Maths Case study (4 Marks)

1. (a) $\text{x}^\text{x}(1+\log\text{x})$

Solution:

Let $\text{y}=\text{x}^\text{x}\Rightarrow\log\text{y}=\text{x}\log\text{x}$

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1\times\log\text{x}+\text{x}\times\frac{1}{\text{x}}]$

$=\text{x}^\text{x}[1+\log\text{x}]$

2. (d) $\text{x}^\text{x}(1+\log\text{x})+\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$

3. (d) $\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$

Solution:

Given $\text{x}=\text{e}^\frac{\text{x}}{\text{y}}\Rightarrow\log\text{x}=\frac{\text{x}}{\text{y}}\log\text{e}\Rightarrow\text{y}\log\text{x}=\text{x}$

$\Rightarrow\text{y}\frac{1}{\text{x}}+(\log\text{x})\frac{\text{dy}}{\text{dx}}=1$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(1-\frac{\text{y}}{\text{x}}\Big)\frac{1}{\log\text{x}}\Rightarrow\frac{1}{\text{x}\log\text{x}}(\text{x}-\text{y})$

4. (c) $(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$

Solution:

$\text{y}=(2-\text{x})^3(3+2\text{x})^5$

$\Rightarrow\log\text{y}=\log(2-\text{x})^3+\log(3+2\text{x})^5$

$=3\log(2-\text{x})+5\log(3+2\text{x})$

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{3\times(-1)}{2-\text{x}}+\frac{5}{3+2\text{x}}\times(2)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=(2-\text{x})^3(3+2\text{x})^5\Big[\frac{10}{3+2\text{x}}-\frac{3}{2-\text{x}}\Big]$

5. (d) $\text{x}^\text{x}\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$

Solution:

$\text{y}=\text{x}^\text{x}\cdot\text{e}^{(2\text{x}+5)}$

$\Rightarrow\log\text{y}=\text{x}\log\text{x}+(2\text{x}+5)$

$\Rightarrow\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\Big(\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\Big)+2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cdot\text{e}^{2\text{x}+5}\cdot(3+\log\text{x})$

1. (b) $\frac{-(3\text{x}^2+2\text{xy}+\text{y}^2)}{\text{x}^2+2\text{xy}+3\text{y}^2}$

Solution:

$\text{x}^3+\text{x}^2\text{y}+\text{xy}^2+\text{y}^3=81$

$\Rightarrow3\text{}^2+\text{x}^2\frac{\text{dy}}{\text{dx}}+2\text{xy}+2\text{xy}\frac{\text{dy}}{\text{dx}}+\text{y}^2+3\text{y}^2\frac{\text{dy}}{\text{dx}}=0$

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

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

2. (c) $\frac{\text{x}-\text{y}}{\text{x}(1+\log\text{x})}$

Solution:

$\text{x}^\text{y}=\text{e}^{\text{x}-\text{y}}\Rightarrow\text{y}\log\text{x}=\text{x}-\text{y}$

$\text{y}\times\frac{1}{\text{x}}+\log\text{x}\cdot\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}[\log\text{x}+1]=1-\frac{\text{y}}{\text{x}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}[1+\log\text{x}]}$

3. (d) $\frac{\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$

Solution:

$\text{e}^{\sin\text{y}}=\text{xy}\Rightarrow\sin\text{y}=\log\text{x}+\log\text{y}$

$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\cos\text{y}-\frac{1}{\text{y}}\Big]=\frac{1}{\text{x}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$

4. (d) $\frac{\sin2\text{x}}{\sin2\text{y}}$

Solution:

$\sin^2\text{x}+\cos^2\text{y}=1$

$\Rightarrow2\sin\text{x}\cos\text{x}+2\cos\text{y}\Big(-\sin\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\sin2\text{x}}{-\sin2\text{y}}=\frac{\sin2\text{x}}{\sin2\text{y}}$

5. (d) $\frac{\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$

Solution:

$\text{y}=(\sqrt{\text{x}})^{\sqrt{\text{x}}^\sqrt{\text{x}}...\infty}\Rightarrow\text{y}=(\sqrt{\text{x}})^\text{y}$

$\Rightarrow\text{y}=\text{y}(\log\sqrt{\text{x}})\Rightarrow\log\text{y}=\frac{1}{2}(\text{y}\log\text{x})$

$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\text{y}\times\frac{1}{\text{x}}+\log\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)\Big]$

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

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{x}}\times\frac{2\text{y}}{(2-\text{y}\log\text{x})}=\frac{\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$

1. (c) 0

Solution:

Given, $\text{y}=\tan^{-1}\Bigg(\frac{\log\big(\frac{\text{e}}{\text{x}^2}\big)}{\log\text{ex}^2}\Bigg)+\tan^{-1}\Big(\frac{3+2\log\text{x}}{1-6\log\text{x}}\Big)$

$=\tan^{-1}\Big(\frac{1-\log\text{x}^2}{1+\log\text{x}^2}\Big)+\tan^{-1}\Big(\frac{3+2\log\text{x}}{1-6\log\text{x}}\Big)$

$=\tan^{-1}(1)-\tan^{-1}(2\log\text{x})+\tan^{-1}(3)+\tan^{-1}(2\log\text{x})$

$\Rightarrow\text{y}=\tan^{-1}(1)+\tan^{-1}(3)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=0\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

2. (d) 10

Solution:

Given, $\text{x}=\text{s}+3\text{t},\text{y}=2\text{s}-\text{t}\Rightarrow\frac{\text{dx}}{\text{ds}}=1,\frac{\text{dy}}{\text{ds}}=2$

Now, $\text{u}=\text{x}^2+\text{y}^2\Rightarrow\frac{\text{du}}{\text{ds}}=2\text{x}\frac{\text{dx}}{\text{ds}}+2\text{y}\frac{\text{dy}}{\text{ds}}=2\text{x}+4\text{y}$

$\Rightarrow\frac{\text{d}^2\text{u}}{\text{ds}^2}=2\Big(\frac{\text{dx}}{\text{ds}}\Big)+4\Big(\frac{\text{dy}}{\text{ds}}\Big)\Rightarrow\frac{\text{d}^2\text{u}}{\text{ds}^2}=2(1)+4(2)=10$

3. (d) $-2\text{cosec}^2\text{x}$

Solution:

We have, $\text{f}(\text{x})=2\log\sin\text{x}$

$\Rightarrow\text{f}'(\text{x})=2\cdot\frac{1}{\sin\text{x}}\cdot\cos\text{x}=2\cot\text{x}\Rightarrow\text{f}''(\text{x})=-2\text{cosec}^2\text{x}$

4. (b) $2\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$

Solution:

We have, $\text{f}(\text{x})=\text{e}^\text{x}\sin\text{x}$

$\Rightarrow\text{f}'(\text{x})=\text{e}^\text{x}\cos\text{x}+\text{e}^\text{x}\sin\text{x}=\text{e}^\text{x}(\cos\text{x}+\sin\text{x})$

$\Rightarrow\text{f}''(\text{x})=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})+\text{e}^\text{x}(\cos\text{x}+\sin\text{x})=2\text{e}^\text{x}\cos\text{x}$

$\Rightarrow\text{f}'''(\text{x})=2[\text{e}^\text{x}\cos\text{x}-\text{e}^\text{x}\sin\text{x}]=2\text{e}^\text{x}[\cos\text{x}-\sin\text{x}]$

5. (d) 0

Solution:

Given, y² = ax² + bx + c

⇒ 2yy₁ = 2ax + b

⇒ 2yy₁ + y₁(2y₁) = 2a

$\Rightarrow\text{yy}_2=\text{a}-\text{y}^2_1\Rightarrow\text{yy}_2=\text{a}-\Big(\frac{2\text{ax}+\text{b}}{2\text{y}}\Big)^2$

$=\frac{4\text{y}^2\text{a}-(4\text{a}^2\text{x}^2+\text{b}^2+4\text{abx})}{4\text{y}^2}$

$\Rightarrow\text{y}^3\text{y}_2=\frac{4\text{a}(\text{ax}^2+\text{bx}+\text{c})-(4\text{a}^2\text{x}^2+\text{b}^2+4\text{abx})}{4}$

$=\frac{4\text{ac}-\text{b}^2}{4}$

$\Rightarrow\frac{\text{d}}{\text{dx}}(\text{y}^3\text{y}_2)=0$

1. (a) f has removable discontinuity.

Solution:

f(3) = 4

$\lim\limits_{\text{x}\rightarrow3}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-9}{\text{x}-3}=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+3)(\text{x}-3)}{(\text{x}-3)}$

$=\lim\limits_{\text{x}\rightarrow3}(\text{x}+3)=6\because\lim\limits_{\text{x}\rightarrow3}\text{f}(\text{x})\neq\text{f}(3)$

$\therefore$ f(x) has removable discontinuity at x = 3.

2. (c) f has irremovable discontinuit.

Solution:

$\lim\limits_{\text{x}\rightarrow4^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow4}(\text{x}+2)=4+2=6$

$\lim\limits_{\text{x}\rightarrow4^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow4}(\text{x}+4)=4+4=8$

$\therefore\lim\limits_{\text{x}\rightarrow4^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow4^+}\text{f}(\text{x})$

$\therefore$ f(x) has an irremovable discontinuity at x = 4.

3. (a) f has removable discontinuity.

Solution:

$\lim\limits_{\text{x}\rightarrow2}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}^2-4)}{(\text{x}-2)}=\lim\limits_{\text{x}\rightarrow2}(\text{x}+2)=4$

and f(2) = 5 (given) $\therefore\lim\limits_{\text{x}\rightarrow2}\text{f}(\text{x})\neq\text{f}(2)$

$\therefore$ f(x) has removable discontinuity at x = 2.

4. (c) f has irremovable discontinuity.

Solution:

f(0) = 2

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

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

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

$\therefore$ f(x) has an irremovable discontinuity at x = 0.

5. (d) f has removable discontinuity.

Solution:

f(0) = 7

$\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\log(1+2\text{x})}=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{\text{e}^\text{x}-1}{\text{x}}\Big)}{\frac{\log(1+2\text{x})}{2\text{x}}\cdot2}=\frac{1}{2}$

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

$\therefore$ f(x) has removable discontinuity at x = 0.

1. (b) -1

Solution:

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

$\lim\limits_{\text{h}\rightarrow0}\frac{3-(1+\text{h})-2}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}-\frac{\text{h}}{\text{h}}=-1$

2. (b) -1

Solution:

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

$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big[\frac{(1-\text{h})^2}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}-2\Big]$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big(\frac{1+\text{h}^2-2\text{h}-6+6\text{h}+13-8}{-4\text{h}}\Big)$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}^2+4\text{h}}{-4\text{h}}\Big)=-1$

3. (c) x = 3

Solution:

Since, R.H.D. at x = 3 is 1

and L.H.D. at x = 3 is - 1

$\therefore$ f(x) is non-differentiable at x = 3.

4. (d) -1

5. (c) -2

Solution:

From above, we have

$\text{f}'(\text{x})=\frac{\text{x}}{2}-\frac{3}{2},\text{x}<1$

$\therefore\text{f}'(-1)=\frac{-1}{2}-\frac{3}{2}=-2$

1. (d) $\text{gof}(\text{x})=\begin{cases}\text{e}^{\sin\text{x}}&,\text{x}\geq0\\\text{e}^{1-\cos\text{x}}&,\text{x}\leq0\end{cases}$

2. (a) $[\text{gof}(\text{x})]'=\begin{cases}\cos\text{x}\cdot\text{e}^{\sin\text{x}}&,\text{x}\geq0\\\text{e}^{1-\cos\text{x}}\cdot\sin\text{x}&,\text{x}\leq0\end{cases}$

3. (b) 1

4. (a) 0

5. (b) $\frac{1}{\sqrt2}$

1. (a) $-\frac{3}{2}$

2. (d) Any real number.

3. (b) $\frac{1}{2}$

4. (c) -1

Solution:

$\text{a}+\text{c}=-\frac{3}{2}+\frac{1}{2}=-1$

5. (d) 2

Solution:

$\text{c}-\text{a}=\frac{1}{2}+\frac{3}{2}=2$

1. (a) $\frac{1}{\sqrt{2}}$

Solution:

Now, $\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}=\frac{\text{f}'(\tan \text{x})\sec^2\text{x}}{\text{g}'(\sec\text{x})\sec\text{x}\tan \text{x}}$

$=\frac{\text{f}'(\tan \text{x})\sec\text{x}}{\text{g}'(\sec\text{x})\tan \text{x}}$

$\therefore\Big[\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}\Big]_{\text{x}=\frac{\pi}{4}}=\frac{\text{f}'(1)\sqrt{2}}{\text{g}'(\sqrt{2})\cdot1}=\frac{2\sqrt{2}}{4\cdot1}=\frac{1}{\sqrt{2}}$

2. (b) 1

3. (c) $3\text{x}^3\text{e}^{\text{x}^3}$

Solution:

Let $\text{y}=\text{e}^{\text{x}^3},\text{z}=\log\text{x}$

Differentiating w.r.t. x, we get

$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^3}(3\text{x}^2)=3\text{x}^2\text{e}^{\text{x}^3}$ and $\therefore\frac{\text{dy}}{\text{dz}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dz}}{\text{dx}}}=\frac{3\text{x}^2\text{e}^{\text{x}^3}}{\Big(\frac{1}{\text{x}}\Big)}=3\text{x}^3\text{e}^{\text{x}^3}$

4. (a) $2$

Solution:

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

Differentiating w.r.t. $\cos^{-1}\text{x},$ we get

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

5. (a) $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$

Solution:

We have, $\text{y}=\frac{1}{4}\text{u}^4\Rightarrow\frac{\text{dy}}{\text{du}}=\frac{1}{4}\cdot4\text{u}^3=\text{u}^3$

and $\text{u}=\frac{2}{3}\text{x}^3+5\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{3}\cdot3\text{x}^2=2\text{x}^2$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{du}}\cdot\frac{\text{du}}{\text{dx}}=\text{u}^3\cdot2\text{x}^2=\Big(\frac{2}{3}\text{x}^3+5\Big)^3(2\text{x})^2$

$=\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$

1. (a) $\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$

Solution:

Let $\text{y}=\cos\sqrt{\text{x}}$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\cos\sqrt{\text{x}})=-\sin\sqrt{\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}})$

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

2. (a) $\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$

Solution:

Let $\text{y}=7^{\text{x}+\frac{1}{\text{x}}}\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(7^{\text{x}+\frac{1}{\text{x}}}\Big)$

$=7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7\cdot\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7\cdot\Big(1-\frac{1}{\text{x}^2}\Big)$

$=\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$

3. (a) $\frac{1}{2}\sec^2\frac{\text{x}}{2}$

Solution:

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

$\therefore\frac{\text{dy}}{\text{dx}}=\sec^2\frac{\text{x}}{2}\cdot\frac{1}{2}=\frac{1}{2}\sec^2\frac{\text{x}}{2}$

4. (b) $\frac{1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$

Solution:

Let $\text{y}=\frac{1}{\text{b}}\tan^{-1}\Big(\frac{\text{x}}{\text{b}}\Big)+\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{b}}\times\frac{1}{1+\frac{\text{x}^2}{\text{b}^2}}\times\frac{1}{\text{b}}+\frac{1}{\text{a}}\times\frac{1}{1+\frac{\text{x}^2}{\text{a}^2}}\times\frac{1}{\text{a}}$

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

5. (d) $\frac{2}{|\text{x}|\sqrt{\text{x}^2-1}}$

Solution:

Let $\text{y}=\sec^{-1}\text{x}+\text{cosec}^{-1}\frac{\text{x}}{\sqrt{\text{x}^2-1}}$

Put $\text{x}=\sec\theta\Rightarrow\theta=\sec^{-1}\text{x}$

$\therefore\text{y}=\sec^{-1}(\sec\theta)+\text{cosec}^{-1}\Big(\frac{\sec\theta}{\sqrt{\sec^2\theta-1}}\Big)$

$=\theta+\sin^{-1}\Big[\sqrt{1-\cos^2\theta}\Big]$

$=\theta+\sin^{-1}(\sin\theta)=\theta+\theta=2\theta=2\sec^{-1}\text{x}$

$\therefore\frac{\text{dy}}{\text{dx}}=2\frac{\text{d}}{\text{dx}}(\sec^{-1}\text{x})=2\times\frac{1}{|\text{x}|\sqrt{\text{x}^2-1}}$

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