Continuity and Differentiability

### Find the values of and so that the function is differentiable at each

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...

### 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 $\mathrm{x}=1$ :...

### 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...

### 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 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...

### 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...

### 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,...