CBSE Class 11 Maths Revision Notes Chapter 13

Q.1

$\begin{array}{l}\mathbf{\text{Find the limit :}}\\ \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{x}}\end{array}$

Ans

$\begin{array}{l}\text{We have,}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{2{\mathrm{x}}^2-\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}(2\mathrm{x}-1)}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\frac{1}{2\mathrm{x}-1}\\ =1\times \frac{1}{2\left(0\right)-1}\left[\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\right]\\ =-1\end{array}$

Q.2

$\begin{array}{l}\mathbf{\text{Free the limit:}}\\ \underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\left[\frac{x^3-4x^2+4x}{x^2-4}\right]\end{array}$

Ans

$\begin{array}{l}\text{We have,}\\ \underset{\mathrm{x}\to 2}{\lim }\left[\frac{{\mathrm{x}}^3-4{\mathrm{x}}^2+4\mathrm{x}}{{\mathrm{x}}^2-4}\right]=\underset{\mathrm{x}\to 2}{\lim }\left[\frac{\mathrm{x}\left({\mathrm{x}}^2-4\mathrm{x}+4\right)}{{\mathrm{x}}^2-4}\right]\\ =\underset{\mathrm{x}\to 2}{\lim }\left[\frac{\mathrm{x}(\mathrm{x}-2)(\mathrm{x}-2)}{(\mathrm{x}-2)(\mathrm{x}+2)}\right]\\ =\underset{\mathrm{x}\to 2}{\lim }\left[\frac{\mathrm{x}(\mathrm{x}-2)}{(\mathrm{x}+2)}\right]\\ =\frac{2(2-2)}{(2+2)}\\ =\frac{0}{4}=0\end{array}$

Q.3

$\mathbf{\text{Compute}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{7}}{\mathbf{lim}}(x+7)$

Ans

$\underset{x\to 7}{\lim }\left(x+7\right)=7+7=14.$

Q.4

$\mathbf{\text{Compute}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\frac{{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{8}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 2}{\lim }\frac{{\mathrm{x}}^3+8}{{\mathrm{x}}^2-1}=\frac{2^3+8}{2^2-1}\\ =\frac{8+8}{4-1}=\frac{16}{3}.\end{array}$

Q.5

$\mathbf{\text{Find the limit:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}}{{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{9}\mathbf{x}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{x}=3,\mathrm{the}\mathrm{expression}\mathrm{reduces}\mathrm{to}\frac{0}{0}.\\ \text{Now,}\\ \underset{\mathrm{x}\to 3}{\lim }\frac{{\mathrm{x}}^2-9}{{\mathrm{x}}^3+9\mathrm{x}-6{\mathrm{x}}^2}=\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}-3)(\mathrm{x}+3)}{\mathrm{x}\left({\mathrm{x}}^2+9-6\mathrm{x}\right)}=\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}-3)(\mathrm{x}+3)}{\mathrm{x}{(\mathrm{x}-3)}^2}\\ =\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}+3)}{\mathrm{x}(\mathrm{x}-3)}=\frac{3+3}{3(3-3)}=\frac{6}{0}=\infty \\ \text{which is not defined.}\end{array}$

Q.6

$\mathbf{\text{Prove that for any positive integer}}\mathbf{n}\mathbf{\text{,}}\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\frac{{\mathbf{x}}^{\mathbf{n}}\mathbf{-}{\mathbf{a}}^{\mathbf{n}}}{\mathbf{x}\mathbf{-}\mathbf{a}}\mathbf{=}{\mathbf{na}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{.}$

Ans

$\begin{array}{l}{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}=\left(\mathrm{x}-\mathrm{a}\right)\left({\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{\mathrm{n}-2}\mathrm{a}+{\mathrm{x}}^{\mathrm{n}-3}{\mathrm{a}}^2+\dots +{\mathrm{xa}}^{\mathrm{n}-2}+{\mathrm{a}}^{\mathrm{n}-1}\right)\\ \Rightarrow \underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}=\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left({\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{\mathrm{n}-2}\mathrm{a}+{\mathrm{x}}^{\mathrm{n}-3}{\mathrm{a}}^2+\dots +{\mathrm{xa}}^{\mathrm{n}-2}+{\mathrm{a}}^{\mathrm{n}-1}\right)\\ ={\mathrm{a}}^{\mathrm{n}-1}+{\mathrm{a}}^{\mathrm{n}-2}\mathrm{a}+{\mathrm{a}}^{\mathrm{n}-3}{\mathrm{a}}^2++{\mathrm{aa}}^{\mathrm{n}-2}+{\mathrm{a}}^{\mathrm{n}-1}\\ ={\mathrm{a}}^{\mathrm{n}-1}+{\mathrm{a}}^{\mathrm{n}-1}+\dots +{\mathrm{a}}^{\mathrm{n}-1}+{\mathrm{a}}^{\mathrm{n}-1}\\ ={\mathrm{na}}^{\mathrm{n}-1}\end{array}$

Q.7

$\mathbf{\text{Compute:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\sqrt{\mathbf{1}\mathbf{+}\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

Ans

$\begin{array}{l}\text{Put}\mathrm{y}=1+\mathrm{x},\text{so that when}\mathrm{x}\to 0\text{then}\mathrm{y}\to 1\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{1+\mathrm{x}}-1}{\mathrm{x}}\underset{\mathrm{y}\to 1}{\lim }\frac{\sqrt{\mathrm{y}}-1}{\mathrm{y}-1}=\underset{\mathrm{y}\to 1}{\lim }\frac{{\mathrm{y}}^{\frac{1}{2}}-1^{\frac{1}{2}}}{\mathrm{y}-1}\\ =\frac{1}{2}{\left(1\right)}^{\frac{1}{2}-1}=\frac{1}{2}\left(\text{as}\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}={\mathrm{na}}^{\mathrm{n}-1}\right)\end{array}$

Q.8

$\mathbf{\text{Compute:}}\underset{\mathbf{r}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{2}\mathbf{\pi r}$

Ans

$\underset{\mathrm{r}\to 1}{\lim }2\mathrm{\pi r}=2\mathrm{\pi }\times 1=2\mathrm{\pi }$

Q.9

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{6}\mathbf{x}}{\mathbf{sin}\mathbf{3}\mathbf{x}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 6\mathrm{x}}{\sin 3\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\left[\frac{\sin 6\mathrm{x}}{6\mathrm{x}}\times \frac{3\mathrm{x}}{\sin 3\mathrm{x}}\times 2\right]\\ =2\cdot \frac{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 6\mathrm{x}}{6\mathrm{x}}}{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\\ =2\cdot \frac{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 6\mathrm{x}}{6\mathrm{x}}}{\underset{3\mathrm{x}\to 0}{\lim }\frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\\ =2\cdot \frac{1}{1}=2\left(\text{as}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\right)\end{array}$

Q.10

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{lim}}\frac{\mathbf{tan}\mathbf{2}\mathbf{x}}{\mathbf{x}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}}\mathbf{.}$

Ans

$\begin{array}{l}\text{When we substitute}\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)\text{as}\mathrm{y}\text{, then}\mathrm{y}\to 0\text{as}\mathrm{x}\to \frac{\mathrm{\pi }}{2}\\ \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\tan 2\mathrm{x}}{\mathrm{x}-\frac{\mathrm{\pi }}{2}}=\underset{\mathrm{y}\to 0}{\lim }\frac{\tan 2\left(\frac{\mathrm{\pi }}{2}-\mathrm{y}\right)}{-\mathrm{y}}\\ =\underset{\mathrm{y}\to 0}{\lim }\frac{\tan (\mathrm{\pi }-2\mathrm{y})}{-\mathrm{y}}=\underset{\mathrm{y}\to 0}{\lim }\frac{-\tan \left(2\mathrm{y}\right)}{-\mathrm{y}}=\underset{\mathrm{y}\to 0}{\lim }\frac{\tan \left(2\mathrm{y}\right)}{\mathrm{y}}\\ =\underset{\mathrm{y}\to 0}{\lim }\frac{\sin \left(2\mathrm{y}\right)}{\mathrm{y}\cos \left(2\mathrm{y}\right)}=\underset{2\mathrm{y}\to 0}{\lim }\frac{\sin \left(2\mathrm{y}\right)}{2\mathrm{y}}\times \frac{2}{\cos \left(2\mathrm{y}\right)}\\ =1\times \frac{2}{1}=2\end{array}$

Q.11

$\mathbf{\text{Find}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{f(x)wheref(x)}}\mathbf{=}\{\frac{1}{\left|x\right|},x\ne 0\}\mathbf{.}$

Ans

$\begin{array}{l}\text{It is given that}=\left\{\frac{1}{\left|\mathrm{x}\right|},\mathrm{x}\ne 0\right\}\\ \text{We observe that when}\mathrm{x}\text{approaches to zero from left side i.e.}\\ \mathrm{x}\text{is negative and close to zero then}\left|\mathrm{x}\right|\text{is close to zero close to zero and positive.}\\ \text{Therefore}1/\left|\mathrm{x}\right|\text{is large and positive.}\\ \underset{\mathrm{x}\to 0^{-}}{\lim }\frac{1}{\left|\mathrm{x}\right|}\to \infty \\ \text{Similarly if}\mathrm{x}\text{approaches to zero from right side, i.e.,}\\ \mathrm{x}\text{is positive and close to zero then}\left|\mathrm{x}\right|\text{is close to zero and positive.}\\ \text{Therefore}1/\left|\mathrm{x}\right|\text{is large and positive.}\\ \underset{\mathrm{x}\to 0^{+}}{\lim }\frac{1}{\left|\mathrm{x}\right|}\to \infty \\ \text{Therefore limits exists and it tends to}\infty \text{.}\end{array}$

Q.12

$\mathbf{\text{Compute:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{x}\mathbf{sec}\mathbf{x}\mathbf{.}$

Ans

$\underset{\mathrm{x}\to 0}{\lim }\mathrm{x}\sec \mathrm{x}=0\times \sec 0=0\times 1=0$

Q.13

$\begin{array}{l}\mathbf{\text{If}}\mathbf{f}\left(x\right)\mathbf{=}\mathbf{\{}\begin{array}{cc}{\mathbf{mx}}^{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{,}& \mathbf{x}\mathbf{<}\mathbf{0}\\ \mathbf{nx}\mathbf{+}\mathbf{m}\mathbf{,}& \mathbf{0}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{1}\\ {\mathbf{nx}}^{\mathbf{3}}\mathbf{+}\mathbf{m}\mathbf{,}& \mathbf{x}\mathbf{>}\mathbf{1}\end{array}\\ \mathbf{\text{For what values of}}\mathbf{m}\mathbf{\text{and}}\mathbf{n}\mathbf{\text{do}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{and}}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{both exist.}}\end{array}$

Ans

$\begin{array}{l}\text{It is given thatf}\left(\text{x}\right)=\left\{\begin{array}{cc}{\mathrm{mx}}^2+\mathrm{n},& \mathrm{x}<0\\ \mathrm{nx}+\mathrm{m},& 0\le \mathrm{x}\le 1\\ {\mathrm{nx}}^3+\mathrm{m},& \mathrm{x}>1\end{array}\right.\\ \underset{\mathrm{x}\to 0}{\lim }\text{f(x)and}\underset{\mathrm{x}\to 1}{\lim }\text{f(x)both exist.}\\ \iff \underset{\mathrm{x}\to 0^{-}}{\lim }\text{f(x)}=\underset{\mathrm{x}\to 0^{+}}{\lim }\text{f(x)}\mathrm{and}\underset{\mathrm{x}\to 1^{-}}{\lim }\text{f(x)}=\underset{\mathrm{x}\to 1^{+}}{\lim }\text{f(x)}\\ \iff \underset{\mathrm{h}\to 0}{\lim }\mathrm{f}\text{(0}-\text{h)}=\underset{\mathrm{h}\to 0}{\lim }\mathrm{f}\text{(0 + h)and}\underset{\mathrm{h}\to 0}{\lim }\text{f(1}-\text{h)}=\underset{\mathrm{h}\to 0}{\lim }\text{f(1+ h)}\\ \iff \underset{\mathrm{h}\to 0}{\lim }\left\{\mathrm{m}{(-\mathrm{h})}^2+\mathrm{n}\right\}=\underset{\mathrm{h}\to 0}{\lim }\text{{nh + m}}\mathrm{and}\underset{\mathrm{h}\to 0}{\lim }\text{{n(1}-\text{h) + m}}=\underset{\mathrm{h}\to 0}{\lim }\left\{{\text{n (1 + h)}}^{\text{3}}\text{+ m}\right\}\\ \iff \mathrm{n}=\mathrm{m}\mathrm{and}\mathrm{n}+\mathrm{m}=\mathrm{n}+\mathrm{m}\\ \text{Hence}\underset{\mathrm{x}\to 0}{\lim }\text{f(x)and}\underset{\mathrm{x}\to 1}{\lim }\text{f(x)both exist for}\mathrm{n}=\mathrm{m}.\end{array}$

Q.14 Compute the derivative of sin x w.r.t. x by first principle.

Ans

$\begin{array}{l}\text{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sinx}\text{.}\\ \text{Then,}\\ \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\sin (\mathrm{x}+\mathrm{h})-\sin \left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{2\cos \left(\frac{2\mathrm{x}+\mathrm{h}}{2}\right)\sin \left(\frac{\mathrm{h}}{2}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\cos \left(\mathrm{x}+\frac{\mathrm{h}}{2}\right)\cdot \underset{\mathrm{h}\to 0}{\lim }\frac{\sin \left(\frac{\mathrm{h}}{2}\right)}{\frac{\mathrm{h}}{2}}=\cos \mathrm{x}\cdot 1=\cos \mathrm{x}\end{array}$

Q.15 Compute the derivative of cos²x.

Ans

$\begin{array}{l}\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x}\cdot \cos \mathrm{x})\\ =\left(\cos \mathrm{x}\right)‘\cos \mathrm{x}+\cos \mathrm{x}\left(\cos \mathrm{x}\right)‘\left(\mathrm{By}\mathrm{using}\mathrm{product}\mathrm{rule}\right)\\ =(-\sin \mathrm{x})\cos \mathrm{x}+\cos \mathrm{x}(-\sin \mathrm{x})\\ =-2\sin \mathrm{x}\cos \mathrm{x}\\ =-\sin 2\mathrm{x}\end{array}$

Q.16

$\begin{array}{l}\mathbf{\text{Find the derivative of :}}\\ \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}(5y^4-2y^2+1)(y-1)\end{array}$

Ans

$\begin{array}{l}\text{(a)}\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}-4{\mathrm{x}}^{2}3\\ \Rightarrow \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\left(3\mathrm{x}-4{\mathrm{x}}^2\right)‘\\ =3-4\cdot 2\mathrm{x}=3-8\mathrm{x}\\ \text{(b)}\mathrm{f}\left(\mathrm{y}\right)=\left(5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\right)\left(\mathrm{y}-1\right)\\ \text{Using the product rule,}\\ \frac{\mathrm{df}\left(\mathrm{y}\right)}{\mathrm{dy}}=\frac{\mathrm{d}}{\mathrm{dy}}\left[\left(5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\right)(\mathrm{y}-1)\right]\\ =\left(5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\right)‘(\mathrm{y}-1)+\left(5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\right)(\mathrm{y}-1)‘\\ =\left(20{\mathrm{y}}^3-4\mathrm{y}\right)(\mathrm{y}-1)+\left(5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\right)\left(1\right)\\ =20{\mathrm{y}}^4-20{\mathrm{y}}^3-4{\mathrm{y}}^2+4\mathrm{y}+5{\mathrm{y}}^4-2{\mathrm{y}}^2+1\\ =25{\mathrm{y}}^4-20{\mathrm{y}}^3-6{\mathrm{y}}^2+4\mathrm{y}+1\end{array}$

Q.17 Find the first derivative of (13t – 1) (5t³ + 2) w.r.t.t.

Ans

$\begin{array}{l}\frac{\mathrm{df}\left(\mathrm{t}\right)}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[(13\mathrm{t}-1)\left(5{\mathrm{t}}^3+2\right)\right]\\ =\left(13\mathrm{t}-1\right)‘\left(5{\mathrm{t}}^3+2\right)+(13\mathrm{t}-1)\left(5{\mathrm{t}}^3+2\right)‘\left(\text{Using the product rule}\right)\\ =13\left(5{\mathrm{t}}^3+2\right)+(13\mathrm{t}-1)\left(15{\mathrm{t}}^2\right)\\ =65{\mathrm{t}}^3+26+195{\mathrm{t}}^3-15{\mathrm{t}}^2\\ =260{\mathrm{t}}^3-15{\mathrm{t}}^2+26\end{array}$

Q.18

$\mathbf{\text{Find the first derivative of}}\frac{\mathbf{x}\mathbf{+}\mathbf{a}}{\mathbf{x}\mathbf{-}\mathbf{a}}\mathbf{\text{by using First Principle.}}$

Ans

$\begin{array}{l}\text{Let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}-\mathrm{a}}.\\ \text{Then,}\\ \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\left(\frac{\mathrm{x}+\mathrm{h}+\mathrm{a}}{\mathrm{x}+\mathrm{h}-\mathrm{a}}\right)-\left(\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}-\mathrm{a}}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{(\mathrm{x}+\mathrm{h}+\mathrm{a})(\mathrm{x}-\mathrm{a})-(\mathrm{x}+\mathrm{a})(\mathrm{x}+\mathrm{h}-\mathrm{a})}{\mathrm{h}(\mathrm{x}+\mathrm{h}-\mathrm{a})(\mathrm{x}-\mathrm{a})}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\mathrm{x}}^2+\mathrm{hx}+\mathrm{ax}-\mathrm{ax}-\mathrm{ah}-{\mathrm{a}}^2\right)-\left({\mathrm{x}}^2+\mathrm{ax}+\mathrm{hx}+\mathrm{ah}-\mathrm{ax}-{\mathrm{a}}^2\right)}{\mathrm{h}(\mathrm{x}+\mathrm{h}-\mathrm{a})(\mathrm{x}-\mathrm{a})}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\mathrm{x}}^2+\mathrm{hx}+\mathrm{ax}-\mathrm{ax}-\mathrm{ah}-{\mathrm{a}}^2\right)-\left({\mathrm{x}}^2+\mathrm{ax}+\mathrm{hx}+\mathrm{ah}-\mathrm{ax}-{\mathrm{a}}^2\right)}{\mathrm{h}(\mathrm{x}+\mathrm{h}-\mathrm{a})(\mathrm{x}-\mathrm{a})}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{{\mathrm{x}}^2+\mathrm{hx}+\mathrm{ax}-\mathrm{ax}-\mathrm{ah}-{\mathrm{a}}^2-{\mathrm{x}}^2-\mathrm{ax}-\mathrm{hx}-\mathrm{ah}+\mathrm{ax}+{\mathrm{a}}^2}{\mathrm{h}(\mathrm{x}+\mathrm{h}-\mathrm{a})(\mathrm{x}-\mathrm{a})}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{-2\mathrm{ah}}{\mathrm{h}(\mathrm{x}+\mathrm{h}-\mathrm{a})(\mathrm{x}-\mathrm{a})}\\ =\frac{-2\mathrm{a}}{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{a})}=\frac{-2\mathrm{a}}{{\mathrm{x}}^2-2\mathrm{ax}+{\mathrm{a}}^2}\end{array}$

Q.19 Compute the derivative of √x from the first principle.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\text{and}\mathrm{f}(\mathrm{x}+\mathrm{h})=\sqrt{(\mathrm{x}+\mathrm{h})}\\ \text{Then by first principle}\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{(\mathrm{x}+\mathrm{h})}-\sqrt{\mathrm{x}}}{\mathrm{h}}\times \frac{\sqrt{(\mathrm{x}+\mathrm{h})}+\sqrt{\mathrm{x}}}{\sqrt{(\mathrm{x}+\mathrm{h})}+\sqrt{\mathrm{x}}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{(\mathrm{x}+\mathrm{h})-\mathrm{x}}{\mathrm{h}(\sqrt{(\mathrm{x}+\mathrm{h})}+\sqrt{\mathrm{x}})}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{h}}{\mathrm{h}(\sqrt{(\mathrm{x}+\mathrm{h})}+\sqrt{\mathrm{x}})}\\ =\frac{1}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}}}\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\frac{1}{2\sqrt{\mathrm{x}}}.\end{array}$

Q.20

$\mathbf{\text{Compute the first derivative of}}\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{tanx}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+\mathrm{tanx}}\\ \text{Using quotient rule for differentiation,}\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\frac{(1+\mathrm{tanx}){\left(\mathrm{x}\right)}^{‘}-\mathrm{x}{(1+\mathrm{tanx})}^{‘}}{{(1+\mathrm{tanx})}^2}\\ =\frac{(1+\mathrm{tanx})\left(1\right)-\mathrm{x}\left({\sec }^2\mathrm{x}\right)}{{(1+\mathrm{tanx})}^2}\\ =\frac{1+\mathrm{tanx}-{\mathrm{xsec}}^2\mathrm{x}}{{(1+\mathrm{tanx})}^2}\end{array}$

Q.21

$\mathbf{\text{Compute the first derivative of}}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{cos}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{)}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \mathrm{x}}{\cos (\mathrm{x}+\mathrm{a})}\\ \text{Using the quotient rule for differentiation we get}\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\frac{\cos (\mathrm{x}+\mathrm{a}){\left(\mathrm{sinx}\right)}^{‘}-{\left\{\cos \right(\mathrm{x}+\mathrm{a}\left)\right\}}^{‘}\sin \mathrm{x}}{{\cos }^2(\mathrm{x}+\mathrm{a})}\\ =\frac{\cos (\mathrm{x}+\mathrm{a})\cdot \mathrm{cosx}-\{-\sin (\mathrm{x}+\mathrm{a}\left)\right\}\cdot \mathrm{sinx}}{{\cos }^2(\mathrm{x}+\mathrm{a})}\\ =\frac{\mathrm{cosx}\cdot \cos (\mathrm{x}+\mathrm{a})+\mathrm{sinx}\cdot \sin (\mathrm{x}+\mathrm{a})}{{\cos }^2(\mathrm{x}+\mathrm{a})}\\ =\frac{\cos (\mathrm{x}+\mathrm{a}-\mathrm{x})}{{\cos }^2(\mathrm{x}+\mathrm{a})}\\ =\frac{\cos \mathrm{a}}{{\cos }^2(\mathrm{x}+\mathrm{a})}\end{array}$

Q.22 Find the derivative of the function f(x) = 2x² + 3x – 3 at x = –2 by first principal method.

Ans

$\begin{array}{l}\text{Wefindthederivativeofthefunctionat}\mathrm{x}=-2.\\ \text{Wehave}\\ {\mathrm{f}}^{‘}(-2)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(-2+\mathrm{h})-\mathrm{f}(-2)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\left[2{(-2+\mathrm{h})}^2+3(-2+\mathrm{h})-3\right]-\left[2{(-2)}^2+3(-2)-3\right]}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\left[2\left(4+{\mathrm{h}}^2-4\mathrm{h}\right)-6+3\mathrm{h}-3\right]-[8-6-3]}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\left[8+2{\mathrm{h}}^2-8\mathrm{h}-6+3\mathrm{h}-3-8+6+3\right]}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{2{\mathrm{h}}^2-5\mathrm{h}}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\lim }2\mathrm{h}-5=2\left(0\right)-5=-5\end{array}$

Q.23

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\mathrm{}\frac{1-\cos 2\mathrm{x}}{\mathrm{x}}=\mathrm{}\underset{\mathrm{x}\to 0}{\lim }\mathrm{}\frac{2{\sin }^2\mathrm{x}}{\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\mathrm{}\left(2\sin \mathrm{x}\times \frac{\sin \mathrm{x}}{\mathrm{x}}\right)\\ =2\times 0\times 1=0\end{array}$

Q.24 Find the derivative of f(x) = x²cosx by first principle.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2\cos \mathrm{x}.\text{Then,}\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)={\left(\mathrm{x}+\mathrm{h}\right)}^2\cos \left(\mathrm{x}+\mathrm{h}\right)\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{(\mathrm{x}+\mathrm{h})}^2\cos (\mathrm{x}+\mathrm{h})-{\mathrm{x}}^2\mathrm{cosx}}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\left({\mathrm{x}}^2+2\mathrm{xh}+{\mathrm{h}}^2\right)\cos (\mathrm{x}+\mathrm{h})-{\mathrm{x}}^2\cos \mathrm{x}}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2\left\{\cos \right(\mathrm{x}+\mathrm{h})-\mathrm{cosx}\}+\mathrm{h}(2\mathrm{x}+\mathrm{h})\cos (\mathrm{x}+\mathrm{h})}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2\left\{\cos \right(\mathrm{x}+\mathrm{h})-\mathrm{cosx}\}}{\mathrm{h}}+\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{h}(2\mathrm{x}+\mathrm{h})\cos (\mathrm{x}+\mathrm{h})}{\mathrm{h}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2\left(-2\sin \frac{\mathrm{h}}{2}\cdot \sin \frac{2\mathrm{x}+\mathrm{h}}{2}\right)}{\mathrm{h}}+\underset{\mathrm{x}\to 0}{\lim }\left(2\mathrm{x}+\mathrm{h}\right)\cos \left(\mathrm{x}+\mathrm{h}\right)\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\times -{\mathrm{x}}^2\sin \frac{2\mathrm{x}+\mathrm{h}}{2}+2\mathrm{x}\cos \mathrm{x}\\ =1\times -{\mathrm{x}}^2\sin \mathrm{x}+2\mathrm{x}\cos \mathrm{x}\\ =-{\mathrm{x}}^2\sin \mathrm{x}+2\mathrm{x}\cos \mathrm{x}\end{array}$

Q.25

$\mathbf{\text{Compute the first derivative of}}\frac{\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}}}{\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{1+\frac{1}{\mathrm{x}}}{1-\frac{1}{\mathrm{x}}}=\frac{\frac{\mathrm{x}+1}{\mathrm{x}}}{\frac{\mathrm{x}-1}{\mathrm{x}}}=\frac{\mathrm{x}+1}{\mathrm{x}-1}\\ \text{Using quotient rule for differentiation,}\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\frac{(\mathrm{x}-1){(\mathrm{x}+1)}^{‘}-{(\mathrm{x}-1)}^{‘}(\mathrm{x}+1)}{{(\mathrm{x}-1)}^2}\\ =\frac{(\mathrm{x}-1)\left(1\right)-\left(1\right)(\mathrm{x}+1)}{{(\mathrm{x}-1)}^2}\\ =\frac{\mathrm{x}-1-\mathrm{x}-1}{{(\mathrm{x}-1)}^2}=\frac{-2}{{(\mathrm{x}-1)}^2},\mathrm{x}\ne 1\end{array}$

Q.26

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sinx}\mathbf{-}\mathbf{2}\mathbf{sin}\mathbf{\left(}\mathbf{3}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{5}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}-2\sin \left(3\mathrm{x}\right)+\sin \left(5\mathrm{x}\right)}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}-\sin \left(3\mathrm{x}\right)-\sin \left(3\mathrm{x}\right)+\sin \left(5\mathrm{x}\right)}{\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{[\mathrm{sinx}-\sin (3\mathrm{x}\left)\right]+\left[\sin \right(5\mathrm{x})-\sin (3\mathrm{x}\left)\right]}{\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{-2\sin \mathrm{x}\cos \left(2\mathrm{x}\right)+2\sin \mathrm{x}\cos \left(4\mathrm{x}\right)}{\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2\sin \mathrm{x}\left[\cos \right(4\mathrm{x})-\cos (2\mathrm{x}\left)\right]}{\mathrm{x}}\\ =2\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)\cdot \underset{\mathrm{x}\to 0}{\lim }\left[\cos \right(4\mathrm{x})-\cos (2\mathrm{x}\left)\right]\\ =2\times 1\times 0=0\end{array}$

Q.27

$\mathbf{\text{Prove that}}\frac{\mathbf{d}}{\mathbf{dx}}\left(x^n\right)\mathbf{=}{\mathbf{nx}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{.}$

Ans

$\begin{array}{l}\text{We can prove it with the help of first principle.}\\ \text{We have}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{n}}\\ \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{{(\mathrm{x}+\mathrm{h})}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}}}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\lim }\frac{{(\mathrm{x}+\mathrm{h})}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}}}{(\mathrm{x}+\mathrm{h})-\mathrm{x}}\\ =\underset{\mathrm{z}\to \mathrm{x}}{\lim }\frac{{\mathrm{z}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}}}{\mathrm{z}-\mathrm{x}}\text{…(i)}\\ \text{where}\mathrm{z}=\mathrm{x}+\mathrm{h}\text{and}\mathrm{z}\to \mathrm{x}\text{as}\mathrm{h}\to 0\\ \text{Using}\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{{\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}={\mathrm{na}}^{\mathrm{n}-1}\text{in(i)we can conclude that}\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1}\\ \text{Hence,}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1}.\end{array}$

Q.28

$\mathbf{\text{Differentiate with respect to}}\mathbf{x}\mathbf{\text{:}}\frac{{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}}{\sqrt{\mathbf{x}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}}{\sqrt{\mathrm{x}}}\right]=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\mathrm{ax}}^2}{\sqrt{\mathrm{x}}}+\frac{\mathrm{bx}}{\sqrt{\mathrm{x}}}+\frac{\mathrm{c}}{\sqrt{\mathrm{x}}}\right]\\ =\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{ax}}^{\frac{3}{2}}+{\mathrm{bx}}^{\frac{1}{2}}+{\mathrm{cx}}^{\frac{-1}{2}}\right]\\ =\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{ax}}^{\frac{3}{2}}\right]+\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{bx}}^{\frac{1}{2}}\right]+\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{cx}}^{\frac{-1}{2}}\right]\\ =\mathrm{a}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\frac{3}{2}}\right]+\mathrm{b}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\frac{1}{2}}\right]+\mathrm{c}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\frac{-1}{2}}\right]\\ =\mathrm{a}\left(\frac{3}{2}{\mathrm{x}}^{\frac{1}{2}}\right)+\mathrm{b}\left(\frac{1}{2}{\mathrm{x}}^{\frac{-1}{2}}\right)+\mathrm{c}\left(-\frac{1}{2}{\mathrm{x}}^{\frac{-3}{2}}\right)\\ =\frac{3\mathrm{a}}{2}{\mathrm{x}}^{\frac{1}{2}}+\frac{\mathrm{b}}{2}{\mathrm{x}}^{\frac{-1}{2}}-\frac{\mathrm{c}}{2}{\mathrm{x}}^{\frac{-3}{2}}.\end{array}$

Q.29 Find the first derivative of x³ – 4 at x = 2.

Ans

f(x) = x³ – 4
f'(x) = 3x²
f'(2) = 3 × 2² = 3 × 4 = 12

Q.30

$\mathbf{\text{For the function,}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{100}}}{\mathbf{100}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{99}}}{\mathbf{99}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{\text{Prove that}}\mathbf{f}\mathbf{‘}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{f}\mathbf{‘}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{100}}{100}+\frac{{\mathrm{x}}^{99}}{99}+\dots \dots .+\frac{{\mathrm{x}}^2}{2}+\mathrm{x}+1\\ {\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\frac{100{\mathrm{x}}^{99}}{100}+\frac{99{\mathrm{x}}^{98}}{99}+\dots \dots +\frac{2\mathrm{x}}{2}+1\\ ={\mathrm{x}}^{99}+{\mathrm{x}}^{98}+\dots \dots .+\mathrm{x}+1\\ \text{LHS-}{\mathrm{f}}^{‘}\left(1\right)=1^{99}+1^{98}+\dots \dots .+1+1\left(100\text{times}\right)=100\\ \text{RHS-}{\mathrm{f}}^{‘}\left(0\right)=0^{99}+0^{98}+\dots \dots .+0+1=1\\ \text{Thus,}{\mathrm{f}}^{‘}\left(1\right)=100{\mathrm{f}}^{‘}\left(0\right)\end{array}$

Q.31

$\mathbf{\text{Evaluate}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{2}}{\mathbf{lim}}\frac{\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}}{\mathbf{x}\mathbf{+}\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\text{On substituting the value}\mathrm{x}=-2\text{, the expression reduces to}0/0.\\ \text{Therefore,}\\ \underset{\mathrm{x}\to -2}{\lim }\frac{\frac{1}{\mathrm{x}}+\frac{1}{2}}{\mathrm{x}+2}=\underset{\mathrm{x}\to -2}{\lim }\frac{\frac{\mathrm{x}+2}{2\mathrm{x}}}{\mathrm{x}+2}\\ =\underset{\mathrm{x}\to -2}{\lim }\frac{1}{2\mathrm{x}}\\ =\frac{1}{2(-2)}=-\frac{1}{4}\end{array}$

Q.32

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 2}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^2}{\mathrm{x}-2}=\underset{\mathrm{x}-2\to 0}{\lim }\frac{{\mathrm{e}}^2\left({\mathrm{e}}^{\mathrm{x}-2}-1\right)}{\mathrm{x}-2}\\ ={\mathrm{e}}^2\underset{\mathrm{y}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{y}}-1}{\mathrm{y}}[\text{Let}\mathrm{x}-2=\mathrm{y}]\\ ={\mathrm{e}}^2\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ ={\mathrm{e}}^2\end{array}$

Q.33

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{3}\mathbf{+}\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{5}}}{\mathbf{x}\mathbf{-}\mathbf{2}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 2}{\lim }\frac{{\mathrm{e}}^{3+\mathrm{x}}-{\mathrm{e}}^5}{\mathrm{x}-2}=\underset{\mathrm{x}-2\to 0}{\lim }\frac{{\mathrm{e}}^5\left({\mathrm{e}}^{\mathrm{x}-2}-1\right)}{\mathrm{x}-2}\\ ={\mathrm{e}}^5\underset{\mathrm{y}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{y}}-1}{\mathrm{y}}[\text{Let}\mathrm{x}-2=\mathrm{y}]\\ ={\mathrm{e}}^5\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ ={\mathrm{e}}^5\end{array}$

Q.34

$\mathbf{\text{Evaluate}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{5}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{5}}}{\mathbf{x}\mathbf{-}\mathbf{5}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 5}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^5}{\mathrm{x}-5}=\underset{\mathrm{x}-5\to 0}{\lim }\frac{{\mathrm{e}}^5\left({\mathrm{e}}^{\mathrm{x}-5}-1\right)}{\mathrm{x}-5}\\ ={\mathrm{e}}^5\underset{\mathrm{y}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{y}}-1}{\mathrm{y}}[\text{Let}\mathrm{x}-5=\mathrm{y}]\\ ={\mathrm{e}}^5\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ ={\mathrm{e}}^5\end{array}$

Q.35

$\mathbf{\text{Evaluate}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{sinx}}\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{.}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\sin \mathrm{x}}-1}{\sin \mathrm{x}}=2\times \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\sin \mathrm{x}}-1}{2\sin \mathrm{x}}\\ =2\times \underset{2\sin \mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\sin \mathrm{x}}-1}{2\sin \mathrm{x}}[\text{as}\mathrm{x}\to 0,2\sin \mathrm{x}\to 0]\\ =2\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ =2\end{array}$

Q.36

$\mathbf{\text{Evaluate}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{x}(e^{2x}-1)}{\mathbf{1}\mathbf{-}\mathbf{cosx}}\mathbf{.}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}\left({\mathrm{e}}^{2\mathrm{x}}-1\right)}{1-\cos \mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2}{1-\cos \mathrm{x}}\\ =2\underset{2\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2}{2{\sin }^2\left(\frac{\mathrm{x}}{2}\right)}\\ =2\times \underset{2\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\frac{2{\left(\frac{\mathrm{x}}{2}\right)}^2}{{\sin }^2\left(\frac{\mathrm{x}}{2}\right)}\\ =4\times \underset{2\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}\times \underset{\mathrm{x}/2\to 0}{\lim }\frac{{\left(\frac{\mathrm{x}}{2}\right)}^2}{{\sin }^2\left(\frac{\mathrm{x}}{2}\right)}\\ =4\times 1\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\text{and}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=1\right]\\ =4\end{array}$

Q.37

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{x}(e^{2x}-1)}{\mathbf{sin}\mathbf{x}}\mathbf{.}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}\left({\mathrm{e}}^{2\mathrm{x}}-1\right)}{\sin \mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{x}}{\sin \mathrm{x}}\right)\times 2\underset{2\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}\\ =1\times 2\left(1\right)\left[\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\text{and}\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ =0\end{array}$

Q.38

$\mathbf{\text{Evaluate}}\mathbf{:}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{x}\mathbf{+}(e^{5x}-1)}{\mathbf{2}\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}+\left({\mathrm{e}}^{5\mathrm{x}}-1\right)}{2\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{2\mathrm{x}}+\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{5\mathrm{x}}-1}{2\mathrm{x}}\\ =\frac{1}{2}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}+\frac{5}{2}\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{5\mathrm{x}}-1}{5\mathrm{x}}\\ =\frac{1}{2}\times 1+\frac{5}{2}\times 1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=1\text{and}\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=1\right]\\ =\frac{6}{2}\\ =3\end{array}$

Q.39

$\mathbf{\text{Evaluate :}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{log}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}{\mathbf{2}\mathbf{x}}$

Ans

$\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}+{\log }_{\mathrm{e}}(1+\mathrm{x})}{2\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{2\mathrm{x}}+\underset{\mathrm{x}\to 0}{\lim }\frac{{\log }_{\mathrm{e}}(1+\mathrm{x})}{2\mathrm{x}}\\ =\frac{1}{2}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}+\frac{1}{2}\underset{\mathrm{x}\to 0}{\lim }\frac{{\log }_{\mathrm{e}}(1+\mathrm{x})}{\mathrm{x}}\\ =\frac{1}{2}\times 1+\frac{1}{2}\times 1\left[\begin{array}{l}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\text{and}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{{\log }_{\mathrm{e}}(1+\mathrm{x})}{\mathrm{x}}=1\end{array}\right]\\ =1\end{array}$

Q.40

$\begin{array}{l}\mathbf{\text{Find the following limit :}}\\ \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\sqrt[\mathbf{3}]{\mathbf{sin}\mathbf{x}}}{\sqrt[\mathbf{3}]{\mathbf{x}}}\end{array}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt[3]{\sin \mathrm{x}}}{\sqrt[3]{\mathrm{x}}}=\underset{\mathrm{x}\to 0}{\lim }\sqrt[3]{\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}\\ =\sqrt[3]{\lim \left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}\\ =\sqrt[3]{1}\left[\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\right]\\ =1\end{array}$

Q.41

$\begin{array}{l}\mathbf{\text{Find the following limit :}}\\ \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left(\frac{x+\sin x}{x}\right)\end{array}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{x}}{\mathrm{x}}+\frac{\sin \mathrm{x}}{\mathrm{x}}\right)\\ =\underset{\mathrm{x}\to 0}{\lim }1+\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}\\ =1+1\left[\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}=1\right]\\ =2\end{array}$

Q.42

$\begin{array}{l}\mathbf{\text{Find the limit :}}\\ \underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}}{{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{15}}\end{array}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \underset{\mathrm{x}\to 3}{\lim }\frac{{\mathrm{x}}^2-9}{{\mathrm{x}}^3+2\mathrm{x}-15}=\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}-3)(\mathrm{x}+3)}{(\mathrm{x}+5)(\mathrm{x}-3)}\\ =\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}+3)}{(\mathrm{x}+5)}\\ =\frac{3+3}{3+5}\\ =\frac{6}{8}\\ =\frac{3}{4}\end{array}$

Q.43 Differentiate x log_ex with respect to x.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}={\mathrm{xlog}}_{\mathrm{e}}\mathrm{x}\\ \text{Differentiating w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\mathrm{x}{\log }_{\mathrm{e}}\mathrm{x}\right\}\\ =\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\log }_{\mathrm{e}}\mathrm{x}+{\log }_{\mathrm{e}}\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\\ =\mathrm{x}\cdot \frac{1}{\mathrm{x}}+{\log }_{\mathrm{e}}\mathrm{x}.1\\ =1+{\log }_{\mathrm{e}}\mathrm{x}\end{array}$

Q.44 Differentiate sin x.e^x with respect to x.

Ans

$\begin{array}{l}\text{Let}\mathrm{y}=\sin \mathrm{x}\cdot {\mathrm{e}}^{\mathrm{x}}\\ \text{Differentiating w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin \mathrm{x}\cdot {\mathrm{e}}^{\mathrm{x}}\right\}=\sin \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\sin \mathrm{x}\\ =\sin \mathrm{x}\cdot {\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}\\ ={\mathrm{e}}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})\end{array}$

Q.45 Differentiate sin²x with respect to x.

Ans

$\begin{array}{l}\text{Let}\mathrm{y}={\sin }^2\mathrm{x}\\ \text{Differentiating w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\{\sin \mathrm{x}\cdot \sin \mathrm{x}\}=\sin \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\sin \mathrm{x}+\sin \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\sin \mathrm{x}\\ =\sin \mathrm{x}\cdot \cos \mathrm{x}+\sin \mathrm{x}\cdot \cos \mathrm{x}\\ =2\sin \mathrm{x}\cos \mathrm{x}\end{array}$

Q.46 Find the derivative of cos x/log x.

Ans

$\begin{array}{l}\text{Let}\mathrm{y}=\frac{\cos \mathrm{x}}{\log \mathrm{x}}\\ \text{Differentiate w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\frac{\cos \mathrm{x}}{\log \mathrm{x}}\right\}\\ =\frac{\log \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\cos \mathrm{x}-\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\log \mathrm{x}}{{\left(\log \mathrm{x}\right)}^2}\\ =\frac{\log \mathrm{x}\{-\mathrm{sinx}\}-\cos \mathrm{x}\frac{1}{\mathrm{x}}}{{\left(\log \mathrm{x}\right)}^2}\\ =\frac{-\sin \mathrm{x}\cdot \log \mathrm{x}-\frac{1}{\mathrm{x}}\cdot \cos \mathrm{x}}{{\left(\log \mathrm{x}\right)}^2}\end{array}$

Q.47 Find the derivative of (x²+1)cos x.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\left({\mathrm{x}}^2-1\right)\cos \mathrm{x}\\ \text{Differentiation w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left({\mathrm{x}}^2-1\right)\cos \mathrm{x}\right\}\\ =\left({\mathrm{x}}^2-1\right)\frac{\mathrm{d}}{\mathrm{dx}}\cos \mathrm{x}+\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2-1\right)\\ =\left({\mathrm{x}}^2-1\right)\times -\sin \mathrm{x}+\cos \mathrm{x}(2\mathrm{x}+0)\\ =-\left({\mathrm{x}}^2-1\right)\sin \mathrm{x}+2\mathrm{x}\cos \mathrm{x}\\ =-{\mathrm{x}}^2\sin \mathrm{x}+\sin \mathrm{x}+2\mathrm{x}\cos \mathrm{x}\end{array}$

Q.48 Find the derivative of (ax + b)(x+1)².

Ans

$\begin{array}{l}\text{Let}\mathrm{y}=\left(\mathrm{ax}+\mathrm{b}\right){\left(\mathrm{x}+1\right)}^2\\ \text{Differentiating w.r.t.}\mathrm{x},\text{we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{(\mathrm{ax}+\mathrm{b}){(\mathrm{x}+1)}^2\right\}\\ =(\mathrm{ax}+\mathrm{b})\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2+2\mathrm{x}+1\right)+{\left(\mathrm{x}+1\right)}^2\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{b}\right)\\ =(\mathrm{ax}+\mathrm{b})(2\mathrm{x}+2)+{\left(\mathrm{x}+1\right)}^2\left(\mathrm{a}+0\right)\\ =2(\mathrm{x}+1)(\mathrm{ax}+\mathrm{b})+\mathrm{a}{\left(\mathrm{x}+1\right)}^2\\ =(\mathrm{x}+1)\left\{2\right(\mathrm{ax}+\mathrm{b})+\mathrm{a}(\mathrm{x}+1\left)\right\}\end{array}$

Q.49

$\begin{array}{l}\mathbf{\text{What is the value of}}\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}(5x^3-4x^2+5)\mathbf{?}\\ \mathbf{\text{(a)29(b)24(c)12(d)60}}\end{array}$

Ans

$\begin{array}{l}\text{(a)29}\\ \text{Explanation:}\\ \underset{\mathrm{x}\to 2}{\lim }\left(5{\mathrm{x}}^3-4{\mathrm{x}}^2+5\right)\\ =\underset{\mathrm{x}\to 2}{\lim }\left(5{\mathrm{x}}^3\right)-\underset{\mathrm{x}\to 2}{\lim }\left(4{\mathrm{x}}^2\right)+\underset{\mathrm{x}\to 2}{\lim }\text{(Addition Rule)}\\ =5\underset{\mathrm{x}\to 2}{\lim }{\mathrm{x}}^3-4\underset{\mathrm{x}\to 2}{\lim }{\mathrm{x}}^2+5\text{(Product Rule)}\\ =5{\left(2\right)}^3-4{\left(2\right)}^2+5\text{(Limit of Polynomials)}\\ =40-16+5=29\end{array}$

Q.50

$\begin{array}{l}\mathbf{\text{The value of}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\sqrt{\mathbf{2}\mathbf{+}\mathbf{x}}\mathbf{-}\sqrt{\mathbf{2}}}{\mathbf{x}}\mathbf{\text{is}}\\ \mathbf{\text{(a)}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\text{(b)}}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\\ \mathbf{\text{(c)}}\frac{\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{2}}}\mathbf{\text{(d)}}\mathbf{2}\sqrt{\mathbf{2}}\end{array}$

Ans

$\begin{array}{l}\text{(c)}\frac{1}{2\sqrt{2}}\\ \text{Explanation:}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sqrt{2+\mathrm{x}}-\sqrt{2}}{\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{(\sqrt{2+\mathrm{x}}-\sqrt{2})(\sqrt{2+\mathrm{x}}+\sqrt{2})}{\mathrm{x}(\sqrt{2+\mathrm{x}}+\sqrt{2})}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2+\mathrm{x}-2}{\mathrm{x}(\sqrt{2+\mathrm{x}}+\sqrt{2})}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{1}{(\sqrt{2+\mathrm{x}}+\sqrt{2})}=\frac{1}{2\sqrt{2}}.\end{array}$

Q.51

$\begin{array}{l}\mathbf{\text{If}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{n}}\mathbf{\text{and if}}{\mathbf{f}}^{\mathbf{‘}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{,}\mathbf{\text{the value of}}\mathbf{n}\mathbf{\text{is}}\\ \left(a\right)\mathbf{9}\left(b\right)\mathbf{10}\left(c\right)\mathbf{89}\left(d\right)\mathbf{90}\end{array}$

Ans

$\begin{array}{l}\text{(b)}10\\ \text{Explanation:}\\ \text{We have,}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{n}}\\ \text{Differentiating both sides w.r.t.}\mathrm{x},\text{we get}{\mathrm{f}}^{‘}\left(\mathrm{x}\right)={\mathrm{nx}}^{\mathrm{n}-1}\\ \text{Putting}\mathrm{x}=1,\text{we get}\\ {\mathrm{f}}^{‘}\left(1\right)=\mathrm{n}\left[{\mathrm{f}}^{‘}\left(1\right)=10\right]\\ \Rightarrow \mathrm{n}=10\end{array}$

Q.52

$\mathbf{\text{Evaluate:}}\underset{\mathbf{x}\mathbf{\to }\sqrt{\mathbf{10}}}{\mathbf{lim}}\frac{\sqrt{\mathbf{7}\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{-}\mathbf{(}\sqrt{\mathbf{5}}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{)}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{10}}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{\sqrt{7-2\mathrm{x}}-(\sqrt{5}-\sqrt{2})}{{\mathrm{x}}^2-10}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{\sqrt{7-2\mathrm{x}}-\sqrt{{(\sqrt{5}-\sqrt{2})}^2}}{{\mathrm{x}}^2-10}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{\sqrt{7-2\mathrm{x}}-\sqrt{(7-2\sqrt{10})}}{{\mathrm{x}}^2-10}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{\sqrt{7-2\mathrm{x}}-\sqrt{(7-2\sqrt{10})}}{{\mathrm{x}}^2-10}\times \frac{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}}{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{(7-2\mathrm{x})-(7-2\sqrt{10})}{(\mathrm{x}-\sqrt{10})(\mathrm{x}+\sqrt{10})\left\{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}\right\}}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{-2\mathrm{x}+2\sqrt{10}}{(\mathrm{x}-\sqrt{10})(\mathrm{x}+\sqrt{10})\left\{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}\right\}}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{-2(\mathrm{x}-\sqrt{10})}{(\mathrm{x}-\sqrt{10})(\mathrm{x}+\sqrt{10})\left\{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}\right\}}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{-2}{(\mathrm{x}+\sqrt{10})\left\{\sqrt{7-2\mathrm{x}}+\sqrt{(7-2\sqrt{10})}\right\}}\\ =\underset{\mathrm{x}\to \sqrt{10}}{\lim }\frac{-2}{2\sqrt{10}\left\{\sqrt{7-2\sqrt{10}}+\sqrt{7-2\sqrt{10}}\right\}}\\ =\frac{-1}{\sqrt{10}\times 2\times \sqrt{7-2\sqrt{10}}}\\ =\frac{-1}{2\sqrt{10}(\sqrt{5}-\sqrt{2})}\\ =\frac{-1}{2\sqrt{10}}\times \frac{(\sqrt{5}+\sqrt{2})}{3}\\ =-\frac{(\sqrt{5}+\sqrt{2})}{6\sqrt{10}}\end{array}$

Q.53

$\begin{array}{l}\mathbf{\text{If}}\mathbf{f}\mathbf{\text{is a real valued function defined by}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{,}\\ \mathbf{\text{then find}}{\mathbf{f}}^{\mathbf{‘}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{and}}{\mathbf{f}}^{\mathbf{‘}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Wehave,}\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+4\mathrm{x}+3\\ \therefore {\mathrm{f}}^{‘}\left(1\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(1+\mathrm{h})-\mathrm{f}\left(1\right)}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(1\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left\{{(1+\mathrm{h})}^2+4(1+\mathrm{h})+3\right\}-\left\{1^2+4\times 1+3\right\}}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(1\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\mathrm{h}}^2+6\mathrm{h}+8\right)-8}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(1\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{{\mathrm{h}}^2+6\mathrm{h}}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(1\right)=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}+6\\ =6\\ \text{and,}\\ {\mathrm{f}}^{‘}\left(3\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(3+\mathrm{h})-\mathrm{f}\left(3\right)}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(3\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left\{{(3+\mathrm{h})}^2+4(3+\mathrm{h})+3\right\}-\left\{3^2+4\times 3+3\right\}}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(3\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\mathrm{h}}^2+10\mathrm{h}+24\right)-24}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(3\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{{\mathrm{h}}^2+10\mathrm{h}}{\mathrm{h}}\\ \Rightarrow {\mathrm{f}}^{‘}\left(3\right)=\underset{\mathrm{h}\to 0}{\lim }\mathrm{h}+10\\ =10\end{array}$

Q.54

$\mathbf{\text{Differentiate}}{\mathbf{e}}^{\sqrt{\mathbf{tanx}}}\mathbf{\text{from first principal.}}$

Ans

$\begin{array}{l}\text{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\\ \text{then,}\mathrm{f}(\mathrm{x}+\mathrm{h})={\mathrm{e}}^{\sqrt{\tan (\mathrm{x}+\mathrm{h})}}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{{\mathrm{e}}^{\sqrt{\tan (\mathrm{x}+\mathrm{h})}}-{\mathrm{e}}^{\sqrt{\mathrm{tanx}}}}{\mathrm{h}}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }{\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\left\{\frac{{\mathrm{e}}^{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\mathrm{tanx}}}-1}{\mathrm{h}}\right\}\\ ={\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\underset{\mathrm{h}\to 0}{\lim }\left\{\frac{{\mathrm{e}}^{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\mathrm{tanx}}}-1}{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}\times \frac{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}{\mathrm{h}}\right\}\\ ={\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\underset{\mathrm{h}\to 0}{\lim }\left\{\frac{{\mathrm{e}}^{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}-1}{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}\right\}\cdot \underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}{\mathrm{h}}\\ ={\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\times 1\times \underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{\tan (\mathrm{x}+\mathrm{h})}-\sqrt{\tan \mathrm{x}}}{\mathrm{h}}\times \frac{\sqrt{\tan (\mathrm{x}+\mathrm{h})}+\sqrt{\tan \mathrm{x}}}{\sqrt{\tan (\mathrm{x}+\mathrm{h})}+\sqrt{\tan \mathrm{x}}}\\ ={\mathrm{e}}^{\sqrt{\mathrm{tanx}}}\times 1\times \underset{\mathrm{h}\to 0}{\lim }\frac{\tan (\mathrm{x}+\mathrm{h})-\tan \mathrm{x}}{\mathrm{h}}\times \frac{1}{\sqrt{\tan (\mathrm{x}+\mathrm{h})}+\sqrt{\tan \mathrm{x}}}\\ ={\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}\times \underset{\mathrm{h}\to 0}{\lim }\frac{\sin \mathrm{h}}{\mathrm{h}\cos (\mathrm{x}+\mathrm{h})\cos \mathrm{x}}\times \frac{1}{\sqrt{\tan (\mathrm{x}+\mathrm{h})}+\sqrt{\tan \mathrm{x}}}\\ ={\mathrm{e}}^{\sqrt{\mathrm{tanx}}}\times \frac{1}{{\cos }^2\mathrm{h}}\times \frac{1}{2\sqrt{\tan \mathrm{x}}}\\ =\frac{{\mathrm{e}}^{\sqrt{\tan \mathrm{x}}}}{2\sqrt{\tan \mathrm{x}}}{\sec }^2\mathrm{x}\end{array}$

Q.55

$\begin{array}{l}\mathbf{\text{Evaluate the following limits :}}\\ \mathbf{\text{(i)}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left(\frac{\cos \mathrm{Ax}-\cos \mathrm{Bx}}{x^2}\right)\\ \mathbf{\text{(ii)}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{3}\mathbf{x}}{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{3}\mathbf{x}}\\ \mathbf{\text{(iii)}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{6}\mathbf{x}}{\mathbf{sin}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{sin}\mathbf{3}\mathbf{x}}\end{array}$

Ans

$\begin{array}{l}\text{(i)Wehave,}\\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{cosAx}-\mathrm{cosBx}}{{\mathrm{x}}^2}\right)\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{xsin}\left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\mathrm{x}}{{\mathrm{x}}^2}\left[\begin{array}{l}\mathrm{cosC}-\mathrm{cosD}\\ =2\sin \frac{\mathrm{C}+\mathrm{D}}{2}\sin \frac{\mathrm{D}-\mathrm{C}}{2}\end{array}\right]\\ =2\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{x}}{\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{x}}\times \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\frac{\sin \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\mathrm{x}}{\left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\times }\times \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\right\}\\ =2\left(\frac{\mathrm{B}+\mathrm{A}}{2}\right)\left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{x}}{\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\mathrm{x}}\right\}\times \left\{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\frac{\mathrm{B}-\mathrm{A}}{2}\right)\mathrm{x}}{2}\right\}\\ =\left(\frac{{\mathrm{B}}^2-{\mathrm{A}}^2}{2}\right)\left(1\right)\left(1\right)=\frac{{\mathrm{B}}^2-{\mathrm{A}}^2}{2}\\ \text{(ii)Wehave,}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sin 2\mathrm{x}+\sin 3\mathrm{x}}{2\mathrm{x}+\sin 3\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{\sin 2\mathrm{x}}{\mathrm{x}}+\frac{\sin 3\mathrm{x}}{\mathrm{x}}}{2+\frac{\sin 3\mathrm{x}}{\mathrm{x}}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2\left(\frac{\sin 2\mathrm{x}}{2\mathrm{x}}\right)+3\left(\frac{\sin 3\mathrm{x}}{3\mathrm{x}}\right)}{2+3\left(\frac{\sin 3\mathrm{x}}{3\mathrm{x}}\right)}\\ =\frac{2\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\sin 2\mathrm{x}}{2\mathrm{x}}\right)+3\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\sin 3\mathrm{x}}{3\mathrm{x}}\right)}{2+3\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\sin 3\mathrm{x}}{3\mathrm{x}}\right)}\\ =\frac{2\times 1+3\times 1}{2+3\times 1}=\frac{5}{5}=1\\ \left(\mathrm{iii}\right)\text{Wehave,}\\ \underset{\mathrm{x}\to 0}{\lim }\frac{\sin 2\mathrm{x}+\sin 6\mathrm{x}}{\sin 5\mathrm{x}-\sin 3\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2\sin 4\mathrm{xcos}2\mathrm{x}}{2\mathrm{sinxcos}4\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{\frac{\sin 4\mathrm{x}}{4\mathrm{x}}\times 4\mathrm{x}\times \cos 2\mathrm{x}}{\frac{\mathrm{sinx}}{\mathrm{x}}\times \mathrm{x}\times \cos 4\mathrm{x}}\right\}\\ =4\frac{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 4\mathrm{x}}{4\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\cos 2\mathrm{x}}{\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}\times \underset{\mathrm{x}\to 0}{\lim }\cos 4\mathrm{x}}\\ =4\left(\frac{1\times 1}{1\times 1}\right)=4\end{array}$

Q.56

$\begin{array}{l}\mathbf{\text{Differentiate the following function with respect to}}\mathbf{x}\mathbf{\text{:}}\\ \mathbf{\text{(i)}}{\mathbf{log}}_{\mathbf{x}}\mathbf{x}\mathbf{\text{(ii)}}{\mathbf{e}}^{\mathbf{3}\mathbf{logx}}\mathbf{\text{(iii)}}{\mathbf{2}}^{{\mathbf{log}}_{\mathbf{2}}\mathbf{x}}\\ \mathbf{\text{(iv)}}\mathbf{5}\left(2^{3{\log }_{2}x}\right)\mathbf{\text{(v)}}\mathbf{5}{\mathbf{e}}^{\mathbf{x}}\mathbf{\text{(vi)}}\mathbf{9}\mathbf{\cdot }\left(3^x\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{We}\mathrm{have},\\ \frac{\mathrm{d}}{\mathrm{dx}}\left({\log }_{\mathrm{x}}\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(1\right)=0.\left[\therefore {\log }_{\mathrm{x}}\mathrm{x}=1\right]\\ \left(\mathrm{ii}\right)\text{Wehave,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{3\mathrm{logx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{{\mathrm{logx}}^3}\right)\\ =\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^3\right)=3{\mathrm{x}}^2\left[{\mathrm{e}}^{\mathrm{logk}}=\mathrm{k}\right]\\ \left(\mathrm{iii}\right)\text{Wehave,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{{\log }_2\mathrm{x}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)=1\left[{\mathrm{a}}^{{\log }_{\mathrm{a}}\mathrm{n}}=\mathrm{n}\right]\\ \left(\mathrm{iv}\right)\text{Wehave,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(5{.2}^{3{\log }_2\mathrm{x}}\right)\\ =5\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{3{\log }_2\mathrm{x}}\right)\\ =5\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(2^{{\log }_2{\mathrm{x}}^3}\right)=5\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^3\right)\\ =5\left(3{\mathrm{x}}^2\right)=15{\mathrm{x}}^2.\\ \left(\mathrm{v}\right)\text{Wehave,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(5{\mathrm{e}}^{\mathrm{x}}\right)=5\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)\\ =5{\mathrm{e}}^{\mathrm{x}}\\ \left(\mathrm{vi}\right)\text{Wehave,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left(9{.3}^{\mathrm{x}}\right)=9\cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(3^{\mathrm{x}}\right)=9\cdot \left(3^{\mathrm{x}}{\log }_{\mathrm{e}}3\right).\end{array}$

Q.57

$\begin{array}{l}\mathbf{\text{(i) The differentiation of}}\mathbf{cosec}\mathbf{x}\mathbf{\text{with respect to}}\mathbf{x}\mathbf{\text{is}}\mathbf{-}\mathbf{cosec}\mathbf{x}\mathbf{cot}\mathbf{x}\mathbf{\text{.}}\\ \mathbf{\text{i.e., prove that}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\mathbf{cosec}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{cosec}\mathbf{x}\mathbf{cot}\mathbf{x}\\ \mathbf{\text{(ii) The differentiation of}}{\mathbf{log}}_{\mathbf{a}}\mathbf{x}\mathbf{(}\mathbf{a}\mathbf{>}\mathbf{0}\mathbf{,}\mathbf{a}\mathbf{\ne }\mathbf{1}\mathbf{)}\mathbf{\text{with respect to}}\mathbf{x}\mathbf{\text{is}}\frac{\mathbf{1}}{\mathbf{x}{\mathbf{log}}_{\mathbf{e}}\mathbf{a}}\\ \mathbf{\text{i.e., prove that}}\frac{\mathbf{d}}{\mathbf{dx}}\left({\log }_{a}x\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{xlog}}_{\mathbf{e}}\mathbf{a}}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\text{(i)Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cosec}\mathrm{x}\text{.}\\ \text{Then,}\\ \mathrm{f}(\mathrm{x}+\mathrm{h})=\mathrm{cosec}(\mathrm{x}+\mathrm{h})\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{cosec}(\mathrm{x}+\mathrm{h})-\mathrm{cosecx}}{\mathrm{h}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\frac{1}{\sin (\mathrm{x}+\mathrm{h})}-\frac{1}{\sin \mathrm{x}}}{\mathrm{h}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{sinx}-\sin (\mathrm{x}+\mathrm{h})}{\mathrm{h}\sin \mathrm{x}\sin (\mathrm{x}+\mathrm{h})}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{2\sin \left(\frac{\mathrm{x}-\mathrm{x}-\mathrm{h}}{2}\right)\cos \left(\frac{\mathrm{x}+\mathrm{x}+\mathrm{h}}{2}\right)}{\mathrm{h}\sin \mathrm{x}\sin (\mathrm{x}+\mathrm{h})}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{2\sin \left(\frac{-\mathrm{h}}{2}\right)\cos (\mathrm{x}+\mathrm{h}/2)}{\mathrm{h}\sin \mathrm{x}\sin (\mathrm{x}+\mathrm{h})}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\sin (\mathrm{h}/2)}{\mathrm{h}/2}\times \underset{\mathrm{h}\to 0}{\lim }\frac{\cos (\mathrm{x}+\mathrm{h}/2)}{\mathrm{sinxsin}(\mathrm{x}+\mathrm{h})}\\ \left[\sin \left(-\frac{\mathrm{h}}{2}\right)=-\sin \frac{\mathrm{h}}{2}\right]\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=(-1)\times \frac{\mathrm{cosx}}{\sin \mathrm{x}\sin \mathrm{x}}\\ =-\cot \mathrm{x}\mathrm{cosec}\mathrm{x}.\\ \text{Hence,}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{cosec}\mathrm{x}\right)=-\mathrm{cosec}\mathrm{x}\cot \mathrm{x}\\ \text{(ii)Let}\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{a}}\mathrm{x}\\ \text{Then,}\\ \mathrm{f}(\mathrm{x}+\mathrm{h})={\log }_{\mathrm{a}}(\mathrm{x}+\mathrm{h})\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}(\mathrm{x}+\mathrm{h})-{\log }_{\mathrm{a}}\left(\mathrm{x}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)}{\mathrm{h}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}\left(1+\frac{\mathrm{h}}{\mathrm{x}}\right)}{\mathrm{h}}\\ =\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}\left(1+\frac{\mathrm{h}}{\mathrm{x}}\right)}{\left({\log }_{\mathrm{e}}\mathrm{a}\right)\cdot \mathrm{h}}\left[{\log }_{\mathrm{a}}\mathrm{\lambda }=\frac{{\log }_{\mathrm{e}}\mathrm{\lambda }}{{\log }_{\mathrm{e}}\mathrm{a}}\right]\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\frac{1}{{\log }_{\mathrm{e}}\mathrm{a}}\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}\left(1+\frac{\mathrm{h}}{\mathrm{x}}\right)}{\mathrm{x}\left(\frac{\mathrm{h}}{\mathrm{x}}\right)}\\ =\frac{1}{{\mathrm{xlog}}_{\mathrm{e}}\mathrm{a}}\left[\underset{\mathrm{h}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}\left(1+\frac{\mathrm{h}}{\mathrm{x}}\right)}{\frac{\mathrm{h}}{\mathrm{x}}}=1\right]\\ \text{Hence,}\\ \frac{\mathrm{d}}{\mathrm{dx}}\left({\log }_{\mathrm{a}}\mathrm{x}\right)=\frac{1}{{\mathrm{xlog}}_{\mathrm{e}}\mathrm{a}}\end{array}$

Q.58

$\mathbf{\text{The value of}}\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\frac{{\mathbf{\text{x}}}^{\mathbf{\text{m}}}\mathbf{-}{\mathbf{\text{a}}}^{\mathbf{\text{m}}}}{\mathbf{\text{x}}\mathbf{-}\mathbf{\text{a}}}\mathbf{\text{is\_\_\_\_\_.}}$

Ans

$\begin{array}{l}{\text{x}}^{\text{m}}–{\text{a}}^{\text{m}}=(\text{x}–\text{a})({\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{2}}\text{a}+{\text{x}}^{\text{m}–\text{3}}{\text{a}}^{\text{2}}+{\text{x}}^{\text{m}–\text{3}}{\text{a}}^{\text{3}}+\dots +{\text{xa}}^{\text{m}–\text{2}}+{\text{a}}^{\text{m}–\text{1}})\\ \underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{{\text{x}}^{\text{m}}–{\text{a}}^{\text{m}}}{\mathrm{x}-\mathrm{a}}=\underset{\mathrm{x}\to \mathrm{a}}{\lim }({\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{2}}\text{a}+{\text{x}}^{\text{m}–\text{3}}{\text{a}}^{\text{2}}+{\text{x}}^{\text{m}–\text{3}}{\text{a}}^{\text{3}}+\dots +{\text{xa}}^{\text{m}–\text{2}}+{\text{a}}^{\text{m}–\text{1}})\\ ={\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{2}}\mathrm{a}+{\text{x}}^{\text{m}–\text{3}}{\mathrm{a}}^2+...+\mathrm{x}{\text{a}}^{\text{m}–\text{2}}+{\text{a}}^{\text{m}–\text{1}}\\ ={\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{1}}+...+{\text{x}}^{\text{m}–\text{1}}+{\text{x}}^{\text{m}–\text{1}}\\ =\mathrm{m}{\text{x}}^{\text{m}–\text{1}}\end{array}$

Q.59 The derivative of x³(x – 3)² is ______.

Ans

$\begin{array}{l}{\text{The given function is x}}^{\text{3}}{(\text{x}-\text{3})}^{\text{2}}.\\ \text{Let y}={\text{x}}^{\text{3}}{(\text{x}-\text{3})}^{\text{2}}\\ \text{Differentiating w.r.t. x, we get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\text{x}}^{\text{3}}{(\text{x}-\text{3})}^{\text{2}}\right\}\\ ={\text{x}}^{\text{3}}\frac{\mathrm{d}}{\mathrm{dx}}{(\text{x}-\text{3})}^{\text{2}}+{(\text{x}-\text{3})}^{\text{2}}\frac{\mathrm{d}}{\mathrm{dx}}{\text{x}}^{\text{3}}\\ ={\text{x}}^{\text{3}}\times 2(\mathrm{x}-3)+{(\text{x}-\text{3})}^{\text{2}}\times 3{\mathrm{x}}^2\\ =2{\text{x}}^{\text{3}}(\mathrm{x}-3)+3{\mathrm{x}}^2{(\text{x}-\text{3})}^{\text{2}}\\ ={\mathrm{x}}^2(\mathrm{x}-3)\{2\mathrm{x}+3(\mathrm{x}-3)\}\\ ={\mathrm{x}}^2(\mathrm{x}-3)(2\mathrm{x}+3\mathrm{x}-9)\\ ={\mathrm{x}}^2(\mathrm{x}-3)(5\mathrm{x}-9)\\ {\text{The derivative of x}}^{\text{3}}{(\text{x}-\text{3})}^{\text{2}}\text{is}\underset{¯}{{\mathrm{x}}^2(\mathrm{x}-3)(5\mathrm{x}-9)}.\end{array}$