Solution: Given that $f(x)$ is differentiable at each $x \in R$ For $x \leq 1$ $f(x)=x^{2}+3 x+a$ i.e. a polynomial for $x>1$ $f(x)=b x+2$, which is also a polynomial As, a polynomial function is...

### If show that

Solution: The given function is $f(x)=|x|$ $\text{Left Hand Derivative}$ at $\mathrm{x}=2$ $\begin{array}{l} \operatorname{Lim}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0}...

### Let Show that is not derivable at .

Solution: The given function $f(x)=\left\{\begin{array}{l}(2+x), \text { if } x \geq 0 ; \\ (2-x), i f x<0\end{array}\right.$ Left Hand Derivative at $x=0$ : $\begin{array}{l}...

### Show that function

Solution: The given function $f(x)=\left\{\begin{array}{l}(1-x), \text { when } x<1 \\ \left(x^{2}-1\right), \text { when } x \geq 1\end{array}\right.$ L.H.L. at $\mathrm{x}=1$ :...

### Show that is neither continuous nor derivable at

Solution: L.H.L. at $x=2$ $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[2-h]=\lim _{h \rightarrow 0} 1=1$ R.H.L. at $x=2$ $\lim _{x \rightarrow 2^{+}}...

### Let Show that is continuous but not differentiable at

Solution: L.H.L. at $x=1$ $\operatorname{Lim}_{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x=1$ $f(x)=x$ is polynomial function and a polynomial function is continuous everywhere R.H.L. at...

### Show that is continuous but not differentiable at

Solution: L.H.L. at $x=5$ $\lim _{x \rightarrow 5^{-}}|x-5|=\lim _{x \rightarrow 5}(5-x)=0$ R.H.L. at $\mathrm{x}=5$ $\lim _{x \rightarrow 5^{+}}|x-5|=\lim _{x \rightarrow 5}(x-5)=0$ Also...

### Show that constant function is always differentiable.

Solution: Asumme that $a$ is any constant number. Therefore, $f(x)=a$ $\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0}...

### Show that is not differentiable at

Solution: The given function $f(x)=(x-1)^{1 / 3}$ $\mathrm{Left Hand Derivative}$ at $\mathrm{x}=1$ $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{h...

### Show that is continuous as well as differentiable at

Solution: It is given that: $f(x)=x^{3}$ If a function is differentiable at a point, it is necessary that it is continuous at that point. Left hand derivative at $x=3$ $\begin{array}{l} \lim _{x...

### Discuss the continuity of the function in the interval of

Solution: The given function $f(x)=|x|+|x-1|$ A function $f(x)$ is said to be continuous on a closed interval $[a, b]$ if and only if, (i) $\mathrm{f}$ is continuous on the open interval...

### Locate the point of discontinuity of the function

Solution: The given function $f(x)=\left\{\begin{array}{c}\left(x^{3}-x^{2}+2 x-2\right), \text { if } x \neq 1 \\ 4, \text { if } x=1\end{array}\right.$ L.H.L. at $\mathrm{x}=1: \lim _{\mathrm{x}...

### Show that is continuous at each point except 0 .

Solution: The given function is $f(x)=\left\{\begin{array}{l}x, \text { if } x \neq 0 \\ 1, \text { if } x=0\end{array}\right.$ L.H.L. at $\mathrm{x}=0$ $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h...

### Show that is continuous.

Solution: The given function $f(x)=\left\{\begin{array}{c}(2 x-1), \text { if } x<2 \\ \frac{3 x}{2}, \text { if } x \geq 2\end{array}\right.$ L.H.L. at $x=2$ $\lim _{x \rightarrow 2^{-}}...

### Discuss the continuity of

Solution: Assume that $n$ be any integer $[\mathrm{x}]=$ Greatest integer less than or equal to $x$ Some values of $[x]$ for specific values of $x$ $\begin{array}{l} {[3]=3} \\ {[4.4]=4} \\...

### Show that function is continuous.

Solution: It is known that $\sin x$ is continuous everywhere Now considering the point $x=0$ L.H.L: $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{\sin x}{x}\right)=\lim...

### Show that sec is a continuous function.

Solution: Assume $f(x)=\sec |x|$ and a be any real number. Then, L.H.L. at $\mathrm{x}=\mathrm{a}$ $\lim _{x \rightarrow \mathrm{a}^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow...

### Show that sec is a continuous function.

Solution: Assume $f(x)=\sec x$ So, $f(x)=\frac{1}{\cos x}$ $f(x)$ is not defined when $\cos x=0$ And $\cos x=0$ when, $x=\frac{\pi}{2}$ and odd multiples of $\frac{\pi}{2}$ like $-\frac{\pi}{2}$...

### Show that function

Solution: It is given that: $f(x)=\left\{\begin{array}{c} \frac{x^{n}-1}{x-1}, \text { when } x \neq 1 \\ n, \text { when } x=1 \end{array}\right.$ L.H.L. and $\mathrm{x}=1$ $\begin{array}{l} \lim...

### Show that function is continuous.

Solution: It is given that: $f(x)=\left\{\begin{array}{c} \sin x, \text { if } x<0 \\ x, \text { if } x \geq 0 \end{array}\right.$ L.H.L. at $\mathrm{x}=0$ $\lim _{x \rightarrow 0^{-}}...

### Show that function is continuous function.

Solution: It is given that: $f(x)=\left\{\begin{array}{l} (7 x+5), \text { when } x \geq 0 \\ (5-3 x), \text { when } x<0 \end{array}\right.$ Let us now calculate the limit of $f(x)$ when $x$...

### Prove that the function given is continuous but not differentiable at

Solution: $f(x)=|x-3|$ As every modulus function is continuous for all real $x, f(x)$ is continuous at $x=3$. $f(x)=f(x)=\left\{\begin{array}{l} 3-x, x<0 \\ x-3, x \geq 0 \end{array}\right.$ In...

### Find the values of a and b such that the following functions , defined as at

Solution: $: \mathrm{f}$ is continuous at $x=0$ $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0+} f(x)$ $\lim _{x \rightarrow 0-}\left(\operatorname{asin} \frac{\pi}{2}(x+1)\right)=\lim _{x...

### Find the values of a and b such that the following functions continuous.

Solution: $f$ is continuous at $x=2$ $\begin{array}{l} \lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2+} f(x)=f(2) \\ \lim _{x \rightarrow 2-}(5)=\lim _{x \rightarrow 2+}[a x+b]=5 \\...

### Show that: is continuous at

Solution: Left Hand Limit: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 1-} x^{3}-3$ $=5$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^{2}+1$ $=5$...

### Show that: is continuous at

Solution: $\begin{array}{l} : \mathrm{Left Hand Limit}: \lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 1-} \mathrm{x}^{2}+1 \\ =2 \end{array}$...

### Show that function:

Solution: $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{2} \sin \frac{1}{x}$ As $\lim _{x \rightarrow 0} x^{2}=0$ and $\sin \left(\frac{1}{x}\right)$ is bounded function between $-1$ and...

### For what valve of is the following function Ans.

Solution: $f$ is continuous at $x=\frac{\pi}{2}$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow-\frac{\pi}{r}} f(x)=f\left(\frac{\pi}{2}\right) \\ \Rightarrow \lim _{x \rightarrow \frac{\pi}{r}}...

### For what value of is the following function is continuous at Ans.

Solution: As, $f(x)$ is continuous at $x=3$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=f(3) \\ \Rightarrow \lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}=f(3) \\...

### For what value of is the following function continuous at

Solution: As, $f(x)$ is continuous at $x=2$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 2-} 2 x+1=\lim _{x \rightarrow 2^{*}} 3 x-1=f(2) \\ \Rightarrow \lim _{x \rightarrow 2^{-}} 2 x+1=f(2)...

### Find the value of for which

; is continuous at x=-1

Solution: As, $f(x)$ is continuous at $x=0$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow-1} \frac{x^{2}-2 x-3}{x+1}=f(0) \\ \Rightarrow \lim _{x \rightarrow-1} \frac{(x-3)(x+1)}{x+1}=\lambda...

### Find the value of for which

Solution: As, $f(x)$ is continuous at $x=0$ $\begin{array}{l} \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 2 \mathrm{x}}{5 \mathrm{x}}=\mathrm{f}(0) \\ \Rightarrow \frac{1}{5} \lim _{\mathrm{x}...

### Prove that

Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=-x \\ =0 \end{array}$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2 *} x$ $=0$...

### Prove that

is discontinuous at x=2

Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{-}} 2 \mathrm{x}$ $=4$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 2^{-}}...

### Prove that

Solution: $\lim _{x \rightarrow 0} \sin \frac{1}{x}=0$ $\sin _{\mathrm{x}}^{1}$ is bounded function between $-1$ and $+1$ Also, $f(0)=0$ As, $\lim _{x \rightarrow 0} f(x)=f(0)$ As a result,...

### Prove that is discontinuous at x=0

Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 0-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0-2} \frac{1}{2}(\mathrm{x}-|\mathrm{x}|)$ $=\lim _{x \rightarrow 0-2}...

### Prove that

Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow \mathrm{a}-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \mathrm{a}-} \frac{|\mathrm{x}-\mathrm{a}|}{\mathrm{x}-\mathrm{a}}$ $=\lim...

### Prove that

Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} \cos x \\ =1 \end{array}$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 0_{-}}...

### Prove that is continuous at

Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{x}-1$ $=1$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim...

### Prove that

Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1-} 5 x-4 \\ =1 \end{array}$ Right Hand Limit: $\lim _{\mathrm{x} \rightarrow 1^{*}}...

### Prove that

Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} 3-x \\ =3 \end{array}$ $\begin{array}{l} \text { Right Hand Limit: } \lim _{x...

### Prove that is discontinuous at

Solution : Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 2-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2-} 2+\mathrm{x}$ $=4$ Right Hand Limit: $\lim _{x \rightarrow 2^{-}} f(x)=\lim...

### Prove that

Solution: $\begin{array}{l} \text { Left Hand Limit: } \lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0-} \frac{1-\operatorname{cosx}}{x^{2}} \\ =\lim _{x \rightarrow 0-} \frac{2 \sin ^{2}...

### Prove that

Solution: Left Hand Limit: $\lim _{\mathrm{x} \rightarrow 0-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0-} \frac{\sin 3 \mathrm{x}}{\mathrm{x}}$ $=3$ $\left[\lim _{x \rightarrow a}...

### Prove that

Solution: Left Hand Limit: $\lim _{x \rightarrow 5-} f(x)=\lim _{x \rightarrow 5-} \frac{x^{2}-25}{x-5}$ $=\lim _{x \rightarrow 5-} \frac{(x+5)(x-5)}{x-5}$ [By middle term splitting]...

### Prove that

Solution: Left Hand Limit: $\lim _{x \rightarrow 3-} f(x)=\lim _{x \rightarrow 3-} \frac{x^{2}-x-6}{x-3}$ $=\lim _{\mathrm{x} \rightarrow 3-} \frac{(\mathrm{x}+2)(\mathrm{x}-3)}{\mathrm{x}-3}$ [By...

### Show that is continuous at

Solution: L.H.L.: $\lim _{\mathrm{x} \rightarrow 1-} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{x}^{2}+3 \mathrm{x}+4$ $=7$ R.H.L.: $\lim _{x \rightarrow 1^{-}} f(x)=\lim...

### Show that is continues at

Solution: L.H.L.: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-} x^{2}$ $=4$ R.H.L.: $\lim _{\mathrm{x} \rightarrow 2^{*}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{*}}...

### Verify Mean Value Theorem if in the interval where and . Find all for which .

Solution: (a) Function is a polynomial as polynomial function is always continuous. As a result,it is continuous in $[1,3]$ (b) $f^{\prime}(x)=3 x^{2}-10 x_{5} f^{\prime}(x)$ exists in $[1,3]$, as a...