Maths M.C.Q (1 Marks)

1. $\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$

Solution:

$\text{x}=\text{f}(\text{t})$ $\text{y}=\text{g}(\text{t}),$

$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{g}'}{\text{f}'}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)\frac{\text{dt}}{\text{dx}}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{\text{f}'^2}\frac{1}{\text{f}'}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$​​​​​​​

4. None of these.

Solution:

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits _{\text{x}\rightarrow0}\frac{1}{1+\text{e}^{\frac{-1}{\text{x}}}}=1\ \Big(\because\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}=0\Big)$

$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\text{f}(0)$

Function is not continuous,

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\text{f}(0)}{-\text{h}}$

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{e}^{\frac{1}{\text{h}}}}-0}{-\text{h}}$

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{-\Big(1+\text{e}^{\frac{1}{\text{h}}}\Big)\text{h}}=-\infty$

Similarly,

$\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\infty$

1. -1

Solution:

Given, f(x) is continuous at every point of its domain. So, it is continuous at x = 1.

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f}\text{(x)}=\text{f}(1)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\text{f}(1)$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Big(4(1+\text{h})^2+3\text{b}(1+\text{h})\Big)=5(1)-4$

$\Rightarrow4+3\text{b}=1$

$\Rightarrow-3=3\text{b}$

$\Rightarrow \text{b} = -1$

3. 0

Solution:

For, x > 10

f(x) = |x - 3| = x - 3

g(x) = fof (x) = |x - 3| - 3

= x - 3 - 3

= x - 6

$\therefore$ g'(x) = 1

2. 1

Solutuion:

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\text{x}}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(1+\cos \text{x}-1)^{\frac{1}{\text{x}}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\frac{\text{x}}{2}\Big)^\frac{1}{\text{x}}$

$\text{f(0)}=\lim\limits\Big(1-2\sin^2\frac{\text{x}}{1}\Big)^{\frac{1}{-2\sin^2\frac{\text{x}}{2}}\times\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$

$\text{f(0)}= \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$

$\text{f(0)}=\lim \limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text{x}}{2}}{\text{x}}\times\sin\frac{\text{x}}{2}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text {x}}{2}}{\frac{\text{x}}{2}}\times\frac{1}{2}\sin\frac{\text{x}}{2}}$

$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{1\times\sin\frac{1}{2}}=\text{e}^0=1$

$\Rightarrow\text{k}=1$

2. 5

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$

If f(x) is continuous at x = 0, then $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f(0)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}}\Big)=\text{k}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3\log(1+3\text{x})}{3\text{x}}-\frac{2\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)-2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{2\text{x}}\Big)=\text{k}$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1+3\text{x})}{3\text{x}}\Big)+2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\log(1-2\text{x})}{-2\text{x}}\Big)=\text{k}$

$\Rightarrow3\times1+2\times1=\text{k}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\Big]$

$\Rightarrow\text{k}=3+2$

$\Rightarrow\text{k}=5$

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

Solution:

Let, $\text{y}=\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$

$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\log\text{e}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$

$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$

$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3}{4\Big(\frac{\text{x}-2}{\text{x}+2}\Big)}\times\frac{(\text{x}+2)\times1-(\text{x}-2)\times1}{(\text{x}+2)^2}$

$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{\text{x}+2-\text{x}+2}{(\text{x}+2)^2}$

$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{4}{(\text{x}+2)}$

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

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-4+3}{\text{x}^2-4}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-1}{\text{x}^2-4}$

3. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$

$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$

$\text{b}=\text{e}^{\text{a}}$

$\text{a}=\log_{\text{e}}\text{b}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$

Here, $\text{c}=1$

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

$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$

$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$

$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\text{a}=\log\text{b}=\log\frac{2}{3}$

1. n²y

Solution:

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

Differentiating both sides we get

$\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}-1}-\text{y}^{\frac{1}{\text{n}}-1}\Big)=2$

$\Rightarrow\text{y}_1\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)=2\text{ny}$

Again differentiating both sides we get

$\text{y}_2\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)+\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}-1}\Big)=2\text{ny}_1$

$\Rightarrow\text{ny}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+\frac{\text{y}^2_1}{\text{y}}\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)=2\text{n}^2\text{y}_1$

$\Rightarrow\text{nyy}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$

$\Rightarrow\text{nyy}_2\frac{2\text{ny}}{\text{y}_1}+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$

$\Rightarrow\frac{\text{n}^2\text{y}^2\text{y}_2}{\text{y}_1^2}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{4\text{x}^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow(\text{x}^2-1)\text{y}_2+\text{xy}_1=\text{n}^2\text{y}$

1. 1

Solution:

Here,

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

$\Rightarrow\sqrt{1-\text{x}}^2\text{f}(\text{x})=\sin^{-1}\text{x}$

Differentiating w.r.t.x, we get

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

$\Rightarrow(1-\text{x}^2)\text{f}'(\text{x})-\text{xf}(\text{x})=1$

2. is not continuous at x = 0

Solution:

Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$

We have

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$

if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then 

$\text{x}\rightarrow0, \text{t}\rightarrow\infty$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$

$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$

Also, f(0) = 0

$\because\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$

Hence, f(x) is discontinuons at x = 0.

3. $\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$

Solution:

Here,

$\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}'(\text{t})\sin\text{t}$

$\text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}'(\text{t})\cos\text{t}$

$ \Rightarrow\frac{\text{dx}}{\text{dt}}=\text{f}'(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}-\text{f}'(\text{t})\cos\text{}\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}'(\text{t})\sin\text)\cos\text{t}+\text{f}'{t}+\text{f}(\text{t})$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}$

Thus

$\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=\{-\text{f}(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\Big\}^2$

$=\{\text{f}(\text{t})\sin\text{t}+\text{f}''(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\}^2$

$=\sin^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2+\cos^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2(\sin^2\text{t}+\cos^2\text{t})$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

1. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$

Solution:

We have, $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}(\text{x}^3-\text{y}^3)$

Putting $\text{x}^3=\sin\text{A}\text{ and y}^3=\sin\text{B}$

$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin2\text{B}}=\text{a}(\sin\text{A}-\sin\text{B})$

$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}(\sin\text{A}-\sin\text{B})$

$\Rightarrow2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{b}}{2}\Big) \\ =2\text{a}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)$

$\Rightarrow\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{a}$

$\Rightarrow\frac{\text{A}+\text{B}}{2}=\cot^{-1}(\text{a})$

$\Rightarrow\text{A}+\text{B}=2\cot^{-1}(\text{a})$

$\Rightarrow\sin^{-1}\text{x}^{3}-\sin^{-1}\text{y}^3=2\cot^{-1}(\text{a})$

$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}\times\frac{\text{d}}{\text{dx}}(\text{x}^3)-\frac{1}{\sqrt{1-\text{y}^6}}\times\frac{\text{d}}{\text{dx}}(\text{y}^3)=0$

$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}3\text{x}^2-\frac{1}{\sqrt{1-\text{y}^6}}\times3\text{y}^2\times\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$

3. Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

Solution:

Given function is

$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$

Consider,

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

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

Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

2. Continuous at x = 0

Solution:

$\text{f(x)}=\sin^{-1}(\cos\text{x})$

$\text{f(x)}=\sin^{-1}\Big[\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$

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

Function is continuous at x = 0.

1. ​​​​​​$\frac{1}{2}\text{a}$

Solution:

Here,

$\text{x}=2\text{ at},\text{y}=\text{at}^2,$

Differentiating w.r.t.x, we get

$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$

Differentiating w.r.t.x, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$

Now, $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$

2. $\frac{\text{y}}{\text{x}}$

Solution:

$\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$

$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\sin(\log\text{a})=\text{k}$

$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$

$\text{x}^2-\text{y}^2=\text{kx}^2+\text{ky}^2$

$\text{x}^2-\text{kx}^2=\text{ky}^2+\text{y}^2$

$(1-\text{k})\text{x}^2=(\text{x}+1)\text{y}^2$

$\frac{1-\text{x}}{\text{k}+1}=\frac{\text{y}^2}{\text{x}^2}\ .....(\text{i})$

Consider,

$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$

Differentaiting with resepect to x,

$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{kx}+2\text{ky}\frac{\text{dy}}{\text{dx}}$

$2\text{x}-2\text{kx}=2\text{ky}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}$

$2\text{x}(1-\text{k})=2\text{y}(\text{k}+1)\frac{\text{dy}}{\text{dx}}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(1-\text{k})}{\text{y}(\text{k}+1)}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{y}^2}{\text{x}^2}\ .....(\because\text{from(i)})$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$

2. $\frac{3}{4\text{t}}$

Solution:

$\text{x}=\text{t}^2\Rightarrow\frac{\text{dx}}{\text{dt}} = 2\text{t}$

$\text{y}=\text{t}^3\Rightarrow\frac{\text{dy}}{\text{dt}} = 3\text{t}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{3\text{t}^2}{2\text{t}} = \frac{3\text{t}}{2}$

Hence, $\frac{\text{d}^2\text{y}}{\text{dx}^2} = \frac{3}{4\text{t}}$

2. 0

Solution:

We have $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1}$

$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}-\text{m})(\text{l}+\text{m})}\times\text{x}^{(\text{m}-\text{n})(\text{m}+\text{n})}\times\text{x}^{(\text{n}-\text{l})(\text{n}-\text{l})}$

$\Rightarrow\text{f}(\text{x})=\text{x}^{\text{l}^2-\text{m}^2}\times\text{x}^{\text{m}^2-\text{n}^2}\times\text{x}^{\text{n}^2-\text{l}^2}$

$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}^2-\text{m}^2+\text{m}^2-\text{n}^2+\text{n}^2-\text{l}^2)}$

$\Rightarrow\text{f}(\text{x})=\text{x}^0$

$\Rightarrow\text{f}(\text{x})=1$

$\Rightarrow\text{f}'(\text{x})=0$

3. $\text{a}<\text{x}_1<\text{b}$

Solution:

We have

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

In the Lagrange's mean value theorem, $\text{c}\in(\text{a},\text{b})$ such that $\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$

So, if there is x₁ such that $\text{f}'(\text{x}_1)=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}},$ then $\text{x}_1\in(\text{a},\text{b})$

$\Rightarrow\text{a}<\text{x}_1<\text{b}$

3. $-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$

Solution:

$\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos\text{x}+\text{i}\sin\text{x})^2...(\cos\text{x}+\text{i}\sin\text{x})^\text{n}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{1+2+3........\text{n}}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{\frac{\text{n}(\text{n}+1)}{2}}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^\text{a}$

$\Rightarrow\text{f}(\text{ax})=(\cos\text{ax}+\text{i}\sin\text{ax})...1$

$\Rightarrow\text{f}(1)=(\cos\text{a}+\text{i}\sin\text{a})$

$\Rightarrow1=(\cos\text{a}+\text{i}\sin\text{a})...2\ [\because\text{f}(1)=1]$

Differentiating eqn.1, we get

$\text{f}'(\text{x})=\text{a}(-\sin\text{ax}+\text{i}\cos\text{ax})$

$\Rightarrow\text{f}''(\text{x})=\text-{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''\text{x}=\text{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''(\text{x})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{ax}+\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{a}+\text{i}\sin\text{a})$

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

1. f(x) is continuous and $\text{f}'(1^+)=\log_{10}\text{e}$

4. f(x) is continuous and $(\text{f}'(1^-)=-\log_{10}\text{e})$

Solution:

Given,

$\text{f(x)}=|\log_{10}\text{x}|=\bigg|\frac{\log_{\text{e}}\text{x}}{\log_{\text{e}}10}\bigg|$

$=|(\log_{\text{e}}\text{x})\times(\log_{10}\text{e})|$

$=(\log_{10}\text{e})|\log_{10}\text{x}|$

$\text{f}'(1^+)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{{10}}\text{e})|\log_{\text{e}}(1+\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1|}{\text{h}}$

$=(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$

$=(\log_{10}\text{e})\times1$

$=(\log_{10}\text{e})$

Also,

$\text{f'(1}^-)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(\log_{10}\text{e})|\log_{\text{e}}(1-\text{h})|-(\log_{10}\text{e})|\log_{\text{e}}1}{\text{h}}$

$=-(\log_{10}\text{e})\lim\limits_{\text{h}\rightarrow0}\frac{|\log_{\text{e}}(1+\text{h})|}{\text{h}}$

$=(\log_{10}\text{e})\times1=-(\log_{10}\text{e})$

2. 6

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$

If f(x) is continuous at x = 0, then

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{\text{x}}=\text{f}(0)$

$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}=\frac{\text{k}}{2}$

$\Rightarrow3\times1=\frac{\text{k}}{2}$

$\Rightarrow\frac{\text{k}}{2}=3$

$\Rightarrow\text{k}=6$

4. None of these.

Solution:

$\text{f}(\text{x})=\sqrt{\text{x}^2-10\text{x}+25}$

$\text{f}(\text{x})=\sqrt{(\text{x}-5)^2}=|\text{x}-5|$

$\text{f}(\text{x})=\text{x}-5\ \text{x}\geq5$

$=-(\text{x}-5)\ \text{x}<5$

$\text{f}'\text{(x)}=1\ \text{x}\geq5$

$=-1\text{ x}<5$

Hence, we can not define derivative of the function on [0, 7].

3. a function of y only

Solution:

$\text{y}^2=\text{ax}^2+\text{bx}+\text{c}$

$\frac{\text{dy}}{\text{dx}}=2\text{ax}+\text{b}$

$\frac{\text{d}^2\text{y}}{\text{d}^2}=2\text{a}$

$=\text{y}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ay}^3$

= A function of y only

1. $(-\infty,\infty)$

Solution:

We have,

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

$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$

When, x < 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(-\infty,0)$

When, x > 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(0,\infty,)$

Thus possible point of non-differentiability of f(x) is x = 0

Now, LHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

And RHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

$\therefore$ LHL (at x = 0) = RHL (at x = 0)

So, f(x) is also differentiable at x = 0

i.e. f(x) is differentiable in $(-\infty,\infty).$

3. $\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$

Solution:

f(x) = 4x⁸

$\Rightarrow\text{f}\Big(\frac{1}{2}\Big)=4\Big(\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$

$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=4\Big(-\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$

$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$

3. $(1+\sin2\text{x})\text{y}_1$

Solution:

$\text{y}=\text{e}^{{\tan}\text{x}},$

$\text{y}_1=\text{sec}^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_1=\text{e}^{\tan\text{x}}$

again differentiating w.r.t.x, we get

$\cos^2\text{xy}_2-2\cos\text{x}\sin\text{xy}_1=\sec^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_2=\text{y}_1\sin2\text{x}+\text{y}_1$

1. $\text{Does not exist.}$

Solution:

Put, $\text{u}=\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$ and $\text{v}=\sqrt{1+3\text{x}}$

$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+3\text{x}}}\times3$

But at $\text{x}=\frac{-1}{3}\frac{\text{dv}}{\text{dx}}$ does not exist

Hence, derivative of $\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$

With respect to $\sqrt{1+3\text{x}}$ does not exist.

2. a + b

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$

$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$

$\text{a}+\text{b}=\text{k}$

1. $2$

Solution:

Let $\text{u}=\cos^{-1}(2\text{x}^2-1)$ and $\text{v}=\cos^{-1}\text{x}$

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

$=\cos^{-1}(2\cos^2\text{v}-1)=\cos^{-1}(\cos2\text{v})=2\text{v}$

$\therefore\ \frac{\text{du}}{\text{dv}}=2$

2. $\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$

1. Continuous as well as differentiable for all $\text{x}\in\text{R}$

Solution:

Here,

$\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2}$

Since, we know that $\pi(\text{x}-\pi)=\text{n}\pi$ and $\sin\text{n}\pi=0.$

$\because4+\text{x}[\text{x}]^2\neq0$

$\therefore\text{f(x)}=0$ for all x

Thus, f(x) is a constant function and it is continuous and differentible everywhere.

3. -1

Solution:

We have, $\sqrt{\text{x}}+\sqrt{\text{y}}=1$

Differentiating with respect to x, we get,

$\frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}\times\frac{2\sqrt{\text{y}}}{1}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\sqrt{\text{y}}}{\sqrt{\text{x}}}$

Now, $\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\Big(\frac{1}{4},\frac{1}{4}\Big)}=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$

4. None of these.

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$

$3(\log4)^3=12(\log4)^3$

$3\text{a}=12$

$\text{a}=12$

Note: The question is incorrect, so it has been modified.

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

Solution:

We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{x}^2}{1-\text{x}^2}\Big[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\Big]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1-\text{x}^2}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)}\Big]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{1-\text{x}^4}$

2.  $\frac{-4\text{x}}{1-\text{x}^4}$

Solution:

We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

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

$=\frac{(1+\text{x}^2)}{(1-\text{x}^2)}\cdot\frac{(1+\text{x}^2)\cdot(-2\text{x})-(1-\text{x}^2)\cdot2\text{x}}{(1+\text{x}^2)^2}$

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

1. $1$

Solution:

f(x) = x³ - 3x in the interval $\big[0,\sqrt3\big]$

$\text{f}(0)=0$ and $\text{f}\big(\sqrt3\big)=0$

f'(x) = 3x² - 3

f'(c) = 3c² - 3

f'(c) = 0

3c² - 3 = 0

3c² = 3

c² = 1

$\text{c}=\pm1$

$\text{x}\in\big[0,\sqrt3\big]$

Hence, x = 1

3. 4a

Solution:

Given that, y = ax² + b

Then, $\frac{\text{dy}}{\text{dx}} = 2\text{ax}$

At x = 2, $\frac{\text{dy}}{\text{dx}}=2(\text{a})(2)=4\text{a}$

1. $(\text{xy}_1-\text{y})^2$

​​​​​​​Solution:

$\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{b}}\Big)^\text{x}$

$\text{y}=\text{x}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$

$\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\text{x}\Big(\frac{1}{\text{x}}-\frac{\text{b}}{\text{a}+\text{bx}}\Big)$

$=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+1-\frac{\text{bx}}{\text{a}+\text{bx}}$

$(\text{a}+\text{bx})\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))(\text{a}+\text{bx})+\text{a}$

Again differentiating w.r.t.x, we get

$(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))\\+(\text{a}+\text{bx})\Big(\frac{1}{\text{x}}\frac{\text{b}}{\text{a}+\text{bx}}\Big)$

$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\frac{\text{a}}{\text{x}}$

$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\frac{\text{by}}{\text{x}}+\frac{\text{a}}{\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\Big(\frac{\text{y}}{\text{x}}+\frac{\text{a}}{\text{a}+\text{bx}}\Big)$

$=\frac{(\text{a}+\text{by})(\text{a}+\text{bx})-\text{b}(\text{ay}+\text{bxy}+\text{ax})}{\text{x}(\text{a}+\text{b})^2}$

$=\frac{\text{a}^2+\text{abx}+\text{aby}+\text{b}^2\text{xy}-\text{bay}-\text{b}^2\text{xy}-\text{abx}}{\text{x}(\text{a}+\text{bx})^2}$

$=\frac{\text{a}^2}{\text{x}(\text{a}+\text{bx})^2}$

$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}$

$(\text{xy}_1-\text{y})=\frac{\text{ax}}{\text{a}+\text{bx}}$

$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}=(\text{xy}_1-\text{y})^2$

1. 0

Solution:

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2}$

$\text{x}\rightarrow\frac{\pi}{2}+\text{h}$

$\text{x}\rightarrow\frac{\pi}{2}\Rightarrow\text{h}\rightarrow0$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\cos\big(\frac{\pi}{2}+\text{h}\big)\Big)-\cos\big(\frac{\pi}{2}+\text{h}\big)}{\Big(\pi-2\big(\frac{\pi}{2}+\text{h}\big)\Big)^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(-\sin\text{h})+\sin\text{h}}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{4\text{h}^2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{\frac{\text{h}-\sin\text{h}}{2}}\times\frac{\text{h}-\sin\text{h}}{2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\text{h}-\sin\text{h}}{2}$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times(\text{h}-\sin\text{h})$

$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{1}{4}\frac{\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{\text{h}^2}\times(\text{h}-\sin\text{h})=0$

$\text{k}=0$

2. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$

Solution:

Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$

We have

$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$

$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$

$=\frac{1}{3}$

$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$

$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$

$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$

$=\frac{\text{b}}{8}\times1$

$=\frac{\text{b}}{8}$

Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$

If f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$

$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$

$\Rightarrow\text{a}=\frac{1}{3}\text{ and }\text{b}=\frac{8}{3}$

2. n(n + 1) y

Solution:

Here

$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$

$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$

$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$

$=\text{n}(\text{n}+1)\text{y}$

2. $\text{x}=\frac{5}{2}$

Solution:

$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}2\sqrt{\text{x}}=2$

$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}4-2\text{x}=2$

Function is continuous at x = 1

$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}4-2\text{x}=-1$

$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}2\text{x}-7=-2$

Function is discontinuous at $\text{x}=\frac{5}{2}$

3. -m²y

Solution:

Here,

$\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}}{\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{mx}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)^\frac{3}{2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xm}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)\times\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xy}_1}{(1-\text{x}^2)}$

$\Rightarrow(1-\text{x}^2)\text{y}_2=-\text{ym}^2+\text{xy}_1$

$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1=-\text{m}^2\text{y}$

3. 1

Solution:

We have, $\text{y}=\tan^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}\times \\ \bigg[\frac{(\cos\text{x}-\sin\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})-(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x}-\sin\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times\\ \bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(-\sin\text{x}-\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times\\ \bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})(-\sin\text{x}+\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$

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

$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$

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

Solution:

Let $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

Differentiating with respect to x using chain rule, we get,

$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{(1+\text{x}^2)^2}\bigg]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\bigg]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{2\text{x}}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)^2}\Big]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{2\text{x}(1+\text{x}^2)}$

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

1. 0

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\times\sin\frac{1}{\text{x}}$

$-1\leq\sin\frac{1}{\text{x}}\leq1$

$\text{x}\times(-1)\leq\text{x}\sin\frac{1}{\text{x}}\leq\text{x}$

$\lim\limits_{\text{x}\rightarrow0}-\text{x}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\frac{1}{\text{x}}\leq\lim\limits_{\text{x}\rightarrow0}\text{x}$

$\text{f}(0)=0$