\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\lim_{x\to\pi}\frac{\sin x}{x-\pi}\text{ is equal to:}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }-1\qquad\text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\pi}\frac{\sin x}{x-\pi}
\displaystyle =\lim_{x\to\pi}\frac{\sin x-\sin\pi}{x-\pi}.
\displaystyle \text{Using }\lim_{x\to a}\frac{\sin x-\sin a}{x-a}=\cos a,
\displaystyle =\cos\pi=-1.
\displaystyle \therefore \text{The correct option is (c) }-1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\lim_{x\to0}\frac{x\sin x}{1-\cos x}\text{ is equal to:}
\displaystyle \text{(a) }2\qquad\text{(b) }\frac{2}{3}\qquad\text{(c) }-\frac{3}{2}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{x\sin x}{1-\cos x}
\displaystyle =\lim_{x\to0}\frac{x\sin x(1+\cos x)}{1-\cos^2x}.
\displaystyle =\lim_{x\to0}\frac{x\sin x(1+\cos x)}{\sin^2x}.
\displaystyle =\lim_{x\to0}\frac{x}{\sin x}(1+\cos x).
\displaystyle =1(1+1)=2.
\displaystyle \therefore \text{The correct option is (a) }2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\lim_{x\to0}\frac{(1+x)^n-1}{x}\text{ is equal to:}
\displaystyle \text{(a) }n\qquad\text{(b) }1\qquad\text{(c) }-n\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{(1+x)^n-1}{x}
\displaystyle =\lim_{x\to0}\frac{1+nx+\frac{n(n-1)}{2!}x^2+\cdots-1}{x}.
\displaystyle =\lim_{x\to0}\left[n+\frac{n(n-1)}{2!}x+\cdots\right].
\displaystyle =n.
\displaystyle \therefore \text{The correct option is (a) }n.
\displaystyle \\

\displaystyle \textbf{Question 4: }\lim_{x\to1}\frac{x^m-1}{x^n-1}\text{ is equal to:}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{m}{n}\qquad\text{(c) }-\frac{m}{n}\qquad\text{(d) }\frac{m^2}{n^2}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{x^m-1}{x^n-1}
\displaystyle =\lim_{x\to1}\frac{(x-1)(x^{m-1}+x^{m-2}+\cdots+x+1)}{(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)}.
\displaystyle =\lim_{x\to1}\frac{x^{m-1}+x^{m-2}+\cdots+x+1}{x^{n-1}+x^{n-2}+\cdots+x+1}.
\displaystyle =\frac{m}{n}.
\displaystyle \therefore \text{The correct option is (b) }\frac{m}{n}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\lim_{\theta\to0}\frac{1-\cos4\theta}{1-\cos6\theta}\text{ is equal to:}
\displaystyle \text{(a) }\frac{4}{9}\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }-\frac{1}{2}\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \lim_{\theta\to0}\frac{1-\cos4\theta}{1-\cos6\theta}
\displaystyle =\lim_{\theta\to0}\frac{2\sin^2 2\theta}{2\sin^2 3\theta}.
\displaystyle =\lim_{\theta\to0}\left(\frac{\sin2\theta}{\sin3\theta}\right)^2.
\displaystyle =\left(\frac{2}{3}\right)^2=\frac{4}{9}.
\displaystyle \therefore \text{The correct option is (a) }\frac{4}{9}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\lim_{x\to0}\frac{\mathrm{cosec}\,x-\cot x}{x}\text{ is equal to:}
\displaystyle \text{(a) }-\frac{1}{2}\qquad\text{(b) }1\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\mathrm{cosec}\,x-\cot x}{x}
\displaystyle =\lim_{x\to0}\frac{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}{x}.
\displaystyle =\lim_{x\to0}\frac{1-\cos x}{x\sin x}.
\displaystyle =\lim_{x\to0}\frac{1-\cos x}{x^2}\cdot\frac{x}{\sin x}.
\displaystyle =\frac{1}{2}\cdot1=\frac{1}{2}.
\displaystyle \therefore \text{The correct option is (c) }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\lim_{x\to0}\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}}\text{ is equal to:}
\displaystyle \text{(a) }2\qquad\text{(b) }0\qquad\text{(c) }1\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}}
\displaystyle =\lim_{x\to0}\frac{\sin x\left(\sqrt{1+x}+\sqrt{1-x}\right)}{(1+x)-(1-x)}.
\displaystyle =\lim_{x\to0}\frac{\sin x\left(\sqrt{1+x}+\sqrt{1-x}\right)}{2x}.
\displaystyle =\frac{1}{2}\lim_{x\to0}\frac{\sin x}{x}\left(\sqrt{1+x}+\sqrt{1-x}\right).
\displaystyle =\frac{1}{2}(1)(1+1)=1.
\displaystyle \therefore \text{The correct option is (c) }1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\lim_{x\to\frac{\pi}{4}}\frac{\sec^2x-2}{\tan x-1}\text{ is equal to:}
\displaystyle \text{(a) }3\qquad\text{(b) }1\qquad\text{(c) }0\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\frac{\pi}{4}}\frac{\sec^2x-2}{\tan x-1}
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{1+\tan^2x-2}{\tan x-1}.
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{\tan^2x-1}{\tan x-1}.
\displaystyle =\lim_{x\to\frac{\pi}{4}}\frac{(\tan x-1)(\tan x+1)}{\tan x-1}.
\displaystyle =\lim_{x\to\frac{\pi}{4}}(\tan x+1).
\displaystyle =1+1=2.
\displaystyle \therefore \text{The correct option is (d) }2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\lim_{x\to1}\frac{(\sqrt{x}-1)(2x-3)}{2x^2+x-3}\text{ is equal to:}
\displaystyle \text{(a) }\frac{1}{10}\qquad\text{(b) }-\frac{1}{10}\qquad\text{(c) }1\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{(\sqrt{x}-1)(2x-3)}{2x^2+x-3}
\displaystyle =\lim_{x\to1}\frac{(\sqrt{x}-1)(2x-3)}{(x-1)(2x+3)}.
\displaystyle \text{Since }x-1=(\sqrt{x}-1)(\sqrt{x}+1),
\displaystyle =\lim_{x\to1}\frac{2x-3}{(\sqrt{x}+1)(2x+3)}.
\displaystyle =\frac{2(1)-3}{(\sqrt{1}+1)(2(1)+3)}.
\displaystyle =\frac{-1}{(2)(5)}=-\frac{1}{10}.
\displaystyle \therefore \text{The correct option is (b) }-\frac{1}{10}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }f(x)=\begin{cases}\dfrac{\sin[x]}{[x]},&[x]\ne0,\\0,&[x]=0,\end{cases}
\displaystyle \text{where }[\,]\text{ denotes the greatest integer function, then }\lim_{x\to0}f(x)\text{ is equal to:}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad\text{(c) }-1\qquad\text{(d) Doesn't exist}
\displaystyle \text{Answer:}
\displaystyle \text{For }x\to0^+,\quad 0<x<1\Rightarrow[x]=0.
\displaystyle \therefore f(x)=0.
\displaystyle \therefore \lim_{x\to0^+}f(x)=0.
\displaystyle \text{For }x\to0^-,\quad -1<x<0\Rightarrow[x]=-1.
\displaystyle \therefore f(x)=\frac{\sin(-1)}{-1}=\sin1.
\displaystyle \therefore \lim_{x\to0^-}f(x)=\sin1.
\displaystyle \text{Since }\lim_{x\to0^-}f(x)\ne\lim_{x\to0^+}f(x),
\displaystyle \therefore \lim_{x\to0}f(x)\text{ does not exist.}
\displaystyle \therefore \text{The correct option is (d) Doesn't exist.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }f(x)=x-[x],\ x\in R,\text{ then }f'\left(\frac{1}{2}\right)\text{ is equal to:}
\displaystyle \text{(a) }\frac{3}{2}\qquad\text{(b) }1\qquad\text{(c) }0\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle f'\left(\frac{1}{2}\right)=\lim_{h\to0}\frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}.
\displaystyle \text{For sufficiently small }h,\quad 0<\frac{1}{2}+h<1.
\displaystyle \therefore \left[\frac{1}{2}+h\right]=0\quad\text{and}\quad\left[\frac{1}{2}\right]=0.
\displaystyle \therefore f\left(\frac{1}{2}+h\right)=\frac{1}{2}+h,\quad f\left(\frac{1}{2}\right)=\frac{1}{2}.
\displaystyle \therefore f'\left(\frac{1}{2}\right)=\lim_{h\to0}\frac{\frac{1}{2}+h-\frac{1}{2}}{h}.
\displaystyle =\lim_{h\to0}\frac{h}{h}=1.
\displaystyle \therefore \text{The correct option is (b) }1.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }y=\sqrt{x}+\frac{1}{\sqrt{x}},\text{ then }\frac{dy}{dx}\text{ at }x=1\text{ is equal to:}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }\frac{1}{\sqrt{2}}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle y=x^{1/2}+x^{-1/2}.
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{-3/2}.
\displaystyle =\frac{1}{2\sqrt{x}}-\frac{1}{2x^{3/2}}.
\displaystyle \text{At }x=1,
\displaystyle \frac{dy}{dx}=\frac{1}{2}-\frac{1}{2}=0.
\displaystyle \therefore \text{The correct option is (d) }0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }f(x)=\frac{x^3}{2\sqrt{x}},\text{ then }f'(1)\text{ is equal to:}
\displaystyle \text{(a) }\frac{5}{4}\qquad\text{(b) }\frac{2}{5}\qquad\text{(c) }1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{x^3}{2\sqrt{x}}.
\displaystyle =\frac{1}{2}x^{3-\frac{1}{2}}=\frac{1}{2}x^{5/2}.
\displaystyle \therefore f'(x)=\frac{1}{2}\cdot\frac{5}{2}x^{3/2}.
\displaystyle =\frac{5}{4}x^{3/2}.
\displaystyle \text{At }x=1,
\displaystyle f'(1)=\frac{5}{4}(1)^{3/2}=\frac{5}{4}.
\displaystyle \therefore \text{The correct option is (a) }\frac{5}{4}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}},\text{ then }\frac{dy}{dx}\text{ is equal to:}
\displaystyle \text{(a) }-\frac{4x}{(x^2-1)^2}\qquad\text{(b) }-\frac{4x}{x^2-1}\qquad\text{(c) }\frac{1-x^2}{4x}\qquad\text{(d) }\frac{4x}{x^2-1}
\displaystyle \text{Answer:}
\displaystyle y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}.
\displaystyle =\frac{x^2+1}{x^2-1}.
\displaystyle \therefore \frac{dy}{dx}=\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}.
\displaystyle =\frac{2x[(x^2-1)-(x^2+1)]}{(x^2-1)^2}.
\displaystyle =\frac{2x(-2)}{(x^2-1)^2}.
\displaystyle =-\frac{4x}{(x^2-1)^2}.
\displaystyle \therefore \text{The correct option is (a) }-\frac{4x}{(x^2-1)^2}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }y=\frac{\sin x+\cos x}{\sin x-\cos x},\text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is equal to:}
\displaystyle \text{(a) }-2\qquad\text{(b) }0\qquad\text{(c) }\lim_{x\to\pi}f(x)\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle y=\frac{\sin x+\cos x}{\sin x-\cos x}.
\displaystyle \therefore \frac{dy}{dx}=\frac{(\sin x-\cos x)(\cos x-\sin x)}{(\sin x-\cos x)^2}
\displaystyle -\frac{(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}.
\displaystyle =\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2}.
\displaystyle =\frac{-2(\sin^2x+\cos^2x)}{(\sin x-\cos x)^2}.
\displaystyle =-\frac{2}{(\sin x-\cos x)^2}.
\displaystyle \text{At }x=0,
\displaystyle \frac{dy}{dx}=-\frac{2}{(\sin0-\cos0)^2}=-\frac{2}{(-1)^2}=-2.
\displaystyle \therefore \text{The correct option is (a) }-2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }f(x)=1+x+\frac{x^2}{2}+\cdots+\frac{x^{100}}{100},\text{ then }f'(1)\text{ is equal to:}
\displaystyle \text{(a) }\frac{1}{100}\qquad\text{(b) }100\qquad\text{(c) }0\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^{100}}{100}.
\displaystyle \therefore f'(x)=1+x+x^2+\cdots+x^{99}.
\displaystyle \text{At }x=1,
\displaystyle f'(1)=1+1+1+\cdots+1.
\displaystyle \text{There are }100\text{ terms.}
\displaystyle \therefore f'(1)=100.
\displaystyle \therefore \text{The correct option is (b) }100.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }f(x)=\frac{x^n-a^n}{x-a}\text{ for some constant }a,\text{ then }f'(a)\text{ is equal to:}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{x^n-a^n}{x-a}.
\displaystyle \text{At }x=a,\text{ the denominator }x-a=0.
\displaystyle \text{Also, the numerator }x^n-a^n=a^n-a^n=0.
\displaystyle \therefore f(a)\text{ is not defined.}
\displaystyle \text{Since }f(a)\text{ is not defined, }f'(a)\text{ does not exist.}
\displaystyle \therefore \text{The correct option is (d) Does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }f(x)=x^{100}+x^{99}+\cdots+x+1,\text{ then }f'(1)\text{ is equal to:}
\displaystyle \text{(a) }5050\qquad\text{(b) }5049\qquad\text{(c) }5051\qquad\text{(d) }50051
\displaystyle \text{Answer:}
\displaystyle f(x)=x^{100}+x^{99}+\cdots+x+1.
\displaystyle \therefore f'(x)=100x^{99}+99x^{98}+\cdots+2x+1.
\displaystyle \text{At }x=1,
\displaystyle f'(1)=100+99+\cdots+2+1.
\displaystyle =\frac{100(100+1)}{2}.
\displaystyle =5050.
\displaystyle \therefore \text{The correct option is (a) }5050.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }y=\frac{\sin(x+9)}{\cos x},\text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is equal to:}
\displaystyle \text{(a) }\cos9\qquad\text{(b) }\sin9\qquad\text{(c) }0\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle y=\frac{\sin(x+9)}{\cos x}.
\displaystyle \therefore \frac{dy}{dx}=\frac{\cos(x+9)\cos x+\sin(x+9)\sin x}{\cos^2x}.
\displaystyle =\frac{\cos[(x+9)-x]}{\cos^2x}.
\displaystyle =\frac{\cos9}{\cos^2x}.
\displaystyle \text{At }x=0,
\displaystyle \frac{dy}{dx}=\frac{\cos9}{\cos^20}=\cos9.
\displaystyle \therefore \text{The correct option is (a) }\cos9.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }f(x)=1-x+x^2-x^3+\cdots-x^{99}+x^{100},\text{ then }f'(1)\text{ is equal to:}
\displaystyle \text{(a) }150\qquad\text{(b) }-50\qquad\text{(c) }-150\qquad\text{(d) }50
\displaystyle \text{Answer:}
\displaystyle f(x)=1-x+x^2-x^3+\cdots-x^{99}+x^{100}.
\displaystyle \therefore f'(x)=-1+2x-3x^2+4x^3-\cdots-99x^{98}+100x^{99}.
\displaystyle \text{At }x=1,
\displaystyle f'(1)=-1+2-3+4-\cdots-99+100.
\displaystyle =(-1+2)+(-3+4)+\cdots+(-99+100).
\displaystyle =1+1+\cdots+1.
\displaystyle \text{There are }50\text{ pairs.}
\displaystyle \therefore f'(1)=50.
\displaystyle \therefore \text{The correct option is (d) }50.
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED QUESTIONS}

\displaystyle \textbf{Question 21: }\text{Indeterminate forms of limits.}
\displaystyle \text{On direct evaluation, if a limit takes an indeterminate form, we use standard results}
\displaystyle \text{for evaluating the limits. Some common indeterminate forms are:}
\displaystyle \frac{0}{0},\quad\frac{\infty}{\infty},\quad 0^0,\quad1^\infty,\quad0\times\infty,\quad\infty-\infty.
\displaystyle \text{Based on the above information, answer any four of the following questions.}

\displaystyle \text{(i) }\lim_{x\to2}\frac{x^6-64}{x-2}\text{ is equal to:}
\displaystyle \text{(a) }0\qquad\text{(b) }80\qquad\text{(c) }192\qquad\text{(d) }\infty
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to2}\frac{x^6-64}{x-2}
\displaystyle =\lim_{x\to2}\frac{x^6-2^6}{x-2}.
\displaystyle =\lim_{x\to2}\left(x^5+2x^4+4x^3+8x^2+16x+32\right).
\displaystyle =32+32+32+32+32+32=192.
\displaystyle \therefore \text{The correct option is (c) }192.

\displaystyle \text{(ii) }\lim_{x\to1}\frac{x^{15}-1}{x^{10}-1}\text{ is equal to:}
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{3}{2}\qquad\text{(c) }\infty\qquad\text{(d) }15
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{x^{15}-1}{x^{10}-1}
\displaystyle =\lim_{x\to1}\frac{\frac{x^{15}-1}{x-1}}{\frac{x^{10}-1}{x-1}}.
\displaystyle =\frac{15}{10}=\frac{3}{2}.
\displaystyle \therefore \text{The correct option is (b) }\frac{3}{2}.

\displaystyle \text{(iii) }\lim_{x\to0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}\text{ is equal to:}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }3\qquad\text{(d) }6
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}
\displaystyle =\lim_{x\to0}\frac{(1+3x)-(1-3x)}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}.
\displaystyle =\lim_{x\to0}\frac{6}{\sqrt{1+3x}+\sqrt{1-3x}}.
\displaystyle =\frac{6}{1+1}=3.
\displaystyle \therefore \text{The correct option is (c) }3.

\displaystyle \text{(iv) }\lim_{x\to0}\frac{8^x-2^x}{x}\text{ is equal to:}
\displaystyle \text{(a) }0\qquad\text{(b) }\log2\qquad\text{(c) }\log4\qquad\text{(d) }\log8
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{8^x-2^x}{x}
\displaystyle =\lim_{x\to0}\left(\frac{8^x-1}{x}-\frac{2^x-1}{x}\right).
\displaystyle =\log8-\log2.
\displaystyle =\log\left(\frac{8}{2}\right)=\log4.
\displaystyle \therefore \text{The correct option is (c) }\log4.

\displaystyle \text{(v) }\lim_{x\to0}\frac{\cos5x-\cos x}{x^2}\text{ is equal to:}
\displaystyle \text{(a) }0\qquad\text{(b) }-12\qquad\text{(c) }1\qquad\text{(d) }12
\displaystyle \text{Answer:}
\displaystyle \cos5x-\cos x=-2\sin3x\sin2x.
\displaystyle \therefore \lim_{x\to0}\frac{\cos5x-\cos x}{x^2}
\displaystyle =-2\lim_{x\to0}\frac{\sin3x}{x}\cdot\frac{\sin2x}{x}.
\displaystyle =-2(3)(2)=-12.
\displaystyle \therefore \text{The correct option is (b) }-12.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The derivative of }y\text{ with respect to }x\text{ is the change in }y\text{ with respect}
\displaystyle \text{to a change in }x.\text{ The derivative of }f(x)\text{ at }x_0\text{ is given by}
\displaystyle f'(x_0)=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.
\displaystyle \text{Based on the above information, answer any four of the following questions.}

\displaystyle \text{(i) The derivative of }\sin x\text{ with respect to }x\text{ is:}
\displaystyle \text{(a) }\sin x\qquad\text{(b) }\cos x\qquad\text{(c) }-\sin x\qquad\text{(d) }-\cos x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\sin x)=\cos x.
\displaystyle \therefore \text{The correct option is (b) }\cos x.

\displaystyle \text{(ii) The derivative of }\cos x\text{ with respect to }x\text{ is:}
\displaystyle \text{(a) }\sin x\qquad\text{(b) }\cos x\qquad\text{(c) }-\sin x\qquad\text{(d) }-\cos x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\cos x)=-\sin x.
\displaystyle \therefore \text{The correct option is (c) }-\sin x.

\displaystyle \text{(iii) The derivative of }\tan x\text{ with respect to }x\text{ is:}
\displaystyle \text{(a) }\sec^2x\qquad\text{(b) }-\sec^2x\qquad\text{(c) }\mathrm{cosec}^2x\qquad\text{(d) }-\mathrm{cosec}^2x
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\tan x)=\sec^2x.
\displaystyle \therefore \text{The correct option is (a) }\sec^2x.

\displaystyle \text{(iv) If }f(x)=x^{100}-x^{50},\text{ then }f'(1)\text{ is:}
\displaystyle \text{(a) }0\qquad\text{(b) }50\qquad\text{(c) }51\qquad\text{(d) }101
\displaystyle \text{Answer:}
\displaystyle f(x)=x^{100}-x^{50}.
\displaystyle \therefore f'(x)=100x^{99}-50x^{49}.
\displaystyle \text{At }x=1,
\displaystyle f'(1)=100-50=50.
\displaystyle \therefore \text{The correct option is (b) }50.

\displaystyle \text{(v) If }y=\frac{x}{\tan x},\text{ then }\frac{dy}{dx}\text{ is:}
\displaystyle \text{(a) }\cos^2x\qquad\text{(b) }\sec^2x\qquad\text{(c) }\frac{\tan x-\sec x}{\tan^2x}
\displaystyle \text{(d) }\frac{\tan x-x\sec^2x}{\tan^2x}
\displaystyle \text{Answer:}
\displaystyle y=\frac{x}{\tan x}.
\displaystyle \therefore \frac{dy}{dx}=\frac{\tan x\cdot1-x\sec^2x}{\tan^2x}.
\displaystyle =\frac{\tan x-x\sec^2x}{\tan^2x}.
\displaystyle \therefore \text{The correct option is (d) }\frac{\tan x-x\sec^2x}{\tan^2x}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Let }f:R\to[0,\infty)\text{ be a function defined as }f(x)=|x|\text{ and}
\displaystyle g(x)=f(x+1)+f(x-1),\quad x\in R.
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) The value of }g(x)\text{ is:}
\displaystyle \text{(a) }g(x)=\begin{cases}-2x,&x<-1,\\2,&-1\le x<1,\\2x,&x\ge1,\end{cases}
\displaystyle \text{(b) }g(x)=\begin{cases}-2x+1,&x<-1,\\-2x,&-1\le x<1,\\2x-1,&x\ge1,\end{cases}
\displaystyle \text{(c) }g(x)=\begin{cases}2x,&x<-1,\\-2,&-1\le x<1,\\-2x,&x\ge1,\end{cases}
\displaystyle \text{(d) }g(x)=\begin{cases}2x-1,&x<-1,\\-2x,&-1\le x<1,\\-2x+1,&x\ge1.\end{cases}
\displaystyle \text{Answer:}
\displaystyle g(x)=|x+1|+|x-1|.
\displaystyle \text{For }x<-1,\quad g(x)=-(x+1)-(x-1)=-2x.
\displaystyle \text{For }-1\le x<1,\quad g(x)=(x+1)-(x-1)=2.
\displaystyle \text{For }x\ge1,\quad g(x)=(x+1)+(x-1)=2x.
\displaystyle \therefore g(x)=\begin{cases}-2x,&x<-1,\\2,&-1\le x<1,\\2x,&x\ge1.\end{cases}
\displaystyle \therefore \text{The correct option is (a).}

\displaystyle \text{(ii) The graph of }g(x)\text{ is:}
\displaystyle \text{Answer:}
\displaystyle g(x)=\begin{cases}-2x,&x<-1,\\2,&-1\le x<1,\\2x,&x\ge1.\end{cases}
\displaystyle \text{For }-1\le x<1,\text{ the graph is the horizontal line segment }y=2.
\displaystyle \text{For }x<-1,\text{ the graph is }y=-2x,\text{ which rises towards the left.}
\displaystyle \text{For }x\ge1,\text{ the graph is }y=2x,\text{ which rises towards the right.}
\displaystyle \therefore \text{The correct graph is (a).}

\displaystyle \text{(iii) If }\lim_{x\to-1}g(x)=a,\text{ then the value of }a\text{ is:}
\displaystyle \text{(a) }0\qquad\text{(b) }-2\qquad\text{(c) }2\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to-1^-}g(x)=\lim_{x\to-1^-}(-2x)=2.
\displaystyle \lim_{x\to-1^+}g(x)=\lim_{x\to-1^+}2=2.
\displaystyle \therefore \lim_{x\to-1}g(x)=2.
\displaystyle \therefore a=2.
\displaystyle \therefore \text{The correct option is (c) }2.

\displaystyle \text{(iv) The value of }\lim_{x\to1^+}g(x)\text{ is:}
\displaystyle \text{(a) }0\qquad\text{(b) }2\qquad\text{(c) }-2\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{For }x>1,\quad g(x)=2x.
\displaystyle \therefore \lim_{x\to1^+}g(x)=\lim_{x\to1^+}2x=2.
\displaystyle \therefore \text{The correct option is (b) }2.

\displaystyle \text{(v) The value of }\lim_{x\to1^-}g(x)\text{ is:}
\displaystyle \text{(a) }0\qquad\text{(b) }2\qquad\text{(c) }-2\qquad\text{(d) Does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{For }-1\le x<1,\quad g(x)=2.
\displaystyle \therefore \lim_{x\to1^-}g(x)=2.
\displaystyle \therefore \text{The correct option is (b) }2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Let }f(x)\text{ be a real function defined as}
\displaystyle f(x)=\begin{cases}\dfrac{\log(1+ax)-\log(1-bx)}{x},&x<0,\\[4pt]5,&x=0,\\[4pt]\dfrac{\sqrt{1+bx}-1}{x},&x>0.\end{cases}
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) }\lim_{x\to0^-}f(x)\text{ is:}
\displaystyle \text{(a) }a+b\qquad\text{(b) }a-b\qquad\text{(c) }b-a\qquad\text{(d) }-a-b
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{\log(1+ax)-\log(1-bx)}{x}.
\displaystyle =\lim_{x\to0^-}\left[\frac{\log(1+ax)}{x}-\frac{\log(1-bx)}{x}\right].
\displaystyle =a\lim_{x\to0^-}\frac{\log(1+ax)}{ax}
\displaystyle +b\lim_{x\to0^-}\frac{\log(1-bx)}{-bx}.
\displaystyle =a+b.
\displaystyle \therefore \text{The correct option is (a) }a+b.

\displaystyle \text{(ii) }\lim_{x\to0^+}f(x)\text{ is:}
\displaystyle \text{(a) }b\qquad\text{(b) }\frac{b}{2}\qquad\text{(c) }\frac{b}{3}\qquad\text{(d) }\frac{b}{4}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0^+}f(x)=\lim_{x\to0^+}\frac{\sqrt{1+bx}-1}{x}.
\displaystyle =\lim_{x\to0^+}\frac{(\sqrt{1+bx}-1)(\sqrt{1+bx}+1)}{x(\sqrt{1+bx}+1)}.
\displaystyle =\lim_{x\to0^+}\frac{bx}{x(\sqrt{1+bx}+1)}.
\displaystyle =\lim_{x\to0^+}\frac{b}{\sqrt{1+bx}+1}=\frac{b}{2}.
\displaystyle \therefore \text{The correct option is (b) }\frac{b}{2}.

\displaystyle \text{(iii) If }\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x),\text{ then a relation between }a\text{ and }b\text{ is:}
\displaystyle \text{(a) }a+2b=0\qquad\text{(b) }2a-b=0\qquad\text{(c) }2a+b=0\qquad\text{(d) }3a+2b=0
\displaystyle \text{Answer:}
\displaystyle a+b=\frac{b}{2}.
\displaystyle 2a+2b=b.
\displaystyle \therefore 2a+b=0.
\displaystyle \therefore \text{The correct option is (c) }2a+b=0.

\displaystyle \text{(iv) The value of }b,\text{ if }\lim_{x\to0^+}f(x)=f(0),\text{ is:}
\displaystyle \text{(a) }5\qquad\text{(b) }15\qquad\text{(c) }20\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0^+}f(x)=\frac{b}{2}\text{ and }f(0)=5.
\displaystyle \therefore \frac{b}{2}=5.
\displaystyle \therefore b=10.
\displaystyle \therefore \text{The correct option is (d) }10.

\displaystyle \text{(v) The values of }a\text{ and }b\text{ if }\lim_{x\to0^-}f(x)=f(0)=\lim_{x\to0^+}f(x)\text{ are:}
\displaystyle \text{(a) }-5,5\qquad\text{(b) }-5,10\qquad\text{(c) }5,10\qquad\text{(d) }10,15
\displaystyle \text{Answer:}
\displaystyle \text{For continuity at }x=0,
\displaystyle \lim_{x\to0^-}f(x)=f(0)=\lim_{x\to0^+}f(x)=5.
\displaystyle \frac{b}{2}=5\Rightarrow b=10.
\displaystyle a+b=5.
\displaystyle \therefore a+10=5\Rightarrow a=-5.
\displaystyle \therefore (a,b)=(-5,10).
\displaystyle \therefore \text{The correct option is (b) }-5,10.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\lim_{x\to c}f(x)^{g(x)}\text{ is of the form }1^\infty,\text{ then}
\displaystyle \lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}g(x)[f(x)-1]}.
\displaystyle \text{For example, }\lim_{x\to0}(1+ax)^{1/x}=e^a.
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) The value of }\lim_{x\to\infty}\left(\frac{x-3}{x+2}\right)^x\text{ is:}
\displaystyle \text{(a) }e\qquad\text{(b) }e^{-1}\qquad\text{(c) }e^{-5}\qquad\text{(d) }e^5
\displaystyle \text{Answer:}
\displaystyle \frac{x-3}{x+2}=1-\frac{5}{x+2}.
\displaystyle \therefore \lim_{x\to\infty}\left(\frac{x-3}{x+2}\right)^x
\displaystyle =e^{\lim_{x\to\infty}x\left(-\frac{5}{x+2}\right)}.
\displaystyle =e^{-5}.
\displaystyle \therefore \text{The correct option is (c) }e^{-5}.

\displaystyle \text{(ii) The value of }\lim_{x\to0}\left\{\tan\left(\frac{\pi}{4}+x\right)\right\}^{1/x}\text{ is:}
\displaystyle \text{(a) }e^2\qquad\text{(b) }e^3\qquad\text{(c) }e^{-2}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{Let }L=\lim_{x\to0}\left\{\tan\left(\frac{\pi}{4}+x\right)\right\}^{1/x}.
\displaystyle L=e^{\lim_{x\to0}\frac{1}{x}\left[\tan\left(\frac{\pi}{4}+x\right)-1\right]}.
\displaystyle \tan\left(\frac{\pi}{4}+x\right)=\frac{1+\tan x}{1-\tan x}.
\displaystyle \therefore \tan\left(\frac{\pi}{4}+x\right)-1=\frac{2\tan x}{1-\tan x}.
\displaystyle \therefore L=e^{\lim_{x\to0}\frac{2\tan x}{x(1-\tan x)}}.
\displaystyle =e^{2}.
\displaystyle \therefore \text{The correct option is (a) }e^2.

\displaystyle \text{(iii) The value of }\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan^2x}\text{ is:}
\displaystyle \text{(a) }e^2\qquad\text{(b) }e^{1/2}\qquad\text{(c) }e\qquad\text{(d) }e^{-1/2}
\displaystyle \text{Answer:}
\displaystyle \text{Let }L=\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan^2x}.
\displaystyle \text{Put }h=x-\frac{\pi}{2}.\text{ Then }h\to0\text{ and }\sin x=\cos h.
\displaystyle L=\lim_{h\to0}(\cos h)^{\cot^2h}.
\displaystyle =e^{\lim_{h\to0}\cot^2h(\cos h-1)}.
\displaystyle =e^{\lim_{h\to0}\frac{\cos^2h}{\sin^2h}(\cos h-1)}.
\displaystyle =e^{-\lim_{h\to0}\frac{\cos^2h}{1+\cos h}}.
\displaystyle =e^{-1/2}.
\displaystyle \therefore \text{The correct option is (d) }e^{-1/2}.

\displaystyle \text{(iv) The value of }\lim_{x\to0}\left(\frac{a^x+b^x+c^x}{3}\right)^{1/x}\text{ is:}
\displaystyle \text{(a) }(abc)^{1/3}\qquad\text{(b) }\frac{1}{3}(abc)\qquad\text{(c) }\log(abc)^{1/3}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{Let }L=\lim_{x\to0}\left(\frac{a^x+b^x+c^x}{3}\right)^{1/x}.
\displaystyle L=e^{\lim_{x\to0}\frac{1}{x}\left(\frac{a^x+b^x+c^x}{3}-1\right)}.
\displaystyle =e^{\frac{1}{3}\lim_{x\to0}\left(\frac{a^x-1}{x}+\frac{b^x-1}{x}+\frac{c^x-1}{x}\right)}.
\displaystyle =e^{\frac{1}{3}(\log a+\log b+\log c)}.
\displaystyle =e^{\frac{1}{3}\log(abc)}=(abc)^{1/3}.
\displaystyle \therefore \text{The correct option is (a) }(abc)^{1/3}.

\displaystyle \text{(v) The value of }\lim_{x\to0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\mathrm{cosec}\,x}\text{ is:}
\displaystyle \text{(a) }e\qquad\text{(b) }\frac{1}{e}\qquad\text{(c) }1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Let }L=\lim_{x\to0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\mathrm{cosec}\,x}.
\displaystyle L=e^{\lim_{x\to0}\mathrm{cosec}\,x\left(\frac{1+\tan x}{1+\sin x}-1\right)}.
\displaystyle =e^{\lim_{x\to0}\frac{\tan x-\sin x}{\sin x(1+\sin x)}}.
\displaystyle \tan x-\sin x=\sin x\left(\frac{1-\cos x}{\cos x}\right).
\displaystyle \therefore L=e^{\lim_{x\to0}\frac{1-\cos x}{\cos x(1+\sin x)}}.
\displaystyle =e^0=1.
\displaystyle \therefore \text{The correct option is (c) }1.
\displaystyle \\

\displaystyle \text{ASSERTION REASONING QUESTIONS}

\displaystyle \textbf{Question 26: }\text{Statement }(S_1):\text{ If }f(x)=\sin^2x+\frac{1}{2}\cos2x+\cot\alpha, \\ \text{then }f'(x)=0.
\displaystyle \text{Reason }(S_2):\text{ The derivative of a constant function is always zero.}
\displaystyle \text{(a) Both }S_1\text{ and }S_2\text{ are true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) Both }S_1\text{ and }S_2\text{ are true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true but }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false but }S_2\text{ is true.}
\displaystyle \text{(e) Both }S_1\text{ and }S_2\text{ are false.}
\displaystyle \text{Answer:}
\displaystyle \sin^2x=\frac{1-\cos2x}{2}.
\displaystyle \therefore \sin^2x+\frac{1}{2}\cos2x=\frac{1}{2}.
\displaystyle \therefore f(x)=\frac{1}{2}+\cot\alpha,
\displaystyle \text{which is a constant. Hence }f'(x)=0.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{Also, the derivative of a constant function is zero, so }S_2\text{ is true.}
\displaystyle \therefore \text{According to the book, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Statement }(S_1):\text{ The derivative of }y=2x-\frac{3}{4}\text{ is }2.
\displaystyle \text{Reason }(S_2):\text{ The derivative of }y=cx\text{ is }c.
\displaystyle \text{(a) Both }S_1\text{ and }S_2\text{ are true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) Both }S_1\text{ and }S_2\text{ are true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true but }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false but }S_2\text{ is true.}
\displaystyle \text{(e) Both }S_1\text{ and }S_2\text{ are false.}
\displaystyle \text{Answer:}
\displaystyle y=2x-\frac{3}{4}.
\displaystyle \therefore \frac{dy}{dx}=\frac{d}{dx}(2x)-\frac{d}{dx}\left(\frac{3}{4}\right).
\displaystyle =2-0=2.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{Also, }\frac{d}{dx}(cx)=c.
\displaystyle \text{Taking }c=2,\quad \frac{d}{dx}(2x)=2.
\displaystyle \therefore S_2\text{ is true and explains }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Statement }(S_1):\text{ The derivative of }f(x)=x^3\text{ is }x^2.
\displaystyle \text{Reason }(S_2):\text{ The derivative of }f(x)=x^n\text{ is }nx^{n-1}.
\displaystyle \text{(a) Both }S_1\text{ and }S_2\text{ are true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) Both }S_1\text{ and }S_2\text{ are true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true but }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false but }S_2\text{ is true.}
\displaystyle \text{(e) Both }S_1\text{ and }S_2\text{ are false.}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(x^3)=3x^{3-1}=3x^2.
\displaystyle \therefore S_1\text{ is false.}
\displaystyle \text{Also, }\frac{d}{dx}(x^n)=nx^{n-1}\text{ is the power rule of differentiation.}
\displaystyle \therefore S_2\text{ is true.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Statement }(S_1):\lim_{x\to0}\frac{\sin ax}{\sin bx}=\frac{a}{b}.
\displaystyle \text{Reason }(S_2):\lim_{x\to0}\frac{\sin x}{x}=1.
\displaystyle \text{(a) Both }S_1\text{ and }S_2\text{ are true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) Both }S_1\text{ and }S_2\text{ are true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true but }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false but }S_2\text{ is true.}
\displaystyle \text{(e) Both }S_1\text{ and }S_2\text{ are false.}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin ax}{\sin bx}
\displaystyle =\lim_{x\to0}\frac{\sin ax}{ax}\cdot\frac{bx}{\sin bx}\cdot\frac{a}{b}.
\displaystyle =1\cdot1\cdot\frac{a}{b}=\frac{a}{b}.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{Also, }\lim_{x\to0}\frac{\sin x}{x}=1,\text{ so }S_2\text{ is true.}
\displaystyle \text{The result in }S_2\text{ is used directly to establish }S_1.
\displaystyle \therefore S_2\text{ is the correct explanation of }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Statement }(S_1):\lim_{x\to0}\frac{\cos2x-1}{\cos x-1}=4.
\displaystyle \text{Reason }(S_2):\lim_{x\to0}\frac{\tan x}{x}=1.
\displaystyle \text{(a) Both }S_1\text{ and }S_2\text{ are true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) Both }S_1\text{ and }S_2\text{ are true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true but }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false but }S_2\text{ is true.}
\displaystyle \text{(e) Both }S_1\text{ and }S_2\text{ are false.}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\cos2x-1}{\cos x-1}
\displaystyle =\lim_{x\to0}\frac{-2\sin^2x}{-2\sin^2(x/2)}.
\displaystyle =\lim_{x\to0}\left(\frac{\sin x}{\sin(x/2)}\right)^2.
\displaystyle =\lim_{x\to0}\left(2\cos\frac{x}{2}\right)^2=4.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{Also, }\lim_{x\to0}\frac{\tan x}{x}=1,\text{ so }S_2\text{ is true.}
\displaystyle \text{However, }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{Find }\lim_{x\to a}\frac{\sqrt{x}+\sqrt{a}}{x+a}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to a}\frac{\sqrt{x}+\sqrt{a}}{x+a}
\displaystyle =\frac{\sqrt{a}+\sqrt{a}}{a+a}.
\displaystyle =\frac{2\sqrt{a}}{2a}=\frac{1}{\sqrt{a}}.
\displaystyle \therefore \text{The required limit is }\frac{1}{\sqrt{a}}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the value of }\lim_{x\to0}\frac{e^x-1}{x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{e^x-1}{x}=1.
\displaystyle \therefore \text{The required limit is }1.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Evaluate }\lim_{x\to0}\frac{\sin nx}{x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin nx}{x}
\displaystyle =n\lim_{x\to0}\frac{\sin nx}{nx}.
\displaystyle =n.
\displaystyle \therefore \text{The required limit is }n.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Evaluate }\lim_{x\to3}\frac{\sqrt{x+1}}{x+1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to3}\frac{\sqrt{x+1}}{x+1}
\displaystyle =\frac{\sqrt{3+1}}{3+1}.
\displaystyle =\frac{2}{4}=\frac{1}{2}.
\displaystyle \therefore \text{The required limit is }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Evaluate }\lim_{x\to1}\frac{\sqrt{1+x}-\sqrt{1-x}}{1+x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{\sqrt{1+x}-\sqrt{1-x}}{1+x}
\displaystyle =\frac{\sqrt{1+1}-\sqrt{1-1}}{1+1}.
\displaystyle =\frac{\sqrt2}{2}=\frac{1}{\sqrt2}.
\displaystyle \therefore \text{The required limit is }\frac{1}{\sqrt2}.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Evaluate }\lim_{n\to\infty}\frac{1+2+3+\cdots+n}{n^2}.
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty}\frac{1+2+3+\cdots+n}{n^2}
\displaystyle =\lim_{n\to\infty}\frac{n(n+1)}{2n^2}.
\displaystyle =\frac{1}{2}\lim_{n\to\infty}\left(1+\frac{1}{n}\right).
\displaystyle =\frac{1}{2}.
\displaystyle \therefore \text{The required limit is }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find }\lim_{x\to0}\frac{e^{-x}-1+x}{x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{e^{-x}-1+x}{x}
\displaystyle =\lim_{x\to0}\frac{e^{-x}-1}{x}+1.
\displaystyle =-1+1=0.
\displaystyle \therefore \text{The required limit is }0.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Evaluate }\lim_{x\to5}\frac{e^x-e^5}{x-5}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to5}\frac{e^x-e^5}{x-5}
\displaystyle =e^5\lim_{x\to5}\frac{e^{x-5}-1}{x-5}.
\displaystyle =e^5.
\displaystyle \therefore \text{The required limit is }e^5.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Find the derivative of the function }y=1+x+x^2+\cdots+x^{50}\text{ at }x=1.
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=1+2x+3x^2+\cdots+50x^{49}.
\displaystyle \text{At }x=1,
\displaystyle \frac{dy}{dx}=1+2+3+\cdots+50.
\displaystyle =\frac{50(51)}{2}=1275.
\displaystyle \therefore \text{The required derivative is }1275.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Find the derivative of }2x^4+x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(2x^4+x)=8x^3+1.
\displaystyle \therefore \text{The required derivative is }8x^3+1.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Find }f'(x),\text{ if }f(x)=(x-2)^2(2x-3).
\displaystyle \text{Answer:}
\displaystyle f'(x)=2(x-2)(2x-3)+2(x-2)^2.
\displaystyle =2(x-2)\left[(2x-3)+(x-2)\right].
\displaystyle =2(x-2)(3x-5).
\displaystyle \therefore f'(x)=2(x-2)(3x-5).
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Find the derivative of }5\sec x+4\cos x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(5\sec x+4\cos x)
\displaystyle =5\sec x\tan x-4\sin x.
\displaystyle \therefore \text{The required derivative is }5\sec x\tan x-4\sin x.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Find the derivative of }\mathrm{cosec}\,x\cdot\cot x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\mathrm{cosec}\,x\cdot\cot x)
\displaystyle =(-\mathrm{cosec}\,x\cot x)\cot x+\mathrm{cosec}\,x(-\mathrm{cosec}^2x).
\displaystyle =-\mathrm{cosec}\,x\cot^2x-\mathrm{cosec}^3x.
\displaystyle \therefore \text{The required derivative is }-\mathrm{cosec}\,x\cot^2x-\mathrm{cosec}^3x.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Find the derivative of }(\sec x-1)(\sec x+1).
\displaystyle \text{Answer:}
\displaystyle (\sec x-1)(\sec x+1)=\sec^2x-1=\tan^2x.
\displaystyle \therefore \frac{d}{dx}(\tan^2x)=2\tan x\sec^2x.
\displaystyle \therefore \text{The required derivative is }2\tan x\sec^2x.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Find the derivative of }\sin(x+a).
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\sin(x+a)=\cos(x+a)\frac{d}{dx}(x+a).
\displaystyle =\cos(x+a).
\displaystyle \therefore \text{The required derivative is }\cos(x+a).
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Find the derivative of }\frac{1}{x^{11}}.
\displaystyle \text{Answer:}
\displaystyle \frac{1}{x^{11}}=x^{-11}.
\displaystyle \therefore \frac{d}{dx}(x^{-11})=-11x^{-12}.
\displaystyle =-\frac{11}{x^{12}}.
\displaystyle \therefore \text{The required derivative is }-\frac{11}{x^{12}}.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Find the derivative of }(x^2+1)\cos x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left[(x^2+1)\cos x\right]
\displaystyle =2x\cos x-(x^2+1)\sin x.
\displaystyle \therefore \text{The required derivative is }2x\cos x-(x^2+1)\sin x.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the derivative of }\cot^3x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(\cot^3x)=3\cot^2x\frac{d}{dx}(\cot x).
\displaystyle =-3\cot^2x\,\mathrm{cosec}^2x.
\displaystyle \therefore \text{The required derivative is }-3\cot^2x\,\mathrm{cosec}^2x.
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 51: }\text{Evaluate }\lim_{x\to1}\frac{x^{15}-1}{x^{10}-1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{x^{15}-1}{x^{10}-1}
\displaystyle =\lim_{x\to1}\frac{\dfrac{x^{15}-1}{x-1}}{\dfrac{x^{10}-1}{x-1}}.
\displaystyle =\frac{15}{10}=\frac{3}{2}.
\displaystyle \therefore \text{The required limit is }\frac{3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Evaluate }\lim_{x\to0}\frac{\sin ax+bx}{ax+\sin bx},\quad a,b,a+b\ne0.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin ax+bx}{ax+\sin bx}
\displaystyle =\lim_{x\to0}\frac{\dfrac{\sin ax}{x}+b}{a+\dfrac{\sin bx}{x}}.
\displaystyle =\frac{a+b}{a+b}=1.
\displaystyle \therefore \text{The required limit is }1.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Find }\lim_{z\to1}\frac{z^{1/3}-1}{z^{1/6}-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=z^{1/6}.
\displaystyle \text{Then }z^{1/3}=t^2\text{ and as }z\to1,\ t\to1.
\displaystyle \therefore \lim_{z\to1}\frac{z^{1/3}-1}{z^{1/6}-1}
\displaystyle =\lim_{t\to1}\frac{t^2-1}{t-1}.
\displaystyle =\lim_{t\to1}(t+1)=2.
\displaystyle \therefore \text{The required limit is }2.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Evaluate the left hand and right hand limits of the following function at }x=2.
\displaystyle \text{Does }\lim_{x\to2}f(x)\text{ exist?}
\displaystyle f(x)=\begin{cases}2x+3,&\text{if }x\le2,\\x+5,&\text{if }x>2.\end{cases}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to2^-}f(x)=\lim_{x\to2^-}(2x+3)=2(2)+3=7.
\displaystyle \lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x+5)=2+5=7.
\displaystyle \therefore \lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=7.
\displaystyle \therefore \lim_{x\to2}f(x)\text{ exists and is equal to }7.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Show that }\lim_{x\to4}\frac{|x-4|}{x-4}\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{For }x<4,\quad |x-4|=-(x-4).
\displaystyle \therefore \lim_{x\to4^-}\frac{|x-4|}{x-4}
\displaystyle =\lim_{x\to4^-}\frac{-(x-4)}{x-4}=-1.
\displaystyle \text{For }x>4,\quad |x-4|=x-4.
\displaystyle \therefore \lim_{x\to4^+}\frac{|x-4|}{x-4}
\displaystyle =\lim_{x\to4^+}\frac{x-4}{x-4}=1.
\displaystyle \text{Since LHL}\ne\text{RHL},
\displaystyle \therefore \lim_{x\to4}\frac{|x-4|}{x-4}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Evaluate }\lim_{x\to1}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}
\displaystyle =\lim_{x\to1}\frac{(2x-3)(\sqrt{x}-1)}{(x-1)(2x+3)}.
\displaystyle =\lim_{x\to1}\frac{2x-3}{(\sqrt{x}+1)(2x+3)}.
\displaystyle =\frac{2-3}{(1+1)(2+3)}.
\displaystyle =-\frac{1}{10}.
\displaystyle \therefore \text{The required limit is }-\frac{1}{10}.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Evaluate }\lim_{x\to1}\frac{x^m-1}{x^n-1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{x^m-1}{x^n-1}
\displaystyle =\lim_{x\to1}\frac{\dfrac{x^m-1}{x-1}}{\dfrac{x^n-1}{x-1}}.
\displaystyle =\frac{m}{n}.
\displaystyle \therefore \text{The required limit is }\frac{m}{n}.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Evaluate }\lim_{x\to0}\frac{\tan2x-\sin2x}{x^3}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\tan2x-\sin2x}{x^3}
\displaystyle =\lim_{x\to0}\frac{\sin2x(1-\cos2x)}{x^3\cos2x}.
\displaystyle =\lim_{x\to0}\frac{\sin2x}{x}\cdot\frac{1-\cos2x}{x^2}\cdot\frac{1}{\cos2x}.
\displaystyle =2\cdot2\cdot1=4.
\displaystyle \therefore \text{The required limit is }4.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Evaluate }\lim_{x\to\frac{\pi}{6}}\frac{\sqrt{3}\sin x-\cos x}{x-\frac{\pi}{6}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\sqrt{3}\sin x-\cos x.
\displaystyle f\left(\frac{\pi}{6}\right)=\sqrt{3}\cdot\frac{1}{2}-\frac{\sqrt{3}}{2}=0.
\displaystyle \therefore \lim_{x\to\frac{\pi}{6}}\frac{\sqrt{3}\sin x-\cos x}{x-\frac{\pi}{6}}
\displaystyle =f'\left(\frac{\pi}{6}\right).
\displaystyle f'(x)=\sqrt{3}\cos x+\sin x.
\displaystyle \therefore f'\left(\frac{\pi}{6}\right)=\sqrt{3}\cdot\frac{\sqrt{3}}{2}+\frac{1}{2}.
\displaystyle =\frac{3}{2}+\frac{1}{2}=2.
\displaystyle \therefore \text{The required limit is }2.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Evaluate }\lim_{x\to0}\frac{\log(6+x)-\log(6-x)}{x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\log(6+x)-\log(6-x)}{x}
\displaystyle =\lim_{x\to0}\frac{1}{x}\log\left(\frac{6+x}{6-x}\right).
\displaystyle =\lim_{x\to0}\frac{1}{x}\log\left(1+\frac{2x}{6-x}\right).
\displaystyle =\lim_{x\to0}\frac{\log\left(1+\frac{2x}{6-x}\right)}{\frac{2x}{6-x}}\cdot\frac{2}{6-x}.
\displaystyle =1\cdot\frac{2}{6}=\frac{1}{3}.
\displaystyle \therefore \text{The required limit is }\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Find the derivative of }\cos x\text{ by first principle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\cos x.
\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.
\displaystyle =\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}.
\displaystyle =\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}.
\displaystyle =\cos x\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\lim_{h\to0}\frac{\sin h}{h}.
\displaystyle =\cos x(0)-\sin x(1).
\displaystyle =-\sin x.
\displaystyle \therefore \frac{d}{dx}(\cos x)=-\sin x.
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Find the derivative of }\sin2x\text{ by first principle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\sin2x.
\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.
\displaystyle =\lim_{h\to0}\frac{\sin(2x+2h)-\sin2x}{h}.
\displaystyle =\lim_{h\to0}\frac{\sin2x(\cos2h-1)+\cos2x\sin2h}{h}.
\displaystyle =\sin2x\lim_{h\to0}\frac{\cos2h-1}{h}
\displaystyle +\cos2x\lim_{h\to0}\frac{\sin2h}{h}.
\displaystyle =\sin2x(0)+2\cos2x(1).
\displaystyle =2\cos2x.
\displaystyle \therefore \frac{d}{dx}(\sin2x)=2\cos2x.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Find the derivative of }f(x)=x\cos x.
\displaystyle \text{Answer:}
\displaystyle f(x)=x\cos x.
\displaystyle f'(x)=x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x).
\displaystyle =-x\sin x+\cos x.
\displaystyle \therefore f'(x)=\cos x-x\sin x.
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{Differentiate }3^x+x^3+4x-5\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(3^x+x^3+4x-5)
\displaystyle =3^x\log3+3x^2+4.
\displaystyle \therefore \text{The required derivative is }3^x\log3+3x^2+4.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Differentiate }e^x\sin x+x^n\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}(e^x\sin x+x^n)
\displaystyle =e^x\sin x+e^x\cos x+nx^{n-1}.
\displaystyle =e^x(\sin x+\cos x)+nx^{n-1}.
\displaystyle \therefore \text{The required derivative is }e^x(\sin x+\cos x)+nx^{n-1}.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Differentiate }\frac{x}{\sin x}\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\frac{x}{\sin x}\right)
\displaystyle =\frac{\sin x-x\cos x}{\sin^2x}.
\displaystyle \therefore \text{The required derivative is }\frac{\sin x-x\cos x}{\sin^2x}.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Differentiate }\frac{x^4+x^3+x^2+1}{x}\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{x^4+x^3+x^2+1}{x}=x^3+x^2+x+\frac{1}{x}.
\displaystyle \therefore \frac{d}{dx}\left(x^3+x^2+x+\frac{1}{x}\right)
\displaystyle =3x^2+2x+1-\frac{1}{x^2}.
\displaystyle \therefore \text{The required derivative is }3x^2+2x+1-\frac{1}{x^2}.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Differentiate }\left(x+\frac{1}{x}\right)^3\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(x+\frac{1}{x}\right)^3
\displaystyle =3\left(x+\frac{1}{x}\right)^2\frac{d}{dx}\left(x+\frac{1}{x}\right).
\displaystyle =3\left(x+\frac{1}{x}\right)^2\left(1-\frac{1}{x^2}\right).
\displaystyle \therefore \text{The required derivative is }3\left(x+\frac{1}{x}\right)^2\left(1-\frac{1}{x^2}\right).
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{Differentiate }(3x+5)(1+\tan x)\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left[(3x+5)(1+\tan x)\right]
\displaystyle =3(1+\tan x)+(3x+5)\sec^2x.
\displaystyle \therefore \text{The required derivative is }3(1+\tan x)+(3x+5)\sec^2x.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Differentiate }\frac{x^2\cos\frac{\pi}{4}}{\sin x}\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle y=\frac{x^2}{\sqrt{2}\sin x}.
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{\sqrt{2}}\frac{d}{dx}\left(\frac{x^2}{\sin x}\right).
\displaystyle =\frac{1}{\sqrt{2}}\left(\frac{2x\sin x-x^2\cos x}{\sin^2x}\right).
\displaystyle =\frac{2x\sin x-x^2\cos x}{\sqrt{2}\sin^2x}.
\displaystyle \therefore \text{The required derivative is }\frac{2x\sin x-x^2\cos x}{\sqrt{2}\sin^2x}.
\displaystyle \\

\displaystyle \text{LONG ANSWER TYPE QUESTIONS}

\displaystyle \textbf{Question 71: }\text{Find }\lim_{x\to1}f(x),\text{ where }
\displaystyle f(x)=\begin{cases}x^2-1,&x\le1,\\-x^2-1,&x>1.\end{cases}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1^-}f(x)=\lim_{x\to1^-}(x^2-1)=1-1=0.
\displaystyle \lim_{x\to1^+}f(x)=\lim_{x\to1^+}(-x^2-1)=-1-1=-2.
\displaystyle \therefore \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x).
\displaystyle \therefore \lim_{x\to1}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{Find the value of }k\text{ if }\lim_{x\to1}\frac{x^4-1}{x-1}=\lim_{x\to k}\frac{x^3-k^3}{x^2-k^2}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{x^4-1}{x-1}
\displaystyle =\lim_{x\to1}\frac{(x-1)(x^3+x^2+x+1)}{x-1}.
\displaystyle =1+1+1+1=4.
\displaystyle \text{Also, }\lim_{x\to k}\frac{x^3-k^3}{x^2-k^2}
\displaystyle =\lim_{x\to k}\frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}.
\displaystyle =\frac{k^2+k^2+k^2}{2k}=\frac{3k}{2}.
\displaystyle \therefore 4=\frac{3k}{2}.
\displaystyle \therefore k=\frac{8}{3}.
\displaystyle \therefore \text{The required value of }k\text{ is }\frac{8}{3}.
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{Find the values of }a\text{ and }b\text{ if }\lim_{x\to2}f(x)\text{ and }\lim_{x\to4}f(x)\text{ exist, where}
\displaystyle f(x)=\begin{cases}x^2+ax+b,&0\le x<2,\\3x+2,&2\le x\le4,\\2ax+5b,&4<x\le8.\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{For }\lim_{x\to2}f(x)\text{ to exist,}
\displaystyle \lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x).
\displaystyle 2^2+2a+b=3(2)+2.
\displaystyle 4+2a+b=8.
\displaystyle \therefore 2a+b=4.\qquad\text{...(1)}
\displaystyle \text{For }\lim_{x\to4}f(x)\text{ to exist,}
\displaystyle \lim_{x\to4^-}f(x)=\lim_{x\to4^+}f(x).
\displaystyle 3(4)+2=2a(4)+5b.
\displaystyle 14=8a+5b.\qquad\text{...(2)}
\displaystyle \text{From (1), }b=4-2a.
\displaystyle \text{Substituting in (2),}
\displaystyle 8a+5(4-2a)=14.
\displaystyle 8a+20-10a=14.
\displaystyle -2a=-6.
\displaystyle \therefore a=3.
\displaystyle \text{From (1), }2(3)+b=4.
\displaystyle \therefore b=-2.
\displaystyle \therefore \text{The required values are }a=3\text{ and }b=-2.
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{Evaluate }\lim_{x\to0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right).
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}
\displaystyle =\lim_{x\to0}\frac{\dfrac{x^2-x}{\sqrt{1+x^2}+\sqrt{1+x}}}{\dfrac{x^3-x}{\sqrt{1+x^3}+\sqrt{1+x}}}.
\displaystyle =\lim_{x\to0}\frac{x(x-1)}{x(x^2-1)}\cdot\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}.
\displaystyle =\lim_{x\to0}\frac{1}{x+1}\cdot\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}.
\displaystyle =1\cdot\frac{1+1}{1+1}=1.
\displaystyle \therefore \text{The required limit is }1.
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{Evaluate }\lim_{x\to0}\frac{\sec4x-\sec2x}{\sec3x-\sec x}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sec4x-\sec2x}{\sec3x-\sec x}
\displaystyle =\lim_{x\to0}\frac{\frac{1}{\cos4x}-\frac{1}{\cos2x}}{\frac{1}{\cos3x}-\frac{1}{\cos x}}.
\displaystyle =\lim_{x\to0}\frac{\cos2x-\cos4x}{\cos x-\cos3x}\cdot\frac{\cos3x\cos x}{\cos4x\cos2x}.
\displaystyle =\lim_{x\to0}\frac{2\sin3x\sin x}{2\sin2x\sin x}\cdot\frac{\cos3x\cos x}{\cos4x\cos2x}.
\displaystyle =\lim_{x\to0}\frac{\sin3x}{\sin2x}\cdot\frac{\cos3x\cos x}{\cos4x\cos2x}.
\displaystyle =\frac{3}{2}\cdot1=\frac{3}{2}.
\displaystyle \therefore \text{The required limit is }\frac{3}{2}.
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{Find the derivative of }f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}.
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}.
\displaystyle =\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2}.
\displaystyle =\frac{-2(\sin^2x+\cos^2x)}{(\sin x-\cos x)^2}.
\displaystyle =-\frac{2}{(\sin x-\cos x)^2}.
\displaystyle \therefore f'(x)=-\frac{2}{(\sin x-\cos x)^2}.
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Differentiate }f(x)=x^5e^x+x^3\log x-2^x\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle f'(x)=\frac{d}{dx}(x^5e^x)+\frac{d}{dx}(x^3\log x)-\frac{d}{dx}(2^x).
\displaystyle =5x^4e^x+x^5e^x+3x^2\log x+x^2-2^x\log2.
\displaystyle =x^4e^x(x+5)+x^2(3\log x+1)-2^x\log2.
\displaystyle \therefore f'(x)=x^4e^x(x+5)+x^2(3\log x+1)-2^x\log2.
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{If }y=\sqrt{x}+\frac{1}{\sqrt{x}},\text{ prove that }2x\frac{dy}{dx}+y=2\sqrt{x}.
\displaystyle \text{Answer:}
\displaystyle y=x^{1/2}+x^{-1/2}.
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{-3/2}.
\displaystyle =\frac{1}{2\sqrt{x}}-\frac{1}{2x^{3/2}}.
\displaystyle \therefore 2x\frac{dy}{dx}=2x\left(\frac{1}{2\sqrt{x}}-\frac{1}{2x^{3/2}}\right).
\displaystyle =\sqrt{x}-\frac{1}{\sqrt{x}}.
\displaystyle \therefore 2x\frac{dy}{dx}+y
\displaystyle =\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}.
\displaystyle =2\sqrt{x}.
\displaystyle \therefore 2x\frac{dy}{dx}+y=2\sqrt{x}.
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{Differentiate }f(x)=\sqrt{\frac{1-\cos2x}{1+\cos2x}}\text{ with respect to }x.
\displaystyle \text{Answer:}
\displaystyle f(x)=\sqrt{\frac{2\sin^2x}{2\cos^2x}}.
\displaystyle =\sqrt{\tan^2x}.
\displaystyle =\tan x\quad\text{(assuming }\tan x>0\text{).}
\displaystyle \therefore f'(x)=\sec^2x.
\displaystyle \therefore \text{The required derivative is }\sec^2x.
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{If }y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}},\text{ prove that }(2xy)\frac{dy}{dx}=\frac{x}{a}-\frac{a}{x}.
\displaystyle \text{Answer:}
\displaystyle y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}.
\displaystyle \text{Squaring both sides,}
\displaystyle y^2=\frac{x}{a}+\frac{a}{x}+2.
\displaystyle \text{Differentiating with respect to }x,
\displaystyle 2y\frac{dy}{dx}=\frac{1}{a}-\frac{a}{x^2}.
\displaystyle \text{Multiplying both sides by }x,
\displaystyle (2xy)\frac{dy}{dx}=\frac{x}{a}-\frac{a}{x}.
\displaystyle \therefore (2xy)\frac{dy}{dx}=\frac{x}{a}-\frac{a}{x}.
\displaystyle \\

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