MATHS M.C.Q (1 Marks)

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

Solution:

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

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

$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^{2}-1).2\text{x}-(\text{x}^{2}+1).2\text{x}}{(\text{x}^{2}-1)^{2}}$

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

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

1. $\cos9$

Solution:

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

$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}.\cos(\text{x}+9)-\sin(\text{x}+9)(-\sin\text{x})}{\cos^{2}\text{x}}$

$=\frac{\cos\text{x}\cos(\text{x}+9)+\sin\text{x}\sin(\text{x}+9)}{\cos^{2}\text{x}}$

$=\frac{\cos(\text{x}+9-\text{x})}{\cos^{2}\text{x}}=\frac{\cos9}{\cos^{2}\text{x}}$

$=\frac{\cos9}{(1)^{2}}=\cos9$ 

4. 0

Solution:

Given that $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ 

$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}^{\frac{3}{2}}}$

$\big(\frac{\text{dy}}{\text{dx}}\big)=\frac{1}{2}-\frac{1}{2}=0$

2. 100

Solution:

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

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

$\therefore \text{f'}(1)=1+11+....+1=100$ 

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

Solution:

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

$\therefore\text{f}(\text{x})=\frac{1}{2}\bigg[\frac{\sqrt{\text{x}.}1-(\text{x}-4).\frac{1}{2\sqrt{\text{x}}}}{\text{x}}\bigg]$

$=\frac{1}{2}\Big[\frac{2\text{x}-\text{x}+4}{2\sqrt{\text{x}.\text{x}}}\Big]$

$=\frac{1}{2}\Bigg[\frac{\text{x}+4}{2(\text{x})^{\frac{3}{2}}}\Bigg]$

$\therefore\text{f}(\text{x})\ \text{x}=1=\frac{1}{2}\Big[\frac{1+4}{2\times1}\Big]=\frac{5}{4}$

3. Does not exist

Solution:

Given $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$

$\text{f'}(\text{x})=\frac{(\text{x}-\text{a})(\text{n.}\text{x}^{\text{n-1}}-(\text{x}^{\text{n}-\text{a}^{\text{n}}})1}{(\text{x}-\text{a})^{2}}$

So, $\text{f}(\text{a})=\frac{0}{0}$ = Does not exist.

4. 50

Solution:

Given that f(x) = 1 - x + x² - x³ + .......-x⁹⁹ + x¹⁰⁰

f'(x) = -1 + 2x - 3x² + ... -99.x⁹⁸ + 100.x⁹⁹

f'(x) = -1 + 2 - 3 + ....-99 + 100

=(- 1 - 3 - 5 ....-99) + (2 + 4 + 6 + ....100)

$=\frac{50}{2}\big[2\times1+(50-1)(-2)\big]+\frac{50}{2}\big[2\times2(50-1)2\big] $

$=25\big[-11+102\big]=25\times2=50$

2. $\frac{\text{m}}{\text{n}}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}=\lim\limits_{\text{x} \rightarrow 1}\frac{\frac{\text{x}^{\text{m}-(1)^\text{m}}}{\text{x}-1}}{\frac{\text{x}^{\text{n}-(1)^{\text{n}}}}{\text{x}-1}}$ 

$=\frac{\text{m}(1)^{\text{m}-1}}{\text{n}(1)^{\text{n}-1}}=\frac{\text{m}}{\text{n}}$

$=\frac{\text{m}}{\text{n}}$

3. $1$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$

$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$

Taking limit, we get

$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]=\frac{1}{2}\times1\times2$

$=1$

1. 5050

Solution:

Given f(x) = x¹⁰⁰ + x⁹⁹ + .... + x + 1

f'(x) = 100.x¹⁰⁰ + 99.x⁹⁸ + .... + 1

S0, f'(1) = 100 + 99 + 98 + ..... + 1

$=\frac{100}{2}[2\times100+(100-1)(-1)]$

$=50[200-99]=50\times101=5050$

$=5050$

1. $\text{n}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$

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

 $=\text{n}$

3. $\frac{1}{2}$

Solution:

Given $​​\lim\limits_{\text{x} \rightarrow 0}\frac{\text{cosec}\text{x}-\cot\text{x}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}{\text{x}}$ 

$ =\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}=\frac{2\sin^{2}\frac{\text{x}}{2}}{\text{x}\cdot\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow1}\frac{\sin\frac{\text{x}}{2}}{\text{x}\cos\frac{\text{x}}{2}} =\frac{\tan\frac{\text{x}}{2}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\frac{\text{x}}{2}}{2\times\frac{\text{x}}{2}}$

$=\frac{1}{2}\times1=\frac{1}{2}$

4. None of these.

Solution:

Given $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ 

$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $

$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}=-1$

$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}=1$

$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$

So, the limit does not exist.

4. $\text{x}^{2}-10\text{x}+21=0$

Solution:

Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ 

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$

$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$

$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$

$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]=7$

Therefore, the quadratic equation whose roots are 3 and 7 is $\text{x}^{2}-10\text{x}+21=0$

1. 2

Solution:

Given $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}^{2}\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\text{x}^{2}}{4}\times4\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\text{x}}{2}\Big)\cdot2\cos\text{x}}{\sin^{2}\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\bigg)\cdot2\cos\text{x}$

$=2\cos\text{x}0=2\times1=2$

2. $\frac{4}{9}$

Solution:

Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$

 $=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$

$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]=\Big[\frac{2\theta}{2\theta}\Big]^{2}=\Big(\frac{2}{3}\Big)^{2}=\frac{4}{9}$

$=\frac{4}{9}$

1. $-2$

Solution:

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

$\frac{\text{dy}}{\text{dx}}=\frac{-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$

$=\frac{-(\sin\text{x}+\cos\text{x})^{2}(\sin\text{x}+\cos\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$

$=\frac{\sin^{2}\text{x}+\cos^{2}\text{x}+2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^{2}}$

$=\frac{-2}{(\sin\text{x}-\cos\text{x})^{2}}$

$\therefore \Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-2}{(\sin\text{0}-\cos0)^{2}}=\frac{-2}{(-1)^{2}}=2$

3. -1

Solution:

Given, $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}=\lim\limits_{\text{x} \rightarrow\pi}\frac{\sin(\pi)-\text{x}}{-(\pi-\text{x})}$ 

$=-1$

3. Does not exist.

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$

$\text{L}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{-\sin\text{x}}{\text{x}}=-1$

$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}=1$

$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$ 

So, the limit does not exist.

2. 1

Solution:

Given f(x) = x - [x]

we have ti first check for differentiability of f(x) at $\text{x}=\frac{1}{2}$

$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}=\frac{-\text{h}}{-\text{h}}=1$

$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}=\frac{\text{h}}{\text{h}}=1$

Since, LHD = RHD

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

4. 2

Solution:

Given $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$ 

$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\tan\text{x}-1}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{(\tan\text{x}+1)(\tan\text{x}-1)}{(\tan\text{x}-1)}$

$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)=\tan\frac{\pi}{4}+1$

$=1+1=2$

$=2$

2. $-\frac{1}{10}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ 

$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$

$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$

Taking limit, we get

$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}=\frac{-1}{2\times5}=\frac{-1}{10}$

2. $\frac{1}{2}$

Solution:

Given $ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}} =\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}\big[\frac{\tan2\text{x}}{\text{x}}-1\big]}{\text{x}\big[3-\frac{\sin\text{x}}{\text{x}}\big]}$

$ =\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\tan2\text{x}}{2\text{x}}\times2-1}{3-\frac{\sin\text{x}}{\text{x}}}=\frac{1.2-1}{3-1}$

$=\frac{2-1}{2}=\frac{1}{2}$