MATHS M.C.Q (1 Marks)

4. $\text{dose not exist}$

Solution:

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).

2. $1$

Solution:

$\text{f(x)}=\text{x}\sin\text{x}$

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)

4. 50

Solution:

$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$

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$)$

$=50$

2. 1

Solution:

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

$\text{f}'(\text{x})=1,\forall\text{x}\in[0,1)$

$\therefore\text{f}'\Big(\frac{1}{2}\Big)=1$

1. $\cos9$ 

Solution:

$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$

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 $\cos9$

1. 5050

Solution:

$\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1$

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 

$\text{f}'(\text{x})=100+99+98+\dots+2+1$

$=\frac{100(100+1)}{2}\ \Big(\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big)$

$=50\times101$

$=5050$​​​​​​​

1. $\frac{5}{4}$

Solution:

$\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$

$=\frac{1}{2}\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}$

$=\frac{1}{2}\text{x}^{\frac{1}{2}}-\text{2x}^{-\frac{1}{2}}$

Differentiate both the sides with respect to x, we get 

$\text{f}'(\text{x})=\frac{1}{2}\times\frac{1}{2}\text{x}^{\frac{1}{2}-1}-2\times\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}\ [\text{f}(\text{x})=\text{x}^\text{n}\Rightarrow\text{f}'(\text{x})=\text{nx}^{\text{n}-1}]$

$\Rightarrow\text{f}'(\text{x})=\frac{1}{4}\text{x}^{-\frac{1}{2}}+\text{x}^{-\frac{3}{2}}$

$\therefore\text{f}'(\text{1})=\frac{1}{4}\times1+1=\frac{5}{4}$

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

Solution:

$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$

$=\frac{\text{x}^2+1}{\text{x}^2-1}$

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)

$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$

$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$

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

4. 0

Solution:

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$

$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$

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.

2. 100

Solution:

$\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}$

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)

$=100$

1. -2

Solution:

$\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$

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

3. y

Solution:

$\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots$

Differentiate both the sides with respect to x, we get 

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\Big)$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^4}{4!}\Big)+\dots$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{1}{1!}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2!}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{1}{3 !}\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{1}{4!}\frac{\text{d}}{\text{dx}}(\text{x}^4)+\dots$

$=0+\frac{1}{1!}\times1+\frac{1}{2!}\times2\text{x}+\frac{1}{3!}\times3\text{x}^2+\frac{1}{4!}\times4\text{x}^3+\dots$

$=1+\frac{1}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\ \Big[\frac{\text{n}}{\text{n}!}=\frac{1}{(\text{n}-1)!}\Big]$

$=\text{y}$