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

Question 25

Examine the continuity of the function

$$f(x)=\{\begin{array}{ll}\sin x-\cos x,& \text{if }x\mathrm{\ne }0\\ -1,& \text{if }x=0\end{array}$$

Answer

We are given the function:

$$f(x)=\{\begin{array}{ll}\sin x-\cos x,& x\mathrm{\ne }0\\ -1,& x=0\end{array}$$

We need to examine the continuity at $x=0$.

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Step 1: Value of the function at $x=0$

$$f(0)=-1$$

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Step 2: Find the limit of $f(x)$as $x\to 0$

$$\underset{x\to 0}{\lim }(\sin x-\cos x)$$

We know standard limits:

$$\sin 0=0,\cos 0=1$$

So,

$$\underset{x\to 0}{\lim }(\sin x-\cos x)=0-1=-1$$

Step 3: Compare Limit and Function Value

Both values are equal.

Conclusion

Since:

$$\underset{x\to 0}{\lim }f(x)=f(0)$$The function $f(x)$ is continuous at $x=0$.

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Question 26

Find the value of $k$ so that the function $f$ is continuous at $x=\frac{\pi }{2}$:

$$f(x)=\{\begin{array}{ll}{\displaystyle \frac{k\cos x}{\pi -2x}},& x\mathrm{\ne }\frac{\pi }{2}\\ 3,& x=\frac{\pi }{2}\end{array}$$

Solution

For continuity at $x=\frac{\pi }{2}$, we need:

$$\underset{x\to \frac{\pi }{2}}{\lim }f(x)=f\left(\frac{\pi }{2}\right)$$

Given:

$$f\left(\frac{\pi }{2}\right)=3$$

Now compute the limit:

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{k\cos x}{\pi -2x}$$

Substitute $x=\frac{\pi }{2}$:

$$\frac{k\cos \left(\frac{\pi }{2}\right)}{\pi -2\left(\frac{\pi }{2}\right)}=\frac{k(0)}{\mathrm{0<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

This is the indeterminate form $\frac{0}{0}$, so apply L’Hôpital’s Rule.

Differentiate numerator and denominator:

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{-k\sin x}{-2}=\underset{x\to \frac{\pi }{2}}{\lim }\frac{k\sin x}{2}$$

Now substitute $x=\frac{\pi }{2}$:

$$\frac{k\sin \left(\frac{\pi }{2}\right)}{2}=\frac{k(1)}{2}=\frac{k}{2}$$

For continuity:$$\frac{k}{2}=3$$
$$k=6$$

Final Answer

$$\boxed{{\displaystyle k=6}}$$

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Question 27

Find the value of $k$ so that the function $f$ is continuous at $x=2$:

$$f(x)=\{\begin{array}{ll}k{x}^2,& x\le 2\\ 3,& x>2\end{array}$$

Solution

For continuity at $x=2$:

$$\underset{x\to 2}{\lim }f(x)=f(2)$$

Step 1: Value at the point

$$f(2)=k(2^2)=4k$$

Step 2: Limit as $x\to 2$

Since the function changes definition at $x=2$, we use left-hand limit (LHL) and right-hand limit (RHL):

Left-hand limit (LHL)

$$\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{-}}{\lim }k{x}^2=k(2^2)=4k$$

Right-hand limit (RHL)

$$\underset{x\to 2^{+}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }3=3$$

Condition for continuity

$$\text{LHL}=\text{RHL}=f(2)$$

So,$$4k=3$$

$$k=\frac{3}{4}$$

Final Answer

$$\boxed{{\displaystyle k=\frac{3}{4}}}$$

Thus, the function is continuous at $x=2$ when $k=\frac{3}{4}$.

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Question 28

Find the value of $k$ so that the function $f$ is continuous at $x=\pi$:

$$f(x)=\{\begin{array}{ll}kx+1,& x\le \pi \\ \cos x,& x>\pi \end{array}$$

Solution

For continuity at $x=\pi$, we require:

$$\underset{x\to \pi }{\lim }f(x)=f(\pi )$$

Step 1: Value of the function at $x=\pi$

Since $x=\pi$ falls in the first part $(x\le \pi )$:

$$f(\pi )=k\pi +1$$Step 2: Left-hand limit (LHL)

$$\underset{x\to {\pi }^{-}}{\lim }f(x)=\underset{x\to {\pi }^{-}}{\lim }(kx+1)=k\pi +1$$

Step 3: Right-hand limit (RHL)

$$\underset{x\to {\pi }^{+}}{\lim }f(x)=\underset{x\to {\pi }^{+}}{\lim }\cos x=\cos \pi$$
$$\cos \pi =-1$$

Condition for Continuity

$$\text{LHL}=\text{RHL}=f(\pi )$$
$$k\pi +1=-1$$

Solve for $k$

$$k\pi =-1-1=-2$$

$$k=\frac{-2}{\pi }$$

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Question 29

Find the value of $k$ so that the function $f$ is continuous at $x=5$:

$$f(x)=\{\begin{array}{ll}kx+1,& x\le 5\\ 3x-5,& x>5\end{array}$$

Solution

For continuity at $x=5$, we need:

$$\underset{x\to 5}{\lim }f(x)=f(5)$$

Step 1: Function value at the point

Since $x\le 5$:

$$f(5)=k(5)+1=5k+1$$

Step 2: Left-hand limit (LHL) as $x\to 5^{-}$

$$\underset{x\to 5^{-}}{\lim }f(x)=\underset{x\to 5^{-}}{\lim }(kx+1)=5k+1$$

Step 3: Right-hand limit (RHL) as $x\to 5^{+}$

$$\underset{x\to 5^{+}}{\lim }f(x)=\underset{x\to 5^{+}}{\lim }(3x-5)=3(5)-5=15-5=10$$

Step 4: Apply continuity condition

$$\text{LHL}=\text{RHL}=f(5)$$
$$5k+1=10$$

$$5k=9$$

$$k=\frac{9}{5}$$

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Question 30

Find the values of $a$ and $b$ such that the function

$$f(x)=\{\begin{array}{ll}5,& x\le 2\\ ax+b,& 2<x<10\\ 21,& x\ge 10\end{array}$$

is continuous.

Solution

For continuity, the function must be continuous at:

• $x=2$ (where left part meets the middle part)

• $x=10$ (where middle part meets the right part)

Continuity at $x=2$

$$\underset{x\to 2^{-}}{\lim }f(x)=f(2)=5$$
$$\underset{x\to 2^{+}}{\lim }f(x)=a(2)+b=2a+b$$

For continuity:

$$2a+b=5\text{(Equation 1)}$$

Continuity at $x=10$

$$\underset{x\to {10}^{-}}{\lim }f(x)=a(10)+b=10a+b$$
$$f(10)=21$$

For continuity:

$$10a+b=21\text{(Equation 2)}$$

Solve Equations (1) and (2)

$$2a+b=5$$
$$10a+b=21$$

Subtract (1) from (2):

$$(10a+b)-(2a+b)=21-5$$
$$8a=16$$

$$a=2$$

Substitute $a=2$ into Equation (1):

$$2(2)+b=5$$

$$4+b=5$$

$$b=1$$

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Question 31

Show that the function

$$f(x)=\cos (x^2)$$

is a continuous function.

Solution

We know that:

• $x^2$ is a polynomial → continuous everywhere

• $\cos x$ is a continuous function for all real numbers

A composition of two continuous functions is also continuous.

Here,

$$f(x)=\cos (x^2)=\cos [g(x)]\text{where }g(x)=x^2$$

Since both $g(x)$ and $\cos x$ are continuous, therefore:

$$f(x)=\cos (x^2)\text{ is continuous for all }x\in \mathbb{R}$$

Answer:

$$\boxed{{\displaystyle \cos (x^2)\text{ is continuous everywhere.}}}$$

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Question 32

Show that the function

$$f(x)=\mathrm{\mid }\cos x\mathrm{\mid }$$

is a continuous function.

Solution

• $\cos x$ is continuous for all real numbers

• The absolute value function $\mathrm{\mid }x\mathrm{\mid }$ is also continuous

• The composition of continuous functions is continuous

$$f(x)=\mathrm{\mid }\cos x\mathrm{\mid }=\mathrm{\mid }g(x)\mathrm{\mid },\text{where }g(x)=\cos x$$

Since both $g(x)$ and $\mathrm{\mid }x\mathrm{\mid }$ are continuous,

$$f(x)=\mathrm{\mid }\cos x\mathrm{\mid }\text{ is continuous for all }x$$

Answer:

$$\boxed{{\displaystyle \mathrm{\mid }\cos x\mathrm{\mid }\text{ is continuous for all real }x\mathrm{.}}}$$

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Question 33

Examine whether

$$f(x)=\sin \mathrm{\mid }x\mathrm{\mid }$$

is a continuous function.

Solution

• $\mathrm{\mid }x\mathrm{\mid }$ is continuous for all real numbers

• $\sin x$ is continuous for all real numbers

• Composition of continuous functions is continuous

Thus:

$$f(x)=\sin (\mathrm{\mid }x\mathrm{\mid })=\sin (g(x)),g(x)=\mathrm{\mid }x\mathrm{\mid }$$

Since both are continuous,

$$\sin \mathrm{\mid }x\mathrm{\mid }\text{ is continuous for all }x$$

Answer:

$$\boxed{{\displaystyle \sin \mathrm{\mid }x\mathrm{\mid }\text{ is continuous everywhere.}}}$$

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Question 34

Find all the points of discontinuity of

$$f(x)=\mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }x+1\mathrm{\mid }$$

Solution

Check where each absolute expression changes form.

Break points occur where inside values become zero:

$$x=0\text{and}x=-1$$

So evaluate function in intervals:

Case 1: $x<-1$

$$\mathrm{\mid }x\mathrm{\mid }=-x,\mathrm{\mid }x+1\mathrm{\mid }=-(x+1)$$
$$f(x)=-x-(-(x+1))=-x+x+1=1$$

Case 2: $-1\le x<0$

$$\mathrm{\mid }x\mathrm{\mid }=-x,\mathrm{\mid }x+1\mathrm{\mid }=x+1$$
$$f(x)=-x-(x+1)=-2x-1$$

Case 3: $x\ge 0$

$$\mathrm{\mid }x\mathrm{\mid }=x,\mathrm{\mid }x+1\mathrm{\mid }=x+1$$
$$f(x)=x-(x+1)=-1$$

Check continuity at critical points

At $x=-1$

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

So continuous at $x=-1$

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At $x=0$

$$\underset{x\to 0^{-}}{\lim }f(x)=-2(0)-1=-1$$
$$\underset{x\to 0^{+}}{\lim }f(x)=-1$$
$$f(0)=-1$$

So continuous at $x=0$

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Final Answer

$$\boxed{{\displaystyle \text{The function }f(x)=\mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }x+1\mathrm{\mid }\text{ is continuous everywhere.}}}$$