MCQ

MCQ 11 Mark

If $\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100},$then $f\ ' (1)$ equals

A
$150$
B
$-50$
C
$-150$
✓
$50$

Answer

Correct option: D.

$50$

$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100})$
$=\frac{\text{d}}{\text{dx}}(1)-\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x}^3)+\dots-\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\frac{\text{d}}{\text{dx}}(\text{x}^{100})$
$=0-1+\text{2x}-\text{3x}^2+\dots-\text{99x}^{98}+100\text{x}^{99}$
Putting $x = 1,$ we get
$\text{f}'(1)=-1+2-3+\dots-99+100$
$=(-1+2)+(-3+4)+(-5+6)+\dots+(-99+100)$
$=1+1+1+\dots+1(50$ terms$)$
$=50$

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MCQ 21 Mark

Let $\text{f}(\text{x})=\text{x}-[\text{x}],\text{x}\in\text{R},$ then $\text{f}'\Big(\frac{1}{2}\Big)$ is :

A
$\frac{3}{2}$
✓
$1$
C
$0$
D
$-1$

Answer

Correct option: B.

$1$

Given : $\text{f}(\text{x})=\text{x}-[\text{x}],\text{x}\in\text{R}$
Now,
For $0\le\text{x}<1,[\text{x}]=0$
$\therefore\text{f}(\text{x})=\text{x}-0=\text{x},\forall\text{x}\in[0,1)$
Differentiate with respect to $x,$ we get
$\text{f}'(\text{x})=1,\forall\text{x}\in[0,1)$
$\therefore\text{f}'\Big(\frac{1}{2}\Big)=1$

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MCQ 31 Mark

If $\text{f(x)}=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}},$ then $\text{f}'\text{(a)}$ is :

A
$1$
B
$0$
C
$\frac{1}{2}$
✓
$\text{dose not exist}$

Answer

Correct option: D.

$\text{dose not exist}$

Given : $\text{f(x)}=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}$
Now $, f(x)$ is not difined at $x = a.$
Therefore $, f(x)$ is not differentiable at $x = a.$
So $, f\ '(a)$ dose not exist.
Hence, the correct answer is option $(d).$

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MCQ 41 Mark

If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is :

✓
$\cos9$
B
$\sin9$
C
$0$
D
$1$

Answer

Correct option: A.

$\cos9$

$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x})\frac{\text{d}}{\text{dx}}\sin(\text{x}+9)-\sin(\text{x}+9)\frac{\text{d}}{\text{dx}}(\cos\text{x})}{\cos^2\text{x}}\ ($Quotient rule$)$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))-(\sin(\text{x}+9))(-\sin\text{x})}{\cos^2\text{x}}$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))+(\sin(\text{x}+9))(\sin\text{x})}{\cos^2\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^2\text{x}}$
$=\frac{\cos9}{\cos^2\text{x}}$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $\cos9$

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MCQ 51 Mark

if $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100},$then $f\, '(1)$ is equal to :

A
$\frac{1}{100}$
✓
$100$
C
$50$
D
$0$

Answer

Correct option: B.

$100$

$\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big(1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2}\Big)+\dots+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\dots+\frac{1}{100}\frac{\text{d}}{\text{dx}}(\text{x}^{100})$
$=0+1+\frac{1}{2}\times\text{2x}+\dots+\frac{1}{100}\times100\text{x}^{99}$
$=1+\text{x}+\text{x}^2+\dots+\text{x}^{99}$
Putting $x = 1,$ we get
$\text{f}'(\text{x})=1+1+1+\dots+1\ (100$ terms$)$
$=100$

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MCQ 61 Mark

if $\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1,$ then $f\ '(1)$ is equal to

✓
$5050$
B
$5049$
C
$5051$
D
$50051$

Answer

Correct option: A.

$5050$

$\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1)$
$=\frac{\text{d}}{\text{dx}}(\text{x}^{100})+\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\dots+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(1)$
$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1+0$
$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1$
Putting $x = 1,$ we get
$\text{f}'(\text{x})=100+99+98+\dots+2+1$
$=\frac{100(100+1)}{2}\ \Big(\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big)$
$=50\times101$
$=5050$

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MCQ 71 Mark

If $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}},$ then $f\ '(1)$ is :

✓
$\frac{5}{4}$
B
$\frac{4}{5}$
C
$1$
D
$0$

Answer

Correct option: A.

$\frac{5}{4}$

$\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$
$=\frac{1}{2}\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}$
$=\frac{1}{2}\text{x}^{\frac{1}{2}}-\text{2x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to x, we get
$\text{f}'(\text{x})=\frac{1}{2}\times\frac{1}{2}\text{x}^{\frac{1}{2}-1}-2\times\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$ [\text{f}(\text{x})=\text{x}^\text{n}\Rightarrow\text{f}'(\text{x})=\text{nx}^{\text{n}-1}]$
$\Rightarrow\text{f}'(\text{x})=\frac{1}{4}\text{x}^{-\frac{1}{2}}+\text{x}^{-\frac{3}{2}}$
$\therefore\text{f}'(\text{1})=\frac{1}{4}\times1+1=\frac{5}{4}$

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MCQ 81 Mark

Mark the correct alternative in each of the following : If $\text{f(x)}=\text{x}\sin\text{x},$ then $\text{f}'\Big(\frac{\text{x}}{2}\Big)=$

A
$0$
✓
$1$
C
$-1$
D
$\frac{1}{2}$

Answer

Correct option: B.

$1$

$\text{f(x)}=\text{x}\sin\text{x}$
Differentiating both sides with respect to $x,$ we get
$\text{f}'\text{(x)}=\text{x}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\times\frac{\text{d}}{\text{dx}}\text{(x)} \ ($Product rule$)$
$=\text{x}\times\cos\text{x}+\sin\text{x}\times1$
$=\text{x}\cos\text{x}+\sin\text{x}$
Putting $\text{x}=\frac\pi{2},$ we get
$=\frac{\pi}{2}\times0+1$
$=1$
Hence, the correct answer is option $(b)$

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MCQ 91 Mark

If $\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}},$ then $\frac{\text{dy}}{\text{dx}}=$

✓
$-\frac{4\text{x}}{(\text{x}^2-1)^2}$
B
$-\frac{4\text{x}}{\text{x}^2-1}$
C
$\frac{1-\text{x}^2}{\text{4x}}$
D
$\frac{4\text{x}}{\text{x}^2-1}$

Answer

Correct option: A.

$-\frac{4\text{x}}{(\text{x}^2-1)^2}$

$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$
$=\frac{\text{x}^2+1}{\text{x}^2-1}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2} \ ($Quotient rule$)$
$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$
$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$
$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$

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MCQ 101 Mark

If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is

A
$1$
B
$\frac{1}{2}$
C
$\frac{1}{\sqrt{\text{2}}}$
✓
$0$

Answer

Correct option: D.

$0$

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x,$ we get
$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$
Putting $x = 1,$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is $0$.

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MCQ 111 Mark

If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is :

✓
$-2$
B
$0$
C
$\frac{1}{2}$
D
does not exist

Answer

Correct option: A.

$-2$

$\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$=\frac{(\sin\text{x}-\cos\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-(\cos^2\text{x}+\sin^2\text{x}-2\cos\text{x}\sin\text{x})-(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-1+2\cos\text{x}\sin\text{x}-1-2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^2}$
Putting $x = 0$ is $-2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=0}=\frac{-2}{(\sin0-\cos0)}=\frac{-2}{(0-1)^2}=-2$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $-2$

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MCQ 121 Mark

If $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots,$ then $\frac{\text{dy}}{\text{dx}}=$

A
$y + 1$
B
$y - 1$
✓
$y$
D
$\text{y}^2$

Answer

Correct option: C.

$y$

$\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^4}{4!}\Big)+\dots$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{1}{1!}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2!}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{1}{3 !}\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{1}{4!}\frac{\text{d}}{\text{dx}}(\text{x}^4)+\dots$
$=0+\frac{1}{1!}\times1+\frac{1}{2!}\times2\text{x}+\frac{1}{3!}\times3\text{x}^2+\frac{1}{4!}\times4\text{x}^3+\dots$
$=1+\frac{1}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\ \Big[\frac{\text{n}}{\text{n}!}=\frac{1}{(\text{n}-1)!}\Big]$
$=\text{y}$

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