{array}{cc} - \cos α & - \sin α \\ - \sin α & \cos α \end{array})$

$= - 1 (\begin{array}{cc} - \cos α & - \sin α \\ - \sin α & \cos α \end{array}) = (\begin{array}{cc} \cos α & \sin α \\ \sin α & - \cos α \end{array}) = M$

Thus

$M^{- 1} = M$

Equivalently

$M^{2} = I_{2}$

Therefore

$A^{- 1} = diag (1, M^{- 1}) = diag (1, M) = A$

Final answer

$A^{- 1} = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{array}) = A$

(Verification:

$A^{2} = diag (1, M^{2}) = diag (1, I_{2}) = I_{3}$ , so indeed

$A^{- 1} = A$

Question 12:

$Let A = (\begin{array}{cc} 3 & 7 \\ 2 & 5 \end{array}) and B = (\begin{array}{cc} 6 & 8 \\ 7 & 9 \end{array}) .$

$Verify that (A B)^{- 1} = B^{- 1} A^{- 1} .$

Let’s verify

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Compute

$A^{- 1}$

and

$B^{- 1}$

(a) $A^{- 1}$ $∣ A ∣ = 3 (5) - 7 (2) = 15 - 14 = 1$

$A^{- 1} = (\begin{array}{cc} 5 & - 7 \\ - 2 & 3 \end{array})$

(b) $B^{- 1}$ $∣ B ∣ = 6 (9) - 8 (7) = 54 - 56 = - 2$

$B^{- 1} = \frac{1}{- 2} (\begin{array}{cc} 9 & - 8 \\ - 7 & 6 \end{array}) = (\begin{array}{cc} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{array})$

Compute

$A B$ $A B = (\begin{array}{cc} 3 & 7 \\ 2 & 5 \end{array}) (\begin{array}{cc} 6 & 8 \\ 7 & 9 \end{array}) = (\begin{array}{cc} 3 \cdot 6 + 7 \cdot 7 & 3 \cdot 8 + 7 \cdot 9 \\ 2 \cdot 6 + 5 \cdot 7 & 2 \cdot 8 + 5 \cdot 9 \end{array}) = (\begin{array}{cc} 67 & 87 \\ 47 & 61 \end{array})$

Compute

$(A B)^{- 1}$ $∣ A B ∣ = 67 (61) - 87 (47) = 4087 - 4089 = - 2$

$(A B)^{- 1} = \frac{1}{- 2} (\begin{array}{cc} 61 & - 87 \\ - 47 & 67 \end{array}) = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array})$

Compute

$B^{- 1} A^{- 1}$ $B^{- 1} A^{- 1} = (\begin{array}{cc} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{array}) (\begin{array}{cc} 5 & - 7 \\ - 2 & 3 \end{array})$

Multiply:

$\begin{aligned} B^{- 1} A^{- 1} & = (\begin{array}{cc} (- \frac{9}{2}) (5) + 4 (- 2) & (- \frac{9}{2}) (- 7) + 4 (3) \\ (\frac{7}{2}) (5) + (- 3) (- 2) & (\frac{7}{2}) (- 7) + (- 3) (3) \end{array}) \\ = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}) \end{aligned}$

Compare

$(A B)^{- 1} = (\begin{array}{cc} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}) = B^{- 1} A^{- 1}$

Hence verified:

$(A B)^{- 1} = B^{- 1} A^{- 1}$

Question 13 :

If $A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$ , show that $A^{2} - 5 A + 7 I = O$ . Hence, find $A^{- 1}$ .

Solution:

Let’s solve step-by-step carefully and clearly.

Given:

$A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$

We have to show that

$A^{2} - 5 A + 7 I = O$

and then use this result to find $A^{- 1}$

Step 1: Compute $A^{2}$

$A^{2} = A \cdot A = [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}] [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}]$

Multiply carefully:

$A^{2} = [\begin{array}{cc} (3) (3) + (1) (- 1) & (3) (1) + (1) (2) \\ (- 1) (3) + (2) (- 1) & (- 1) (1) + (2) (2) \end{array}] = [\begin{array}{cc} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{array}] = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}]$

So,

$A^{2} = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}]$

Step 2: Compute $5 A$ and $7 I$

$5 A = 5 [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}]$ $7 I = 7 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}]$

Step 3: Compute $A^{2} - 5 A + 7 I$

$A^{2} - 5 A + 7 I = [\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}] - [\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}] + [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}]$

First subtract $A^{2} - 5 A$ :

$[\begin{array}{cc} 8 - 15 & 5 - 5 \\ - 5 - (- 5) & 3 - 10 \end{array}] = [\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}]$

Now add $7 I$ :

$[\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}] + [\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Hence proved:

$A^{2} - 5 A + 7 I = O$

Step 4: Finding $A^{- 1}$

From the equation:

$A^{2} - 5 A + 7 I = 0$

Rearrange it as:

$A^{2} - 5 A = - 7 I$ Multiply both sides by $A^{- 1}$ (on the right):

$A - 5 I = - 7 A^{- 1}$

So, $A^{- 1} = \frac{1}{7} (5 I - A)$

Step 5: Substitute values

$A^{- 1} = \frac{1}{7} (5 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] - [\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}])$ $A^{- 1} = \frac{1}{7} [\begin{array}{cc} 5 - 3 & 0 - 1 \\ 0 - (- 1) & 5 - 2 \end{array}] = \frac{1}{7} [\begin{array}{cc} 2 & - 1 \\ 1 & 3 \end{array}]$

Click Here For Question 14 to 18 of Exercise 4.4
October 29, 2025
Exercise-4.1, Class 12th, Maths, Chapter 4, NCERT

DETERMINANTS

Question 1

Evaluate $∣ \begin{array}{cc} 2 & 4 \\ - 5 & - 1 \end{array} ∣$

Solution.
$\det = 2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

Question 2

(i) $∣ \begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array} ∣$

Solution 2 (i):

$\det = \cos θ \cdot \cos θ - (- \sin θ) \cdot \sin θ = \cos^{2} θ + \sin^{2} θ = 1 .$

Question 2 (ii):

$∣ \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} ∣$ Solution:

We know that for a $2 \times 2$ determinant.

Now compute:

$\det = (x^{2} - x + 1) (x + 1) - (x - 1) (x + 1) .$

Step 1: Expand both terms

$(x^{2} - x + 1) (x + 1) = x^{3} + x^{2} - x^{2} - x + x + 1 = x^{3} + 1$ $(x - 1) (x + 1) = x^{2} - 1$

Step 2: Substitute back

$\det = (x^{3} + 1) - (x^{2} - 1) = x^{3} - x^{2} + 2$

Question 3

If $A = (\begin{array}{cc} 1 & 2 \\ 4 & 2 \end{array})$ , show $∣ 2 A ∣ = 4 ∣ A ∣$

Solution.
For an $n \times n$ matrix scaling by scalar $k$ multiplies the determinant by $k^{n}$ . Here $n = 2$ and $k = 2$ . So $∣ 2 A ∣ = 2^{2} ∣ A ∣ = 4 ∣ A ∣$

(Direct check: $∣ A ∣ = 1 \cdot 2 - 2 \cdot 4 = 2 - 8 = - 6$ . Then $∣ 2 A ∣ = \det (\begin{array}{cc} 2 & 4 \\ 8 & 4 \end{array}) = 2 \cdot 4 - 4 \cdot 8 = 8 - 32 = - 24 = 4 (- 6)$

Question 4

If $A = (\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array})$ , show $∣ 3 A ∣ = 27 ∣ A ∣$

Solution.
$A$ is $3 \times 3$ . Scaling by 3 multiplies determinant by $3^{3} = 27$ . Hence $∣ 3 A ∣ = 27 ∣ A ∣$ .

(One can check: $∣ A ∣$ is product of diagonal entries $1 \cdot 1 \cdot 4 = 4$ .

$∣ 3 A ∣$ has diagonal $3, 3, 12$ so determinant $3 \cdot 3 \cdot 12 = 108 = 27 \cdot 4$

Question 5 — Evaluate the following determinants

(i) $∣ \begin{array}{ccc} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{array} ∣$

Solution (5.i).
Expand along second row (only one nonzero entry $a_{23} = - 1$ ):

Minor for $a_{23}$ is $∣ \begin{array}{cc} 3 & - 1 \\ 3 & - 5 \end{array} ∣ = 3 (- 5) - (- 1) 3 = - 15 + 3 = - 12$

Cofactor $C_{23} = (- 1)^{2 + 3} \cdot (- 12) = - 1 \cdot (- 12) = 12$ . Then determinant = $a_{23} C_{23} = (- 1) \cdot 12 = - 12$

(ii) $∣ \begin{array}{ccc} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{array} ∣$

Solution (5.ii).
Compute by expansion / rule of Sarrus:

$\det = 3 ∣ \begin{array}{cc} 1 & - 2 \\ 3 & 1 \end{array} ∣ - (- 4) ∣ \begin{array}{cc} 1 & - 2 \\ 2 & 1 \end{array} ∣ + 5 ∣ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} ∣ .$

Evaluate minors: $\cdot$
So $\det = 3 \cdot 7 + 4 \cdot 5 + 5 \cdot 1 = 21 + 20 + 5 = 46$

(iii) $∣ \begin{array}{ccc} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{array} ∣$

Solution (5.iii).
Compute (expanding first row):

$\det = 0 \cdot (\dots) - 1 \cdot ∣ \begin{array}{cc} - 1 & - 3 \\ - 2 & 0 \end{array} ∣ + 2 \cdot ∣ \begin{array}{cc} - 1 & 0 \\ - 2 & 3 \end{array} ∣$

First minor: $(- 1) \cdot 0 - (- 3) (- 2) = 0 - 6 = - 6$ ; the corresponding contribution is $- 1 \cdot (- 6) = 6$
Second minor: $(- 1) \cdot 3 - 0 \cdot (- 2) = - 3$ ; contribution $2 \cdot (- 3) = - 6$ . Sum $6 - 6 = 0$

(iv) $∣ \begin{array}{ccc} 2 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{array} ∣$

Solution (5.iv).
Expand along first row:

$\det = 2 ∣ \begin{array}{cc} 2 & - 1 \\ - 5 & 0 \end{array} ∣ - (- 1) ∣ \begin{array}{cc} 0 & - 1 \\ 3 & 0 \end{array} ∣ + (- 2) ∣ \begin{array}{cc} 0 & 2 \\ 3 & - 5 \end{array} ∣$

Compute minors: first $= 2 \cdot 0 - (- 1) (- 5) = 0 - 5 = - 5$ ⇒ $2 (- 5) = - 10$
second minor $= 0 \cdot 0 - (- 1) \cdot 3 = 3$ ⇒ sign gives $+ 1 \cdot 3 = + 3$
third minor $= 0 \cdot (- 5) - 2 \cdot 3 = - 6$ ⇒ multiplied by $- 2$ gives $+ 12$ .
Sum $- 10 + 3 + 12 = 5 .$

Question 6

If $A = (\begin{array}{ccc} 1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9 \end{array})$ , find $∣ A ∣$

Solution.
Expand along first row:

$∣ A ∣ = 1 ∣ \begin{array}{cc} 1 & - 3 \\ 4 & - 9 \end{array} ∣ - 1 ∣ \begin{array}{cc} 2 & - 3 \\ 5 & - 9 \end{array} ∣ + (- 2) ∣ \begin{array}{cc} 2 & 1 \\ 5 & 4 \end{array} ∣$

Compute minors: first = $1 (- 9) - (- 3) 4 = - 9 + 12 = 3$
second = $2 (- 9) - (- 3) 5 = - 18 + 15 = - 3$ . With the minus sign yields $- 1 \cdot (- 3) = + 3$ .
third minor = $2 \cdot 4 - 1 \cdot 5 = 8 - 5 = 3$ , times $- 2$ gives $- 6$ . Sum $3 + 3 - 6 = 0$

So $∣ A ∣ = 0$

Question 7 (i):

$Find the value of x if ∣ \begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array} ∣ = ∣ \begin{array}{cc} 2 x & 4 \\ 6 & x \end{array} ∣$

Solution:

For any $2 \times 2$ matrix
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c$ Step 1: Compute the determinant on the left-hand side (LHS)

$∣ \begin{array}{cc} 2 & 4 \\ 5 & 1 \end{array} ∣ = (2) (1) - (4) (5) = 2 - 20 = - 18$ Step 2: Compute the determinant on the right-hand side (RHS)

$∣ \begin{array}{cc} 2 x & 4 \\ 6 & x \end{array} ∣ = (2 x) (x) - (4) (6) = 2 x^{2} - 24$

Step 3: Set them equal

$- 18 = 2 x^{2} - 24.$

Simplify:

$2 x^{2} = - 18 + 24 = 6.$ $x^{2} = 3.$ $x = \pm \sqrt{3}$

Question 7 (ii):

$Find the value of x if ∣ \begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array} ∣ = ∣ \begin{array}{cc} x & 3 \\ 2 x & 5 \end{array} ∣$

Solution:

For any $2 \times 2$ matrix
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c .$

Step 1: Compute LHS

$∣ \begin{array}{cc} 2 & 3 \\ 4 & 5 \end{array} ∣ = (2) (5) - (3) (4) = 10 - 12 = - 2$

Step 2: Compute RHS

$∣ \begin{array}{cc} x & 3 \\ 2 x & 5 \end{array} ∣ = (x) (5) - (3) (2 x) = 5 x - 6 x = - x$

Step 3: Equate LHS and RHS

$- 2 = - x \Rightarrow x = 2$

Question 8:

$If ∣ \begin{array}{cc} x & 2 \\ 18 & x \end{array} ∣ = ∣ \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} ∣, then find x$

Options:
(A) $6$ (B) $\pm 6$ (C) $- 6$ (D) $0$

Solution:

For any $2 \times 2$ determinant
$(\begin{array}{cc} a & b \\ c & d \end{array})$ ,

$\det = a d - b c .$

Step 1: Compute LHS

$∣ \begin{array}{cc} x & 2 \\ 18 & x \end{array} ∣ = x \cdot x - 2 \cdot 18 = x^{2} - 36$ Step 2: Compute RHS

$∣ \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} ∣ = 6 \cdot 6 - 2 \cdot 18 = 36 - 36 = 0$

Step 3: Equate LHS and RHS

$x^{2} - 36 = 0$

$x^{2} = 36$

$x = \pm 6$

Final Answer:

$(B) x = \pm 6$

October 28, 2025
Miscellaneous Exercise on Chapter 3, Class 12th, Maths, NCERT
Question 1:

$If A and B are symmetric matrices, prove that A B - B A is a skew-symmetric matrix.$

Answer (proof):

Recall a matrix $M$ is skew-symmetric if $M^{T} = - M$

Given $A$ and $B$ are symmetric, so

$A^{T} = A, B^{T} = B .$

Consider the transpose of $A B - B A$ :

$(A B - B A)^{T} = (A B)^{T} - (B A)^{T} = B^{T} A^{T} - A^{T} B^{T} .$

Using symmetry of $A$ and $B$ we get

$(A B - B A)^{T} = B A - A B = - (A B - B A) .$

Thus $(A B - B A)^{T} = - (A B - B A)$ , so $A B - B A$ is skew-symmetric.

(As a remark, every skew-symmetric matrix has zeros on its diagonal, so the diagonal entries of $A B - B A$ are all zero.)

Question 2:

$Show that the matrix B^{'} A B is symmetric according as A is symmetric or skew-symmetric.$

Answer (proof):

Let $A$ and $B$ be square matrices of the same order, and let $B^{'}$ denote the transpose of $B$ .
We need to show that:
- If $A$ is symmetric, then $B^{'} A B$ is symmetric.
- If $A$ is skew-symmetric, then $B^{'} A B$ is skew-symmetric.
Case 1: $A$ is symmetric

If $A$ is symmetric, then $A^{'} = A$ .

Consider $(B^{'} A B)^{'}$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B .$

But since $A^{'} = A$ ,

$(B^{'} A B)^{'} = B^{'} A B .$

Hence, $B^{'} A B$ is symmetric.

Case 2: $A$ is skew-symmetric

If $A$ is skew-symmetric, then $A^{'} = - A$ .

Now, take the transpose of $B^{'} A B$ :

$(B^{'} A B)^{'} = B^{'} A^{'} B = B^{'} (- A) B = - B^{'} A B .$

Thus, $(B^{'} A B)^{'} = - B^{'} A B$ ,
which means $B^{'} A B$ is skew-symmetric.

Hence proved:
The matrix $B^{'} A B$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

Question 3:

$Find the values of x, y, z if A = (\begin{array}{ccc} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}) satisfies A^{T} A = I .$

Answer (solution):

Compute $A^{T} A$ . A straightforward multiplication gives

$A^{T} A = (\begin{array}{ccc} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{array}) .$

For $A^{T} A = I$ we must have the diagonal entries equal to 1, hence

$2 x^{2} = 1, 6 y^{2} = 1, 3 z^{2} = 1$

Therefore

$x^{2} = \frac{1}{2} ⟹ x = \pm \frac{1}{\sqrt{2}}, y^{2} = \frac{1}{6} ⟹ y = \pm \frac{1}{\sqrt{6}}, z^{2} = \frac{1}{3} ⟹ z = \pm \frac{1}{\sqrt{3}} .$

All choices of independent signs are allowed, so the solutions are

$(x, y, z) = (\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{3}}),$

(8 sign-combinations in total).

Question 4:

$For what values of x does [1 2 1] (\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = 0 ?$

Answer (solution):

First compute the product of the matrix with the column vector:

$(\begin{array}{ccc} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}) (\begin{array}{c} 0 \\ 2 \\ x \end{array}) = (\begin{array}{c} 1 \cdot 0 + 2 \cdot 2 + 0 \cdot x \\ 2 \cdot 0 + 0 \cdot 2 + 1 \cdot x \\ 1 \cdot 0 + 0 \cdot 2 + 2 \cdot x \end{array}) = (\begin{array}{c} 4 \\ x \\ 2 x \end{array})$

Now left-multiply by $[1 2 1]$ :

$[1 2 1] (\begin{array}{c} 4 \\ x \\ 2 x \end{array}) = 1 \cdot 4 + 2 \cdot x + 1 \cdot 2 x = 4 + 4 x$

Set equal to zero:

$4 + 4 x = 0 \Rightarrow x = - 1$

Question 5:

$If A = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}), show that A^{2} - 5 A + 7 I = 0$

Answer (solution):

Given

$A = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}), I = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) .$

Step 1: Compute $A^{2}$

$A^{2} = (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 3 (3) + 1 (- 1) & 3 (1) + 1 (2) \\ (- 1) (3) + 2 (- 1) & (- 1) (1) + 2 (2) \end{array}) = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) .$

Step 2: Compute $5 A$

$5 A = 5 (\begin{array}{cc} 3 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array})$

Step 3: Compute $7 I$

$7 I = 7 (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Step 4: Substitute into $A^{2} - 5 A + 7 I$

$A^{2} - 5 A + 7 I = (\begin{array}{cc} 8 & 5 \\ - 5 & 3 \end{array}) - (\begin{array}{cc} 15 & 5 \\ - 5 & 10 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array})$

Compute step-by-step:

$A^{2} - 5 A = (\begin{array}{cc} 8 - 15 & 5 - 5 \\ - 5 - (- 5) & 3 - 10 \end{array}) = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array})$

Now add $7 I$ :

$A^{2} - 5 A + 7 I = (\begin{array}{cc} - 7 & 0 \\ 0 & - 7 \end{array}) + (\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}) = (\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$

Hence proved:

$A^{2} - 5 A + 7 I = 0$

Question 6:

$Find x if [x - 5 - 1] (\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = 0.$

Solution:

We will simplify step-by-step.

Step 1: Multiply the matrix with the column vector

$(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}) (\begin{array}{c} x \\ 4 \\ 1 \end{array}) = (\begin{array}{c} 1 \cdot x + 0 \cdot 4 + 2 \cdot 1 \\ 0 \cdot x + 2 \cdot 4 + 1 \cdot 1 \\ 2 \cdot x + 0 \cdot 4 + 3 \cdot 1 \end{array}) = (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array})$

Step 2: Multiply the row vector $[x - 5 - 1]$ with the result

$[x - 5 - 1] (\begin{array}{c} x + 2 \\ 9 \\ 2 x + 3 \end{array}) = x (x + 2) + (- 5) (9) + (- 1) (2 x + 3)$

Simplify:

$= x^{2} + 2 x - 45 - 2 x - 3 = x^{2} - 48.$

Step 3: Set equal to zero

$x^{2} - 48 = 0 \Rightarrow x^{2} = 48$

$x= \pm 4 \sqrt{3}$

Question 7

$A manufacturer produces three products x, y, z sold in two markets with annual sales$ $S = (\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}) (rows: Market I, Market II; columns: x, y, z).$ $(a) Unit sale prices p = (\begin{array}{c} 2.50 \\ 1.50 \\ 1.00 \end{array}) . (b) Unit costs c = (\begin{array}{c} 2.00 \\ 1.00 \\ 0.50 \end{array}) .$ $Find (a) total revenue in each market, (b) gross profit in each market.$

Solution

Use matrix multiplication. Revenue vector $R$ (marketwise) is

$R = S p .$

Compute component-wise:

Market I revenue

$10000 (2.5) + 2000 (1.5) + 18000 (1) = 25000 + 3000 + 18000 = 46000.$

Market II revenue

$6000 (2.5) + 20000 (1.5) + 8000 (1) = 15000 + 30000 + 8000 = 53000.$

So

$R = (\begin{array}{c} 46000 \\ 53000 \end{array}) rupees.$

(b) Total cost vector $C = S c$ :

Market I cost

$10000 (2) + 2000 (1) + 18000 (0.5) = 20000 + 2000 + 9000 = 31000.$

Market II cost

$6000 (2) + 20000 (1) + 8000 (0.5) = 12000 + 20000 + 4000 = 36000.$

So

$C = (\begin{array}{c} 31000 \\ 36000 \end{array}) rupees.$

Gross profit for each market $G = R - C$ :

$G = (\begin{array}{c} 46000 - 31000 \\ 53000 - 36000 \end{array}) = (\begin{array}{c} 15000 \\ 17000 \end{array}) rupees$ Final answers

(a) Total revenue — Market I: Rs. 46,000; Market II: Rs. 53,000.
(b) Gross profit — Market I: Rs. 15,000; Market II: Rs. 17,000.

(Also note matrix forms: $R = S p, C = S c, G = S (p - c)$

Question 8:

$Find the matrix X such that X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

Solution:

We are given:

$X (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array})$

Let:

$A = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}), B = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) .$

We need $X$ such that $X A = B$ .

To isolate $X$ , multiply both sides on the right by $A^{T} (A A^{T})^{- 1}$ :
(since $A$ is $2 \times 3$ , not square, we use this formula)

$X = B A^{T} (A A^{T})^{- 1} .$

Step 1: Compute $A A^{T}$

$A A^{T} = (\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}) = (\begin{array}{cc} 1^{2} + 2^{2} + 3^{2} & 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 \\ 4 \cdot 1 + 5 \cdot 2 + 6 \cdot 3 & 4^{2} + 5^{2} + 6^{2} \end{array}) = (\begin{array}{cc} 14 & 32 \\ 32 & 77 \end{array})$

Step 2: Compute $(A A^{T})^{- 1}$

First find determinant:

$∣ A A^{T} ∣ = (14) (77) - (32) (32) = 1078 - 1024 = 54$

So,

$(A A^{T})^{- 1} = \frac{1}{54} (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Step 3: Compute $B A^{T}$

$B A^{T} = (\begin{array}{ccc} - 7 & - 8 & - 9 \\ 2 & 4 & 6 \end{array}) (\begin{array}{cc} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array})$

$= (\begin{array}{cc} (- 7) (1) + (- 8) (2) + (- 9) (3) & (- 7) (4) + (- 8) (5) + (- 9) (6) \\ (2) (1) + (4) (2) + (6) (3) & (2) (4) + (4) (5) + (6) (6) \end{array})$

Compute each entry:

$B A^{T} = (\begin{array}{cc} - 7 - 16 - 27 & - 28 - 40 - 54 \\ 2 + 8 + 18 & 8 + 20 + 36 \end{array}) = (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array})$

Step 4: Compute $X = B A^{T} (A A^{T})^{- 1}$

$X = \frac{1}{54} (\begin{array}{cc} - 50 & - 122 \\ 28 & 64 \end{array}) (\begin{array}{cc} 77 & - 32 \\ - 32 & 14 \end{array}) .$

Compute the product:

First row:

$(- 50) (77) + (- 122) (- 32) = - 3850 + 3904 = 54,$ $(- 50) (- 32) + (- 122) (14) = 1600 - 1708 = - 108.$

Second row:

$(28) (77) + (64) (- 32) = 2156 - 2048 = 108,$ $(28) (- 32) + (64) (14) = - 896 + 896 = 0$

So:

$X = \frac{1}{54} (\begin{array}{cc} 54 & - 108 \\ 108 & 0 \end{array}) = (\begin{array}{cc} 1 & - 2 \\ 2 & 0 \end{array}) .$

Choose the correct answer in the following questions:

Question 9:

$If A = (\begin{array}{cc} α & β \\ γ & - α \end{array}) is such that A^{2} = I, then find the correct relation among α, β, γ .$

Options:
(A) $1 + α^{2} + β γ = 0$
(B) $1 - α^{2} + β γ = 0$
(C) $1 - α^{2} - β γ = 0$
(D) $1 + α^{2} - β γ = 0$

Solution:

Given

$A = (\begin{array}{cc} α & β \\ γ & - α \end{array})$

Compute $A^{2}$ :

$A^{2} = (\begin{array}{cc} α & β \\ γ & - α \end{array}) (\begin{array}{cc} α & β \\ γ & - α \end{array}) = (\begin{array}{cc} α^{2} + β γ & α β - α β \\ α γ - α γ & γ β + α^{2} \end{array}) = (\begin{array}{cc} α^{2} + β γ & 0 \\ 0 & α^{2} + β γ \end{array})$

Thus,

$A^{2} = (α^{2} + β γ) I .$

We are told $A^{2} = I$ , so:

$(α^{2} + β γ) I = I$

That means:

$α^{2} + β γ = 1$

Rearranging:

$1 - α^{2} - β γ = 0$

Correct Option:

$(C) 1 - α^{2} - β γ = 0.$

Question 10:

$If the matrix A is both symmetric and skew-symmetric, then:$

Options:
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these

Solution:

By definition:
- $A$ is symmetric if $A^{T} = A$
- $A$ is skew-symmetric if $A^{T} = - A$
If $A$ is both, then:

$A = A^{T} = - A$

This implies:

$A = - A \Rightarrow 2 A = 0 \Rightarrow A = 0$

That means all elements of $A$ are zero.

Correct Option:

$(B) A is a zero matrix.$

Question 11:

$If A is a square matrix such that A^{2} = A, then find (I + A)^{3} - 7 A .$

Options:
(A) $A$
(B) $I - A$
(C) $I$
(D) $3 A$

Solution:

We are given that:

$A^{2} = A .$

Let’s expand $(I + A)^{3}$ :

$(I + A)^{3} = I^{3} + 3 I^{2} A + 3 I A^{2} + A^{3}$

Simplify each term using $I^{2} = I$ and $A^{2} = A$ :

$(I + A)^{3} = I + 3 A + 3 A + A^{3}$

Now, compute $A^{3}$ :

$A^{3} = A^{2} A = A \cdot A = A^{2} = A$

Substitute this back:

$(I + A)^{3} = I + 3 A + 3 A + A = I + 7 A$

Now compute:

$(I + A)^{3} - 7 A = (I + 7 A) - 7 A = I$

Final Answer:

$(C) I$
October 28, 2025
Exercise-3.1, Class 12th, Maths, Chapter 3- Matrices, NCERT

Exercise 3.1

Question 1:

In the matrix

$A = [\begin{array}{cccc} 2 & 5 & 19 & - 7 \\ 1 & - 5 & 17 & 3 \\ 1 & - 2 & 12 & 5 \end{array}]$

write:
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Answer:
(i) Order = 3 × 4
(ii) Number of elements = 3 × 4 = 12
(iii) a₁₃ = 19, a₂₁ = 1, a₃₃ = 12, a₂₄ = 3, a₂₃ = 17.

Question 2:

If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements?

Answer:
We know that if a matrix has m × n elements, then total elements = m × n.

(i) For 24 elements:
Possible orders = (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1)

(ii) For 13 elements (prime number):
Possible orders = (1,13), (13,1)

Question 3:

If a matrix has 18 elements, what are its possible orders? What if it has 5 elements?

Answer:
(i) For 18 elements: $m \times n = 18$
Possible orders = (1,18), (2,9), (3,6), (6,3), (9,2), (18,1)

(ii) For 5 elements (prime): (1,5), (5,1)

Question 4:

Construct a 2 × 2 matrix A = [aᵢⱼ] where:

(i) aᵢⱼ = 2i − j
(ii) aᵢⱼ = i² − 3j
(iii) aᵢⱼ = i² / 2j

Answer:
For i, j = 1, 2

(i)

$A = [\begin{array}{cc} 2 (1) - 1 & 2 (1) - 2 \\ 2 (2) - 1 & 2 (2) - 2 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 3 & 2 \end{array}]$

(ii)

$A = [\begin{array}{cc} 1^{2} - 3 (1) & 1^{2} - 3 (2) \\ 2^{2} - 3 (1) & 2^{2} - 3 (2) \end{array}] = [\begin{array}{cc} - 2 & - 5 \\ 1 & - 2 \end{array}]$

(iii)

$A = [\begin{array}{cc} \frac{1^{2}}{2 \times 1} & \frac{1^{2}}{2 \times 2} \\ \frac{2^{2}}{2 \times 1} & \frac{2^{2}}{2 \times 2} \end{array}] = [\begin{array}{cc} \frac{1}{2} & \frac{1}{4} \\ 2 & 1 \end{array}]$

Question 5:

Construct a 3 × 4 matrix where:
(i) aᵢⱼ = ½ (i − 3j)
(ii) aᵢⱼ = 2i − j

Answer:

(i)

$A = [\begin{array}{cccc} - 1 & - 2.5 & - 4 & - 5.5 \\ - 0.5 & - 2 & - 3.5 & - 5 \\ 0 & - 1.5 & - 3 & - 4.5 \end{array}]$

(ii)

$A = [\begin{array}{cccc} 1 & 0 & - 1 & - 2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{array}]$

Question 6:

Find the values of x, y, and z from the following equations:

(i)

$[\begin{array}{ccc} x & 5 & z \\ y & 1 & 5 \end{array}] = [\begin{array}{ccc} 2 & 6 & 2 \\ 1 & 5 & 8 \end{array}]$

(ii)

$[\begin{array}{ccc} 5 & z & x + y \\ 5 & 8 & x + y \end{array}] = [\begin{array}{ccc} x + z & y & z \\ x z & y & z \end{array}]$

Answer:

(i) Comparing elements:
x = 2, y = 1, z = 2.

(ii) On comparing:
x = 3, y = 1, z = 2.

Question 7:

Find a, b, c, d from the equation:

$[\begin{array}{cc} a - b & 2 a - b \\ 2 a + c & 3 c - d \end{array}] = [\begin{array}{cc} 0 & - 15 \\ 0 & 13 \end{array}]$

Answer:
Equating elements:
a − b = 0
2a − b = −15
2a + c = 0
3c − d = 13

Solving, we get
a = −15, b = −15, c = 30, d = 77.

Question 8:

A = [aᵢⱼ] is a square matrix if:
(A) m < n (B) m > n (C) m = n (D) None of these

Answer:
(C) m = n

Question 9:

Which values of x and y make the following matrices equal?

$[\begin{array}{cc} 3 x + 7 & 5 \\ y + 1 & 2 \end{array}] = [\begin{array}{cc} - 3 x & 0 \\ - 2 & 8 \end{array}]$

Answer:
3x + 7 = −3x ⇒ 6x = −7 ⇒ x = −7/6
y + 1 = −2 ⇒ y = −3

x = −7/6, y = −3

Question 10:

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512

Answer:
Each entry can be 0 or 1, i.e., 2 possibilities per element.
Total = $2^{3 \times 3} = 2^{9} = 512$

Answer: (D) 512

October 26, 2025
Exercise-2.1, Class 12th, Maths, Chapter – 2, NCERT
EXERCISE 2.1 — Solutions

1. Find the principal value of $\sin^{- 1} ⁣ (\frac{1}{2})$ .

Solution. Principal branch of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ .
$\sin \frac{π}{6} = \frac{1}{2}$ . $\frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ .

$\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$

2. Find the principal value of $\cos^{- 1} ⁣ (\frac{\sqrt{3}}{2})$ .

Solution. Principal branch of $\cos^{- 1}$ is $[0, π]$ .
$\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\frac{π}{6} \in [0, π]$ .

$\cos^{- 1} ⁣ (\frac{\sqrt{3}}{2}) = \frac{π}{6}$

3. Find the principal value of ${cosec}^{- 1} (2)$ .

Solution. If $y = {cosec}^{- 1} 2$ then $cosec y = 2 \Rightarrow \sin y = \frac{1}{2}$ . Principal branch for ${cosec}^{- 1}$ (as used in the book) corresponds to $y \in [- \frac{π}{2}, \frac{π}{2}]$ excluding 0 (or equivalent principal branch placing the value at $π / 6$ or $π - π / 6$ ; standard choice giving principal value in $[- \frac{π}{2}, \frac{π}{2}] ∖ {0}$ yields $y = \frac{π}{6}$ . So

${cosec}^{- 1} (2) = \frac{π}{6}$

(Interpretation note: many texts give ${cosec}^{- 1} 2 = π - \frac{π}{6} = \frac{5 π}{6}$ if using a different branch — but the book’s principal branch choice yields $π / 6$ .)

4. Find the principal value of $\tan^{- 1} (\sqrt{3})$ .

Solution. $\tan \frac{π}{3} = \sqrt{3}$ . Principal branch of $\tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , and $\frac{π}{3} \in (- \frac{π}{2}, \frac{π}{2})$ .

$\tan^{- 1} (\sqrt{3}) = \frac{π}{3}$

5. Find the principal value of $\cos^{- 1} ⁣ (- \frac{1}{2})$ .

Solution. $\cos \frac{2 π}{3} = - \frac{1}{2}$ , and $\frac{2 π}{3} \in [0, π]$ . So

$\cos^{- 1} ⁣ (- \frac{1}{2}) = \frac{2 π}{3}$

6. Find the principal value of $\tan^{- 1} (- 1)$ .

Solution. $\tan (- \frac{π}{4}) = - 1$ . Principal branch is $(- \frac{π}{2}, \frac{π}{2})$ .

$\tan^{- 1} (- 1) = - \frac{π}{4}$

7. Find the principal value of $\sec^{- 1} ⁣ (\frac{2}{\sqrt{3}})$ .

Solution. $\sec θ = \frac{2}{\sqrt{3}}$ ⇒ $\cos θ = \frac{\sqrt{3}}{2}$ ⇒ $θ = \frac{π}{6}$ (principal $\sec^{- 1}$ branch is $[0, π] ∖ {\frac{π}{2}}$ ).

$\sec^{- 1} ⁣ (\frac{2}{\sqrt{3}}) = \frac{π}{6}$

8. Find the principal value of $\cot^{- 1} (\sqrt{3})$ .

Solution. $\cot θ = \sqrt{3} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6}$ . Principal branch of $\cot^{- 1}$ in the book is $(0, π)$ , and $\frac{π}{6} \in (0, π)$ .

$\cot^{- 1} (\sqrt{3}) = \frac{π}{6}$

9. Find the principal value of $\cos^{- 1} ⁣ (- \frac{1}{2})$ .

(This repeats Q5 — same answer.)

Solution. As in Q5,

$\frac{2 π}{3}$

10. Find the principal value of ${cosec}^{- 1} (- 2)$ .

Solution. $cosec y = - 2 \Rightarrow \sin y = - \frac{1}{2}$ . Principal value of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ , and $\sin (- \frac{π}{6}) = - \frac{1}{2}$ . So choose $y = - \frac{π}{6}$ in the principal branch for cosec $^{- 1}$ .

${cosec}^{- 1} (- 2) = - \frac{π}{6}$

11. Evaluate $\tan^{- 1} (1) + \cos^{- 1} ⁣ (\frac{1}{2}) + \sin^{- 1} ⁣ (\frac{1}{2})$

Solution.
$\tan^{- 1} (1) = \frac{π}{4}$
$\cos^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{3}$
$\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$
Sum: $\frac{π}{4} + \frac{π}{3} + \frac{π}{6} = \frac{3 π + 4 π + 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$

$\frac{3 π}{4}$

12. Evaluate $\cos^{- 1} ⁣ (\frac{1}{2}) + 2 \sin^{- 1} ⁣ (\frac{1}{2})$ .

Solution. $\cos^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{3}$ , $\sin^{- 1} ⁣ (\frac{1}{2}) = \frac{π}{6}$ .
So $\frac{π}{3} + 2 \cdot \frac{π}{6} = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3}$

$\frac{2 π}{3}$

13. If $\sin^{- 1} x = y$ , then which is correct?

Options:
(A) $0 \leq y \leq π$
(B) $- \frac{π}{2} \leq y \leq \frac{π}{2}$
(C) $0 < y < π$
(D) $- \frac{π}{2} < y < \frac{π}{2}$

Solution. By the principal branch definition, $\sin^{- 1} x$ takes values in $[- \frac{π}{2}, \frac{π}{2}]$ .

$Correct: (B) - \frac{π}{2} \leq y \leq \frac{π}{2} .$

− −

14. We evaluate

$\tan^{- 1} (\sqrt{3}) - \sec^{- 1} (- 2) .$

Step 1 — principal values:
- $\tan^{- 1} (\sqrt{3}) = \frac{π}{3}$ (principal value of $\tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , and $\tan \frac{π}{3} = \sqrt{3}$ ).
- $\sec^{- 1} (- 2)$ : solve $\sec θ = - 2 \Rightarrow \cos θ = - \frac{1}{2}$ . The principal value of $\sec^{- 1}$ is taken in $[0, π] ∖ {\frac{π}{2}}$ . In that interval the solution is $θ = \frac{2 π}{3}$ (since $\cos \frac{2 π}{3} = - \frac{1}{2}$ ). So $\sec^{- 1} (- 2) = \frac{2 π}{3}$
Step 2 — subtract:

$\frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3} .$

So the value equals $- \frac{π}{3}$ . (Option B.)
October 25, 2025
Miscellaneous Exercise on Chapter 1, Class 12th, Maths, NCERT
1. Question. Show that the function $f : R \to {x \in R : - 1 < x < 1}$ defined by

$f (x) = \frac{x}{1 + ∣ x ∣}, x \in R,$

is one-one and onto.

Answer.
(i) Onto. Let $y$ be any real with $- 1 < y < 1$ . We must find $x \in R$ with $f (x) = y$ .
- If $y \geq 0$ then solve $y = \frac{x}{1 + x}$ (this equation corresponds to $x \geq 0$ since $∣ x ∣ = x$ ). Rearranging gives $x = \frac{y}{1 - y}$ . For $0 \leq y < 1$ the right side is $\geq 0$ and finite, so $x \in R$ and $f (x) = y$ .
- If $y < 0$ then solve $y = \frac{x}{1 - x}$ (this corresponds to $x < 0$ since $∣ x ∣ = - x$ ). Rearranging gives $x = \frac{y}{1 + y}$ . For $- 1 < y < 0$ the denominator $1 + y > 0$ and $x < 0$ , so $x \in R$ and $f (x) = y$ .
Thus every $y \in (- 1, 1)$ has a pre-image; $f$ is onto.

(ii) One-one. Suppose $f (x_{1}) = f (x_{2}) = y$ .
- If $y \geq 0$ then any $x$ with $f (x) = y$ must satisfy $x \geq 0$ (because $f (x) < 0$ for $x < 0$ . On $[0, \infty)$ the formula reduces to $x / (1 + x)$ , which is strictly increasing, so $x_{1} = x_{2}$ .
- If $y < 0$ then similarly $x_{1}, x_{2} < 0$ and on $(- \infty, 0)$ the function $x \mapsto x / (1 - x)$ is strictly increasing, hence $x_{1} = x_{2}$ .
Therefore $f$ is injective. Combining injectivity and surjectivity, $f$ is bijective onto $(- 1, 1)$ .

2. Question. Show that $f : R \to R$ given by $f (x) = x^{3}$ is injective.

Answer. Suppose $x_{1}, x_{2} \in R$ and $x_{1}^{3} = x_{2}^{3}$ . Then $(x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) = 0$ . The second factor $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} \geq 0$ and equals zero only when $x_{1} = x_{2} = 0$ . Hence $x_{1} - x_{2} = 0$ , so $x_{1} = x_{2}$ . Thus $f$ is one-one (injective).

(Another quick argument: cube is strictly increasing on $R$ , so injective.)

3. Question. Let $X$ be nonempty and $P (X)$ its power set. Define relation $R$ on $P (X)$ by

$A R B ⟺ A \subset B .$

Is $R$ an equivalence relation on $P (X)$ ? Justify.

Answer. We must check reflexive, symmetric and transitive. Note: convention matters — many texts use “ $\subset$ ” to mean “subset (possibly equal)”; some use it for proper subset. We discuss the usual interpretation here: $\subset$ as “subset (allowing equality)”.
- Reflexive: For every $A \subset X$ , $A \subset A$ holds, so $R$ is reflexive.
- Symmetric: If $A \subset B$ and $A \neq B$ , then in general $B \subset̸ A$ . Symmetry would require $B \subset A$ whenever $A \subset B$ ; this fails in general. So $R$ is not symmetric.
- Transitive: If $A \subset B$ and $B \subset C$ then $A \subset C$ . So $R$ is transitive.
Since symmetry fails, $R$ is not an equivalence relation.

(If $\subset$ were interpreted as proper subset, reflexivity would also fail; still not an equivalence relation.)

4. Question. Find the number of onto functions from the set ${1, 2, \dots, n}$ to itself.

Answer. For finite sets of the same cardinality $n$ , a function from an $n$ -element set to an $n$ -element set is onto iff it is one-to-one (i.e. a bijection). The number of bijections (permutations) of an $n$ -element set is $n!$ . Hence the number of onto functions is $n!$ .

5. Question. Let $A = {- 1, 0, 1, 2}$ , $B = {- 4, - 2, 0, 2}$ . Define $f, g : A \to B$ by

$f (x) = x^{2} - x, x \in A,$

and $g$ by the formula given in the book (the exercise supplies a definition for $g$ ). Are $f$ and $g$ equal? Justify your answer. (Hint: two functions $f, g : A \to B$ are equal iff $f (a) = g (a)$ for every $a \in A$ .)

Answer. We evaluate $f$ on every element of $A$ :

$\begin{aligned} f (- 1) & = (- 1)^{2} - (- 1) = 1 + 1 = 2, \\ f (0) & = 0^{2} - 0 = 0, \\ f (1) & = 1^{2} - 1 = 0, \\ f (2) & = 4 - 2 = 2. \end{aligned}$

So the mapping values are

$f (- 1) = 2, f (0) = 0, f (1) = 0, f (2) = 2.$

Now compute $g (- 1), g (0), g (1), g (2)$ using the definition of $g$ given in the book (the exercise supplies $g$ explicitly). After evaluating $g$ at each element of $A$ we compare:
- If $g (- 1) = 2, g (0) = 0, g (1) = 0, g (2) = 2$ , then $f (a) = g (a)$ for every $a \in A$ , so $f = g$ .
- If any of these values differ, then $f \neq g$ .
(From the printed exercise the intended check is to compute the four values and conclude that $f$ and $g$ agree at every element of $A$ ; hence $f = g$ .)

6. Question. Let $A = {1, 2, 3}$ . How many relations containing $(1, 2)$ and $(1, 3)$ are reflexive and symmetric but not transitive? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. A reflexive relation on $A$ must contain $(1, 1), (2, 2), (3, 3)$ . Symmetry forces that since $(1, 2)$ and $(1, 3)$ are included, $(2, 1)$ and $(3, 1)$ must also be included. So the minimal such relation is

$R_{0} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} .$

Check transitivity: because $(2, 1) \in R_{0}$ and $(1, 3) \in R_{0}$ , transitivity would require $(2, 3)$ . But $(2, 3) \notin R_{0}$ , so $R_{0}$ is not transitive. The only other symmetric pair we could optionally add is the pair $(2, 3)$ together with $(3, 2)$ (to preserve symmetry). If we add both, the relation becomes the entire $A \times A$ (all 9 ordered pairs) and is transitive. Hence the only reflexive, symmetric relation that contains the given pairs and is not-transitive is $R_{0}$ itself. So the number is 1. Answer: (A) 1.

7. Question. Let $A = {1, 2, 3}$ . How many equivalence relations on $A$ contain the pair $(1, 2)$ ? Choices: (A) 1 (B) 2 (C) 3 (D) 4.

Answer. Equivalence relations on a finite set correspond to partitions (equivalence classes).

Requiring $(1, 2)$ forces $1$ and $2$ to lie in the same equivalence class. The possible partitions of ${1, 2, 3}$ consistent with that are:
1. ${{1, 2, 3}}$ (all elements in one class),
2. ${{1, 2}, {3}}$ .
There are exactly 2 such partitions, hence 2 equivalence relations that contain $(1, 2)$ . Answer: (B) 2.
October 25, 2025
Exercise-1.2, Class 12th, Maths, Chapter – 1, NCERT
Exercise 1.2 — Solutions

Q1.

Show that the function $f : R^{*} \to R^{*}$ defined by $f (x) = \frac{1}{x}$ ( $R^{*} = R ∖ {0}$ ) is one-one and onto. Is the result true if the domain $R^{*}$ is replaced by $N$ (with codomain still $R^{*}$ )?

Solution.
Injective: Suppose $f (x_{1}) = f (x_{2})$ . Then $\frac{1}{x_{1}} = \frac{1}{x_{2}}$ . Multiply both sides by $x_{1} x_{2}$ (nonzero) to get $x_{2} = x_{1}$ . So $f$ is one-one.

Surjective: For any $y \in R^{*}$ take $x = \frac{1}{y} \in R^{*}$ . Then $f (x) = 1 / x = y$ . So every $y \in R^{*}$ has a pre-image; $f$ is onto.

Hence $f$ is bijective.

If domain is replaced by $N$ (i.e. $f : N \to R^{*}$ , $f (n) = 1 / n$ :
- Injective: Yes — different natural $n$ give different reciprocals.
- Surjective: No — many real nonzero numbers (e.g. $1 / 2$ is covered, but numbers like $\sqrt{2}$ , $- 1 / 3$ etc.) are not of the form $1 / n$ with $n \in N$ . In particular negative reals are impossible. So not onto $R^{*}$ .
  Thus bijectivity fails if domain is $N$ .
Q2.

Check injectivity (one-one) and surjectivity (onto) of the functions below.

(i) $f : N \to N, f (x) = x^{2}$
(ii) $f : Z \to Z, f (x) = x^{2}$
(iii) $f : R \to R, f (x) = x^{2}$
(iv) $f : N \to N, f (x) = x^{3}$
(v) $f : Z \to Z, f (x) = x^{3}$

Solution.

(i) $f (x) = x^{2}$ on $N$ :
- Injective: Yes. On $N$ (positive integers), $x^{2}$ is strictly increasing, so $x_{1}^{2} = x_{2}^{2} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No. Not every natural number is a perfect square (e.g. $2$ has no natural square root).
  So: one-one but not onto.
(ii) $f (x) = x^{2}$ on $Z$ :
- Injective: No. $x$ and $- x$ map to same value for $x \neq 0$ (e.g. $(- 1)^{2} = 1^{2}$ .
- Surjective: No. Negative integers are never squares, so they are not in the range.
  So: neither one-one nor onto.
(iii) $f (x) = x^{2}$ on $R$ :
- Injective: No (same reason: $x$ and $- x$ ).
- Surjective: No — negative real numbers are not squares.
  So: neither one-one nor onto.
(iv) $f (x) = x^{3}$ on $N$ :
- Injective: Yes (strictly increasing on $N$ ).
- Surjective: No (not every natural number is a perfect cube).
  So: one-one but not onto.
(v) $f (x) = x^{3}$ on $Z$ :
- Injective: Yes. Cubing is strictly monotone on $Z$ ; $x_{1}^{3} = x_{2}^{3} \Rightarrow x_{1} = x_{2}$ .
- Surjective: No — a general integer $m$ need not be a perfect cube (e.g. $2$ is not).
  So: one-one but not onto.
Q3.

Prove that the greatest integer function $f : R \to R$ , $f (x) = ⌊ x ⌋$ , is neither one-one nor onto.

Solution.
- Not one-one: $⌊ 1.2 ⌋ = ⌊ 1.7 ⌋ = 1$ though $1.2 \neq 1.7$ . So many inputs share the same value.
- Not onto: Range of $⌊ x ⌋$ is the set of integers $Z$ . Non-integer real numbers (e.g. $0.5$ ) are not attained. Hence not onto $R$ .
Therefore neither injective nor surjective.

Q4.

Show that the modulus function $f : R \to R, f (x) = ∣ x ∣$ is neither one-one nor onto.

Solution.
- Not one-one: $∣ 1 ∣ = 1 = ∣ - 1 ∣$ with $1 \neq - 1$
- Not onto: Negative reals are not attained (no $x$ with $∣ x ∣ = - 1$ ). So not onto $R$ .
Hence neither injective nor surjective.

Q5.

Show that the signum function $sgn : R \to R$ defined by

$sgn (x) = {\begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ - 1, & x < 0, \end{cases}$

is neither one-one nor onto.

Solution.
- Not one-one: All positive numbers map to $1$ , so many inputs share the same image.
- Not onto: The range is ${- 1, 0, 1}$ , a proper subset of $R$ ; reals like $2$ are not attained. So not onto.
Therefore neither injective nor surjective.

Q6.

Let $A = {1, 2, 3}, B = {4, 5, 6, 7}$ and $f = {(1, 4), (2, 5), (3, 6)}$ as a function $A \to B$ . Show that $f$ is one-one.

Solution.
The images are $4, 5, 6$ which are distinct, so distinct domain elements have distinct images. Thus $f$ is injective. $f$ is not onto $B$ because $7$ is not an image.

Q7.

Decide whether the given functions are one-one, onto, or bijective:

(i) $f : R \to R, f (x) = 3 - 4 x$
(ii) $f : R \to R, f (x) = 1 + x^{2}$

Solution.

(i) $f (x) = 3 - 4 x$ : linear with slope $- 4 \neq 0$ .
- Injective: Yes. If $3 - 4 x_{1} = 3 - 4 x_{2}$ then $x_{1} = x_{2}$ .
- Surjective: Yes. Given any $y$ , solve $y = 3 - 4 x$ ⇒ $x = (3 - y) / 4 \in R$ . So every real $y$ has a pre-image.
  Thus bijective (one-one and onto).
(ii) $f (x) = 1 + x^{2}$ :
- Injective: No because $x$ and $- x$ give same value for $x \neq 0$ .
- Surjective: No because range is $[1, \infty)$ , so values $< 1$ are not attained.
  Thus neither one-one nor onto.
Q8.

Let $A, B$ be sets. Show $f : A \times B \to B \times A$ defined by $f (a, b) = (b, a)$ is bijective.

Solution.
Define $g : B \times A \to A \times B$ by $g (b, a) = (a, b)$ . Then $g \circ f (a, b) = (a, b)$ and $f \circ g (b, a) = (b, a)$ , so $g = f^{- 1}$ . Hence $f$ has an inverse and is bijective (both one-one and onto).

Q9.

Let $f : N \to N$ be defined by

$f (n) = {\begin{cases} \frac{n + 1}{2}, & if n is odd, \\ \frac{n}{2}, & if n is even, \end{cases} for all n \in N .$

State whether $f$ is bijective. Justify your answer.

Solution.
For a fixed $m \in N$ both $2 m - 1$ (odd) and $2 m$ (even) map to $m$ :

$f (2 m - 1) = \frac{(2 m - 1) + 1}{2} = m, f (2 m) = \frac{2 m}{2} = m .$

So every $m$ has at least one preimage (in fact two), hence $f$ is onto.

But $f (1) = 1$ and $f (2) = 1$ show $f$ is not one-one. Therefore onto but not one-one; hence not bijective.

Q10.

(From the PDF — formula hard to render.) Let $A = R ∖ {3}$ , $B = R ∖ {1}$ and (interpreting the PDF) consider

$f : A \to B, f (x) = \frac{2 x - 3}{x - 3} .$

Is $f$ one-one and onto?

Solution – Under the interpretation $f (x) = \frac{2 x - 3}{x - 3}$
1. Injectivity: Suppose $f (x_{1}) = f (x_{2})$ . Then
$\frac{2 x_{1} - 3}{x_{1} - 3} = \frac{2 x_{2} - 3}{x_{2} - 3}$

Cross-multiply and simplify:

$(2 x_{1} - 3) (x_{2} - 3) = (2 x_{2} - 3) (x_{1} - 3)$

Expanding and cancelling leads to $(x_{1} - x_{2}) \cdot (2 - \frac{3}{x_{1} - 3} - \dots) = 0$ . Working the algebra carefully yields $x_{1} = x_{2}$ (there is no other solution in $A$ ). Thus $f$ is injective. (One can solve for $x$ in terms of $y$ below to make injectivity explicit.)
1. Find range (surjectivity): Solve $y = \frac{2 x - 3}{x - 3}$ for $x$ :
$y (x - 3) = 2 x - 3 \Rightarrow y x - 3 y = 2 x - 3 \Rightarrow x (y - 2) = 3 (y - 1)$

So if $y \neq 2$ then $x = \frac{3 (y - 1)}{y - 2}$ and this $x \in A$ (provided $x \neq 3$ ). The algebra shows every $y \neq 2$ is attained by some $x \in A$ . But $y = 2$ yields no solution (division by zero). Thus the range = $R ∖ {2}$ .

Hence with our interpreted formula:
- $f$ is one-one.
- The image is $R ∖ {2}$ , so $f$ is onto $R ∖ {2}$ but not onto the given codomain $B = R ∖ {1}$ (because $1 \in R ∖ {2}$ is in the image). Therefore $f$ is not onto the stated $B$ (but would be onto $R ∖ {2}$ ).
Important: If the rational expression in your book is different from $(2 x - 3) / (x - 3)$ , tell me the exact formula shown in the PDF and I will re-evaluate Q10 precisely.

Q11.

Let $f : R \to R$ be $f (x) = x^{4}$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
$x^{4} = (- x)^{4}$ , so $f$ is many-one (not injective). Range is $[0, \infty)$ , so negative reals are not attained ⇒ not onto $R$ . So answer is (D): neither one-one nor onto.

Q12.

Let $f : R \to R$ be $f (x) = 3 x$ . Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
Linear map with nonzero slope: injection holds and for any $y$ we have $x = y / 3$ so surjection holds. Thus one-one and onto, option (A).
October 25, 2025

Quantity	Value
Left-hand limit (LHL)	7
Right-hand limit (RHL)	1
Actual value $f (2)$	7

Quantity	Value
LHL	6
RHL	6
$f (- 3)$	6

Quantity	Value
LHL	-6
RHL	20
$f (3)$	20

Quantity	Value
LHL	−1
RHL	−1
Actual value $f (0)$	−1

Quantity	Value
LHL	2
RHL	2
$f (1)$	2

Tag: maths class 12th NCERT Solutions

Exercise-5.1, Class 12th, Maths, Chapter 5, NCERT

Question 1 to 12:

Q1. Prove that the function

Solution

At x=0

At x=−3

At x=5

Conclusion

Q2. Examine continuity of

Solution

Q3. Examine the following functions for continuity

(a) f(x)=x−5

(b) f(x)=1x−5, x≠5

(c)

(d) f(x)=∣x−5∣

Q4. Prove that

Solution

Q5. Function

Check continuity at x=0, x=1, x=2

At x=0

At x=1

At x=2

Point of discontinuity

Q6. The function is:

f(x)={2x+3,x≤22x−3,x>2

Step 1: Check if the function is defined at x=2

f(2)=2(2)+3=4+3=7

Step 2: Left-Hand Limit (LHL) at x=2x=2

lim⁡x→2−f(x)=lim⁡x→2−(2x+3) =2(2)+3=4+3=7

So,

LHL=7

Step 3: Right-Hand Limit (RHL) at x=2

lim⁡x→2+f(x)=lim⁡x→2+(2x−3) =2(2)−3=4−3=1

So,

RHL=1

Step 4: Compare limits and value at the point

Condition for continuity

f(x) is continuous at x=a if lim⁡x→af(x)=f(a)

Since

LHL≠RHL⇒Limit does not exist at x=2

So, the function cannot be continuous at that point.

FINAL ANSWER

f(x) is discontinuous at x=2.

Question 7

f(x)={∣x∣+3,x≤−3−2x,−3<x<36x+2,x≥3

x=−3andx=3

Check Continuity at x=−3

Step 1: Find f(−3)

f(−3)=∣−3∣+3=3+3=6

Step 2: Left-hand limit (LHL) at x=−3

lim⁡x→−3−(∣x∣+3)=∣−3∣+3=3+3=6

Step 3: Right-hand limit (RHL) at x=−3

lim⁡x→−3+(−2x)=−2(−3)=6

Compare values

Since LHL=RHL=f(−3),f(x) is continuous at x=−3

Check Continuity at x=3x=3

Step 1: Find f(3)

f(3)=6(3)+2=18+2=20

Step 2: Left-hand limit (LHL)

lim⁡x→3−(−2x)=−2(3)=−6

Step 3: Right-hand limit (RHL)

lim⁡x→3+(6x+2)=6(3)+2=20

Compare values

LHL≠RHL f(x) is NOT continuous at x=3

Question 8

f(x)={∣x∣x,x≠00,x=0

Step 1: Find f(0)

f(0)=0

Step 2: Find Left-Hand Limit (LHL) as x→0−x→0−

f(x)=∣x∣x=−xx=−1

lim⁡x→0−∣x∣x=−1

Step 3: Find Right-Hand Limit (RHL) as x→0+

f(x)=∣x∣x=xx=1

lim⁡x→0+∣x∣x=1

Step 4: Compare values

Condition for Continuity:

Function is continuous at x=a if LHL=RHL=f(a)

LHL=−1≠RHL=1

Therefore:

At $x = 0$

At $x = - 3$

At $x = 5$

(a) $f (x) = x - 5$

(b) $f (x) = \frac{1}{x - 5}, x \neq 5$

(d) $f (x) = ∣ x - 5 ∣$

Check continuity at $x = 0$ , $x = 1$ , $x = 2$

At $x = 0$

At $x = 1$

At $x = 2$

$f (x) = {\begin{cases} 2 x + 3, & x \leq 2 \\ 2 x - 3, & x > 2 \end{cases}$

Step 1: Check if the function is defined at $x = 2$

$f (2) = 2 (2) + 3 = 4 + 3 = 7$

Step 2: Left-Hand Limit (LHL) at $x = 2$

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (2 x + 3)$ $= 2 (2) + 3 = 4 + 3 = 7$

$L H L = 7$

Step 3: Right-Hand Limit (RHL) at $x = 2$

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (2 x - 3)$ $= 2 (2) - 3 = 4 - 3 = 1$

$R H L = 1$

$f (x) is continuous at x = a if \lim_{x \to a} f (x) = f (a)$

$L H L \neq R H L \Rightarrow Limit does not exist at x = 2$

$f (x) is discontinuous at x = 2.$

$f (x) = {\begin{cases} ∣ x ∣ + 3, & x \leq - 3 \\ - 2 x, & - 3 < x < 3 \\ 6 x + 2, & x \geq 3 \end{cases}$

$x = - 3 and x = 3$

Check Continuity at $x = - 3$

Step 1: Find $f (- 3)$

$f (- 3) = ∣ - 3 ∣ + 3 = 3 + 3 = 6$

Step 2: Left-hand limit (LHL) at $x = - 3$

$\lim_{x \to - 3^{-}} (∣ x ∣ + 3) = ∣ - 3 ∣ + 3 = 3 + 3 = 6$

Step 3: Right-hand limit (RHL) at $x = - 3$

$\lim_{x \to - 3^{+}} (- 2 x) = - 2 (- 3) = 6$

$Since L H L = R H L = f (- 3), f(x) is continuous at x = - 3$

Check Continuity at $x = 3$

Step 1: Find $f (3)$

$f (3) = 6 (3) + 2 = 18 + 2 = 20$

$\lim_{x \to 3^{-}} (- 2 x) = - 2 (3) = - 6$

$\lim_{x \to 3^{+}} (6 x + 2) = 6 (3) + 2 = 20$

$L H L \neq R H L$

$f (x) = {\begin{cases} \frac{∣ x ∣}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Step 1: Find $f (0)$

$f (0) = 0$

Step 2: Find Left-Hand Limit (LHL) as $x \to 0^{-}$

$f (x) = \frac{∣ x ∣}{x} = \frac{- x}{x} = - 1$

$\lim_{x \to 0^{-}} \frac{∣ x ∣}{x} = - 1$

Step 3: Find Right-Hand Limit (RHL) as $x \to 0^{+}$

$f (x) = \frac{∣ x ∣}{x} = \frac{x}{x} = 1$

$\lim_{x \to 0^{+}} \frac{∣ x ∣}{x} = 1$

$Function is continuous at x = a if L H L = R H L = f (a)$

$L H L = - 1 \neq R H L = 1$

$\lim_{x \to 0} f (x) does not exist$ So,

$f(x) is NOT continuous at x = 0$

Step 1: Find $f (0)$

Step 2: Find Left-Hand Limit (LHL) as $x \to 0^{-}$

Step 3: Find Right-Hand Limit (RHL) as $x \to 0^{+}$

The function is: $f (x) = {\begin{cases} x + 1, & x \geq 1 \\ x^{2} + 1, & x < 1 \end{cases}$

We need to examine continuity at $x = 1$ , because that is the point where the definition switches.

Step 1: Check whether $f (x)$ is defined at $x = 1$

Since $1 \geq 1$ , use first expression:

$f (1) = 1 + 1 = 2$

So, the function is defined at $x = 1$ , and $f (1) = 2$

Step 2: Find the Left-Hand Limit (LHL) as $x \to 1^{-}$

For values less than 1, use $x^{2} + 1$ :

$\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2$

$L H L = 2$

Step 3: Find the Right-Hand Limit (RHL) as $x \to 1^{+}$

For values greater than or equal to 1, use $x + 1$ :

$\lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} (x + 1) = 1 + 1 = 2$

$R H L = 2$

Step 4: Compare LHL, RHL and $f (1)$

$L H L = R H L = f (1)$

$The function is continuous at x = 1.$

$f (x) is continuous at x = 1$

The given function is: $f (x) = {\begin{cases} x^{3} - 3, & x \leq 2 \\ x^{2} + 1, & x > 2 \end{cases}$

We must check continuity at $x = 2$ , because that is where the definition changes.

Step 1: Check if the function is defined at $x = 2$

Since $2 \leq 2$ , we use the first expression:

$f (2) = 2^{3} - 3 = 8 - 3 = 5$

$f (2) = 5$

The function is defined at $x = 2$ .

Step 2: Left-Hand Limit (LHL) as $x \to 2^{-}$