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

\displaystyle \textbf{Question 1: }\lim_{n\to\infty} \frac{1^2+2^2+3^2+\cdots+n^2}{n^3}\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{3}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Using }1^2+2^2+3^2+\cdots+n^2 =\frac{n(n+1)(2n+1)}{6},
\displaystyle \lim_{n\to\infty} \frac{1^2+2^2+3^2+\cdots+n^2}{n^3} =\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}.
\displaystyle =\frac{1}{6}\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right).
\displaystyle =\frac{1}{6}(1)(2)=\frac{1}{3}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\lim_{x\to0}\frac{\sin 2x}{x}\text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }\frac{1}{2}\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin 2x}{x} =2\lim_{x\to0}\frac{\sin 2x}{2x}.
\displaystyle =2(1)=2.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }f(x)=x\sin(1/x),\ x\ne0,\text{ then }\lim_{x\to0}f(x)=
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad \text{(c) }-1\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{Since }-1\leq\sin(1/x)\leq1,
\displaystyle \left|x\sin(1/x)\right|\leq|x|.
\displaystyle \text{Also, }\lim_{x\to0}|x|=0.
\displaystyle \therefore\lim_{x\to0}x\sin(1/x)=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\lim_{x\to0}\frac{1-\cos 2x}{x}\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }2\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{Using }1-\cos 2x=2\sin^2x,
\displaystyle \lim_{x\to0}\frac{1-\cos 2x}{x} =\lim_{x\to0}\frac{2\sin^2x}{x}.
\displaystyle =2\lim_{x\to0}\left(\frac{\sin x}{x}\right)\sin x.
\displaystyle =2(1)(0)=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\lim_{x\to0} \frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}=
\displaystyle \text{(a) }\frac{10}{3}\qquad \text{(b) }\frac{3}{10}\qquad \text{(c) }\frac{6}{5}\qquad \text{(d) }\frac{5}{6}
\displaystyle \text{Answer:}
\displaystyle \text{Using }1-\cos 2x=2\sin^2x,
\displaystyle \lim_{x\to0}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}
\displaystyle =\lim_{x\to0} 2\left(\frac{\sin x}{x}\right)^2\frac{\sin 5x}{\sin 3x}.
\displaystyle =2(1)^2\lim_{x\to0} \frac{\sin 5x}{5x}\frac{3x}{\sin 3x}\frac{5}{3}.
\displaystyle =2(1)(1)\frac{5}{3}=\frac{10}{3}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\lim_{x\to0}\frac{x}{\tan x}\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }4\qquad\text{(d) not defined}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{x}{\tan x} =\lim_{x\to0}\frac{x\cos x}{\sin x}.
\displaystyle =\lim_{x\to0}\frac{x}{\sin x}\cos x.
\displaystyle =1\times1=1.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\lim_{n\to\infty} \left\{\frac{1}{1-n^2}+\frac{2}{1-n^2}+\cdots+ \frac{n}{1-n^2}\right\}\text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }-\frac{1}{2}\qquad \text{(c) }\frac{1}{2}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty} \left\{\frac{1+2+\cdots+n}{1-n^2}\right\}
\displaystyle =\lim_{n\to\infty} \frac{\frac{n(n+1)}{2}}{1-n^2}.
\displaystyle =\lim_{n\to\infty} \frac{n(n+1)}{2(1-n)(1+n)}.
\displaystyle =\lim_{n\to\infty}\frac{n}{2(1-n)}.
\displaystyle =\lim_{n\to\infty} \frac{1}{2\left(\frac{1}{n}-1\right)}=-\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\lim_{x\to\infty}\frac{\sin x}{x}\text{ equals}
\displaystyle \text{(a) }0\qquad\text{(b) }\infty\qquad \text{(c) }1\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{Since }-1\leq\sin x\leq1,
\displaystyle -\frac{1}{x}\leq\frac{\sin x}{x}\leq\frac{1}{x}.
\displaystyle \text{As }x\to\infty,\quad \lim_{x\to\infty}-\frac{1}{x}=0=\lim_{x\to\infty}\frac{1}{x}.
\displaystyle \therefore\lim_{x\to\infty}\frac{\sin x}{x}=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\lim_{x\to0}\frac{\sin x^\circ}{x} \text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }\pi\qquad \text{(c) }x\qquad\text{(d) }\frac{\pi}{180}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x^\circ=\frac{\pi x}{180}\text{ radians,}
\displaystyle \lim_{x\to0}\frac{\sin x^\circ}{x} =\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x}.
\displaystyle =\frac{\pi}{180}\lim_{x\to0} \frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}.
\displaystyle =\frac{\pi}{180}(1)=\frac{\pi}{180}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\lim_{x\to3}\frac{x-3}{|x-3|} \text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }0\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{For }x<3,\quad |x-3|=-(x-3).
\displaystyle \therefore\lim_{x\to3^-}\frac{x-3}{|x-3|} =\lim_{x\to3^-}\frac{x-3}{-(x-3)}=-1.
\displaystyle \text{For }x>3,\quad |x-3|=x-3.
\displaystyle \therefore\lim_{x\to3^+}\frac{x-3}{|x-3|} =\lim_{x\to3^+}\frac{x-3}{x-3}=1.
\displaystyle \text{Since LHL}\ne\text{RHL, the given limit does not exist.}
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\lim_{x\to a}\frac{x^n-a^n}{x-a} \text{ is equal to}
\displaystyle \text{(a) }na^n\qquad\text{(b) }na^{n-1}\qquad \text{(c) }na\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle x^n-a^n=(x-a) \left(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}\right).
\displaystyle \therefore\lim_{x\to a}\frac{x^n-a^n}{x-a}
\displaystyle =\lim_{x\to a} \left(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}\right).
\displaystyle =a^{n-1}+a^{n-1}+\cdots+a^{n-1}.
\displaystyle =na^{n-1}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\lim_{x\to\pi/4} \frac{\sqrt{2}\cos x-1}{\cot x-1}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{\sqrt{2}}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{2\sqrt{2}}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}
\displaystyle =\lim_{x\to\pi/4} \frac{(\sqrt{2}\cos x-1)\sin x}{\cos x-\sin x}.
\displaystyle \text{Since }1=\sqrt{2}\cos\frac{\pi}{4},
\displaystyle =\sqrt{2}\lim_{x\to\pi/4} \frac{\left(\cos x-\cos\frac{\pi}{4}\right)\sin x} {\cos x-\sin x}.
\displaystyle \text{Using }\cos x-\cos\frac{\pi}{4} =-2\sin\left(\frac{x+\pi/4}{2}\right) \sin\left(\frac{x-\pi/4}{2}\right),
\displaystyle \text{and }\cos x-\sin x =-2\sin\left(\frac{x-\pi/4}{ }\right) \sin\frac{\pi}{4},
\displaystyle \text{the limit simplifies to }\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\lim_{x\to\infty} \frac{\sqrt{x^2-1}}{2x+1}\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad \text{(c) }-1\qquad\text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\infty}\frac{\sqrt{x^2-1}}{2x+1} =\lim_{x\to\infty} \frac{x\sqrt{1-\frac{1}{x^2}}}{x\left(2+\frac{1}{x}\right)}.
\displaystyle =\lim_{x\to\infty} \frac{\sqrt{1-\frac{1}{x^2}}}{2+\frac{1}{x}}.
\displaystyle =\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\lim_{h\to0}2\left\{ \frac{\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) -\cos\left(\frac{\pi}{6}+h\right)} {\sqrt{3}h(\sqrt{3}\cos h-\sin h)}\right\}\text{ is equal to}
\displaystyle \text{(a) }\frac{2}{3}\qquad \text{(b) }\frac{4}{3}\qquad \text{(c) }-2\sqrt{3}\qquad \text{(d) }-\frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle \sqrt{3}\sin\left(\frac{\pi}{6}+h\right) =\sqrt{3}\left(\frac{1}{2}\cos h+\frac{\sqrt{3}}{2}\sin h\right).
\displaystyle =\frac{\sqrt{3}}{2}\cos h+\frac{3}{2}\sin h.
\displaystyle \cos\left(\frac{\pi}{6}+h\right) =\frac{\sqrt{3}}{2}\cos h-\frac{1}{2}\sin h.
\displaystyle \therefore\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) -\cos\left(\frac{\pi}{6}+h\right)=2\sin h.
\displaystyle \therefore\lim_{h\to0}2\left\{ \frac{2\sin h}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}\right\}
\displaystyle =\frac{4}{\sqrt{3}}\lim_{h\to0} \frac{\sin h}{h}\frac{1}{\sqrt{3}\cos h-\sin h}.
\displaystyle =\frac{4}{\sqrt{3}}\times1\times\frac{1}{\sqrt{3}} =\frac{4}{3}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\lim_{h\to0}\left\{ \frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}=
\displaystyle \text{(a) }-\frac{1}{12}\qquad \text{(b) }-\frac{4}{3}\qquad \text{(c) }-\frac{16}{3}\qquad \text{(d) }-\frac{1}{48}
\displaystyle \text{Answer:}
\displaystyle \lim_{h\to0}\left\{ \frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}
\displaystyle =\lim_{h\to0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}.
\displaystyle \text{Let }a=\sqrt[3]{8+h}.\text{ Then }a^3=8+h.
\displaystyle 8-a^3=-h.
\displaystyle (2-a)(4+2a+a^2)=-h.
\displaystyle \therefore 2-a=\frac{-h}{4+2a+a^2}.
\displaystyle \therefore\lim_{h\to0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}
\displaystyle =\lim_{h\to0} \frac{-1}{2a(4+2a+a^2)}.
\displaystyle \text{As }h\to0,\quad a\to2.
\displaystyle \therefore\text{the limit} =-\frac{1}{2(2)(4+4+4)}=-\frac{1}{48}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\lim_{n\to\infty}\left\{ \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+ \frac{1}{(2n+1)(2n+3)}\right\}\text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{9}\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \frac{1}{(2r+1)(2r+3)} =\frac{1}{2}\left(\frac{1}{2r+1}-\frac{1}{2r+3}\right).
\displaystyle \therefore\frac{1}{1\cdot3}+\frac{1}{3\cdot5} +\cdots+\frac{1}{(2n+1)(2n+3)}
\displaystyle =\frac{1}{2}\left\{\left(1-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+ \left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\right\}.
\displaystyle =\frac{1}{2}\left(1-\frac{1}{2n+3}\right).
\displaystyle \therefore\lim_{n\to\infty} \frac{1}{2}\left(1-\frac{1}{2n+3}\right)=\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\lim_{x\to1}\frac{\sin\pi x}{x-1} \text{ is equal to}
\displaystyle \text{(a) }-\pi\qquad\text{(b) }\pi\qquad \text{(c) }-\frac{1}{\pi}\qquad\text{(d) }\frac{1}{\pi}
\displaystyle \text{Answer:}
\displaystyle \text{Put }x=1+h.\text{ Then, as }x\to1,\ h\to0.
\displaystyle \therefore\lim_{x\to1}\frac{\sin\pi x}{x-1} =\lim_{h\to0}\frac{\sin\pi(1+h)}{h}.
\displaystyle =\lim_{h\to0}\frac{-\sin\pi h}{h}.
\displaystyle =-\pi\lim_{h\to0}\frac{\sin\pi h}{\pi h}=-\pi.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\lim_{x\to1} \frac{x+x^2+x^3+\cdots+x^n-n}{x-1}=5050,\text{ then }n\text{ equals}
\displaystyle \text{(a) }10\qquad\text{(b) }100\qquad \text{(c) }150\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x+x^2+\cdots+x^n-n =(x-1)+(x^2-1)+\cdots+(x^n-1).
\displaystyle \therefore\lim_{x\to1} \frac{x+x^2+\cdots+x^n-n}{x-1}
\displaystyle =\lim_{x\to1}\left\{ \frac{x-1}{x-1}+\frac{x^2-1}{x-1}+\cdots+ \frac{x^n-1}{x-1}\right\}.
\displaystyle =1+2+3+\cdots+n=\frac{n(n+1)}{2}.
\displaystyle \therefore\frac{n(n+1)}{2}=5050.
\displaystyle \Rightarrow n(n+1)=10100=100\times101.
\displaystyle \therefore n=100.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The value of }\lim_{x\to\infty} \frac{\sqrt{1+x^4}+(1+x^2)}{x^2}\text{ is}
\displaystyle \text{(a) }-1\qquad\text{(b) }1\qquad \text{(c) }2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to\infty} \frac{\sqrt{1+x^4}+(1+x^2)}{x^2}
\displaystyle =\lim_{x\to\infty} \left\{\frac{\sqrt{1+x^4}}{x^2}+\frac{1+x^2}{x^2}\right\}.
\displaystyle =\lim_{x\to\infty} \left\{\sqrt{1+\frac{1}{x^4}}+1+\frac{1}{x^2}\right\}.
\displaystyle =1+1+0=2.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\lim_{x\to0} \frac{\sqrt{1+x}-1}{x}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad\text{(b) }2\qquad \text{(c) }0\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sqrt{1+x}-1}{x} =\lim_{x\to0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)} {x(\sqrt{1+x}+1)}.
\displaystyle =\lim_{x\to0} \frac{x}{x(\sqrt{1+x}+1)}.
\displaystyle =\lim_{x\to0}\frac{1}{\sqrt{1+x}+1}.
\displaystyle =\frac{1}{1+1}=\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\lim_{x\to\pi/3} \frac{\sin\left(\frac{\pi}{3}-x\right)}{2\cos x-1}\text{ is equal to}
\displaystyle \text{(a) }\sqrt{3}\qquad\text{(b) }\frac{1}{2}\qquad \text{(c) }\frac{1}{\sqrt{3}}\qquad\text{(d) }\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle \text{Put }h=\frac{\pi}{3}-x.\text{ Then, as }x\to\frac{\pi}{3},\ h\to0.
\displaystyle \therefore x=\frac{\pi}{3}-h.
\displaystyle 2\cos x-1=2\cos\left(\frac{\pi}{3}-h\right)-1.
\displaystyle =2\left(\frac{1}{2}\cos h+\frac{\sqrt{3}}{2}\sin h\right)-1.
\displaystyle =\cos h+\sqrt{3}\sin h-1.
\displaystyle \therefore\lim_{x\to\pi/3} \frac{\sin\left(\frac{\pi}{3}-x\right)}{2\cos x-1}
\displaystyle =\lim_{h\to0}\frac{\sin h} {\cos h-1+\sqrt{3}\sin h}.
\displaystyle =\lim_{h\to0}\frac{\frac{\sin h}{h}} {\frac{\cos h-1}{h}+\sqrt{3}\frac{\sin h}{h}}.
\displaystyle =\frac{1}{0+\sqrt{3}}=\frac{1}{\sqrt{3}}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\lim_{x\to3} \frac{\displaystyle\sum_{r=1}^{n}x^r-\displaystyle\sum_{r=1}^{n}3^r}{x-3} \text{ is equal to}
\displaystyle \text{(a) }\frac{(2n-1)3^n}{4}\qquad \text{(b) }\frac{(2n-1)3^n+1}{4}
\displaystyle \text{(c) }(2n-1)3^n+1\qquad \text{(d) }\frac{(2n-1)3^n-1}{4}
\displaystyle \text{Answer:}
\displaystyle \sum_{r=1}^{n}x^r=x+x^2+x^3+\cdots+x^n.
\displaystyle \therefore\lim_{x\to3} \frac{\sum_{r=1}^{n}(x^r-3^r)}{x-3}
\displaystyle =\sum_{r=1}^{n}\lim_{x\to3}\frac{x^r-3^r}{x-3}.
\displaystyle =\sum_{r=1}^{n}r3^{r-1}.
\displaystyle =1+2(3)+3(3^2)+\cdots+n3^{n-1}.
\displaystyle \text{Let }S=1+2(3)+3(3^2)+\cdots+n3^{n-1}.
\displaystyle 3S=3+2(3^2)+3(3^3)+\cdots+n3^n.
\displaystyle \therefore 2S=1+3+3^2+\cdots+3^{n-1}-n3^n.
\displaystyle \text{Taking the equivalent positive form,}
\displaystyle 2S=n3^n-\frac{3^n-1}{2}.
\displaystyle \Rightarrow S=\frac{2n3^n-3^n+1}{4}.
\displaystyle =\frac{(2n-1)3^n+1}{4}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\lim_{n\to\infty} \frac{1-2+3-4+5-6+\cdots+(2n-1)-2n} {\sqrt{n^2+1}+\sqrt{n^2-1}}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }-\frac{1}{2}\qquad \text{(c) }1\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle 1-2+3-4+\cdots+(2n-1)-2n
\displaystyle =(1-2)+(3-4)+\cdots+\{(2n-1)-2n\}.
\displaystyle =(-1)+(-1)+\cdots+(-1)=-n.
\displaystyle \therefore\lim_{n\to\infty} \frac{-n}{\sqrt{n^2+1}+\sqrt{n^2-1}}
\displaystyle =\lim_{n\to\infty} \frac{-1}{\sqrt{1+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n^2}}}.
\displaystyle =\frac{-1}{1+1}=-\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }f(x)= \begin{cases}x\sin\frac{1}{x},&x\ne0\\0,&x=0\end{cases}, \text{ then }\lim_{x\to0}f(x)\text{ equals}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad \text{(c) }-1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since }-1\leq\sin\frac{1}{x}\leq1,
\displaystyle \left|x\sin\frac{1}{x}\right|\leq|x|.
\displaystyle \text{Also, }\lim_{x\to0}|x|=0.
\displaystyle \therefore\lim_{x\to0}x\sin\frac{1}{x}=0.
\displaystyle \therefore\lim_{x\to0}f(x)=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\lim_{n\to\infty} \frac{n!}{(n+1)!+n!}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}\qquad\text{(b) }0\qquad \text{(c) }2\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty}\frac{n!}{(n+1)!+n!} =\lim_{n\to\infty}\frac{n!}{(n+1)n!+n!}.
\displaystyle =\lim_{n\to\infty}\frac{n!}{(n+2)n!}.
\displaystyle =\lim_{n\to\infty}\frac{1}{n+2}=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\lim_{x\to\pi/4} \frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2x}\text{ is equal to}
\displaystyle \text{(a) }5\sqrt{2}\qquad\text{(b) }3\sqrt{2}\qquad \text{(c) }\sqrt{2}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\cos x+\sin x.
\displaystyle y^2=1+\sin2x\Rightarrow\sin2x=y^2-1.
\displaystyle \therefore1-\sin2x=2-y^2=(\sqrt2-y)(\sqrt2+y).
\displaystyle 4\sqrt2=(\sqrt2)^5.
\displaystyle \therefore4\sqrt2-y^5=(\sqrt2-y) \left(4+2\sqrt2y+2y^2+\sqrt2y^3+y^4\right).
\displaystyle \therefore\lim_{x\to\pi/4} \frac{4\sqrt2-(\cos x+\sin x)^5}{1-\sin2x}
\displaystyle =\lim_{y\to\sqrt2} \frac{4+2\sqrt2y+2y^2+\sqrt2y^3+y^4}{\sqrt2+y}.
\displaystyle =\frac{4+4+4+4+4}{2\sqrt2} =\frac{20}{2\sqrt2}=5\sqrt2.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\lim_{x\to2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt3}{x-2}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{8\sqrt3}\qquad \text{(b) }\frac{1}{\sqrt3}\qquad \text{(c) }8\sqrt3\qquad\text{(d) }\sqrt3
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt3}{x-2}
\displaystyle =\lim_{x\to2} \frac{\sqrt{2+x}-2} {(x-2)\left(\sqrt{1+\sqrt{2+x}}+\sqrt3\right)}.
\displaystyle =\lim_{x\to2} \frac{1}{(\sqrt{2+x}+2) \left(\sqrt{1+\sqrt{2+x}}+\sqrt3\right)}.
\displaystyle =\frac{1}{(2+2)(\sqrt3+\sqrt3)} =\frac{1}{8\sqrt3}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\lim_{x\to\infty} a^x\sin\left(\frac{b}{a^x}\right),\ a,b>1\text{ is equal to}
\displaystyle \text{(a) }b\qquad\text{(b) }a\qquad \text{(c) }a\log_e b\qquad\text{(d) }b\log_e a
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to\infty,\quad a^x\to\infty \text{ and }\frac{b}{a^x}\to0.
\displaystyle \lim_{x\to\infty}a^x\sin\left(\frac{b}{a^x}\right)
\displaystyle =b\lim_{x\to\infty} \frac{\sin\left(\frac{b}{a^x}\right)}{\frac{b}{a^x}}.
\displaystyle =b(1)=b.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\lim_{x\to0}\frac{8}{x^8} \left\{1-\cos\frac{x^2}{2}-\cos\frac{x^2}{4} +\cos\frac{x^2}{2}\cos\frac{x^2}{4}\right\}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{16}\qquad \text{(b) }-\frac{1}{16}\qquad \text{(c) }\frac{1}{32}\qquad \text{(d) }-\frac{1}{32}
\displaystyle \text{Answer:}
\displaystyle 1-\cos\frac{x^2}{2}-\cos\frac{x^2}{4} +\cos\frac{x^2}{2}\cos\frac{x^2}{4}
\displaystyle =\left(1-\cos\frac{x^2}{2}\right) \left(1-\cos\frac{x^2}{4}\right).
\displaystyle \text{Using }1-\cos\theta=2\sin^2\frac{\theta}{2},
\displaystyle \therefore\lim_{x\to0}\frac{8}{x^8} \left(1-\cos\frac{x^2}{2}\right) \left(1-\cos\frac{x^2}{4}\right)
\displaystyle =\lim_{x\to0}\frac{32}{x^8} \sin^2\frac{x^2}{4}\sin^2\frac{x^2}{8}.
\displaystyle =\frac{32}{4^2\cdot8^2} \lim_{x\to0}\left(\frac{\sin(x^2/4)}{x^2/4}\right)^2 \left(\frac{\sin(x^2/8)}{x^2/8}\right)^2.
\displaystyle =\frac{32}{16\cdot64}(1)(1)=\frac{1}{32}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }\alpha\text{ is a repeated root of }ax^2+bx+c=0,\text{ then}
\displaystyle \lim_{x\to\alpha}\frac{\tan(ax^2+bx+c)}{(x-\alpha)^2}\text{ is}
\displaystyle \text{(a) }a\qquad\text{(b) }b\qquad \text{(c) }c\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Since }\alpha\text{ is a repeated root of }ax^2+bx+c=0,
\displaystyle ax^2+bx+c=a(x-\alpha)^2.
\displaystyle \therefore\lim_{x\to\alpha} \frac{\tan(ax^2+bx+c)}{(x-\alpha)^2}
\displaystyle =\lim_{x\to\alpha} \frac{\tan\{a(x-\alpha)^2\}}{(x-\alpha)^2}.
\displaystyle =a\lim_{x\to\alpha} \frac{\tan\{a(x-\alpha)^2\}}{a(x-\alpha)^2}.
\displaystyle =a(1)=a.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{The value of }\lim_{x\to0} \frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}} {\sqrt{a+x}-\sqrt{a-x}}\text{ is}
\displaystyle \text{(a) }a\qquad\text{(b) }\sqrt{a}\qquad \text{(c) }-a\qquad\text{(d) }-\sqrt{a}
\displaystyle \text{Answer:}
\displaystyle \text{Rationalising the numerator,}
\displaystyle \sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}
\displaystyle =\frac{-2ax} {\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}.
\displaystyle \text{Also, rationalising the denominator,}
\displaystyle \sqrt{a+x}-\sqrt{a-x} =\frac{2x}{\sqrt{a+x}+\sqrt{a-x}}.
\displaystyle \therefore\text{the given limit}
\displaystyle =\lim_{x\to0} \frac{-a\left(\sqrt{a+x}+\sqrt{a-x}\right)} {\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}.
\displaystyle =\frac{-a(2\sqrt{a})}{2a}=-\sqrt{a}.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The value of }\lim_{x\to0} \frac{1-\cos x+2\sin x-\sin^3x-x^2+3x^4} {\tan^3x-6\sin^2x+x-5x^3}\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }-1\qquad\text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x,
\displaystyle =\lim_{x\to0} \frac{\frac{1-\cos x}{x}+2\frac{\sin x}{x} -\frac{\sin^3x}{x}-x+3x^3} {\frac{\tan^3x}{x}-6\frac{\sin^2x}{x}+1-5x^2}.
\displaystyle \text{Now, }\lim_{x\to0}\frac{1-\cos x}{x}=0,\qquad \lim_{x\to0}\frac{\sin x}{x}=1.
\displaystyle \text{Also, }\lim_{x\to0}\frac{\sin^3x}{x} =\lim_{x\to0}\sin^2x\frac{\sin x}{x}=0.
\displaystyle \lim_{x\to0}\frac{\tan^3x}{x} =\lim_{x\to0}\tan^2x\frac{\tan x}{x}=0,
\displaystyle \text{and }\lim_{x\to0}\frac{\sin^2x}{x} =\lim_{x\to0}\sin x\frac{\sin x}{x}=0.
\displaystyle \therefore\text{the required limit} =\frac{0+2(1)-0-0+0}{0-0+1-0}=2.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\lim_{\theta\to\pi/2} \frac{1-\sin\theta}{(\pi/2-\theta)\cos\theta}\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }\frac{1}{2}\qquad\text{(d) }-\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Put }h=\frac{\pi}{2}-\theta.\text{ Then, as }\theta\to\frac{\pi}{2},\ h\to0.
\displaystyle \sin\theta=\sin\left(\frac{\pi}{2}-h\right)=\cos h,
\displaystyle \cos\theta=\cos\left(\frac{\pi}{2}-h\right)=\sin h.
\displaystyle \therefore\lim_{\theta\to\pi/2} \frac{1-\sin\theta}{(\pi/2-\theta)\cos\theta} =\lim_{h\to0}\frac{1-\cos h}{h\sin h}.
\displaystyle =\lim_{h\to0} \frac{2\sin^2(h/2)}{h\sin h}.
\displaystyle =\frac{1}{2}\lim_{h\to0} \left(\frac{\sin(h/2)}{h/2}\right)^2\frac{h}{\sin h}.
\displaystyle =\frac{1}{2}(1)^2(1)=\frac{1}{2}.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The value of }\lim_{x\to\pi/2} (\sec x-\tan x)\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }-1\qquad \text{(c) }1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sec x-\tan x =\frac{1-\sin x}{\cos x}.
\displaystyle =\frac{(1-\sin x)(1+\sin x)} {\cos x(1+\sin x)}.
\displaystyle =\frac{1-\sin^2x}{\cos x(1+\sin x)} =\frac{\cos x}{1+\sin x}.
\displaystyle \therefore\lim_{x\to\pi/2}(\sec x-\tan x) =\lim_{x\to\pi/2}\frac{\cos x}{1+\sin x}.
\displaystyle =\frac{0}{1+1}=0.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The value of }\lim_{n\to\infty} \frac{n!}{(n+1)!-n!}\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }-1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty}\frac{n!}{(n+1)!-n!} =\lim_{n\to\infty}\frac{n!}{(n+1)n!-n!}.
\displaystyle =\lim_{n\to\infty}\frac{n!}{n\cdot n!} =\lim_{n\to\infty}\frac{1}{n}=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The value of }\lim_{n\to\infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }-1\qquad \text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}
\displaystyle =\lim_{n\to\infty} \frac{(n+1)!\{(n+2)+1\}}{(n+1)!\{(n+2)-1\}}.
\displaystyle =\lim_{n\to\infty}\frac{n+3}{n+1}.
\displaystyle =\lim_{n\to\infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}}=1.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{The value of }\lim_{x\to\infty} \frac{(x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}} {x^{10}+10^{10}}\text{ is}
\displaystyle \text{(a) }10\qquad\text{(b) }100\qquad \text{(c) }10^{10}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Dividing the numerator and denominator by }x^{10},
\displaystyle \text{the required limit}
\displaystyle =\lim_{x\to\infty} \frac{\left(1+\frac1x\right)^{10}+ \left(1+\frac2x\right)^{10}+\cdots+ \left(1+\frac{100}{x}\right)^{10}} {1+\frac{10^{10}}{x^{10}}}.
\displaystyle \text{As }x\to\infty,\text{ each term in the numerator tends to }1.
\displaystyle \text{There are }100\text{ such terms.}
\displaystyle \therefore\text{the required limit}=\frac{100}{1}=100.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{The value of }\lim_{n\to\infty} \left\{\frac{1+2+3+\cdots+n}{n+2}-\frac{n}{2}\right\}\text{ is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }1\qquad \text{(c) }-1\qquad\text{(d) }-\frac12
\displaystyle \text{Answer:}
\displaystyle \text{Using }1+2+3+\cdots+n=\frac{n(n+1)}{2},
\displaystyle \lim_{n\to\infty} \left\{\frac{n(n+1)}{2(n+2)}-\frac{n}{2}\right\}
\displaystyle =\lim_{n\to\infty} \frac{n}{2}\left(\frac{n+1}{n+2}-1\right).
\displaystyle =\lim_{n\to\infty} \frac{n}{2}\left(\frac{-1}{n+2}\right).
\displaystyle =-\frac12\lim_{n\to\infty}\frac{n}{n+2} =-\frac12.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 39: }\lim_{x\to1}[x-1],\text{ where }[\,\cdot\,]\text{ is the greatest integer function, is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }0\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to1^-,\quad -1<x-1<0.
\displaystyle \therefore [x-1]=-1.
\displaystyle \therefore\lim_{x\to1^-}[x-1]=-1.
\displaystyle \text{As }x\to1^+,\quad 0<x-1<1.
\displaystyle \therefore [x-1]=0.
\displaystyle \therefore\lim_{x\to1^+}[x-1]=0.
\displaystyle \text{Since LHL}\ne\text{RHL, the given limit does not exist.}
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\lim_{x\to\infty}\frac{|x|}{x} \text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }0\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to\infty,\quad x>0.
\displaystyle \therefore |x|=x.
\displaystyle \therefore\lim_{x\to\infty}\frac{|x|}{x} =\lim_{x\to\infty}\frac{x}{x}=1.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 41: }\lim_{x\to0}\frac{|\sin x|}{x}\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }0\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \text{For }x\to0^+,\quad \sin x>0\Rightarrow|\sin x|=\sin x.
\displaystyle \therefore\lim_{x\to0^+}\frac{|\sin x|}{x} =\lim_{x\to0^+}\frac{\sin x}{x}=1.
\displaystyle \text{For }x\to0^-,\quad \sin x<0\Rightarrow|\sin x|=-\sin x.
\displaystyle \therefore\lim_{x\to0^-}\frac{|\sin x|}{x} =-\lim_{x\to0^-}\frac{\sin x}{x}=-1.
\displaystyle \text{Since LHL}\ne\text{RHL, the given limit does not exist.}
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }f(x)= \begin{cases}\dfrac{\sin[x]}{[x]},&[x]\ne0\\0,&[x]=0\end{cases}, \text{ where }[\,\cdot\,]\text{ denotes the greatest integer function, then}
\displaystyle \lim_{x\to0^-}f(x)\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad \text{(c) }-1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^-,\quad -1<x<0.
\displaystyle \therefore[x]=-1.
\displaystyle \therefore f(x)=\frac{\sin(-1)}{-1}=\sin1.
\displaystyle \therefore\lim_{x\to0^-}f(x)=\sin1.
\displaystyle \text{Since }\sin1\text{ is not among the given options,}
\displaystyle \text{the correct option is (d), none of these.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Let }f(x)=x-[x],\ x\in R,\text{ then }f'\left(\frac{1}{2}\right)\text{ is}
\displaystyle \text{(a) }\frac{3}{2}\qquad\text{(b) }1\qquad \text{(c) }0\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \text{For }0<x<1,\quad [x]=0.
\displaystyle \therefore f(x)=x.
\displaystyle \therefore f'(x)=1.
\displaystyle \therefore f'\left(\frac{1}{2}\right)=1.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{If }f(x)=\frac{x-4}{2\sqrt{x}}, \text{ then }f'(1)\text{ is}
\displaystyle \text{(a) }\frac{5}{4}\qquad\text{(b) }\frac{4}{5}\qquad \text{(c) }1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{1}{2}\left(x^{1/2}-4x^{-1/2}\right).
\displaystyle \therefore f'(x)=\frac{1}{2} \left(\frac{1}{2}x^{-1/2}+2x^{-3/2}\right).
\displaystyle =\frac{1}{4\sqrt{x}}+\frac{1}{x^{3/2}}.
\displaystyle \therefore f'(1)=\frac{1}{4}+1=\frac{5}{4}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{If }y=1+\frac{x}{1!} +\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots,\text{ then }\frac{dy}{dx}=
\displaystyle \text{(a) }y+1\qquad\text{(b) }y-1\qquad \text{(c) }y\qquad\text{(d) }y^2
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=0+\frac{1}{1!} +\frac{2x}{2!}+\frac{3x^2}{3!}+\cdots
\displaystyle =1+\frac{x}{1!}+\frac{x^2}{2!} +\frac{x^3}{3!}+\cdots
\displaystyle =y.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{If }f(x)=1-x+x^2-x^3+\cdots -x^{99}+x^{100},\text{ then }f'(1)\text{ equals}
\displaystyle \text{(a) }150\qquad\text{(b) }-50\qquad \text{(c) }-150\qquad\text{(d) }50
\displaystyle \text{Answer:}
\displaystyle f'(x)=-1+2x-3x^2+4x^3-\cdots -99x^{98}+100x^{99}.
\displaystyle \therefore 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 \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }y= \frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}},\text{ then }\frac{dy}{dx}=
\displaystyle \text{(a) }-\frac{4x}{(x^2-1)^2}\qquad \text{(b) }-\frac{4x}{x^2-1}
\displaystyle \text{(c) }\frac{1-x^2}{4x}\qquad \text{(d) }\frac{4x}{x^2-1}
\displaystyle \text{Answer:}
\displaystyle y=\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{4x}{(x^2-1)^2}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If }y=\sqrt{x}+ \frac{1}{\sqrt{x}},\text{ then }\frac{dy}{dx}\text{ at }x=1\text{ is}
\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 \therefore\left.\frac{dy}{dx}\right|_{x=1} =\frac{1}{2}-\frac{1}{2}=0.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 49: }\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)=100x^{99}+99x^{98}+\cdots+2x+1.
\displaystyle \therefore f'(1)=100+99+\cdots+2+1.
\displaystyle =\frac{100(100+1)}{2}=5050.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 50: }\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) }50\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle f'(x)=1+x+x^2+\cdots+x^{99}.
\displaystyle \therefore f'(1)=1+1+1+\cdots+1.
\displaystyle \text{There are }100\text{ terms.}
\displaystyle \therefore f'(1)=100.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{If }y=\frac{\sin x+\cos x}{\sin x-\cos x}, \text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is}
\displaystyle \text{(a) }-2\qquad\text{(b) }0\qquad \text{(c) }\frac{1}{2}\qquad\text{(d) does not exist}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}= \frac{(\sin x-\cos x)(\cos x-\sin x) -(\sin x+\cos x)(\cos x+\sin x)} {(\sin x-\cos x)^2}.
\displaystyle \text{At }x=0,\quad \sin0=0,\quad\cos0=1.
\displaystyle \therefore\left.\frac{dy}{dx}\right|_{x=0} =\frac{(-1)(1)-(1)(1)}{(-1)^2}=-2.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{If }y=\frac{\sin(x+9)}{\cos x}, \text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is}
\displaystyle \text{(a) }\cos9\qquad\text{(b) }\sin9\qquad \text{(c) }0\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \sin(x+9)=\sin x\cos9+\cos x\sin9.
\displaystyle \therefore y=\tan x\cos9+\sin9.
\displaystyle \therefore\frac{dy}{dx}=\sec^2x\cos9.
\displaystyle \therefore\left.\frac{dy}{dx}\right|_{x=0} =\sec^20\cos9=\cos9.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{If }f(x)=\frac{x^n-a^n}{x-a}, \text{ then }f'(a)\text{ is}
\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{ both numerator and denominator are zero.}
\displaystyle \therefore f(a)\text{ is not defined.}
\displaystyle \text{Since }f(a)\text{ does not exist, }f'(a)\text{ does not exist.}
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{If }f(x)=x\sin x, \text{ then }f'\left(\frac{\pi}{2}\right)=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }-1\qquad\text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle f'(x)=x\cos x+\sin x.
\displaystyle \therefore f'\left(\frac{\pi}{2}\right) =\frac{\pi}{2}\cos\frac{\pi}{2}+\sin\frac{\pi}{2}.
\displaystyle =\frac{\pi}{2}(0)+1=1.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the value of }\lim_{x\to0} \frac{\sqrt{1-\cos 2x}}{x}.
\displaystyle \text{Answer:}
\displaystyle \text{Using }1-\cos 2x=2\sin^2x,
\displaystyle \lim_{x\to0}\frac{\sqrt{1-\cos 2x}}{x} =\lim_{x\to0}\frac{\sqrt{2\sin^2x}}{x}.
\displaystyle =\sqrt2\lim_{x\to0}\frac{|\sin x|}{x}.
\displaystyle \text{For }x\to0^+,\quad |\sin x|=\sin x.
\displaystyle \therefore\lim_{x\to0^+}\frac{\sqrt{1-\cos 2x}}{x} =\sqrt2\lim_{x\to0^+}\frac{\sin x}{x}=\sqrt2.
\displaystyle \text{For }x\to0^-,\quad |\sin x|=-\sin x.
\displaystyle \therefore\lim_{x\to0^-}\frac{\sqrt{1-\cos 2x}}{x} =-\sqrt2\lim_{x\to0^-}\frac{\sin x}{x}=-\sqrt2.
\displaystyle \text{Since LHL}\ne\text{RHL, the given limit does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the value of }\lim_{x\to0^-}[x].
\displaystyle \text{Answer:}
\displaystyle \text{Here, }[x]\text{ denotes the greatest integer function.}
\displaystyle \text{As }x\to0^-,\quad -1<x<0.
\displaystyle \therefore [x]=-1.
\displaystyle \therefore\lim_{x\to0^-}[x]=-1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the value of }\lim_{x\to0^+}[x].
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^+,\quad 0<x<1.
\displaystyle \therefore [x]=0.
\displaystyle \therefore\lim_{x\to0^+}[x]=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Write the value of }\lim_{x\to1^-}(x-[x]).
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to1^-,\quad 0<x<1.
\displaystyle \therefore [x]=0.
\displaystyle \therefore\lim_{x\to1^-}(x-[x]) =\lim_{x\to1^-}x=1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the value of }\lim_{x\to0^-} \frac{\sin[x]}{[x]}.
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^-,\quad -1<x<0.
\displaystyle \therefore [x]=-1.
\displaystyle \therefore\lim_{x\to0^-}\frac{\sin[x]}{[x]} =\frac{\sin(-1)}{-1}=\sin1.
\displaystyle \therefore\text{the value of the limit is }\sin1.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the value of }\lim_{x\to\pi} \frac{\sin x}{x-\pi}.
\displaystyle \text{Answer:}
\displaystyle \text{Put }x-\pi=h.\text{ Then, as }x\to\pi,\ h\to0.
\displaystyle \therefore\lim_{x\to\pi}\frac{\sin x}{x-\pi} =\lim_{h\to0}\frac{\sin(\pi+h)}{h}.
\displaystyle =-\lim_{h\to0}\frac{\sin h}{h}=-1.
\displaystyle \therefore\text{the value of the limit is }-1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the value of }\lim_{x\to\infty} \frac{\sin x}{x}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }-1\leq\sin x\leq1,
\displaystyle -\frac{1}{x}\leq\frac{\sin x}{x}\leq\frac{1}{x}.
\displaystyle \text{As }x\to\infty,\quad \lim_{x\to\infty}-\frac{1}{x}=0=\lim_{x\to\infty}\frac{1}{x}.
\displaystyle \therefore\lim_{x\to\infty}\frac{\sin x}{x}=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the value of }\lim_{x\to2} \frac{|x-2|}{x-2}.
\displaystyle \text{Answer:}
\displaystyle \text{For }x<2,\quad |x-2|=-(x-2).
\displaystyle \therefore\lim_{x\to2^-}\frac{|x-2|}{x-2} =\lim_{x\to2^-}\frac{-(x-2)}{x-2}=-1.
\displaystyle \text{For }x>2,\quad |x-2|=x-2.
\displaystyle \therefore\lim_{x\to2^+}\frac{|x-2|}{x-2} =\lim_{x\to2^+}\frac{x-2}{x-2}=1.
\displaystyle \text{Since LHL}\ne\text{RHL, the given limit does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the value of }\lim_{x\to0} \frac{\sin x^\circ}{x}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }x^\circ=\frac{\pi x}{180}\text{ radians,}
\displaystyle \lim_{x\to0}\frac{\sin x^\circ}{x} =\lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x}.
\displaystyle =\frac{\pi}{180}\lim_{x\to0} \frac{\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}.
\displaystyle =\frac{\pi}{180}.
\displaystyle \therefore\text{the value of the limit is }\frac{\pi}{180}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the value of }\lim_{x\to0^-} \frac{\sin x}{\sqrt{x}}.
\displaystyle \text{Answer:}
\displaystyle \text{As }x\to0^-,\quad x<0.
\displaystyle \text{But }\sqrt{x}\text{ is not defined for }x<0\text{ in the real number system.}
\displaystyle \therefore\frac{\sin x}{\sqrt{x}}\text{ is not defined to the left of }x=0.
\displaystyle \therefore\lim_{x\to0^-}\frac{\sin x}{\sqrt{x}}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the value of }\lim_{x\to0} \frac{\sin x}{\sqrt{1+x}-1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to0}\frac{\sin x}{\sqrt{1+x}-1} =\lim_{x\to0}\frac{\sin x(\sqrt{1+x}+1)} {(\sqrt{1+x}-1)(\sqrt{1+x}+1)}.
\displaystyle =\lim_{x\to0}\frac{\sin x(\sqrt{1+x}+1)}{x}.
\displaystyle =\lim_{x\to0}\frac{\sin x}{x} \left(\sqrt{1+x}+1\right).
\displaystyle =1(1+1)=2.
\displaystyle \therefore\text{the value of the limit is }2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write the value of }\lim_{x\to-\infty} \left(3x+\sqrt{9x^2-x}\right).
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to-\infty}\left(3x+\sqrt{9x^2-x}\right)
\displaystyle =\lim_{x\to-\infty} \frac{(3x+\sqrt{9x^2-x})(\sqrt{9x^2-x}-3x)} {\sqrt{9x^2-x}-3x}.
\displaystyle =\lim_{x\to-\infty} \frac{-x}{\sqrt{9x^2-x}-3x}.
\displaystyle \text{Since }x<0,\quad\sqrt{x^2}=-x.
\displaystyle \therefore\sqrt{9x^2-x} =(-x)\sqrt{9-\frac{1}{x}}.
\displaystyle \therefore\lim_{x\to-\infty} \frac{-x}{(-x)\sqrt{9-\frac{1}{x}}+(-x)3}
\displaystyle =\lim_{x\to-\infty} \frac{1}{\sqrt{9-\frac{1}{x}}+3}.
\displaystyle =\frac{1}{3+3}=\frac{1}{6}.
\displaystyle \therefore\text{the value of the limit is }\frac{1}{6}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Write the value of }\lim_{n\to\infty} \frac{n!+(n+1)!}{(n+1)!+(n+2)!}.
\displaystyle \text{Answer:}
\displaystyle \lim_{n\to\infty}\frac{n!+(n+1)!}{(n+1)!+(n+2)!}
\displaystyle =\lim_{n\to\infty} \frac{n![1+(n+1)]}{(n+1)![1+(n+2)]}.
\displaystyle =\lim_{n\to\infty} \frac{n!(n+2)}{(n+1)n!(n+3)}.
\displaystyle =\lim_{n\to\infty}\frac{n+2}{(n+1)(n+3)}.
\displaystyle =\lim_{n\to\infty} \frac{1+\frac{2}{n}}{n\left(1+\frac{1}{n}\right) \left(1+\frac{3}{n}\right)}=0.
\displaystyle \therefore\text{the value of the limit is }0.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Write the value of }\lim_{x\to\pi/2} \frac{2x-\pi}{\cos x}.
\displaystyle \text{Answer:}
\displaystyle \text{Put }x=\frac{\pi}{2}+h.\text{ Then, as }x\to\frac{\pi}{2},\ h\to0.
\displaystyle \therefore 2x-\pi=2\left(\frac{\pi}{2}+h\right)-\pi=2h.
\displaystyle \text{Also, }\cos x=\cos\left(\frac{\pi}{2}+h\right)=-\sin h.
\displaystyle \therefore\lim_{x\to\pi/2}\frac{2x-\pi}{\cos x} =\lim_{h\to0}\frac{2h}{-\sin h}.
\displaystyle =-2\lim_{h\to0}\frac{h}{\sin h}=-2.
\displaystyle \therefore\text{the value of the limit is }-2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the value of }\lim_{n\to\infty} \frac{1+2+3+\cdots+n}{n^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Using }1+2+3+\cdots+n=\frac{n(n+1)}{2},
\displaystyle \lim_{n\to\infty}\frac{1+2+3+\cdots+n}{n^2} =\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}(1+0)=\frac{1}{2}.
\displaystyle \therefore\text{the value of the limit is }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Write the value of } \lim_{x\to c}\frac{f(x)-f(c)}{x-c}.
\displaystyle \text{Answer:}
\displaystyle \text{By the definition of derivative,}
\displaystyle f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}.
\displaystyle \therefore\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f'(c).
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the value of } \lim_{x\to a}\frac{x f(a)-a f(x)}{x-a}.
\displaystyle \text{Answer:}
\displaystyle xf(a)-af(x)=(x-a)f(a)-a\{f(x)-f(a)\}.
\displaystyle \therefore\lim_{x\to a}\frac{xf(a)-af(x)}{x-a}
\displaystyle =\lim_{x\to a}\left\{f(a) -a\frac{f(x)-f(a)}{x-a}\right\}.
\displaystyle =f(a)-af'(a).
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }x<2,\text{ then write the value of } \frac{d}{dx}\left(\sqrt{x^2-4x+4}\right).
\displaystyle \text{Answer:}
\displaystyle \sqrt{x^2-4x+4}=\sqrt{(x-2)^2}=|x-2|.
\displaystyle \text{Since }x<2,\quad |x-2|=2-x.
\displaystyle \therefore\frac{d}{dx}\left(\sqrt{x^2-4x+4}\right) =\frac{d}{dx}(2-x)=-1.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\frac{\pi}{2}<x<\pi,\text{ then find } \frac{d}{dx}\left(\sqrt{\frac{1+\cos2x}{2}}\right).
\displaystyle \text{Answer:}
\displaystyle \frac{1+\cos2x}{2}=\cos^2x.
\displaystyle \therefore\sqrt{\frac{1+\cos2x}{2}} =\sqrt{\cos^2x}=|\cos x|.
\displaystyle \text{Since }\frac{\pi}{2}<x<\pi,\quad \cos x<0.
\displaystyle \therefore|\cos x|=-\cos x.
\displaystyle \therefore\frac{d}{dx}\left(\sqrt{\frac{1+\cos2x}{2}}\right) =\frac{d}{dx}(-\cos x)=\sin x.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the value of } \frac{d}{dx}(x|x|).
\displaystyle \text{Answer:}
\displaystyle \text{For }x>0,\quad |x|=x.
\displaystyle \therefore x|x|=x^2.
\displaystyle \therefore\frac{d}{dx}(x|x|)=2x.
\displaystyle \text{For }x<0,\quad |x|=-x.
\displaystyle \therefore x|x|=-x^2.
\displaystyle \therefore\frac{d}{dx}(x|x|)=-2x.
\displaystyle \therefore\frac{d}{dx}(x|x|) =\begin{cases}2x,&x>0\\-2x,&x<0.\end{cases}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Write the value of } \frac{d}{dx}\{(x+|x|)|x|\}.
\displaystyle \text{Answer:}
\displaystyle \text{For }x>0,\quad |x|=x.
\displaystyle \therefore(x+|x|)|x|=(x+x)x=2x^2.
\displaystyle \therefore\frac{d}{dx}\{(x+|x|)|x|\}=4x=4|x|.
\displaystyle \text{For }x<0,\quad |x|=-x.
\displaystyle \therefore(x+|x|)|x|=(x-x)(-x)=0.
\displaystyle \therefore\frac{d}{dx}\{(x+|x|)|x|\}=0.
\displaystyle \text{At }x=0,\text{ the derivative is }0.
\displaystyle \therefore\frac{d}{dx}\{(x+|x|)|x|\} =\begin{cases}0,&x\leq0\\4x,&x>0.\end{cases}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }f(x)=\frac{x^2}{|x|}, \text{ write }\frac{d}{dx}(f(x)).
\displaystyle \text{Answer:}
\displaystyle \text{For }x>0,\quad |x|=x.
\displaystyle \therefore f(x)=\frac{x^2}{x}=x.
\displaystyle \therefore f'(x)=1.
\displaystyle \text{For }x<0,\quad |x|=-x.
\displaystyle \therefore f(x)=\frac{x^2}{-x}=-x.
\displaystyle \therefore f'(x)=-1.
\displaystyle \therefore\frac{d}{dx}(f(x)) =\begin{cases}1,&x>0\\-1,&x<0.\end{cases}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Write the value of } \frac{d}{dx}(\log|x|).
\displaystyle \text{Answer:}
\displaystyle \text{For }x>0,\quad \log|x|=\log x.
\displaystyle \therefore\frac{d}{dx}(\log|x|)=\frac{1}{x}.
\displaystyle \text{For }x<0,\quad \log|x|=\log(-x).
\displaystyle \therefore\frac{d}{dx}(\log|x|) =\frac{1}{-x}(-1)=\frac{1}{x}.
\displaystyle \therefore\frac{d}{dx}(\log|x|)=\frac{1}{x}, \quad x\ne0.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }f(x)=|x|+|x-1|, \text{ write the value of }\frac{d}{dx}(f(x)).
\displaystyle \text{Answer:}
\displaystyle \text{For }x<0,\quad |x|=-x,\quad |x-1|=1-x.
\displaystyle \therefore f(x)=1-2x\quad\Rightarrow\quad f'(x)=-2.
\displaystyle \text{For }0<x<1,\quad |x|=x,\quad |x-1|=1-x.
\displaystyle \therefore f(x)=1\quad\Rightarrow\quad f'(x)=0.
\displaystyle \text{For }x>1,\quad |x|=x,\quad |x-1|=x-1.
\displaystyle \therefore f(x)=2x-1\quad\Rightarrow\quad f'(x)=2.
\displaystyle \text{At }x=0\text{ and }x=1,\text{ the derivative does not exist.}
\displaystyle \therefore f'(x)= \begin{cases} -2,&x<0\\ 0,&0<x<1\\ 2,&x>1 \end{cases}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Write the value of the derivative of } f(x)=|x-1|+|x-3|\text{ at }x=2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }1<x<3,\quad |x-1|=x-1,\quad |x-3|=3-x.
\displaystyle \therefore f(x)=(x-1)+(3-x)=2.
\displaystyle \therefore f'(x)=0\text{ for }1<x<3.
\displaystyle \therefore f'(2)=0.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }f(1)=1,\ f'(1)=2,\text{ then write the value of}
\displaystyle \lim_{x\to1}\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}.
\displaystyle \text{Answer:}
\displaystyle \lim_{x\to1}\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}
\displaystyle =\lim_{x\to1} \frac{f(x)-1}{x-1}\cdot \frac{\sqrt{x}+1}{\sqrt{f(x)}+1}.
\displaystyle =f'(1)\cdot\frac{1+1}{\sqrt{f(1)}+1}.
\displaystyle =2\cdot\frac{2}{1+1}=2.
\displaystyle \therefore\text{the required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Write the derivative of } f(x)=3|2+x|\text{ at }x=-3.
\displaystyle \text{Answer:}
\displaystyle \text{Since }x=-3<-2,\quad |2+x|=-(2+x).
\displaystyle \therefore f(x)=3\{-2-x\}=-6-3x.
\displaystyle \therefore f'(x)=-3.
\displaystyle \therefore f'(-3)=-3.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }|x|<1\text{ and } y=1+x+x^2+x^3+\cdots,\text{ then write the value of }\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }|x|<1,\quad y=1+x+x^2+x^3+\cdots=\frac{1}{1-x}.
\displaystyle \therefore\frac{dy}{dx} =\frac{d}{dx}\left(\frac{1}{1-x}\right).
\displaystyle =\frac{d}{dx}(1-x)^{-1} =(1-x)^{-2}.
\displaystyle \therefore\frac{dy}{dx}=\frac{1}{(1-x)^2}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }f(x)=\log_{x^2}x^3, \text{ write the value of }f'(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\log_{x^2}x^3 =\frac{\log x^3}{\log x^2}.
\displaystyle =\frac{3\log x}{2\log x}=\frac{3}{2}.
\displaystyle \therefore f'(x)=\frac{d}{dx}\left(\frac{3}{2}\right)=0.
\displaystyle \\

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