\displaystyle \textbf{Question 1. }\text{Let }f(x)=x-[x],\ x\in R,\text{ then }f'\left(\frac{1}{2}\right)\text{ is}
\displaystyle \text{Answer:}
\displaystyle f(x)=x-[x]
\displaystyle \text{At }x=\frac{1}{2},\ [x]=0
\displaystyle \therefore f(x)=x
\displaystyle \therefore f'\left(\frac{1}{2}\right)=1
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 2. }\text{If }f(x)=\frac{x-4}{2\sqrt{x}},\text{ then }f'(1)\text{ is}
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{x-4}{2\sqrt{x}}=\frac{1}{2}\left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}\right)
\displaystyle f'(x)=\frac{1}{2}\left(\frac{1}{2}x^{-\frac{1}{2}}+2x^{-\frac{3}{2}}\right)
\displaystyle =\frac{1}{4\sqrt{x}}+\frac{1}{x^{\frac{3}{2}}}
\displaystyle \therefore f'(1)=\frac{1}{4}+1=\frac{5}{4}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 3. }\text{If }y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots,\text{ then }\frac{dy}{dx}=
\displaystyle \text{Answer:}
\displaystyle y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x
\displaystyle \therefore \frac{dy}{dx}=e^x
\displaystyle \therefore \frac{dy}{dx}=y
\displaystyle \therefore \text{Correct option is (c)}
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\displaystyle \textbf{Question 4. }\text{If }f(x)=1-x+x^2-x^3+\cdots-x^{99}+x^{100},\text{ then }f'(1)\text{ equals}
\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 =(2-1)+(4-3)+\cdots+(100-99)
\displaystyle =1+1+\cdots+1\text{ }(50\text{ times})
\displaystyle \therefore f'(1)=50
\displaystyle \therefore \text{Correct option is (d)}
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\displaystyle \textbf{Question 5. }\text{If }y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}},\text{ then }\frac{dy}{dx}=
\displaystyle \text{Answer:}
\displaystyle y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}=\frac{x^2+1}{x^2-1}
\displaystyle \frac{dy}{dx}=\frac{(x^2-1)\cdot2x-(x^2+1)\cdot2x}{(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{Correct option is (a)}
\\

\displaystyle \textbf{Question 6. }\text{If }y=\sqrt{x}+\frac{1}{\sqrt{x}},\text{ then }\frac{dy}{dx}\text{ at }x=1\text{ is}
\displaystyle \text{Answer:}
\displaystyle y=x^{\frac{1}{2}}+x^{-\frac{1}{2}}
\displaystyle \frac{dy}{dx}=\frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{3}{2}}
\displaystyle \therefore \left(\frac{dy}{dx}\right)_{x=1}=\frac{1}{2}-\frac{1}{2}=0
\displaystyle \therefore \text{Correct option is (d)}
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\displaystyle \textbf{Question 7. }\text{If }f(x)=x^{100}+x^{99}+\cdots+x+1,\text{ then }f'(1)\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle f'(x)=100x^{99}+99x^{98}+\cdots+1
\displaystyle \therefore f'(1)=100+99+\cdots+1
\displaystyle =\frac{100\times101}{2}=5050
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 8. }\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{Answer:}
\displaystyle f'(x)=1+x+x^2+\cdots+x^{99}
\displaystyle \therefore f'(1)=1+1+1+\cdots+1
\displaystyle =100
\displaystyle \therefore \text{Correct option is (b)}
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\displaystyle \textbf{Question 9. }\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{Answer:}
\displaystyle y=\frac{\sin x+\cos x}{\sin x-\cos x}
\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,\ \sin0=0,\ \cos0=1
\displaystyle \therefore \left(\frac{dy}{dx}\right)_{x=0}=\frac{(-1)(1)-(1)(1)}{(-1)^2}=-2
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 10. }\text{If }y=\frac{\sin(x+9)}{\cos x},\text{ then }\frac{dy}{dx}\text{ at }x=0\text{ is}
\displaystyle \text{Answer:}
\displaystyle y=\frac{\sin(x+9)}{\cos x}
\displaystyle \frac{dy}{dx}=\frac{\cos x\cos(x+9)-\sin(x+9)(-\sin x)}{\cos^2x}
\displaystyle =\frac{\cos x\cos(x+9)+\sin x\sin(x+9)}{\cos^2x}
\displaystyle \text{At }x=0,\ \sin0=0,\ \cos0=1
\displaystyle \therefore \left(\frac{dy}{dx}\right)_{x=0}=\cos9
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 11. }\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{This is not defined at }x=a
\displaystyle \therefore f'(a)\text{ does not exist}
\displaystyle \therefore \text{Correct option is (d)}
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\displaystyle \textbf{Question 12. }\text{If }f(x)=x\sin x,\text{ then }f'\left(\frac{\pi}{2}\right)=
\displaystyle \text{Answer:}
\displaystyle f(x)=x\sin x
\displaystyle \therefore 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 =0+1=1
\displaystyle \therefore \text{Correct option is (b)}
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