Question 13.

Is the function defined by

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

a continuous function?

Answer

To check the continuity of the function at $x=1$, evaluate the following:

Value of the function at $x=1$

Since $x\le 1$:

$$f(1)=1+5=6$$

Left-hand limit (LHL) as $x\to 1^{-}$

$$\underset{x\to 1^{-}}{\lim }(x+5)=6$$

Right-hand limit (RHL) as $x\to 1^{+}$

$$\underset{x\to 1^{+}}{\lim }(x-5)=-4$$

Comparison

$$\text{LHL}=6,\text{RHL}=-4,f(1)=6$$

Since

$$\text{LHL}\mathrm{\ne }\text{RHL}$$

Final Answer

The function is not continuous at $x=1$ because the left-hand limit and right-hand limit are not equal.

---

Discuss the continuity of the function f, where f is defined by

Question 14.

Discuss the continuity of the function

$$f(x)=\{\begin{array}{ll}3,& \text{if }0\le x\le 1\\ 4,& \text{if }1<x<3\\ 5,& \text{if }3\le x\le 10\end{array}$$

Answer

To check continuity, we examine the points where the definition of the function changes:
at $x=1$ and $x=3$.

Continuity on the intervals

• On $[0,1]$: $f(x)=3$ (a constant function → continuous).

• On $(1,3)$: $f(x)=4$ (constant function → continuous).

• On $[3,10]$: $f(x)=5$ (constant function → continuous).

So the only possible discontinuities are at the endpoints where the pieces join.

---

Check continuity at $x=1$

Left-hand limit (as $x\to 1^{-}$)

$$\underset{x\to 1^{-}}{\lim }f(x)=3$$

Right-hand limit (as $x\to 1^{+}$)

$$\underset{x\to 1^{+}}{\lim }f(x)=4$$

Value of the function

$$f(1)=3$$

Since

$$\underset{x\to 1^{-}}{\lim }f(x)=3\text{and}\underset{x\to 1^{+}}{\lim }f(x)=4$$

Left-hand limit ≠ Right-hand limit
→ function is discontinuous at $x=1$

---

Check continuity at $x=3$

Left-hand limit (as $x\to 3^{-}$)

$$\underset{x\to 3^{-}}{\lim }f(x)=4$$

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

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

Value of the function

$$f(3)=5$$

Since

$$4\mathrm{\ne }5$$

→ function is discontinuous at $x=3$.

---

Final Conclusion

The function $f(x)$ is:

• Continuous on each open interval $(0,1)$, $(1,3)$, and $(3,10)$
  

• Discontinuous at the points $x=1$ and $x=3$

---

Question 15

Discuss the continuity of the function $f$ defined by:

$$f(x)=\{\begin{array}{ll}2x,& \text{if }x<0\\ 0,& \text{if }0<x<1\\ 4x,& \text{if }x>1\end{array}$$

Answer

This is a piecewise function, and we need to check continuity at the points where the definition changes, i.e., at $x=0$ and $x=1$.

---

Continuity for $x<0$, $0<x<1$, and $x>1$

In each interval, the function is a polynomial (linear function), and polynomials are continuous everywhere.
So, $f(x)$ is continuous within each interval.

---

Check continuity at $x=0$

Left-hand limit (LHL) as $x\to 0^{-}$

Using $2x$:

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

Right-hand limit (RHL) as $x\to 0^{+}$

Using 0:

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

Value of the function at $x=0$

Notice: The function definition does not include $x=0$ in any case, so:

$$f(0)\text{ does not exist}$$

Conclusion at $x=0$

Even though

$$\text{LHL}=\text{RHL}=0,$$

the value of the function at $x=0$ is not defined.
So, the function is not continuous at $x=0$.

---

Check continuity at $x=1$

Left-hand limit (LHL) as $x\to 1^{-}$

Using 0:

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

Right-hand limit (RHL) as $x\to 1^{+}$

Using $4x$:

$$\underset{x\to 1^{+}}{\lim }f(x)=4(1)=4$$

Value of the function at $x=1$

Not defined in any case, so:

$$f(1)\text{ does not exist}$$

Conclusion at $x=1$

Since

$$\text{LHL}=0\mathrm{\ne }4=\text{RHL}$$

The limit does not exist. Therefore, the function is not continuous at $x=1$.

Final Conclusion

The function $f(x)$ is:

• Continuous within each open interval $(-\mathrm{\infty },0)$, $(0,1)$, and $(1,\mathrm{\infty })$

• Not continuous at $x=0$ (because $f(0)$ is not defined)

• Not continuous at $x=1$ (because left-hand and right-hand limits are not equal and $f(1)$ is not defined)

---

Question 16

Discuss the continuity of the function $f(x)$, where

$$f(x)=\{\begin{array}{ll}-2,& \text{if }x\le -1\\ 2x,& \text{if }-1<x\le 1\\ 2,& \text{if }x>1\end{array}$$

Answer:

To check continuity, we must examine the possible points of discontinuity—here the function changes its definition at $x=-1$ and $x=1$.
So, we check continuity at these points.

---

Continuity at $x=-1$

Value of the function at $x=-1$

Since $x\le -1$:

$$f(-1)=-2$$

Left-hand limit (LHL) as $x\to -1^{-}$

Using $-2$:

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

Right-hand limit (RHL) as $x\to -1^{+}$

Using $2x$:

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

Conclusion

$$LHL=RHL=f(-1)=-2$$

So, the function is continuous at $x=-1$.

---

Continuity at $x=1$

Value of the function at $x=1$

Using $2x$:

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

Left-hand limit (LHL) as $x\to 1^{-}$

Using $2x$:

$$\underset{x\to 1^{-}}{\lim }2x=2$$

Right-hand limit (RHL) as $x\to 1^{+}$

Using constant $2$:

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

Conclusion

$$LHL=RHL=f(1)=2$$

Thus, the function is continuous at $x=1$.

---

Final Conclusion

Since the function is continuous at both points where it changes its definition ($x=-1$ and $x=1$), and there are no other breaks, gaps, or jumps:

$$\boxed{{\displaystyle \text{The function }f(x)\text{ is continuous for all real values of }x\mathrm{.}}}$$

Question 17

Find the relationship between $a$ and $b$ so that the function $f(x)$ defined by

$$f(x)=\{\begin{array}{ll}ax+1,& \text{if }x\le 3\\ bx+3,& \text{if }x>3\end{array}$$

is continuous at $x=3$.

Answer

To ensure continuity at $x=3$, we require:

$$\text{Left-hand limit}=\text{Right-hand limit}=f(3)$$

Value of the function at $x=3$

Using the first expression since $x\le 3$:

$$f(3)=a(3)+1=3a+1$$

Left-hand Limit (LHL) as $x\to 3^{-}$

$$\underset{x\to 3^{-}}{\lim }f(x)=3a+1$$

Right-hand Limit (RHL) as $x\to 3^{+}$

Using $bx+3$:

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

Condition for continuity

$$3a+1=3b+3$$

Solving

$$3a-3b=2$$
$$a-b=\frac{2}{3}$$

Final Answer

$$\boxed{{\displaystyle a-b=\frac{2}{3}}}$$

This is the required relationship between $a$ and $b$ for $f(x)$ to be continuous at $x=3$.

---

Question 18

For what value of $\lambda$ is the function defined by

$$f(x)=\{\begin{array}{ll}\lambda (x^2-2x),& \text{if }x\le 0\\ 4x+1,& \text{if }x>0\end{array}$$

continuous at $x=0$? What about at $x=1$?

Answer

Checking continuity at $x=0$

Value of the function at $x=0$

Since $x\le 0$:

$$f(0)=\lambda (0^2-2\cdot 0)=\lambda (0)=0$$

Left-hand limit (LHL)

$$\underset{x\to 0^{-}}{\lim }\lambda (x^2-2x)=\lambda (0-0)=0$$

Right-hand limit (RHL)

$$\underset{x\to 0^{+}}{\lim }(4x+1)=4(0)+1=1$$

Condition for continuity

$$LHL=RHL=f(0)$$

So,

$$0=1$$

This statement is never true, so no real value of $\lambda$ can make the function continuous at $x=0$.

---

Checking continuity at $x=1$

At $x=1$, the value is taken from the second piece $4x+1$, and there is no matching left-hand expression at 1, since the first part ends at $x=0$.
Thus, the function is not defined in a neighborhood around 1 from the left side.

Therefore, the function cannot be continuous at $x=1$.

---

Final Conclusion

$$\boxed{{\displaystyle \text{There is no value of }\lambda \text{ that makes the function continuous at }x=0.}}$$
$$\boxed{{\displaystyle \text{The function is not continuous at }x=1\text{ because continuity cannot be checked.}}}$$

---

Question 19

Show that the function $g(x)=x-[x]$ is discontinuous at all integral points.
Here $[x]$ denotes the greatest integer less than or equal to $x$.

Answer

Given Function

$$g(x)=x-[x]$$

The expression $x-[x]$ represents the fractional part of $x$, denoted as $\{x\}$.
Thus,

$$g(x)=\{x\}$$

The fractional part function always satisfies:

$$0\le \{x\}<1$$

To show discontinuity at integer points

Let $x=n$, where $n$ is any integer.

---

Left-hand limit (LHL) as $x\to n^{-}$

When $x$ approaches $n$ from the left, $x=n-h$ where $h\to 0^{+}$.
Then the greatest integer function gives:

$$[x]=n-1$$

So,

$$g(x)=x-[x]=(n-h)-(n-1)=1-h$$

Taking limit,

$$\underset{x\to n^{-}}{\lim }g(x)=\underset{h\to 0^{+}}{\lim }(1-h)=1$$

Right-hand limit (RHL) as $x\to n^{+}$

When approaching from the right, $x=n+h$, $h\to 0^{+}$, then

$$[x]=n$$

So,

$$g(x)=(n+h)-n=h$$

Taking limit,

$$\underset{x\to n^{+}}{\lim }g(x)=\underset{h\to 0^{+}}{\lim }h=0$$

---

Value of the function at $x=n$

$$g(n)=n-[n]=n-n=0$$

Comparison

$$\underset{x\to n^{-}}{\lim }g(x)=1,\underset{x\to n^{+}}{\lim }g(x)=0,g(n)=0$$

Since,

$$\underset{x\to n^{-}}{\lim }g(x)\mathrm{\ne }\underset{x\to n^{+}}{\lim }g(x)$$

The limit does not exist at $x=n$, and therefore, the function is discontinuous at every integer $n$.

Final Conclusion

$$\boxed{{\displaystyle g(x)=x-[x]\text{ is discontinuous at all integral points.}}}$$

---

Question 20

Is the function defined by

$$f(x)=x^2-\sin x+5$$

continuous at $x=\pi$?

Answer

The function $f(x)=x^2-\sin x+5$ is composed of the following functions:

• $x^2$ → a polynomial function (continuous for all real $x$)

• $\sin x$ → a trigonometric function (continuous for all real $x$)

• Constant $5$ → continuous everywhere

The sum or difference of continuous functions is also continuous for all real numbers.

Therefore, $f(x)$ is continuous everywhere, including at $x=\pi$.

---

Check using limits

Value at $x=\pi$:

$$f(\pi )={\pi }^2-\sin \pi +5={\pi }^2-0+5={\pi }^2+5$$

Limit as $x\to \pi$:

$$\underset{x\to \pi }{\lim }f(x)=\underset{x\to \pi }{\lim }(x^2-\sin x+5)={\pi }^2-0+5={\pi }^2+5$$

Comparison

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

So the function is continuous at $x=\pi$.

Final Answer

$$\boxed{{\displaystyle \text{Yes, the function is continuous at }x=\pi \mathrm{.}}}$$

---

Question 21

Discuss the continuity of the following functions:

(a) $f(x)=\sin x+\cos x$
(b) $f(x)=\sin x-\cos x$
(c) $f(x)=\sin x\cdot \cos x$

---

Answer

To discuss continuity, recall that:

• $\sin x$ and $\cos x$ are continuous functions for all real values of $x$.

• Sum, difference, and product of continuous functions are also continuous.

So, we analyze each function:

(a) $f(x)=\sin x+\cos x$

• $\sin x$ is continuous for all $x\in \mathbb{R}$

• $\cos x$ is continuous for all $x\in \mathbb{R}$

The sum of two continuous functions is continuous.

Conclusion

$$\boxed{{\displaystyle f(x)=\sin x+\cos x\text{ is continuous for all real }x\mathrm{.}}}$$

(b) $f(x)=\sin x-\cos x$

• Difference of continuous functions remains continuous.

Conclusion

$$\boxed{{\displaystyle f(x)=\sin x-\cos x\text{ is continuous for all real }x\mathrm{.}}}$$

(c) $f(x)=\sin x\cdot \cos x$

• Product of two continuous functions is continuous.

Conclusion

$$\boxed{{\displaystyle f(x)=\sin x\cdot \cos x\text{ is continuous for all real }x\mathrm{.}}}$$

---

Question 22

Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Answer

To discuss the continuity of these trigonometric functions, recall:

• A function is continuous at a point if the limit exists and equals the value of the function at that point.

• Discontinuity occurs where the function is not defined.

---

Cosine Function

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

• $\cos x$ is defined for all real values of $x$.

• It is smooth and has no breaks or gaps.

Conclusion

$$\boxed{{\displaystyle \cos x\text{ is continuous for all }x\in \mathbb{R}\mathrm{.}}}$$

---

Cosecant Function

$$f(x)=\csc x=\frac{1}{\sin x}$$

• $\csc x$ is not defined where $\sin x=0$.

• $\sin x=0$ at $x=n\pi$, where $n$ is any integer.

Conclusion

$$\boxed{{\displaystyle \csc x\text{ is discontinuous at }x=n\pi \mathrm{.}}}$$
$$\boxed{{\displaystyle \text{It is continuous for all other real values of }x\mathrm{.}}}$$

---

Secant Function

$$f(x)=\sec x=\frac{1}{\cos \mathrm{x<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

• $\sec x$ is not defined where $\cos x=0$.

• $\cos x=0$ at $x=\frac{(2n+1)\pi }{2}$, where $n$ is any integer.

Conclusion

$$\boxed{{\displaystyle \sec x\text{ is discontinuous at }x=\frac{(2n+1)\pi }{2}\mathrm{.}}}$$
$$\boxed{{\displaystyle \text{It is continuous everywhere else.}}}$$

Cotangent Function

$$f(x)=\cot x=\frac{\cos x}{\sin x}$$

• $\cot x$ is not defined where $\sin x=0$.

• Similar to cosecant, discontinuity occurs at $x=n\pi$.

Conclusion

$$\boxed{{\displaystyle \cot x\text{ is discontinuous at }x=n\pi \mathrm{.}}}$$
$$\boxed{{\displaystyle \text{It is continuous for all other real values of }x\mathrm{.}}}$$

---

Question 23

Find all points of discontinuity of

$$f(x)=\{\begin{array}{ll}{\displaystyle \frac{\sin x}{x}},& \text{if }x<0\\ x+1,& \text{if }x\ge 0\end{array}$$

Answer

The function changes its definition at $x=0$, so discontinuity (if any) must be checked at $x=0$.

Step 1: Compute limit from the left side as $x\to 0^{-}$

Consider

$$\underset{x\to 0^{-}}{\lim }\frac{\sin x}{x}$$

We know the standard limit identity:

$$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$$

So,

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

Step 2: Compute right-hand limit as $x\to 0^{+}$

From the second part $f(x)=x+1$ when $x\ge 0$:

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

Step 3: Value of the function at $x=0$

Since $x\ge 0$,

$$f(0)=0+1=1$$

Comparison

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

Since all three are equal:

$$\underset{x\to 0}{\lim }f(x)=f(0)$$

Final Conclusion

$$\boxed{{\displaystyle \text{The function is continuous at }x=0.}}$$

There are no other points where the definition changes, and both components of the function are continuous in their respective intervals.

$$\boxed{{\displaystyle \text{Therefore, }f(x)\text{ is continuous for all }x\mathrm{.}}}$$

---

Question 24

Determine if the function defined by

$$f(x)=\{\begin{array}{ll}{x}^2\sin \left(\frac{1}{x}\right),& \text{if }x\mathrm{\ne }0\\ 0,& \text{if }x=0\end{array}$$

is continuous at $x=0$.

---

Answer

To check continuity at $x=0$, we need to verify:

$$\underset{x\to 0}{\lim }f(x)=f(0)$$

Given:

$$f(0)=0$$

Find the limit of $f(x)$ as $x\to 0$

$$f(x)=x^2\sin \left(\frac{1}{x}\right)$$

We know that:

$$-1\le \sin \left(\frac{1}{x}\right)\le 1$$

Multiplying the entire inequality by $x^2\ge 0$:

$$-x^2\le x^2\sin \left(\frac{1}{x}\right)\le x^2$$

Now take the limit as $x\to 0$:

$$\underset{x\to 0}{\lim }-x^2=0\text{and}\underset{x\to 0}{\lim }{x}^2=0$$

By Squeeze (Sandwich) Theorem:

$$\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{1}{x}\right)=0$$

Therefore:

$$\underset{x\to 0}{\lim }f(x)=0=f(0)$$

Final Conclusion

$$\boxed{{\displaystyle \text{Yes, the function }f(x)\text{ is continuous at }x=0.}}$$

---