Question 3015 Marks
$\text{If y}=3\cos(\log \text{x})+4\sin(\log\text{x}),\text{ show that }\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
AnswerGiven: $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x})\ \dots\text{(i)}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}_1=-3\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}+4\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{y}_1=-3\sin(\log\text{x})\frac{1}{\text{x}}+4\cos(\log\text{x})\frac{1}{\text{x}}$ $=\frac{1}{\text{x}}[-3\sin(\log\text{x})+4\cos(\log\text{x})]$
$\Rightarrow\ \text{xy}_1=-3\sin(\log\text{x})+4\cos(\log\text{x})$
Now $\frac{\text{d}}{\text{dx}}(\text{xy}_1)=-3\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}-4\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{x}\frac{\text{d}}{\text{dx}}(\text{y}_1)+\text{y}_1\frac{\text{d}}{\text{dx}}\text{x}$ $=-3\cos(\log\text{x})\frac{1}{\text{x}}-4\sin(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\ \text{xy}_2+\text{y}_1=-\frac{[3\cos(\log\text{x})+4\sin(\log\text{x})]}{\text{x}}$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-[3\cos(\log\text{x})+4\sin(\log\text{x})]$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-\text{y}\ \ [\text{From eq.(i)}]$
$\Rightarrow\ \text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0 \ \text{ Hence proved}.$
View full question & answer→Question 3025 Marks
Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$
AnswerSince f(x) is continuous at $\text{x}=\frac{\pi}{2},$ L.H.Limit = R.H.Limit.
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi^+}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}=3$
$\Rightarrow\text{k}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
View full question & answer→Question 3035 Marks
If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ prove that $2\text{x}\frac{\text{dy}}{\text{dx}}=\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}$
Answer$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{x}^{-1\frac{1}{2}}\Big)$
$=\frac{1}{2\sqrt{\text{x}}}+\Big(-\frac{1}{2}\times\text{x}^{-\frac{1}{2}-1}\Big)$
$=\frac{2}{2\sqrt{\text{x}}}-\frac{1}{2\sqrt[\text{x}]{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}\sqrt{\text{x}}}$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
Hence, the solution is, $2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
View full question & answer→Question 3045 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{If x}=\sqrt{\text{a}^{\sin^{-1}}\text{t}},\text{y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}},\text{ Show that}\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
AnswerThe given equations are $\text{x}=\sqrt{\text{a}^{\sin{-1}}\text{t}}\text{ and y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}}$
$$$\text{x}=\sqrt{\text{a}^{\sin^{-1}}\text{t}}\text{ and y}=\sqrt{\text{a}^{\cos^{-1}}\text{t}}$
$\Rightarrow\ \text{x}=\Big({\text{a}^{\sin^{-1}}\text{t}\Big)}^{\frac{1}{2}}\text{ and y}=\Big({\text{a}^{\cos^{-1}}\text{t}\Big)^{\frac{1}{2}}}$
$\Rightarrow\ \text{x}=\text{a}^{\frac{1}{2}\sin^{-1}\text{t}}\text{ and y}=\text{a}^{\frac{1}{2}\cos^{-1}\text{t}}$
Consider $\text{x}=\text{a}^{\frac{1}{2}\sin^{-1}\text{t}}$
Taking logarithm on both the sides, we obtain
$\log\text{x}=\frac{1}{2}\sin^{-1}\text{t}\log\text{a}$
$\therefore \frac{1}{\text{x}}.\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\log\text{a}.\frac{\text{d}}{\text{dt}}(\sin^{-1}\text{t})$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{x}}{2}\log\text{a}.\frac{1}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{x}\log\text{a}}{2\sqrt{1-\text{t}^2}}$
Then, consider $\text{a}^{\frac{1}{2}\cos^{-1}\text{t}}$
Taking logarithm on both the sides, we obtain
$\log\text{y}=\frac{1}{2}\cos^{-1}\text{t}\log\text{a}$
$\therefore \frac{1}{\text{y}}.\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\log\text{a}.\frac{\text{d}}{\text{dt}}(\cos^{-1}\text{t})$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{-\text{y}\log\text{a}}{2}.\Big(\frac{1}{\sqrt{1-\text{t}^2}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}=\frac{-\text{y}\log\text{a}}{2\sqrt{1-\text{t}^2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dx}}\Big)}=\frac{\Big(\frac{-\text{y}\log\text{a}}{2\sqrt{1-\text{t}^2}}\Big)}{\Big(\frac{\text{x}\log\text{a}}{2\sqrt{1-\text{t}^2}}\Big)}=-\frac{\text{y}}{\text{x}}.$
Hence, proved.
View full question & answer→Question 3055 Marks
Differentiate $\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$ w.r.t. $\tan^{-1}\text{x}$ when $\text{x}\neq0.$
AnswerLet $\text{u}=\tan^{-1}\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}$ and $\text{y}=\tan^{-1}\text{x}$
Put $\text{x}=\tan\theta$
$\Rightarrow\ \text{u}=\tan^{-1}\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}$
$=\tan^{-1}\frac{\sec\theta-1}{\tan\theta}=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg[\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\bigg]$ $=\tan^{-1}\Big[\tan\frac{\theta}{2}\Big]=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}\text{x}$
$\therefore\ \frac{\text{du}}{\text{dx}}=\frac{1}{2}\cdot\frac{1}{1+\text{x}^2}\ \ \dots(\text{i})$
and $\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\ \ \dots(\text{ii})$
$\therefore\ \frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{2}$
View full question & answer→Question 3065 Marks
Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\sqrt{4-\text{x}^2}\text{ in }[-2,2].$
AnswerWe have, $\text{f(x)}=\sqrt{4-\text{x}^2}=(4-\text{x}^2)^{\frac{1}{2}}$
- $\text{f(x)}=\sqrt{4-\text{x}^2}$ is continuous function.
[since every polynomial function is a continuous function]
Hence, f(x) is continuous in [-2, 2]
- $\text{f}'(\text{x})=\frac{1}{2}(4-\text{x}^2)^{\frac{-1}{2}}\cdot(-2\text{x})$
$=-\text{x}\cdot\frac{1}{\sqrt{4-\text{x}^2}},$ which exists everywhere except at $\text{x}=\pm2.$
Hence, f(x) is differentiable in (-2, 2).
- $\text{f}(-2)=\sqrt{(4-4)}=0$ and $\text{f}(2)=\sqrt{(4-4)}=0$
$\Rightarrow\ \text{f}(-2)=\text{f}(2)$
Conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that f'(c) = 0
$\Rightarrow\ -\text{c}\frac{1}{\sqrt{4-\text{c}^2}}=0$
$\Rightarrow\ \text{c}=0\in(-2,2)$
Hence, Rolle’s theorem has been verified. View full question & answer→Question 3075 Marks
If $\text{x}=\cos\theta,\text{y}=\sin^3$ prove that $\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=3\sin^2\theta(5\cos^2\theta-1)$
AnswerHere$\text{x}=\cos\theta,\text{y}=\sin^3$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=3\sin^2\theta\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}=-3\sin\theta\cos\theta$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-3\cos^2\theta+3\sin^2\theta)\frac{\text{d}\theta}{\text{dx}}\frac{)-3\cos^2\theta+3\sin^2\theta)}{-\sin\theta}$
Now,
$\text{LHS}=\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\sin^3\theta\times\frac{(-3\cos^2\theta+3\sin^2\theta)}{\sin\theta}+(-3\sin\theta\cos\theta)^2$
$=3\sin^2\theta\cos^2\theta-3\sin^4\theta+9\sin^2\theta\cos^2\theta$
$=12\sin^2\theta\cos^2\theta-3\sin^4\theta$
$=3\sin^2\theta(4\cos^2\theta-\sin^2\theta)$
$=3\sin^2\theta(5\cos^2\theta-1)$
$=\text{RHS}$
View full question & answer→Question 3085 Marks
If $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}},$ prove that $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}$
AnswerGivne, $\text{y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Differentiate with respect to x,
$\frac{\text{d}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\bigg[\frac{\sqrt{1-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x}\sin^{-1}\text{x})-(\text{x}\sin^{-1}\text{x})\frac{\text{d}}{\text{dx}}(\sqrt{1-\text{x}^2})}{(\sqrt{1-\text{x}^2})^2}\bigg]$
[Using quotient rule, product rule, chain rule]
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}-\big(\text{x}\sin^{-1}\text{x}\big)\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(1-\text{x}^2\big)}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\sqrt{1-\text{x}^2}\Big\{\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big\}-\frac{\text{x}\sin{-1}\text{x}(-2\text{x})}{2\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$=\begin{bmatrix}\frac{\text{x}+\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}}{\big(1-\text{x}^2\big)} \end{bmatrix}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\sqrt{1-\text{x}^2}\sin^{-1}}{1}+\frac{\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{(1-\text{x}^2)\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}-\text{x}^2\sin^{-1}\text{x}+\text{x}^2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow\big(1-\text{x}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}+\bigg(\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\bigg)$
$\Rightarrow(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}=\text{x}+\frac{\text{y}}{\text{x}}\ \Big\{\text{Since, given y}=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\Big\}$
View full question & answer→Question 3095 Marks
Differentiate the following functions with respect to x:
$\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
AnswerLet $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)-(\text{x}^2+2)\frac{\text{d}}{\text{dx}}(\sqrt{\cos\text{x}})}{(\sqrt{\cos\text{x}})^2}$
[Using quotient rule and chain rule]
$=\frac{2\text{x}\sqrt{\cos\text{x}}-(\text{x}^2+2)\Big(-\frac{1}{2}\frac{\sin\text{x}}{\sqrt{\cos\text{x}}}\Big)}{\cos\text{x}}$
$=\frac{2\text{x}\sqrt{\cos\text{x}}+\frac{(\text{x}^2+2)\sin\text{x}}{2\sqrt{\cos\text{x}}}}{\cos\text{x}}$
$=\frac{4\text{x}\cos\text{x}+(\text{x}^2+2)\sin\text{x}}{2(\cos\text{x})^\frac{3}{2}}$
$=\frac{2\text{x}}{\sqrt{\cos\text{x}}}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{(\cos\text{x})^\frac{3}{2}}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{1}{2}\frac{(\text{x}^2+2)\sin\text{x}}{\cos\text{x}}\Big\}$
$=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\tan\text{x}}{2}\Big\}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}\Big)=\frac{1}{\sqrt{\cos\text{x}}}\Big\{2\text{x}+\frac{(\text{x}^2+2)\sin\text{x}}{2}\Big\}$
View full question & answer→Question 3105 Marks
If $\text{y}=\tan^{-1}\text{x},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y alone.
AnswerWe have, $\text{y}=\tan^{-1}\text{x}$ [ondifferentiating w.r.t. x]
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$ [again differentiating w.r.t. x]
Now, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(1+\text{x}^2)^{-1}$
$=-1(1+\text{x}^2)^{-2}\cdot\frac{\text{d}}{\text{dx}}(1+\text{x}^2)$
$=-\frac{1}{(1+\text{x}^2)^2}\cdot2\text{x}$
$=\frac{-2\tan\text{y}}{(1+\tan^2\text{y})^2}$ $[\because\text{y}=\tan^{-1}\text{x}\Rightarrow\tan\text{y}=\text{x}]$
$=\frac{-2\tan\text{y}}{(\sec^2\text{y})^2}$
$=-2\frac{\sin\text{y}}{\cos\text{y}}\cdot\cos^2\text{y}\cdot\cos^2\text{y}$
$=-\sin2\text{y}\cdot\cos^2\text{y}$ $[\because\sin2\text{x}=2\sin\text{x}\cos\text{x}]$
View full question & answer→Question 3115 Marks
Prove that $\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
Answer$\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}=\sqrt{\text{a}^2-\text{x}^2}$
$\text{L.H.S}=\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big\}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\text{x}\frac{\text{d}}{\text{dx}}\sqrt{\text{a}^2-\text{x}^2}+\sqrt{\text{a}^2-\text{x}^2}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\frac{\text{a}^2}{2}\times\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{\text{x}}\Big)^2}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$
[Using product rule, chain rule]
$=\frac{1}{2}\bigg[\text{x}\times\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)+\sqrt{\text{a}^2-\text{x}^2}\Big] \\ +\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2}}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\Big[\frac{\text{x}(-2\text{x})}{2\sqrt{\text{a}^2-\text{x}^2}}+\sqrt{\text{a}^2-\text{x}^2}\Big]+\Big(\frac{\text{a}^2}{2}\Big)\times\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\times\Big(\frac{1}{\text{a}}\Big)$
$=\frac{1}{2}\bigg[\frac{-2\text{x}^2+2\big(\text{a}^2-\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{1}{2}\bigg[\frac{2\big(\text{a}^2-2\text{x}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}\bigg]+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}+\frac{\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{\text{a}^2-2\text{x}^2+\text{a}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\text{a}^2-2\text{x}^2}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{2\big(\text{a}^2-\text{a}^2\big)}{2\sqrt{\text{a}^2-\text{x}^2}}$
$=\frac{(\text{a}^2-\text{x}^2)}{\sqrt{\text{a}^2-\text{x}^2}}$
$=\sqrt{\text{a}^2-\text{x}^2}$
$=\text{R.H.S}$
View full question & answer→Question 3125 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
AnswerWe have, $\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
Differentiating it with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{2}{3}}\Big)+\frac{\text{d}}{\text{dx}}\Big(\text{y}^{\frac{2}{3}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\text{a}^{\frac{2}{3}}\Big)$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{2}{3}-1}+\frac{2}{3}\big(\text{y}\big)^{\frac{2}{3}-1}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}+\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}\times\frac{3}{2\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{-1}{3}}}{\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{1}{3}}}{\text{y}^{\frac{1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\big(\frac{\text{x}}{\text{y}}\big)^{\frac{1}{3}}$
View full question & answer→Question 3135 Marks
Show that $\text{f}(\text{x})=\text{x}^\frac{1}{3}$ is not differentible at x = 0.
Answer$\text{f}(\text{x})=\text{x}^\frac{1}{3}$
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-\text{h})^\frac{1}{3}}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-1)^\frac{1}{3}\text{h}^\frac{1}{3}}{(-1)-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}(-1)^\frac{-2}{3}\text{h}^\frac{-2}{3}$
= Not defined
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(\text{h})^\frac{1}{3}-0}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\text{h}^\frac{-2}{3}$
= Not defined
Since,
LHL and RHL does not exists at x = 0
$\therefore$ f(x) is not differentiable at x = 0
View full question & answer→Question 3145 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\big\{2\text{x}\sqrt{1-\text{x}^2}\big\},\frac{1}{\sqrt{2}}<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{2\text{x}\sqrt{1-\text{x}^2}\Big\}$
Put $\text{x}=\cos\theta$
$\text{y}=\cos^{-1}\Big\{2\cos\sqrt{1-\cos^2\theta}\Big\}$
$=\cos^{-1}\big\{2\cos\theta\sin\theta\big\}$
$\text{y}=\cos^{-1}\big\{\sin2\theta\big\}$
$\big[\text{Since}, \sin2\theta=2\sin\theta\cos\theta,\sin^2\theta+\cos^2\theta=1\big]$
$\text{y}=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)\Big]\ .....(\text{i})$
Now,
$\frac{1}{\sqrt{2}}<\text{x}<1$
$\Rightarrow\frac{1}{\sqrt{2}}<\cos\theta<1$
$\Rightarrow 0<\theta<\frac{\pi}{4}$
$\Rightarrow 0<2\theta<\frac{\pi}{2}$
$\Rightarrow 0>-2\theta>-\frac{\pi}{2}$
$\Rightarrow\frac{\pi}{2}>\Big(\frac{\pi}{2}-2\theta\Big)>0$
Hence, from equation (i),
$\text{y}=\frac{\pi}{2}-2\theta$
$\big[\text{Sicne}, \cos^{-1}(\cos\theta)=\theta,\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{\pi}{2}-2\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)-2\frac{\text{d}}{\text{dx}}\big(\cos^{-1}\text{x}\big)$
$=0-2\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 3155 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}$
AnswerWe have, $\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{e}^{3\text{x}}+\log\sin4\text{x}+\log2^\text{x}$
$\Rightarrow\log\text{y}=3\text{x}\log\text{e}+\log\sin4\text{x}+\text{x}\log2$
$\Rightarrow\log\text{y}=3\text{x}+\log\sin4\text{x}+\text{x}\log2$
Differentiating with resepect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log2)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\log2(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\cot4\text{x}(4)+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+4\cot4\text{x}+\log2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[3+4\cot4\text{x}+\log2\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{3\text{x}}\sin4\text{x}2^\text{x}\big[3+4\cot4\text{x}+\log2\big]$
[Using equation (i)]
View full question & answer→Question 3165 Marks
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
AnswerGiven,
$\text{x}=\text{a}\cos\theta\dots\text{ eq. }1$
$\text{y}=\text{b}=\sin\theta\dots\text{ eq. }2$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, let's first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and defferentiate it again.
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}\cos\theta=-\text{a}\sin\theta\dots\text{ eq. 3}$
Similarly, $\frac{\text{dy}}{\text{d}\theta}=\text{b}\cos\theta\dots\text{ eq. 4}$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x}\tan\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=-\frac{\text{b}\cos\theta}{\text{a}\sin\theta}=-\frac{\text{b}}{\text{a}}\cot\theta$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{b}}{\text{a}}\cot\theta\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\text{cosec}^2\theta\frac{\text{d}\theta}{\text{dx}}\dots\text{ eq. 5}$
$\Big[$Using chain rule and $\frac{\text{d}}{\text{dx}}\cot\text{x}=-\text{cosec}^2\text{x}\Big]$
From equation 3:
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{d}\theta}{\text{dx}}=\frac{-1}{\text{a}\sin\theta}$
Putting the value in equation 5:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}}{a}\text{cosec}^2\theta\frac{1}{\text{a}\sin\theta}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\text{a}^2\sin^3\theta}$
From equation 1:
$\text{y}=\text{b}\sin\theta$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}\dots\text{proved.}$
View full question & answer→Question 3175 Marks
Differentiate the following functions with respect to x:
$\sin(2\sin^{-1}\text{x})$
AnswerLet, $\text{y}=\sin(2\sin^{-1}\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sin(2\sin^{-1}\text{x})\Big)$
$=\cos\big(2\sin^{-1}\text{x}\big)\frac{\text{d}}{\text{dx}}\big(2\sin^{-1}\text{x}\big)$
[Using chain rule]
$=\cos\big(2\sin^{-1}\text{x}\big)\times2\frac{1}{\sqrt{1-\text{x}^2}}$
$=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\sin\big(2\sin^{-1}\text{x}\big)\Big)=\frac{2\cos\big(2\sin^{-1}\text{x}\big)}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 3185 Marks
If $\text{x}=\text{a}\Big(\frac{1+\text{t}^2}{1-\text{t}^2}\Big)\text{ and y}=\frac{2\text{t}}{1-\text{t}^2},$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\text{x}=\text{a}\Big(\frac{1+\text{t}^{2}}{1-\text{t}^{2}}\Big)$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})-(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})}\bigg]$[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{(1-\text{t}^{2})(2\text{t)}-(1+\text{t}^{2})(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{a}\bigg[\frac{2\text{t}-2\text{t}^{3}+2\text{t}+2\text{t}^{3}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{4\text{a}\text{t}}{(1-\text{t}^{2})^{2}}\ .....(\text{i})$ and, $\text{y}=\frac{2\text{t}}{1-\text{t}^{2}}$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(\text{t})-\text{t}\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})(1)-\text{t}(-2\text{t})}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dx}}{\text{dt}}=2\bigg[\frac{(1-\text{t}^{2})+2\text{t}^{2}}{(1-\text{t}^{2})^{2}}\bigg]$ $\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{2(1-\text{t}^{2})}{(1-\text{t}^{2})^{2}}\ .....(\text{ii})$ Dividing equation (ii) by (i), $\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2(1+\text{t}^{2})}{(1-\text{t}^{2})^{2}}\times\frac{(1-\text{t}^{2})^{2}}{4\text{a}\text{t}}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{t}^{2})}{2\text{a}\text{t}}$
View full question & answer→Question 3195 Marks
If $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2},$ prove that $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
AnswerWe have, $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]$
$=\frac{\text{d}}{\text{dx}}\big(\text{x}\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^2}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$=\Big[\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big]-\frac{2\text{x}}{2\sqrt{1-\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}$
$=\sin^{-1}\text{x}$
View full question & answer→Question 3205 Marks
Differentiate the following functions with respect to x:
$\text{e}^{3\text{x}}\cos2\text{x}$
AnswerConsider $\text{y}=\text{e}^{3\text{x}}\cos2\text{x}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}\times\frac{\text{d}}{\text{dx}}(\cos2\text{x})+\cos2\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})$
[Using chain rule]
$=\text{e}^{3\text{x}}\times(-\sin2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x})+\cos2\text{xe}^{3\text{x}}\frac{\text{d}}{\text{dx}}(3\text{x})$
[Using chain rule]
$=-2\text{e}^{3\text{x}}\sin2\text{x}+3\text{e}^{3\text{x}}\cos2\text{x}$
$=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}}\cos2\text{x})=\text{e}^{3\text{x}}(3\cos2\text{x}-2\sin2\text{x})$
View full question & answer→Question 3215 Marks
Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\text{x}(\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}\text{ in }[-3,0].$
AnswerWe have, $\text{f(x)}=\text{x}(\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}$
Since polynomial function x(x + 3) and exponential funxtion $\text{e}^{-\frac{\text{x}}{2}}$ are continuous and diffferentiable in R, given function f(x) is also continuous and differentiable in R
Also f(0) = f(-3) = 0
So, conditions of Rolle's theorem are satisfied.
Hence, there exists a real number $\text{c}\in(-3,0)$ such that f'(c) = 0
Now $\text{f(x)}=(\text{x}^2+3\text{x})\text{e}^{-\frac{\text{x}}{2}}$
$\therefore\ \text{f}'(\text{x})=(2\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2+3\text{x})$
$=-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2+3\text{x}-4\text{x}-6)=-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2-\text{x}-6)$
So, f'(c) = 0
$\Rightarrow\ -\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{c}+2)(\text{c}-3)=0$
$\Rightarrow\ \text{c}=-2\in(-3,0)$
Therefore, Rolle's theorem has been verified.
View full question & answer→Question 3225 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
AnswerLet, $\text{y}=\text{e}^{\tan^{-1}\sqrt{\text{x}}}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\big)$
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\sqrt{\text{x}}\big)$
[Using chain rule]
$=\text{e}^{\tan^{-1}\sqrt{\text{x}}}\times\frac{1}{1+(\sqrt{\text{x}})^2}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{1+\text{x}}\times\frac{1}{2\sqrt{\text{x}}}$
$=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\tan^{-1}\sqrt{\text{x}}}\Big)=\frac{\text{e}^{\tan^{-1}\sqrt{\text{x}}}}{2\sqrt{\text{x}}(1+\text{x})}$
View full question & answer→Question 3235 Marks
Differentiate w.r.t. x the function in Exercise:
$(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})},\ \frac{\pi}{4}<\text{x}<\frac{3\pi}{4}$
AnswerLet $\text{y}=(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}$
Tanking logarithm on both the sides, we obtain
$\log\text{y}=\log\big[(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}\big]$
$\Rightarrow\ \log\text{y}=(\sin\text{x}-\cos\text{x)}.\log(\sin\text{x}-\cos\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[(\sin\text{x}-\cos\text{x})\log(\sin\text{x}-\cos\text{x})]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x}).\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})+(\sin\text{x}-\cos\text{x}).\frac{\text{d}}{\text{dx}}\log(\sin\text{x}-\cos\text{x})$
$\Rightarrow\ \ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\sin\text{x}-\cos\text{x}).(\cos\text{x}+\sin\text{x})+(\sin\text{x}-\cos\text{x}).\frac{1}{(\sin\text{x}-\cos\text{x})}.\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x)}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x)}^{(\sin\text{x}-\cos\text{x})}[(\cos\text{x}+\sin\text{x}).\log(\sin\text{x}-\cos\text{x})+(\cos\text{x}+\sin\text{x})]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\sin\text{x}-\cos\text{x})^{(\sin\text{x}-\cos\text{x})}(\cos\text{x}+\sin\text{x})[1+\log(\sin\text{x}-\cos\text{x})]$
View full question & answer→Question 3245 Marks
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
Taking log on both sides we get
$\text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}-\text{y}\tan\text{x}=\log\cos\text{y}-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}+\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\cos\text{x}+\text{x}\tan\text{y})=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}+\text{y}\tan\text{x}}{\log\cos\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 3255 Marks
Differentiate the following w.r.t. x:
$\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big),\frac{-\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\cos^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$=\frac{-1}{\sqrt{1-\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)^2}}\cdot\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big)$
$\Big[\because\ \frac{\text{d}}{\text{dx}}(\cos\text{x})=-\frac{1}{\sqrt{1-\text{x}^2}}\Big]$
$=\frac{-1}{\sqrt{4-\frac{(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cdot\cos\text{x})}{2}}}\cdot\frac{1}{\sqrt{2}}(\cos\text{x}-\sin\text{x})$
$=\frac{-1\cdot\sqrt{2}}{\sqrt{1-\sin2\text{x}}}\cdot\frac{1}{\sqrt{2}}(\cos\text{x}-\sin\text{x})$
$\big[\because1-\sin2\text{x}=(\cos\text{x}-\sin\text{x})^2=\cos^2\text{x}+\sin^2\text{x}-2\sin\text{x}\cos\text{x}\big]$
$=\frac{-1(\cos\text{x}-\sin\text{x})}{(\cos\text{x}-\sin\text{x})}=-1$
View full question & answer→Question 3265 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\text{e}^\theta\Big(\theta+\frac{1}{\theta}\Big),\text{ y}=\text{e}^{-\theta}\Big(\theta-\frac{1}{\theta}\Big)$
AnswerWe have, $\text{x}=\text{e}^\theta\Big(\theta+\frac{1}{\theta}\Big)$ and $\text{y}=\text{e}^{-\theta}\Big(\theta-\frac{1}{\theta}\Big)$
$\therefore\ \frac{\text{dx}}{\text{d}}=\frac{\text{d}}{\text{d}\theta}\bigg[\text{e}^\theta\Big(\theta+\frac{1}{\theta}\Big)\bigg]$
$=\text{e}^\theta\frac{\text{d}}{\text{d}\theta}\Big(\theta+\frac{1}{\theta}\Big)+\Big(\theta+\frac{1}{\theta}\Big)\frac{\text{d}}{\text{d}\theta}\text{e}^\theta$ $=\text{e}^\theta\Big(1-\frac{1}{\theta^2}\Big)+\Big(\theta+\frac{1}{\theta}\Big)\text{e}^\theta$
$=\text{e}^\theta\Big(1-\frac{1}{\theta^2}+\theta+\frac{1}{\theta}\Big)$ $=\text{e}^\theta\Big(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\Big)\ \ \dots(\text{i})$
and $\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\bigg[\text{e}^{-\theta}\cdot\Big(\theta-\frac{1}{\theta}\Big)\bigg]$ $=\text{e}^{-\theta}\frac{\text{d}}{\text{d}\theta}\Big(\theta-\frac{1}{\theta}\Big)+\Big(\theta-\frac{1}{\theta}\Big)\frac{\text{d}}{\text{d}\theta}\text{e}^{-\theta}$
$=\text{e}^{-\theta}\Big(1+\frac{1}{\theta^2}\Big)-\Big(1-\frac{1}{\theta}\Big)\text{e}^{-\theta}$ $=\text{e}^{\theta}\Big[\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\Big]\ \ \dots(\text{ii})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{e}^{-\text{e}\Big(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\Big)}}{\text{e}^\theta\Big(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\Big)}$ $=\text{e}^{-2\theta}\Big(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\Big)$
View full question & answer→Question 3275 Marks
Given the function $\text{f(x)}=\frac{1}{\text{x}+2}.$ Find the points of discontinuity of the composite function y = f(f(x)).
AnswerWe have, $\text{f(x)}=\frac{1}{\text{x}+2}$
$\therefore\ \text{y}=\text{f}\big\{\text{f(x)}\big\}$
$=\text{f}\Big(\frac{1}{\text{x}+2}\Big)=\frac{1}{\frac{1}{\text{x}+2}+2}$
$=\frac{1}{1+2\text{x}+4}\cdot(\text{x}+2)=\frac{(\text{x}+2)}{(2\text{x}+5)}$
So, the function y will not be continuous at those points, where it is not defined as it is a rational function.
Therefore, $\text{y}=\frac{(\text{x}+2)}{(2\text{x}+5)}$ is not defined, when 2x + 5 = 0
$\therefore\ \text{x}=\frac{-5}{2}$
Hence, y is discontinuous at $\text{x}=\frac{-5}{2}$
View full question & answer→Question 3285 Marks
If $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}},$ then find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}}$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\big]$
$=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\Big]$
$=\frac{1}{\sqrt{1-\big(6\text{x}\sqrt{1-9\text{x}^2}\big)^2}}\times\frac{\text{d}}{\text{dx}}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)$
$=\frac{1}{\sqrt{1-[36\text{x}^2(1-9\text{x}^2)]}}\times\Big(6\text{x}\frac{\text{d}}{\text{dx}}\sqrt{1-9\text{x}^2}+\sqrt{1-9\text{x}^2}\frac{\text{d}}{\text{dx}}(6\text{x})\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-9\text{x}^2)+\sqrt{1-9\text{x}^2}(6)\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}(-18\text{x}^2)+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2}{\sqrt{1-9\text{x}^2}}+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2+6\sqrt{1-9\text{x}^2}}{\sqrt{1-9\text{x}^2}}\Big)$
$=\frac{-54\text{x}^2+6-54\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
$=\frac{6-108\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
View full question & answer→Question 3295 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$
AnswerHere $\text{f}(\text{x})=\cos2\text{x}\text{ on }\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$ We know that $\cos\text{x}$ is continuous and differentiable everywhere. So, f(x) is continuous in $\Big[\frac{-\pi}{4},\frac{\pi}{4}\Big]$ and differentiable is $\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$. Now, $\text{f}\Big(-\frac{\pi}{4}\Big)=\cos2\Big(-\frac{\pi}{4}\Big)=\cos\Big(-\frac{\pi}{2}\Big)=0$ $\text{f}\Big(\frac{\pi}{4}\Big)=\cos2\Big(\frac{\pi}{4}\Big)=\cos\Big(\frac{\pi}{2}\Big)=0$ $\Rightarrow\text{f}\Big(-\frac{\pi}{4}\Big)=\text{f}\Big(\frac{\pi}{4}\Big)$ So, Rolle's theorem is applicable, so, there must exist a $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0 Now, $\text{f}'(\text{x})=2\sin2\text{x}$ $\text{f}'(\text{c})=2\sin2\text{c}=0$ $\Rightarrow\sin2\text{c}=0$ $\Rightarrow2\text{c}=0$$\Rightarrow\text{c}=0\in\Big(\frac{-\pi}{4},\frac{\pi}{4}\Big)$
Thus, Rolle's theorem verified.
View full question & answer→Question 3305 Marks
If $\text{y}=\sqrt{\text{a}^2-\text{x}^2},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
AnswerDIfferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{a}^2-\text{x}^2}\big)$
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$
$=\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}$
$\big[\text{Since},\sqrt{\text{a}^2-\text{x}^2}=\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}$
Hence, the solution is, $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
View full question & answer→Question 3315 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}$
We have,
$(\text{LHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{k}(\pi-\text{h})+1=\text{k}\pi+1$
$(\text{RHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\cos(\pi+\text{h})=\cos\pi=-1$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}$
$\Rightarrow\text{k}\pi+1=-1$
$\Rightarrow\text{k}=\frac{-2}{\pi}$
View full question & answer→Question 3325 Marks
Find all points of discontinuity of f, where f is defined by: $\text{f(x)}= \begin{cases}\text{x}^{10} - 1,\ \ \text{if x}\leq 1 \\\text{x}^2,\ \ \ \ \ \ \ \ \ \ \text{if x}>1\end{cases}$
AnswerHere $\text{f(x)}= \begin{cases}\text{x}^{10} - 1,\ \ \text{if x}\leq 1 \\\text{x}^2,\ \ \ \ \ \ \ \ \ \ \text{if x}>1\end{cases}$ Function f is defined at all points of the real line. Let c be any real number. Three cases arise: Case I: c < 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(x}^{10} - 1) = \text{c}^{10}- 1f(c) = c^{10} - 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x < 1. Case II: c > 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{x}^{2} = \text{c}^{2} f(c) = c^2$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x > 1. Case III: c = 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{(x}^{10} - 1) = 1^{10}- 1 = 1 - 1 =0$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{x}^2 = (1)^2 = 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = 1.
View full question & answer→Question 3335 Marks
Discuss the continuity of the function $\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
AnswerWhen x < 2, we have
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
Hence, f(x) is everywhere continuous.
View full question & answer→Question 3345 Marks
If $\text{x}-\text{e}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=\text{x}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}}$
$\text{y}=\text{e}^{\tan\text{x}\log\text{x}}+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\text{x})+\text{e}^{\frac{1}{2}\log\big(\frac{\text{x}^2+1}{2}\big)}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2}\log\Big(\frac{\text{x}^2+1}{2}\Big)\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{1}{2}\times\frac{2}{\text{x}^2+1}\times(\text{x})\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\sqrt{\frac{\text{x}^2+1}{2}}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\tan\text{x}}\Big[\frac{\tan\text{x}}{\text{x}}+\sec^3\text{x}\log\text{x}\Big]+\frac{\text{x}}{\sqrt{2(\text{x}^2+1)}}$
View full question & answer→Question 3355 Marks
If $\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$ show that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
AnswerHere,
$\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$
$\Rightarrow\frac{1}{\text{a}}\log\text{y}=\sin^{-1}\text{x}$
$\Rightarrow\text{y}=\text{e}^\text{a}\sin^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\text{y}_1=\text{e}^{\text{a} \sin^{-1}\text{x}}\times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{\text{ay}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\text{ay}_1\sqrt{1-\text{x}^2}+\frac{\text{x}\text{ay}}{\sqrt{1-\text{x}^2}}}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1(1-\text{x}^2)+\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1}{\sqrt{1-\text{x}^2}}+\frac{\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{a}^2\text{y}}{1-\text{x}^2}+{\frac{\text{xy}_1}{(1-\text{x}^2)}}$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
View full question & answer→Question 3365 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\big(2\text{x}\sqrt{1-\text{x}^{2}}\big), -\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerThe given relationship is,$\text{y}=\sin^{-1}\big(2\text{x}\sqrt{1-\text{x}^{2}}\big)$
$\text{y}=\sin^{-1}\big(2\text{x}\sqrt{1-\text{x}^{2}}\big)$
$\Rightarrow\sin\text{y}=2\text{x}\sqrt{1-\text{x}^{2}}$
Differentiating this relationship with respect to x, we obtain
$\cos\text{y}\frac{\text{dy}}{\text{dx}}=2\bigg[\text{x}\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^{2}}\big)+\sqrt{1-\text{x}^{2}}\frac{\text{d}}{\text{dx}}\text{x}\bigg]$
$\Rightarrow\sqrt{1-\sin^{2}\text{y}\frac{\text{dy}}{\text{dx}}}=2\bigg[\frac{\text{x}}{2}.\frac{-2\text{x}}{\sqrt{1-\text{x}^{2}}}+\sqrt{1-\text{x}^{2}}\bigg]$
$\Rightarrow\sqrt{1-\big(2\text{x}\sqrt{1-\text{x}^{2}}\big)^{2}}\frac{\text{dy}}{\text{dx}}=2\bigg[\frac{-\text{x}^{2}+1-\text{x}^{2}}{\sqrt{1-\text{x}^{2}}}\bigg]$
$\Rightarrow\sqrt{1-4\text{x}^{2}\big({1-\text{x}^{2}}\big)}\frac{\text{dy}}{\text{dx}}=2\bigg[\frac{1-2\text{x}^{2}}{\sqrt{1-\text{x}^{2}}}\bigg]$
$\sqrt{1-4x^2+4x^4}\frac{\text{dy}}{\text{dx}}=2\bigg[\frac{1-2\text{x}^{2}}{\sqrt{1-\text{x}^{2}}}\bigg]$
$\Rightarrow\sqrt{(1-2\text{x}^{2}})^{2}\frac{\text{dy}}{\text{dx}}=2\bigg[\frac{1-2\text{x}^{2}}{\sqrt{1-\text{x}^{2}}}\bigg]$
$\Rightarrow\big(1-2\text{x}^{2}\big)\frac{\text{dy}}{\text{dx}}=2\bigg[\frac{1-2\text{x}^{2}}{\sqrt{1-\text{x}^{2}}}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 3375 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
AnswerHere,$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
We know that, exponantial function is continuous and differentiable everywhere. So, f(x) is continuous IS $[-1,1]$ and differentiable is $(-1,1).$ Now, $\text{f}(-1)=\text{e}^{1-1}=1$ $\text{f}(1)=\text{e}^{1-1}=1$ $\Rightarrow\text{f}(-1)=1$ So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(-1,1)$ such that f'(c) = 0. Now,$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}$
$\text{f}'(\text{x})=\text{e}^{1-\text{x}^2}(-2\text{x})$ Now, $\text{f}'(\text{c})=0$ $-2\text{ce}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0$ or $\text{e}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0\in(-1,1)$ Hence, Rolle's theorem is verified.
View full question & answer→Question 3385 Marks
If $\text{xy}=1,$ prove that $\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
AnswerHere, xy = 1 .....(i)
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$ \Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)=0$
$\Big[\text{Put x}=\frac{1}{\text{y}}\text{ from equation (i)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\frac{1}{\text{y}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$
View full question & answer→Question 3395 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\cos^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\cos^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\cos^{-1}\text{x}}$
$\Rightarrow\log\text{y}=\cos^{-1}\text{x}\log\text{x}$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos^{-1}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\cos^{-1}\text{x}}\Big[\frac{\cos^{-1}\text{x}}{\text{x}}-\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
View full question & answer→Question 3405 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$
AnswerThe given function f is $\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$ It is evident that f is defind at all points of the real line. Let c be a real number.Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\sin\text{c}-\cos\text{c}$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\sin\text{x}-\cos\text{x})=\sin\text{c}-\cos\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that $\text{x}\neq0$Case II:
If c = 0, then f(0) = -1 $\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ $\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ Therefore, f is continuousb at x = 0 From tha above observations, it can be continuous that f is continuous at every point of the real line. Thus, f is a continuous function.
View full question & answer→Question 3415 Marks
Examine the differentiability of f, where f is defined by: $\text{f(x)}=\begin{cases}\text{x[x]},&\text{if }0\leq\text{x}<2(\text{x}-1)\text{x},&\text{if }2\leq\text{x}<3\end{cases}$ at $x = 2.$
AnswerWe have, $\text{f(x)}=\begin{cases}\text{x[x]},&\text{if }0\leq\text{x}<2(\text{x}-1)\text{x},&\text{if }2\leq\text{x}<3\end{cases}$ at x = 2 Let us first check continuity at x = 2 $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow2^-}\text{x[x]}$ $=\lim\limits_{\text{h}\rightarrow0}(2-\text{h})[2-\text{h}]=2\times1=2$ $\text{R.H.L}=\lim\limits_{\text{x}\rightarrow2^+}(\text{x}-1)\text{x}$ $=\lim\limits_{\text{h}\rightarrow0}(2+\text{h}-1)(2+\text{h})=1\times2=2$ Also, f(2) = (2 - 1)2 = 2 Thus, f(x) is continuous at x = 2. Now $\text{Lf}'(2)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(2-\text{h})[2-\text{h}]-2}{-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(2-\text{h}).1-2}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{-\text{h}}=1$ $\text{Rf}'(2)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(2+\text{h}-1)(2+\text{h})-2}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2+3\text{h}}{\text{h}}=3$ Clearly, Lf'(2) ≠ Rf'(2) Thus, f(x) is not diffcrentiable at x = 2.
View full question & answer→Question 3425 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}\text{ on }\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
Since $\cos\text{x}\ \&\ \text{e}^{\text{x}}$ are everywhere continuous and differentiable.
Therefore, f(x) being a product of these two, is continuous on $\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$ and differentiable on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
Also,
$\text{f}\Big(\frac{-\pi}{2}\Big)=\text{f}\Big(\frac{\pi}{2}\Big)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\text{e}^{\text{x}}\cos{\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\text{e}^{\text{x}}(\cos\text{x}+\sin\text{x})=0$
$\Rightarrow\sin\text{x}+\cos\text{x}=0$
$\Rightarrow\tan\text{x}=-1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Since, $\text{c}=\frac{\pi}{4}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$ such that f'(c) = 0
Hence, Rolle's theorem verified.
View full question & answer→Question 3435 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3 - 2x^2 - x + 3$ on $[0, 1]$
AnswerHere,$f(x) = x^3 - 2x^2 - x + 3$
Since f(x) is polynomial function. So, f(x) is continuous in [0, 1] and differentiable in (0, 1).
Thus, both conditions of Lagrange's mean value theorem is appplicable.
Thus, there exist a point $\text{c}\in(0,1)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$
$\Rightarrow 3c^2 - 4c - 1 = 1^{-2}$
$\Rightarrow 3c^2 - 4c + 1 = 0$
$\Rightarrow 3c^2- 3c - c + 1 = 0$
$\Rightarrow 3c(c - 1) - 1(c - 1) = 0$
$\Rightarrow (3c - 1)(c - 1) = 0$
$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 3445 Marks
Find the values of p and q so that $\text{f(x)}=\begin{cases}\text{x}^2+3\text{x}+\text{p},&\text{if x}\leq1\\\text{qx}+2,&\text{if x}>1\end{cases}$ is differentiable at x = 1.
AnswerWe have, $\text{f(x)}=\begin{cases}\text{x}^2+3\text{x}+\text{p},&\text{if x}\leq1\\\text{qx}+2,&\text{if x}>1\end{cases}$ is differentiable at x = 1.
$\text{Lf}'(1)=\lim\limits_{\text{x}\rightarrow1^-}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^-}\frac{\big(\text{x}^2+3\text{x}+\text{p}\big)-\big(1+3+\text{p}\big)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(1-\text{h})^2+3(1-\text{h})+\text{p}\big]-[4+\text{p}]}{(1-\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[1+\text{h}^2-2\text{h}+3-3\text{h}+\text{p}\big]-[4+\text{p}]}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[\text{h}^2-5\text{h}+\text{p}+4-4-\text{p}\big]}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}[\text{h}-5]}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}-[\text{h}-5]=5$
$\text{Rf}'(1)=\lim\limits_{\text{x}\rightarrow1^+}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1^+}\frac{(\text{qx}+2)-(\text{q}+2)}{\text{x}-1}$
$\lim\limits_{\text{h}\rightarrow0}\frac{\big[\text{q}+\text{qh}+2-2-\text{q}\big]}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{qh}}{\text{h}}$
$\Rightarrow\ \text{q}$
Since, f(x) is differentiable at x = 1
$\therefore\ \text{Lf}'(1)=\text{Rf}'(1),$
$\therefore\ 5=\text{q}$
Substituting q = 5, we get
$\text{p}=3$
$\therefore\ \text{p}=3\text{ and q}=5$
View full question & answer→Question 3455 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{1+\sqrt{1-\text{x}^3}}\Big\}$
Put $\text{x}=\sin\theta$
$\text{y}=\tan^{-1}\Big\{\frac{\sin\theta}{1+\sqrt{1-\sin^2\theta}}\Big\}$
$\text{y}=\tan^{-1}\Big(\frac{\sin\theta}{1+\cos\theta}\Big)$
$\text{y}=\tan^{-1}\bigg\{\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\bigg\}$
$\text{y}=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)\ .....(\text{i})$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\sin\theta<1$
$\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\frac{\theta}{2}\ \Big[\text{Since}, \tan^{-1}(\tan\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\sin^{-1}\text{x}\big[\text{Since, x}=\sin\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 3465 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{ y}=\frac{3+2\log\text{t}}{\text{t}}.$
AnswerConsider, $\text{x}=\frac{1+\log\text{t}}{\text{t}^2}$ and $\text{y}=\frac{3+2\log\text{t}}{\text{t}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{t}^2\cdot\frac{\text{d}}{\text{dt}}(1+\log\text{t})-(1+\log\text{t})\cdot\frac{\text{d}}{\text{dt}}\text{t}^2}{(\text{t}^2)^2}$
$=\frac{\text{t}^2\cdot\frac{1}{\text{t}}-(1+\log\text{t})\cdot2\text{t}}{\text{t}^4}$
$=\frac{\text{t}-(1+\log\text{t})\cdot2\text{t}}{\text{t}^4}$
$=\frac{-1-2\log\text{t}}{\text{t}^3}$
and $\frac{\text{dy}}{\text{dt}}=\frac{\text{t}\cdot\frac{\text{d}}{\text{dt}}(3+2\log\text{t})-(3+2\log\text{t})\cdot\frac{\text{d}}{\text{dt}}\text{t}}{\text{t}^2}$
$=\frac{\text{t}\cdot2\cdot\frac{1}{\text{t}}-(3+2\log\text{t})\cdot1}{\text{t}^2}$a
$=\frac{-1-2\log\text{t}}{\text{t}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-1-2\log\frac{\text{t}}{\text{t}^2}}{-1-2\log\frac{\text{t}}{\text{t}^3}}=\text{t}$
View full question & answer→Question 3475 Marks
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$
AnswerConsider, $\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\sin\text{xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)=\frac{\text{d}}{\text{dx}}\text{x}^2-\frac{\text{d}}{\text{dx}}\text{y}$
$\Rightarrow\ \cos\text{xy}\cdot\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{y}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}}{\text{y}^2}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \cos\text{xy}\cdot\Big[\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\cdot\text{x}\Big]+\frac{\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}=2\text{x}-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{xy}+\frac{\text{y}}{\text{y}^2}\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\text{x}\cos\text{xy}-\frac{\text{x}}{\text{y}^2}+1\Big]=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\Big[\frac{2\text{xy}-\text{y}^2\cos\text{xy}-1}{\text{y}}\Big]\Big[\frac{\text{y}^2}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}\Big]$
$=\frac{2\text{xy}^2-\text{y}^3\cos\text{xy}-\text{y}}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}$
View full question & answer→Question 3485 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{if }\text{ x}<0\\2\text{x}+3,&\text{ x}\geq0\end{cases}$
AnswerWhen x < 0, we have $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ and the identity function continuous for x < 0, so the quotient function
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous for x < 0
When x > 0 f(x) = 2x + 3, which is a polynomial of degree 1. So, f(x) 2x + 3 is continuous for x > 0
Now, consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(-\text{h})}{-\text{h}}=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin\text{h}}{-\text{h}}=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
$\text{f}(0)=2\times0+3=3$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence, f(x) is discontinuous at x = 0
View full question & answer→Question 3495 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta),\text{y}=\text{ae}^\theta(\sin\theta+\cos\theta)$
AnswerWe have, $\text{x}=\text{ae}^{\theta}(\sin\theta-\cos\theta)$ and $\text{y}=\text{ae}^{\theta}(\sin\theta+\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta-\cos\theta)+(\sin\theta-\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta+\cos\theta)+(\sin\theta+\cos\theta)\frac{\text{d}}{\text{d}\theta}(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta+\sin\theta)+(\sin\theta-\cos\theta)(\text{e}^\theta)\Big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\text{e}^\theta(\cos\theta-\sin\theta)+(\sin\theta+\cos\theta)(\text{e}^\theta)\Big]$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\sin\theta)\big]$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[2\text{e}^\theta(\cos\theta)\big]$
$\therefore\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}(2\text{e}^\theta\cos\theta)}{\text{a}(2\text{e}^\theta\sin\theta)}=\cot\theta$
View full question & answer→Question 3505 Marks
If $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}}$ for $\text{x}\neq\frac{\pi}{4},$ find the value which can be assigned to f(x) at $\text{x}=\frac{\pi}{4}$ so that the function f(x) becomes continuous every where in $\Big[0,\frac{\pi}{2}\Big]$
AnswerWhere $\text{x}\neq\frac{\pi}{4},\ \tan\big(\frac{\pi}{4}-\text{x}\big)$ and $\cot2\text{x}$ are continuous in $\Big[0,\frac{\pi}{2}\Big]$
Thus, the quotient function $\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$ is continuous in $\Big[0,\frac{\pi}{2}\Big]$ for each $\text{x}\neq\frac{\pi}{4}$
So, if f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then it will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
Now,
Let us consider the point $\text{x}=\frac{\pi}{4}$
Given, $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}},\text{x}\neq\frac{\pi}{4}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}+\text{h}\big)}{\cot\big(\frac{\pi}{2}-2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
$\Big(\text{RHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}-\text{h}\big)}{\cot\big(\frac{\pi}{2}+2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(-\text{h})}{-\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)\\=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
If f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\text{f}\Big(\frac{\pi}{4}\Big)$
$\therefore\ \text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2}$
Hence, for $\text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2},$ the function f(x) will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
View full question & answer→