\displaystyle \textbf{Question 1. }\text{If }y=\log x^{x},\text{ then the value of }\frac{dy}{dx}\text{ is}
\displaystyle \text{(a) }x^{x}(1+\log x)\qquad\text{(b) }\log ex
\displaystyle \text{(c) }\log\frac{e}{x}\qquad\text{(d) }\log\left(\frac{x}{e}\right)
\displaystyle \text{Answer:}
\displaystyle y=\log x^{x}
\displaystyle =x\log x
\displaystyle \therefore \frac{dy}{dx}=\frac{d}{dx}(x\log x)
\displaystyle =\log x+x\cdot\frac{1}{x}
\displaystyle =1+\log x
\displaystyle =\log e+\log x
\displaystyle =\log(ex)
\displaystyle \therefore \frac{dy}{dx}=\log(ex)
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If }f(x)=\begin{cases}mx+1,&x\leq\frac{\pi}{2}\\ \sin x+n,&x>\frac{\pi}{2}\end{cases}\text{ is continuous at }x=\frac{\pi}{2},\text{ then}
\displaystyle \text{(a) }m=1,\ n=0\qquad\text{(b) }m=\frac{n\pi}{2}+1
\displaystyle \text{(c) }n=\frac{m\pi}{2}\qquad\text{(d) }m=n=\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Since }f(x)\text{ is continuous at }x=\frac{\pi}{2},
\displaystyle \lim_{x\to(\pi/2)^-}f(x)=\lim_{x\to(\pi/2)^+}f(x)=f\!\left(\frac{\pi}{2}\right)
\displaystyle \therefore m\left(\frac{\pi}{2}\right)+1=\sin\frac{\pi}{2}+n
\displaystyle \therefore \frac{m\pi}{2}+1=1+n
\displaystyle \therefore n=\frac{m\pi}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 3. }\text{If }y=\log_{7}(\log x),\text{ then }\frac{dy}{dx}\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{x\log x\log 7}\qquad\text{(b) }-\frac{1}{x\log x\log 7}
\displaystyle \text{(c) }\frac{1}{x\log x}\qquad\text{(d) None\ of\ these}
\displaystyle \text{Answer:}
\displaystyle y=\log_{7}(\log x)
\displaystyle \therefore y=\frac{\log(\log x)}{\log 7}
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{\log 7}\cdot\frac{1}{\log x}\cdot\frac{1}{x}
\displaystyle =\frac{1}{x\log x\log 7}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 4. }\text{If }f(x)=|\cos x|,\text{ then}
\displaystyle \text{(a) }f\text{ is everywhere differentiable}
\displaystyle \text{(b) }f\text{ is everywhere continuous but not differentiable at }x=n\pi,\ n\in Z
\displaystyle \text{(c) }f\text{ is everywhere continuous but not differentiable at }x=(2n+1)\frac{\pi}{2},\ n\in Z
\displaystyle \text{(d) None\ of\ the\ above}
\displaystyle \text{Answer:}
\displaystyle f(x)=|\cos x|
\displaystyle \text{Since }|\cdot|\text{ preserves continuity, }f(x)\text{ is continuous for all }x.
\displaystyle \text{Differentiability fails where }\cos x=0.
\displaystyle \cos x=0\implies x=(2n+1)\frac{\pi}{2},\ n\in Z
\displaystyle \therefore f(x)\text{ is everywhere continuous but not differentiable at }x=(2n+1)\frac{\pi}{2},\ n\in Z
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 5. }\text{The function }f(x)=\begin{cases}1,&x\neq0\\2,&x=0\end{cases}\text{ is not continuous at}
\displaystyle \text{(a) }x=0\qquad\text{(b) }x=1
\displaystyle \text{(c) }x=-1\qquad\text{(d) None\ of\ these}
\displaystyle \text{Answer:}
\displaystyle \text{For }x\neq0,\ f(x)=1
\displaystyle \therefore \lim_{x\to0}f(x)=1
\displaystyle \text{But }f(0)=2
\displaystyle \therefore \lim_{x\to0}f(x)\neq f(0)
\displaystyle \therefore f(x)\text{ is not continuous at }x=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 6. }\text{The point of discontinuity of the function }f(x)=\begin{cases}2x+3,&x\leq2\\2x-3,&x>2\end{cases}\text{ is}
\displaystyle \text{(a) }x=0\qquad\text{(b) }x=1
\displaystyle \text{(c) }x=2\qquad\text{(d) None\ of\ these}
\displaystyle \text{Answer:}
\displaystyle \text{At }x=2,
\displaystyle \lim_{x\to2^-}f(x)=2(2)+3=7
\displaystyle \lim_{x\to2^+}f(x)=2(2)-3=1
\displaystyle \therefore \lim_{x\to2^-}f(x)\neq\lim_{x\to2^+}f(x)
\displaystyle \therefore f(x)\text{ is discontinuous at }x=2
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 7. }\text{The function }f(x)=\begin{cases}\frac{k\cos x}{\pi-2x},&x\neq\frac{\pi}{2}\\3,&x=\frac{\pi}{2}\end{cases}\text{ is continuous at }x=\frac{\pi}{2},\text{ when }k\text{ equals}
\displaystyle \text{(a) }-6\qquad\text{(b) }6
\displaystyle \text{(c) }5\qquad\text{(d) }-5
\displaystyle \text{Answer:}
\displaystyle \text{Since }f(x)\text{ is continuous at }x=\frac{\pi}{2},
\displaystyle \lim_{x\to\pi/2}f(x)=f\left(\frac{\pi}{2}\right)
\displaystyle \therefore \lim_{x\to\pi/2}\frac{k\cos x}{\pi-2x}=3
\displaystyle =k\lim_{x\to\pi/2}\frac{\cos x}{\pi-2x}
\displaystyle \text{Let }t=x-\frac{\pi}{2}
\displaystyle \therefore \cos x=\cos\left(\frac{\pi}{2}+t\right)=-\sin t
\displaystyle \pi-2x=\pi-2\left(\frac{\pi}{2}+t\right)=-2t
\displaystyle \therefore \lim_{x\to\pi/2}\frac{\cos x}{\pi-2x}=\lim_{t\to0}\frac{-\sin t}{-2t}=\frac{1}{2}
\displaystyle \therefore \frac{k}{2}=3
\displaystyle \therefore k=6
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 8. }\text{If }f(x)=\begin{cases}mx+1,&x\leq\frac{\pi}{2}\\ \sin x+n,&x>\frac{\pi}{2}\end{cases}\text{ is continuous at }x=\frac{\pi}{2},\text{ then}
\displaystyle \text{(a) }m=1,\ n=0\qquad\text{(b) }m=\frac{n\pi}{2}+1
\displaystyle \text{(c) }n=\frac{m\pi}{2}\qquad\text{(d) }m=n=\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Since }f(x)\text{ is continuous at }x=\frac{\pi}{2},
\displaystyle \lim_{x\to(\pi/2)^-}f(x)=\lim_{x\to(\pi/2)^+}f(x)=f\left(\frac{\pi}{2}\right)
\displaystyle \therefore m\left(\frac{\pi}{2}\right)+1=\sin\frac{\pi}{2}+n
\displaystyle \therefore \frac{m\pi}{2}+1=1+n
\displaystyle \therefore n=\frac{m\pi}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 9. }\text{Find the derivative of }xe^{x}\sin x\text{ w.r.t. }x.
\displaystyle \text{Answer:}
\displaystyle y=xe^{x}\sin x
\displaystyle \frac{dy}{dx}=x\frac{d}{dx}(e^{x}\sin x)+e^{x}\sin x
\displaystyle =x(e^{x}\sin x+e^{x}\cos x)+e^{x}\sin x
\displaystyle =e^{x}\left[x(\sin x+\cos x)+\sin x\right]
\\

\displaystyle \textbf{Question 10. }\text{Differentiate }\frac{a^{x}}{1+a^{x}}\text{ w.r.t. }a^{x}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=a^{x}
\displaystyle y=\frac{u}{1+u}
\displaystyle \frac{dy}{du}=\frac{(1+u)-u}{(1+u)^{2}}
\displaystyle =\frac{1}{(1+u)^{2}}
\displaystyle \therefore \frac{d}{d(a^{x})}\left(\frac{a^{x}}{1+a^{x}}\right)=\frac{1}{(1+a^{x})^{2}}
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\displaystyle \textbf{Question 11. }\text{Find the second derivative of function }y=\log\log x.
\displaystyle \text{Answer:}
\displaystyle y=\log\log x
\displaystyle \frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}
\displaystyle =\frac{1}{x\log x}
\displaystyle \therefore \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(x\log x)^{-1}
\displaystyle =-\frac{\log x+1}{(x\log x)^{2}}
\displaystyle =-\frac{\log x+1}{x^{2}(\log x)^{2}}
\\

\displaystyle \textbf{Question 12. }\text{Is the function }f\text{ defined by }f(x)=\begin{cases}x,&x\leq1\\5,&x>1\end{cases}\text{ continuous at } \\ x=0,\text{ at }x=1\text{ and at }x=2?
\displaystyle \text{Answer:}
\displaystyle \text{For }x\leq1,\ f(x)=x\text{ and for }x>1,\ f(x)=5
\displaystyle \text{At }x=0,\ f(x)=x
\displaystyle \therefore f(0)=0\text{ and }\lim_{x\to0}f(x)=0
\displaystyle \therefore f\text{ is continuous at }x=0
\displaystyle \text{At }x=1,\ f(1)=1
\displaystyle \lim_{x\to1^-}f(x)=1
\displaystyle \lim_{x\to1^+}f(x)=5
\displaystyle \therefore \lim_{x\to1^-}f(x)\neq\lim_{x\to1^+}f(x)
\displaystyle \therefore f\text{ is not continuous at }x=1
\displaystyle \text{At }x=2,\ f(x)=5
\displaystyle \therefore f(2)=5\text{ and }\lim_{x\to2}f(x)=5
\displaystyle \therefore f\text{ is continuous at }x=2
\\

\displaystyle \textbf{Question 13. }\text{Find the value of }P\text{ for which the function }f(x)=\begin{cases}\frac{1-\cos4x}{x^{2}},&x\neq0\\P,&x=0\end{cases} \\ \text{ is continuous at }x=0
\displaystyle \text{Answer:}
\displaystyle \text{For continuity at }x=0,
\displaystyle \lim_{x\to0}f(x)=f(0)
\displaystyle \therefore \lim_{x\to0}\frac{1-\cos4x}{x^{2}}=P
\displaystyle \text{Now, }1-\cos4x=2\sin^{2}2x
\displaystyle \therefore \lim_{x\to0}\frac{1-\cos4x}{x^{2}}=\lim_{x\to0}\frac{2\sin^{2}2x}{x^{2}}
\displaystyle =2\lim_{x\to0}\left(\frac{\sin2x}{x}\right)^{2}
\displaystyle =2\lim_{x\to0}\left(2\cdot\frac{\sin2x}{2x}\right)^{2}
\displaystyle =2(2)^{2}
\displaystyle =8
\displaystyle \therefore P=8
\\

\displaystyle \textbf{Question 14. }\text{If }f(x)=\sin2x-\cos2x,\text{ then find }f'\left(\frac{\pi}{6}\right)
\displaystyle \text{Answer:}
\displaystyle f(x)=\sin2x-\cos2x
\displaystyle \therefore f'(x)=2\cos2x+2\sin2x
\displaystyle \therefore f'\left(\frac{\pi}{6}\right)=2\cos\frac{\pi}{3}+2\sin\frac{\pi}{3}
\displaystyle =2\cdot\frac{1}{2}+2\cdot\frac{\sqrt{3}}{2}
\displaystyle =1+\sqrt{3}
\\

\displaystyle \textbf{Question 15. }\text{Find }\frac{dy}{dx},\text{ when }\sin(x+y)=x^{2}+y^{2}
\displaystyle \text{Answer:}
\displaystyle \sin(x+y)=x^{2}+y^{2}
\displaystyle \text{Differentiating w.r.t. }x,
\displaystyle \cos(x+y)\left(1+\frac{dy}{dx}\right)=2x+2y\frac{dy}{dx}
\displaystyle \cos(x+y)+\cos(x+y)\frac{dy}{dx}=2x+2y\frac{dy}{dx}
\displaystyle \left(\cos(x+y)-2y\right)\frac{dy}{dx}=2x-\cos(x+y)
\displaystyle \therefore \frac{dy}{dx}=\frac{2x-\cos(x+y)}{\cos(x+y)-2y}
\\

\displaystyle \textbf{Question 16. }\text{If }y=ax-\frac{(ax)^{2}}{2}+\frac{(ax)^{3}}{3}-\cdots,\text{ then find }\frac{dy}{dx}
\displaystyle \text{Answer:}
\displaystyle y=ax-\frac{(ax)^{2}}{2}+\frac{(ax)^{3}}{3}-\cdots
\displaystyle \text{We know that }\log(1+t)=t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots
\displaystyle \therefore y=\log(1+ax)
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{1+ax}\cdot a
\displaystyle =\frac{a}{1+ax}
\\

\displaystyle \textbf{Question 17. }\text{Determine the value of }k\text{ for which the following function}
\displaystyle \text{is continuous at }x=3: \ \ \ f(x)=\begin{cases}\frac{(x+3)^{2}-36}{x-3},&x\neq3\\k,&x=3\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{For continuity at }x=3,
\displaystyle \lim_{x\to3}f(x)=f(3)=k
\displaystyle \therefore k=\lim_{x\to3}\frac{(x+3)^{2}-36}{x-3}
\displaystyle =\lim_{x\to3}\frac{x^{2}+6x+9-36}{x-3}
\displaystyle =\lim_{x\to3}\frac{x^{2}+6x-27}{x-3}
\displaystyle =\lim_{x\to3}\frac{(x-3)(x+9)}{x-3}
\displaystyle =\lim_{x\to3}(x+9)
\displaystyle =12
\displaystyle \therefore k=12
\\

\displaystyle \textbf{Question 18. }\text{Find the values of }a\text{ and }b\text{ such that the function defined as follows is continuous.}
\displaystyle f(x)=\begin{cases}x+2,&x\leq2\\ax+b,&2<x<5\\3x-2,&x\geq5\end{cases}
\displaystyle \text{Answer:}
\displaystyle \text{For continuity at }x=2,
\displaystyle \lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)
\displaystyle \therefore 2+2=2a+b
\displaystyle \therefore 2a+b=4 \qquad ...(1)
\displaystyle \text{For continuity at }x=5,
\displaystyle \lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)
\displaystyle \therefore 5a+b=3(5)-2
\displaystyle \therefore 5a+b=13 \qquad ...(2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle 3a=9
\displaystyle \therefore a=3
\displaystyle \text{Substituting in (1),}
\displaystyle 2(3)+b=4
\displaystyle \therefore b=-2
\displaystyle \therefore a=3,\ b=-2
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\displaystyle \textbf{Question 19. }\text{If }y=3\cos(\log x)+4\sin(\log x),\text{ then show that } \\ x^{2}y_{2}+xy_{1}+y=0.
\displaystyle \text{Answer:}
\displaystyle y=3\cos(\log x)+4\sin(\log x)
\displaystyle y_{1}=\frac{dy}{dx}=-\frac{3}{x}\sin(\log x)+\frac{4}{x}\cos(\log x)
\displaystyle =\frac{-3\sin(\log x)+4\cos(\log x)}{x}
\displaystyle \therefore xy_{1}=-3\sin(\log x)+4\cos(\log x)
\displaystyle \text{Differentiating again,}
\displaystyle y_{2}=\frac{d}{dx}\left[\frac{-3\sin(\log x)+4\cos(\log x)}{x}\right]
\displaystyle =\frac{-3\cos(\log x)-4\sin(\log x)}{x^{2}}-\frac{-3\sin(\log x)+4\cos(\log x)}{x^{2}}
\displaystyle \therefore x^{2}y_{2}=-3\cos(\log x)-4\sin(\log x)+3\sin(\log x)-4\cos(\log x)
\displaystyle =-7\cos(\log x)+3\sin(\log x)-4\sin(\log x)
\displaystyle =-7\cos(\log x)-\sin(\log x)
\displaystyle \text{Now,}
\displaystyle x^{2}y_{2}+xy_{1}+y
\displaystyle =[-7\cos(\log x)-\sin(\log x)]+[-3\sin(\log x)+4\cos(\log x)]
\displaystyle \qquad +[3\cos(\log x)+4\sin(\log x)]
\displaystyle =0
\displaystyle \therefore x^{2}y_{2}+xy_{1}+y=0
\\

\displaystyle \textbf{Question 20. }\text{Show that the function }f(x)=\begin{cases}\frac{\sin x}{x}+\cos x,&x\neq0\\2,&x=0\end{cases} \\ \text{ is continuous at }x=0
\displaystyle \text{Answer:}
\displaystyle f(0)=2
\displaystyle \lim_{x\to0}f(x)=\lim_{x\to0}\left(\frac{\sin x}{x}+\cos x\right)
\displaystyle =\lim_{x\to0}\frac{\sin x}{x}+\lim_{x\to0}\cos x
\displaystyle =1+1=2
\displaystyle \therefore \lim_{x\to0}f(x)=f(0)
\displaystyle \therefore f(x)\text{ is continuous at }x=0
\\

\displaystyle \textbf{Question 21. }\text{Show that the function }f(x)=2x-|x|\text{ is continuous at } \\ x=0
\displaystyle \text{Answer:}
\displaystyle f(x)=2x-|x|
\displaystyle f(0)=0
\displaystyle \lim_{x\to0^-}f(x)=\lim_{x\to0^-}(2x+x)=\lim_{x\to0^-}3x=0
\displaystyle \lim_{x\to0^+}f(x)=\lim_{x\to0^+}(2x-x)=\lim_{x\to0^+}x=0
\displaystyle \therefore \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)
\displaystyle \therefore f(x)\text{ is continuous at }x=0
\\

\displaystyle \textbf{Question 22. }\text{Prove that the function }f(x)=|x-1|,\ x\in R\text{ is not differentiable at } \\ x=1.
\displaystyle \text{Answer:}
\displaystyle f(x)=|x-1|=\begin{cases}1-x,&x<1\\x-1,&x\geq1\end{cases}
\displaystyle \text{LHD at }x=1=\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to0^-}\frac{|h|-0}{h}=\lim_{h\to0^-}\frac{-h}{h}=-1
\displaystyle \text{RHD at }x=1=\lim_{h\to0^+}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to0^+}\frac{|h|-0}{h}=\lim_{h\to0^+}\frac{h}{h}=1
\displaystyle \therefore \text{LHD}\neq\text{RHD}
\displaystyle \therefore f(x)=|x-1|\text{ is not differentiable at }x=1
\\

\displaystyle \textbf{Question 23. }\text{Find }\frac{dy}{dx}\text{ at }x=1,\ y=\frac{\pi}{4},\text{ if }\sin^{2}y+\cos xy=K
\displaystyle \text{Answer:}
\displaystyle \sin^{2}y+\cos xy=K
\displaystyle \text{Differentiating w.r.t. }x,
\displaystyle 2\sin y\cos y\frac{dy}{dx}-\sin xy\left(y+x\frac{dy}{dx}\right)=0
\displaystyle \sin2y\frac{dy}{dx}=\sin xy\left(y+x\frac{dy}{dx}\right)
\displaystyle \text{At }x=1,\ y=\frac{\pi}{4},
\displaystyle 1\cdot\frac{dy}{dx}=\frac{1}{\sqrt{2}}\left(\frac{\pi}{4}+\frac{dy}{dx}\right)
\displaystyle \sqrt{2}\frac{dy}{dx}=\frac{\pi}{4}+\frac{dy}{dx}
\displaystyle (\sqrt{2}-1)\frac{dy}{dx}=\frac{\pi}{4}
\displaystyle \therefore \frac{dy}{dx}=\frac{\pi}{4(\sqrt{2}-1)}=\frac{\pi(\sqrt{2}+1)}{4}
\\

\displaystyle \textbf{Question 24. }\text{Find the derivative of }\tan^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\text{ with respect to }x
\displaystyle \text{Answer:}
\displaystyle y=\tan^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)
\displaystyle =\tan^{-1}\left(\frac{1+\tan x}{1-\tan x}\right)
\displaystyle =\tan^{-1}\left(\tan\left(\frac{\pi}{4}+x\right)\right)
\displaystyle =\frac{\pi}{4}+x
\displaystyle \therefore \frac{dy}{dx}=1
\\

\displaystyle \textbf{Question 25. }\text{Differentiate the following function w.r.t. }x,
\displaystyle \tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)
\displaystyle \text{Put }x=\sin\theta
\displaystyle y=\tan^{-1}\left(\frac{\sqrt{1+\sin\theta}-\sqrt{1-\sin\theta}}{\sqrt{1+\sin\theta}+\sqrt{1-\sin\theta}}\right)
\displaystyle =\tan^{-1}\left(\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}-\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}+\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}\right)
\displaystyle =\tan^{-1}\left(\tan\frac{\theta}{2}\right)
\displaystyle =\frac{\theta}{2}=\frac{1}{2}\sin^{-1}x
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}
\\

\displaystyle \textbf{Question 26. }\text{If }y=(\tan^{-1}x)^{2},\text{ then show that } \\ (x^{2}+1)^{2}\frac{d^{2}y}{dx^{2}}+2x(x^{2}+1)\frac{dy}{dx}=2.
\displaystyle \text{Answer:}
\displaystyle y=(\tan^{-1}x)^{2}
\displaystyle \therefore \frac{dy}{dx}=2\tan^{-1}x\cdot\frac{1}{1+x^{2}}
\displaystyle =\frac{2\tan^{-1}x}{1+x^{2}}
\displaystyle \therefore \frac{d^{2}y}{dx^{2}}=\frac{2(1+x^{2})\cdot\frac{1}{1+x^{2}}-2\tan^{-1}x(2x)}{(1+x^{2})^{2}}
\displaystyle =\frac{2-4x\tan^{-1}x}{(1+x^{2})^{2}}
\displaystyle \text{Now,}
\displaystyle (x^{2}+1)^{2}\frac{d^{2}y}{dx^{2}}+2x(x^{2}+1)\frac{dy}{dx}
\displaystyle =(2-4x\tan^{-1}x)+2x(1+x^{2})\cdot\frac{2\tan^{-1}x}{1+x^{2}}
\displaystyle =2-4x\tan^{-1}x+4x\tan^{-1}x
\displaystyle =2
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 27. }\text{If }\sin y=x\sin(a+y),\text{ then prove that }\frac{dy}{dx}=\frac{\sin^{2}(a+y)}{\sin a}.
\displaystyle \text{Answer:}
\displaystyle \sin y=x\sin(a+y)
\displaystyle \text{Differentiating w.r.t. }x,
\displaystyle \cos y\frac{dy}{dx}=\sin(a+y)+x\cos(a+y)\frac{dy}{dx}
\displaystyle \left[\cos y-x\cos(a+y)\right]\frac{dy}{dx}=\sin(a+y)
\displaystyle \therefore \frac{dy}{dx}=\frac{\sin(a+y)}{\cos y-x\cos(a+y)}
\displaystyle \text{From }\sin y=x\sin(a+y),
\displaystyle x=\frac{\sin y}{\sin(a+y)}
\displaystyle \therefore \frac{dy}{dx}=\frac{\sin(a+y)}{\cos y-\frac{\sin y}{\sin(a+y)}\cos(a+y)}
\displaystyle =\frac{\sin^{2}(a+y)}{\cos y\sin(a+y)-\sin y\cos(a+y)}
\displaystyle =\frac{\sin^{2}(a+y)}{\sin(a+y-y)}
\displaystyle =\frac{\sin^{2}(a+y)}{\sin a}
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 28. }\text{Show that the function }f(x)=x-[x]\text{ is discontinuous at all integral points.}
\displaystyle \text{Answer:}
\displaystyle f(x)=x-[x]
\displaystyle \text{Let }n\in Z
\displaystyle f(n)=n-[n]=n-n=0
\displaystyle \lim_{x\to n^-}f(x)=\lim_{x\to n^-}(x-(n-1))=n-(n-1)=1
\displaystyle \lim_{x\to n^+}f(x)=\lim_{x\to n^+}(x-n)=n-n=0
\displaystyle \therefore \lim_{x\to n^-}f(x)\neq\lim_{x\to n^+}f(x)
\displaystyle \therefore f(x)=x-[x]\text{ is discontinuous at all integral points.}
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\displaystyle \textbf{Question 29. }\text{Show that the function }f(x)=\begin{cases}\frac{e^{1/x}-1}{e^{1/x}+1},&x\neq0\\0,&x=0\end{cases} \\ \text{ is discontinuous at }x=0
\displaystyle \text{Answer:}
\displaystyle f(0)=0
\displaystyle \lim_{x\to0^+}f(x)=\lim_{x\to0^+}\frac{e^{1/x}-1}{e^{1/x}+1}
\displaystyle \text{As }x\to0^+,\ \frac{1}{x}\to\infty,\text{ so }e^{1/x}\to\infty
\displaystyle \therefore \lim_{x\to0^+}f(x)=1
\displaystyle \lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{e^{1/x}-1}{e^{1/x}+1}
\displaystyle \text{As }x\to0^-,\ \frac{1}{x}\to-\infty,\text{ so }e^{1/x}\to0
\displaystyle \therefore \lim_{x\to0^-}f(x)=\frac{0-1}{0+1}=-1
\displaystyle \therefore \lim_{x\to0^+}f(x)\neq\lim_{x\to0^-}f(x)
\displaystyle \therefore f(x)\text{ is discontinuous at }x=0
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\displaystyle \textbf{Question 30. }\text{If }y=\left[x+\sqrt{x^{2}+a^{2}}\right]^{n},\text{ then prove that }\frac{dy}{dx}=\frac{ny}{\sqrt{x^{2}+a^{2}}}.
\displaystyle \text{Answer:}
\displaystyle y=\left[x+\sqrt{x^{2}+a^{2}}\right]^{n}
\displaystyle \log y=n\log\left(x+\sqrt{x^{2}+a^{2}}\right)
\displaystyle \text{Differentiating w.r.t. }x,
\displaystyle \frac{1}{y}\frac{dy}{dx}=n\cdot\frac{1}{x+\sqrt{x^{2}+a^{2}}}\left(1+\frac{x}{\sqrt{x^{2}+a^{2}}}\right)
\displaystyle =n\cdot\frac{1}{x+\sqrt{x^{2}+a^{2}}}\cdot\frac{x+\sqrt{x^{2}+a^{2}}}{\sqrt{x^{2}+a^{2}}}
\displaystyle =\frac{n}{\sqrt{x^{2}+a^{2}}}
\displaystyle \therefore \frac{dy}{dx}=\frac{ny}{\sqrt{x^{2}+a^{2}}}
\displaystyle \text{Hence proved.}
\\

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