Question 3515 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=3\cos\theta-2\cos^3\theta,\text{ y}=3\sin\theta-2\sin^3\theta.$
Answer$\because\ \text{x}=3\cos\theta-2\cos^3\theta$ and $\text{y}=3\sin\theta-2\sin^3\theta$
$\therefore\ \frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(3\cos\theta)-\frac{\text{d}}{\text{d}\theta}(2\cos^3\theta)$
$=3.(-\sin\theta)-2.3\cos^2\theta.\frac{\text{d}}{\text{d}\theta}.\cos\theta$
$=-3\sin\theta+6\cos^2\theta\sin\theta$
and $\frac{\text{dy}}{\text{d}\theta}=3\cos\text{A}-2.3\sin^2\theta.\frac{\text{d}}{\text{d}\theta}.\sin\theta$
$=3\cos\theta-6\sin^2\theta.\cos\theta$
Now, $\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{3\cos\theta-6\sin^2\theta\cos\theta}{-3\sin\theta+6\cos^2\theta\sin\theta}$
$=\frac{3\cos\theta(1-2\sin^2\theta)}{3\sin\theta(-1+2\cos^2\theta)}$ $=\cot\theta.\frac{\cos2\theta}{\cos2\theta}=\cot\theta$
View full question & answer→Question 3525 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{\text{e}^\text{t}+\text{e}^{-\text{t}}}{2}\text{ and y}=\frac{\text{e}^\text{t}-\text{e}^\text{-t}}{2}$
AnswerWe have, $ \text{x}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$ and $ \text{y}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})+\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})-\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}+\text{e}^{\text{-t} \frac{\text{d}}{\text{dt}}}(\text{-t})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}-\text{e}^{\text{-t}}\frac{\text{d}}{\text{dt}}({\text{e}^{\text{-t}}})\bigg]$
$\Rightarrow\frac{1}{2}(\text{e}^{\text{t}}-\text{e}^\text{-t})=\text{y}$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}(\text{e}^\text{t}+\text{e}^{\text{-t}})=\text{x}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{x}}{\text{y}}$
View full question & answer→Question 3535 Marks
It is given that the Rolle's theorem holds for the function $f(x) = x^3 + bx^2 + cx, \text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of b and c.
AnswerSo, $f(1) = f(2)$
$\Rightarrow (1)^3 + b(1)^2 + c(1) = (2)^3 + b(2)^2 + c(2)$
$\Rightarrow 1 + b + c = 8 + 4b + 2c$
$\Rightarrow 3b + c + 7 = 0$ ....(i)
And $\text{f}'\Big(\frac{4}{3}\Big)=0$
$\Rightarrow3\Big(\frac{4}{3}\Big)^2+2\text{b}\Big(\frac{4}{3}\Big)+\text{c}=0$ [As, $f'(x) = 3x^2 + 2bx + c$]
$\Rightarrow\frac{16}{3}+\frac{8\text{b}}{3}+\text{c}=0$
$\Rightarrow8\text{b}+3\text{c}+16=0\ ....(\text{ii})$
(ii) - (i) × 3, we get
$8b - 9b + 16 - 21 = 0$
$\Rightarrow -b - 5 = 0$
$\Rightarrow b = -5$
Substituting $b = -5$ in (i), we get
$3(-5) + c + 7 = 0$
$\Rightarrow -15 + c + 7 = 0$
$\Rightarrow c = 8$
View full question & answer→Question 3545 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{2^\text{z+2}-16}{4^\text{x}-16}, &\text{if x} \neq 2\\\text{k}, & \text{x} = 2\end{cases}$
is continuous at x = 2, Find k.
AnswerWe are given that the function is continuous at x = 2
$\therefore$ LHL = RHL = f(2)...(i)
Now,
f(2) = k...(A)
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{(2-\text{h)+2}}-16}{4^{(2-\text{h)}}-16}=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{2-\text{h}}-16}{4^{2-\text{h}}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^4-2^\text{-h}-16}{4^2.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16.2^\text{-h}-16}{16.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16\Big(2^\text{-h}-1\Big)}{16\Big(4^\text{-h}-1\Big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}\Big)-1^2}$ $\Big[\because2^{-2\text{h}}=\Big(2^{-\text{h}}\Big)^2=4^{-\text{h}}\Big]$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}-1\Big)\Big(2^\text{-h}+1\Big)}=\frac{1}{2}\dots(\text{B})$
$\therefore$ Using (i) from (A) & (B)
$\text{k}=\frac{1}{2}$
View full question & answer→Question 3555 Marks
If f is defined by $\text{f(x)}=\text{x}^2-4\text{x}+7,$ show that $\text{f}'(5)=2\text{f}'\Big(\frac{7}{2}\Big).$
Answer$f(x) = x^2 - 4x + 7$ is a polynomial function, So it is differentiable everywhere.
$\text{f}'(5)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(5+\text{h})-\text{f}(5)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{(5+\text{h})^2-4(5+\text{h})+7\big\}-[25-20+7]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+25+10\text{h}-20-4\text{h}+7-12}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+6\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+6)$
$=6$
$\text{f}'\Big(\frac{7}{2}\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{7}{2}+\text{h}\Big)^2-\text{f}\Big(\frac{7}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\Big(\frac{7}{2}+\text{h}\Big)^2-4\Big(\frac{7}{2}+\text{h}\Big)+7\Big]-\Big[\Big(\frac{7}{2}\Big)^2-4\Big(\frac{7}{2}\Big)+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7\Big]-\Big[\frac{49}{2}-14+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7-\frac{49}{2}-14+7}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+3\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+3)$
$=3$
$\text{f}'(5)=6$
$=2(3)$
$=\text{f}'(5)=2\text{f}'\Big(\frac{7}{2}\Big)$
View full question & answer→Question 3565 Marks
Find the value of k in this question, so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}},&\text{if}-1\leq0\\\frac{2\text{x}+1}{\text{x}-1},&\text{if }0\leq\text{x}\leq1\end{cases}$ at x = 0.
AnswerWe have, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}},&\text{if}-1\leq\text{x<0}\\\frac{2\text{x}+1}{\text{x}-1},&\text{if }0\leq\text{x}\leq1\end{cases}$ at x = 0
$\text{L.H.L}=\lim\limits_{\text{h}\rightarrow0^-}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0^-}\bigg(\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}}\bigg)\cdot\bigg(\frac{\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}{\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0^-}\frac{1+\text{kx}-1+\text{kx}}{\text{x}\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}$
$=\lim\limits_{\text{h}\rightarrow0^-}\frac{2\text{kx}}{\text{x}\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{k}}{\sqrt{1+\text{k}(0-\text{h})}+\sqrt{1-\text{k}(0-\text{h})}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{k}}{\sqrt{1-\text{kh}}+\sqrt{1+\text{kh}}}$
$=\frac{2\text{k}}{2}=\text{k}$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow0^+}\frac{2\text{x}+1}{\text{x}-1}=\lim\limits_{\text{h}\rightarrow0}\frac{2(0+\text{h})+1}{(0+\text{h})-1}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}+1}{\text{h}-1}=-1$
Also $\text{f}(0)=\frac{2\times0+1}{0-1}=-1$
We must have L.H.L = R.H.L = f(0)
⇒ k = -1
View full question & answer→Question 3575 Marks
If $\text{xy}=4,$ prove that $\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
AnswerWe have, $\text{xy}=4$
$\Rightarrow\text{y}=\frac{4}{\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\frac{4}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4(-1\times\text{x}^{-1-1})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4}{\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4}{\big(\frac{4}{\text{y}}\big)^2}\ \Big[\because\text{x}=\frac{4}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4\text{y}^2}{16}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2}{4}$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}+4\text{y}^2=3\text{y}^2$
$\Rightarrow4\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}^2$
Dividing both side by x,
$\Rightarrow\frac{4}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
View full question & answer→Question 3585 Marks
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
AnswerWe have, $\text{y}=\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}}\Big)$
$=\frac{1}{\sqrt[2]{\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)}}\times\frac{\text{d}}{\text{dx}}\Big(\frac{1+\text{e}^\text{x}}{1-\text{e}^\text{x}}\Big)$
[Using chain rule]
$=\frac{1}{2}\times\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Bigg[\frac{\big(1-\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(1+\text{e}^\text{x})-(1+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1-\text{e}^\text{x})\big)}{(1-\text{e}^\text{x})^2}\Bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\bigg[\frac{(1-\text{e}^\text{x})\text{e}^\text{x}+(1+\text{e}^\text{x})\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\bigg]$
$=\frac{1}{2}\sqrt{\frac{1-\text{e}^\text{x}}{1+\text{e}^\text{x}}}\Big[\frac{2\text{e}^\text{x}}{(1-\text{e}^\text{x})^2}\Big]$
$=\frac{\text{e}^\text{x}}{\sqrt{(1+\text{e}^\text{x})\sqrt{(1+\text{e}^\text{x})}}}\frac{1}{(1-\text{e}^\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}}{(1-\text{e}^\text{x})\sqrt{1-\text{e}^{2\text{x}}}}$
View full question & answer→Question 3595 Marks
If $x^m + y^n = 1$, Prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
AnswerWe have, $x^m + y^n = 1$
Taking log on both side,
$\log(\text{x}^\text{m}\text{y}^{\text{n}})=\log(1)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}(\text{m}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{n}\log\text{y})=\frac{\text{d}}{\text{dx}}\big\{\log(1)\big\}$
$\Rightarrow\frac{\text{m}}{\text{n}}+\frac{\text{n}}{\text{y}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{m}}{\text{x}}\times\frac{\text{y}}{\text{n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{my}}{\text{nx}}$
View full question & answer→Question 3605 Marks
Discuss the continuity of the f(x) at the indicated points f(x) = |x| + |x - 1| at x = 0, 1.
AnswerGiven,
$\text{f(x)}=|\text{x}|+|\text{x}-1|$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0-\text{h}|+|0-\text{h}-1|\big]=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0+\text{h}|+|0+\text{h}-1|\big]=1$
Also, $\text{f}(1)=|1|+|1-1|=1+0=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$ and $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=1$
Hence, f(x) is continuous at x = 0, 1
View full question & answer→Question 3615 Marks
Differentiate the following functions with respect to x:
$(\sin\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\sin\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\sin\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\sin\text{x}}(\cos\text{x})-\sin\text{x}\log\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}[\cos\text{x}\cot\text{x}-\sin\text{x}\log\sin\text{x}]$
View full question & answer→Question 3625 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}\text{ on }0\leq\text{x}\leq\pi$
AnswerThe given function is $\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}.$
Since $\cos\text{x}$ and $\text{e}^\text{x}$ are everywhere continuous and differentiable, being a quotient of these two, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi).$
Also,
$\text{f}(\pi)=\text{f}(0)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in(0,\pi)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}$
$\Rightarrow \text{f}'(\text{x})=\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}=0$
$\Rightarrow\cos\text{x}-\sin\text{x}=0$
$\Rightarrow\tan\text{x}=1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Thus, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.
View full question & answer→Question 3635 Marks
Explain if Rolle's theorem is applicable to any one of the following functions.
- $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
- $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$
Can you say something about the converse of Rolle's Theorem from these functions? AnswerBy Rolle’s theorem, for a function $\text{f}:[\text{a},\text{b}]\rightarrow\text{R},$ if
- f is continuous on [a, b],
- f is differentiable on (a, b) and
- f(a) = f(b)
Then there exists some $\text{c}\in(\text{a},\text{b})$ such that f'(c) = 0
Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
- $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9.
Thus, f(x) is not continuous on [5, 9].
Also, f(5) = [5] = 5 and f(9) = [9] = 9
$\therefore\ \text{f}(5)\neq\text{f}(9)$
The differentiability of f on (5, 9) is checked in the following way.
Let n be an integer such that $\text{n}\in(5,9).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
- $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = −2 and x = 2.
Thus, f (x) is not continuous on [−2, 2].
Also, f(-2) = [-2] = -2 and f(2) = [2] = 2
$\therefore\ \text{f}(-2)\neq\text{f}(2)$
The differentiability of f on (-2, 2) is checked in the following way.
Let n be an integer such that $\text{n}\in(-2,2).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (-2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$ View full question & answer→Question 3645 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\tan^{-1}\text{x}}$
AnswerLet $\text{y}=\text{x}^{\tan^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\tan^{-1}\text{x}}$
$\log\text{y}=\tan^{-1}\text{x}\log\text{x}\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{1}{1+\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\tan^{-1}\text{x}}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 3655 Marks
Differentiate $\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$ with respect to $\sqrt{1-4\text{x}^2},$ if:
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
AnswerLet $\text{u}=\sin^{-1}\Big(4\text{x}\sqrt{1-4\text{x}^2}\Big)$
Put $2\text{x}=\cos\theta \text{ So},$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\times\cos\theta\sqrt{1-\cos^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\cos\theta\sin\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let, $\text{v}=\sqrt{1-4\text{x}^2}\ .....(\text{ii})$
Here,
$\text{x}\in\Big(-\frac{1}{2\sqrt{2}},\frac{1}{2}\Big)$
$\Rightarrow2\text{x}\in\Big(-\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\cos\theta\in\Big(\frac{1}{\sqrt{2}},1\Big)$
$\Rightarrow\theta\in\Big(0,\frac{\pi}{4}\Big)$
So, from equation (i),
$\text{u}=2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dv}}{\text{dx}}=2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{-2}{\sqrt{1-4\text{x}^2}}(2)\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{v})$
Dividing equation (v) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{-4\text{x}}$
$\frac{\text{du}}{\text{dv}}=\frac{1}{\text{x}}$
View full question & answer→Question 3665 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\},-\text{a}<\text{x}<\text{a}$
AnswerLet $\text{y}=\tan^{-1}\Big\{\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big\}$ Put $\text{x}=\text{a}\sin\theta$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2-\text{a}^2\sin^2\theta}}\Big\}$ $\text{y}=\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{\sqrt{\text{a}^2(1-\sin^2\theta)}}\bigg\}$ $\text{y}=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}\cos\theta}\Big\}$ $\text{y}=\tan^{-1}(\tan\theta)\ .....(\text{i})$ Here, $-\text{a}<\text{x}<\text{a}$ $\Rightarrow-1<\frac{\text{x}}{\text{a}}<1$ $\Rightarrow -\frac{\pi}{2}<\theta<\frac{\pi}{2}$ From equation (i), $\text{y}=\theta$ $\Big[\text{Since},\tan^{-1}(\tan\theta)=\theta,\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$ $\text{y}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\big[\text{Since x}=\text{a }\sin \theta\big]$ Differentiating it with respect to x, $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\big(\frac{\text{x}}{\text{a}}\big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{a}}\Big)$ $=\frac{\text{a}}{\sqrt{\text{a}^2-\text{x}^2}}\times\big(\frac{1}{\text{a}}\big)$$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}$
View full question & answer→Question 3675 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big), -\frac{1}{\sqrt{6}}<\text{x}<\frac{1}{\sqrt{6}}$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{5\text{x}}{1+6\text{x}^3}\Big)$
$=\tan^{-1}\Big(\frac{3\text{x}+2\text{x}}{1-(3\text{x})(2\text{x})}\Big)$
$\text{y}=\tan^{-1}(3\text{x})+\tan^{-1}(2\text{x})$
$\Big[\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{1+(2\text{x})^2}\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{1}{1+9\text{x}^2}(3)+\frac{1}{1+4\text{x}^2}(2)$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}+\frac{2}{1+4\text{x}^2}$
View full question & answer→Question 3685 Marks
Find all points of discontinuity of f, where f is defined by:
$\text{f(x)}= \begin{cases}\text{x} + 1,\ \ \text{if x}\geq 1 \\\text{x}^2 + 1,\ \text{if x}<1\end{cases}$
AnswerHere $\text{f(x)}= \begin{cases}\text{x} + 1,\ \ \text{if x}\geq 1 \\\text{x}^2 + 1,\ \text{if x}<1\end{cases}$
Function f is defined at all points of the real line.
Let c be any real number.
Three cases arise:
Case I: c < 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}^2 + 1) = \text{c}^2 + 1$
$f(x) = c^2 + 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x < 1.
Case II: c > 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x} + 1) = \text{c} + 1$
f(c) = c + 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 III: c = 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{(x}^2 + 1) = 1 + 1 = 2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{(x}^2 + 1) = 1 + 1 = 2$
f(1) = 1 + 1 = 2
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = \text{f(1)}$
$\therefore$ f is continuous at all points x < 1.
$\therefore$ f is continuous at all point of domain.
View full question & answer→Question 3695 Marks
If $\text{x}=\text{a}(\theta+\sin\theta),\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{x}=\text{a}(\theta+\sin\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big(\frac{\text{d}}{\text{d}\theta}(\theta)+\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta)\ .....(\text{i})$
And, $\text{y}=\text{a}(1+\cos\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(0-\sin\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sin\theta\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1+\cos\theta)}$
$=\frac{-\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\tan\theta}{2}$
View full question & answer→Question 3705 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\Bigg(\frac{{1}-\text{x}^{2}}{1+\text{x}^{2}}\Bigg), 0 <\text{x}<1$
AnswerThe given relationship is,$\text{y}=\sin^{-1}\Bigg(\frac{{1}-\text{x}^{2}}{1+\text{x}^{2}}\Bigg)$
$\text{y}=\sin^{-1}\Bigg(\frac{{1}-\text{x}^{2}}{1+\text{x}^{2}}\Bigg)$
$\Rightarrow\sin\text{y}=\frac{{1}-\text{x}^{2}}{1+\text{x}^{2}}$
Differentiating this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}\sin\text{y}=\frac{\text{d}}{\text{dx}}\Big[\frac{1-\text{x}^{2}}{1+\text{x}^{2}}\Big]$
$\cos\text{y}.\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)\cdot2\text{x}}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{1-\sin^2\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{1-\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^2}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{1-\frac{(1-\text{x}^2)^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{1-\frac{(1-\text{x}^2)^2-(1-\text{x}^2)^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{\frac{1+\text{x}^4+2\text{x}^2-1-\text{x}^4+2\text{x}^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\sqrt{\frac{4\text{x}^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{2\text{x}}{1+\text{x}^2}\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}^2}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{(1+\text{x}^2)2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{1+\text{x}^2}$
View full question & answer→Question 3715 Marks
Verify the Rolle’s theorem for each of the functions:
$f(x) = x(x - 1)^2$ in $[0, 1].$
AnswerConsider, $f ( x )= x ( x -1)^2$ in $[0,1]$
i. Since, $f(x)=x(x-1)^2$ is a polynomial function
So, it is continuous in $[0,1]$.
ii. Now, $f(0)=0$ and $f(1)=0 \Rightarrow f(0)=1$ f satisfies the above conditions of Rolle’s theorem.
Hence, by Rolle’s theorem $\exists\text{ c}\in(0,1)$ such that
$f(c) = 0$
$\Rightarrow 3c^2 - 4c + 1 = 0$
$\Rightarrow 3c^2 - 3c - c + 1 = 0$
$\Rightarrow 3c(c - 1) -1(c - 1) = 0$
$\Rightarrow (3c - 1)(c - 1) = 0$
$\Rightarrow\ \text{c}=\frac{1}{3},1\Rightarrow\frac{1}{3}\in(0,1)$
Thus, we see that there exists a real number c in the open interval $(0, 1).$
Hence, Rolle’s theorem has been verified.
View full question & answer→Question 3725 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$ is continuous at x = 0. find k.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)\dots(\text{i})$
Consider:
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}\Bigg)=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}\sin\text{x}}\Bigg)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\frac{2\text{k}^2}{4}\Big(\sin\frac{\text{kx}}{\text{x}}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\begin{pmatrix}\frac{\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\sin\frac{\text{kx}}{2}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2}}{\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\times1=\frac{\text{k}^2}{2}$
From equation (i), we have
$\frac{\text{k}^2}{2}=\text{f}(0)$
$\Rightarrow\frac{\text{k}^2}{2}=\frac{1}{2}$
$\Rightarrow\text{k}=\pm1$
View full question & answer→Question 3735 Marks
If $\text{x}=\cos\text{t}(3-2\cos^2\text{t}),\text{y}\sin\text{t}(3-2\sin^2\text{t})$ find the value of $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{4}$
Answer$\text{x}=\cos\text{t}(3-2\cos^2\text{t})\text{ and }\text{y}\sin\text{t}(3-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}(3-2\cos^2\text{t})+\cos\text{t}(4\cos\text{t}\sin\text{t})$ and $\frac{\text{dy}}{\text{dx}}=\cos\text{t}(3-2\sin^2\text{t})+\sin\text{t}(-4\sin\text{t}\cos\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+6\sin\text{t}\cos^2\text{t}$ and $ \frac{\text{dy}}{\text{dx}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sin\text{t}+(1-2\cos^2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\sin\text{t}\cos2\text{t}\cos2\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\cos\text{t}(1-2\sin^2\text{t})$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\cos\text{t}\cos2\text{t}}{3\sin\text{t}\cos2\text{t}}=\cot\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1$
View full question & answer→Question 3745 Marks
Is the function f defined by
$\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$
continuous at x = 0? At x = 1? At x = 2?
AnswerHere $\text{f(x)} = \begin{cases}\text{x}, \text{if}\ \text{x}\leq1\\5, \text{if}\ \text{x} > 1\end{cases}$At x = 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 - \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^-]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0-\text{h})$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(-h)} = (0) = 0$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{(x)}\ [ \text{Put}\ \text{x} = 0 + \text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow 0\ \text{as}\ \text{x}\rightarrow 0^+]$
$=\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(0 + \text{h})$
= 0 + 0 = 0
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = 0$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\text{f(x)} = 0$
Also f(0) = value of x at x = 0
= 0
$\therefore$ f is continous at x = 0.
At x = 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{x}\ [ \text{Put}\ \text{x} = 1 - \text{h}, \text{h}>0,\ \text{so that}\ \text{h}\rightarrow 0\ \text{on}\ \text{x}\rightarrow 1^-]$
$^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}(1 - \text{h}) = 1 - 0 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}(5) = 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}\neq\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)}$ does not exist
$\therefore$ f is discontinuous at x =1.
At x = 2
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}(5) = 5$
Also f(2) = 5
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}}\text{f(x)} = \text{f(2)} = 5$
$\therefore$ f is continous at x = 2.
View full question & answer→Question 3755 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$x^5 + y^5 = 5xy$
AnswerWe Heve, $x^5 + y^5 = 5xy$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^5\big)+\frac{\text{d}}{\text{dx}}\big(\text{y}^5\big)=\frac{\text{d}}{\text{dx}}\big(5\text{xy}\big)$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{dy}}{\text{dx}}\big(\text{x}\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\big(1\big)\Big]$
$\Rightarrow5\text{x}^4+5\text{y}^4\frac{\text{dy}}{\text{dx}}=5\text{x}\frac{\text{dy}}{\text{dx}}+5\text{y}$
$\Rightarrow5\text{y}^4\frac{\text{dy}}{\text{dx}}-5\text{x}\frac{\text{dy}}{\text{dx}}=5\text{y}-5\text{x}^4$
$\Rightarrow5\frac{\text{dy}}{\text{dx}}\big(\text{y}^4-\text{x}\big)=5\big(\text{y}-\text{x}^4\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{5(\text{y}-\text{x}^4)}{5(\text{y}^4-\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}^4}{\text{y}^4-\text{x}}$
View full question & answer→Question 3765 Marks
Differentiate $\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$ with respect to x:
AnswerWe have $\text{y}=\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\Big\{\frac{2\times6^\text{x}}{1+6^{2\text{x}}}\Big\}$
Put $6^\text{x}=\tan\theta$
$\Rightarrow \theta=\tan^{-1}(6^\text{x})$
Now, $\text{y}=\sin^{-1}\Big\{\frac{2\tan\theta}{1+\tan^2\theta}\Big\}$
$\Rightarrow \text{y}=\sin^{-1}\big\{\sin2\theta\big\}$
$\Rightarrow \text{y}=2\theta$
$\Rightarrow \text{y}=2\tan^{-1}(6^\text{x})$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=2\times\frac{1}{(6^\text{x})^2}\times6^\text{x}\log6$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2(\log6)6^\text{x}}{36^\text{x}}$
View full question & answer→Question 3775 Marks
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
AnswerConsider $\text{y}=\cos(\log\text{ x})^2$
Differentiating it with respect to x and applying the chain and the product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}+\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}$
$=\frac{1}{2}(\text{x}+1)^\frac{-1}{2}+\frac{1}{2}(\text{x}-1)^\frac{-1}{2}$
$=\frac{1}{2}\Big(\frac{1}{\sqrt{\text{x}+1}}\frac{1}{\sqrt{\text{x}-1}}\Big)$
$=\frac{1}{2}\bigg(\frac{\sqrt{\text{x}-1}+\sqrt{\text{x}+1}}{\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{y}}{\sqrt{\text{x}^2-1}}\Big)$
So,
$\sqrt{\text{x}^2}-\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}$
View full question & answer→Question 3785 Marks
If $\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$ find whethe f is continuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$
We observe
$(\text{LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{e}^\frac{-1}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\bigg(\frac{1}{\text{e}^{\frac{1}{\text{h}}}}\bigg)=\frac{1}{\lim\limits_{\text{h} \rightarrow 0}\text{e}^{\frac{1}{\text{h}}}}=0$
$(\text{RHL at x}=0)=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}(\text{x})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0^+}\text{e}^\frac{1}{\text{h}}=\infty$
Given,
$\text{f}(0)=1$
It is known for a function f(x) to be continuous at x = a,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$
But here,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{0}^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 3795 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\},-1<\text{x}<1$
AnswerLet $\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\cos^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\cos^{-1}\Big\{\sqrt{\frac{2\cos^2\theta}{2}}\Big\}$
$\text{y}=\cos^{-1}\{\cos\theta\}$
Here, $-1<\text{x}<1$
$\Rightarrow\ -1<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\pi$
$\Rightarrow\ 0<\theta<\frac{\pi}{2}$
So, equation (i),
$\text{y}=\theta$
$\big[\text{Since}, \cos^{-1}(\cos\theta)=\theta\text{ if }\theta\in[0,\pi]\big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 3805 Marks
Differentiate the following functions with respect to x:
$\sin(\text{x}^\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Taking log on both sides,
$\log(\sin^{-1}\text{y})=\log\text{x}^\text{x}$
$\Rightarrow\ \log(\sin^{-1}\text{y})=\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\frac{\text{dy}}{\text{dy}}(\sin^{-1}\text{y})=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\sin^{-1}\text{y}}\times\Big(\frac{1}{\sqrt{1-\text{y}^2}}\Big)\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{y}\sqrt{1-\text{y}^2}(1+\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin^{-1}(\sin\text{x}^\text{x})\sqrt{1-(\sin\text{x}^\text{x})^2}(1+\log\text{x})$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\cos\text{x}^\text{x}(1+\log\text{x})$
[Using equation (i)]
View full question & answer→Question 3815 Marks
Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.
AnswerA function f is a differentiable function if and only if $\text{Lf (c)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a}-\text{h})-\text{f}(\text{a})}{-\text{h}}$ and $\text{Rf}'(\text{c})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{a}-\text{h})-\text{f(a)}}{\text{h}}$ are equal.
Consider, f(x) = |x - 5|
$\therefore\ \text{f(x)}=\begin{cases}-(\text{x}-5),&\text{if x}<5\\\text{x}-5,&\text{if x}\geq5\end{cases}$
For continuity at x = 5,
$\text{L.H.L}=\lim\limits_{\text{x}\rightarrow5^-}(-\text{x}+5)$
$=\lim\limits_{\text{h}\rightarrow0}\big[-(5-\text{h})+5\big]=\lim\limits_{\text{h}\rightarrow0}\text{h}=0$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow5^+}(\text{x}-5)$
$=\lim\limits_{\text{h}\rightarrow0}(5+\text{h}-5)=\lim\limits_{\text{h}\rightarrow0}\text{h}=0$
$\text{f}(5)=5-5=0$
$\Rightarrow\ \text{L.H.L}=\text{R.H.L}=\text{f}(5)$
Hence, f(x) is continuous at x = 5
Now, $\text{Lf}'(5)=\lim\limits_{\text{x}\rightarrow5^-}\frac{\text{f(x)}-\text{f}(5)}{\text{x}-5}$
$=\lim\limits_{\text{x}\rightarrow5^-}\frac{-\text{x}+5-0}{\text{x}-5}=-1$
$\text{Rf}'(5)=\lim\limits_{\text{x}\rightarrow5^+}\frac{\text{f(x)}-\text{f}(5)}{\text{x}-5}$
$=\lim\limits_{\text{x}\rightarrow5^+}\frac{\text{x}-5-0}{\text{x}-5}=1$
Lf'(5) ≠ Rf'(5)
So, f(x) = |x - 5| is not differentiable at x = 5.
View full question & answer→Question 3825 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
AnswerHere,
$\text{y}=\text{x}^{\text{x}}+\text{x}^\frac{1}{\text{x}}$
$=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{x}^\frac{1}{\text{x}}}$
$\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\big(\frac{1}{\text{x}}\log\text{x}\big)}$
$\big[\text{Since, e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{e}^{\frac{1}{\text{x}}\log\text{x}}\Big)$
$=\text{e}^{\text{x}\log\text{x}}+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\frac{1}{\text{x}}\log\text{x}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\log\text{x}\Big)$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{x}^\frac{1}{\text{x}}}\Big[\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\frac{1}{\text{x}}\big)\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +\text{x}^\frac{1}{\text{x}}\Big[\Big(\frac{1}{\text{x}}\Big)\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(-\frac{1}{\text{x}^2}\Big)\Big]$
$=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\Big(\frac{1}{\text{x}^2}-\frac{1}{\text{x}^2}\log\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}[1+\log\text{x}]+\text{x}^{\frac{1}{\text{x}}}\frac{(1-\log\text{x})}{\text{x}^2}$
View full question & answer→Question 3835 Marks
If $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big),\text{x}>0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\sec^{-1}\Big(\frac{1+\text{x}^2}{1-\text{x}^2}\Big)$
$\Rightarrow \text{y}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Put $\text{x}=\tan\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)+\cos^{-1}\Big(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\Big)$
$\Rightarrow \text{y}=\tan^{-1}(\tan2\theta)+\cos^{-1}(\cos2\theta)$
$\Rightarrow \text{y}=2\theta+2\theta$
$\Rightarrow \text{y}=4\theta$
$\Rightarrow \text{y}=4\tan^{-1} \text{x} \big[\text{using, x}=\tan\theta\big]$
Differentiate it with respect to x,
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{4}{1+\text{x}^2}$
View full question & answer→Question 3845 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
AnswerWe have, $\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}+\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\cos\text{x}\log\sin\text{x}}+\text{e}^{\sin\text{x}\log\cos\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}\log\cos\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}\big(\cos\text{x}\log\sin\text{x})+\text{e}^{\sin\text{x}\log\cos\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\cos\text{x})$
$=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}\Big[\sin\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\Big[\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\times(-\sin\text{x})\Big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\sin\text{x}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\times(\cos\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\tan\text{x}(-\sin\text{x})+\cos\text{x}\log\cos\text{x}\big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}\big]$
View full question & answer→Question 3855 Marks
Find the points on the curve $\text{y}=(\cos\text{x}-1)$ in $[0, 2\pi],$ where the tangent is parallel to x-axis.
AnswerThe equation of the curve is $\text{y}=\cos\text{x}-1.$
Now, we have to find a point on the curve in $[0,2\pi].$
where the tangent is parallel to X-axis i.e., the tangent to the curve at x = c has a slope o,
where $\text{c}\in[0,2\pi].$
Let us apply Rolle’s theorem to get the point.
- $\text{y}=\cos\text{x}-1$ is a continuous function in $[0,2\pi].$
[since it is a combination of cosine function and a constant function]
- $\text{y}'=-\sin\text{x},$ which exists in $(0,2\pi).$
Hence, y is differentiable in $(0,2\pi).$
- $\text{y}(0)=\cos0-1=0$ and $\text{y}(2\pi)=\cos2\pi-1=0$
$\therefore\ \text{y}(0)=\text{y}(2\pi)$
Since, conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that
$\text{f}'(\text{c})=0$
$\Rightarrow-\sin\text{c}=0$
$\Rightarrow\ \text{c}=\pi$ or 0, where $\pi\in(0,2\pi)$
$\Rightarrow\ \text{x}=\pi$
$\therefore\ \text{y}=\cos\pi-1=-2$
Hence, the required point on the curve, where the tangent drawn is parallel to the X-axis is $(\pi,2).$ View full question & answer→Question 3865 Marks
Find A and B so that $\text{y}=\text{A}\sin3\text{x}+\text{B}\cos3\text{x}$ satisfy the equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\frac{\text{dy}}{\text{dx}}+3\text{y}=10\cos3\text{x}.$
Answer$\text{y}=\text{Ae}^{-kt}\cos(\text{pt}+\text{c})$ Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\text{A}\Big\{\text{e}^{-\text{kt}}(-\sin(\text{pt}+\text{c})\times\text{p})+(\cos(\text{pt}+{c}))(-\text{re}^{-\text{kt}})\Big\}$
$\Rightarrow-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}$ Differentiating w.r.t.x, $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{Ap}\Big\{\text{e}^{-\text{kt}}(\cos(\text{pt}+\text{c})\times\text{p})+(\sin(\text{pt}+\text{c}))(\text{e}^{-\text{kt}}\times-\text{R})-\text{ky}^1\Big\}=-\text{p}^2\text{y}+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}^1$
Adding & substracting $ky_1$ on RHS $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1+\text{ky}^1$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1-\text{k}\text{ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{k}^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-(\text{p}+\text{k}^2)\text{y}-2\text{k}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0$
View full question & answer→Question 3875 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
AnswerLet $\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}-\text{a}}{\text{x}}}{\frac{\text{x}+\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{\frac{\text{x}}{\text{x}}-\frac{\text{x}}{\text{x}}}{\frac{\text{x}}{\text{x}}+\frac{\text{a}}{\text{x}}}\bigg)$
$=\tan^{-1}\bigg(\frac{1-\frac{\text{x}}{\text{x}}}{1+1\times\frac{\text{a}}{\text{x}}}\bigg)$
$\text{y}=\tan^{-1}(1)-\tan^{-1}\big(\frac{\text{a}}{\text{x}}\big)$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{1+\big(\frac{\text{a}}{\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\big(\frac{\text{a}}{\text{x}}\big)$
$=-\frac{\text{x}^2}{\text{x}^2+\text{a}^2}\Big(\frac{-\text{a}}{\text{x}^2}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}}{\text{a}^2+\text{x}^2}$
View full question & answer→Question 3885 Marks
Differentiate the following functions with respect to x:
$(\text{x}^\text{x})\sqrt{\text{x}}$
AnswerLet $\text{y}=(\text{x}^\text{x})\sqrt{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\text{x}^\text{x}\sqrt{\text{x}})$
$\log\text{y}=\text{x}\log\text{x}+\frac{1}{2}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)+\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1+\log\text{x}+\frac{1}{2\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}\sqrt{\text{x}}\Big[1+\log\text{x}+\frac{1}{2\text{x}}\Big]$
[Using equation (i)]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}+\frac{1}{2}}\Big[\Big(\frac{2\text{x}+1}{2\text{x}}\Big)+\log\text{x}\Big]$
View full question & answer→Question 3895 Marks
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big), \frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\tan^{-1}\Big[\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big]$
$=\tan\bigg[\frac{\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\bigg[\frac{\frac{\cos\text{x}}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}}\bigg]$
$=\tan^{-1}\Big[\frac{1+\tan\text{x}}{1-\tan\text{x}}\Big]$
$=\tan^{-1}\bigg[\frac{\frac{\tan\pi}{4}+\tan\text{x}}{1-\frac{\tan\pi}{4}\tan\text{x}}\bigg]$
$=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\text{x}\Big)\Big]$
$\text{y}=\frac{\pi}{4}+\text{x}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+1$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 3905 Marks
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
Answerconsider the given function,
$\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1$
Differentiating both sides w.r.t. 'x' we get
$-\sin\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(-\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}\Big)+\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos(\text{a}+\text{y})}{\text{x}\sin(\text{a}+\text{y})-\sin\text{y}}$
Multiplying the numerator and the denominator
by $\cos(\text{a}+\text{y})$ on the R.H.S., we have,
$\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$\big[\because\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{given function}\big]$
$=\frac{\cos^2(\text{a}+\text{y})}{\sin\big[(\text{a}+\text{y})-\text{y}\big]}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
View full question & answer→Question 3915 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
AnswerLet, $\text{y}=\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\Big]$
$=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)$
[Using chain rule and quotient rule]
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{\text{x}^2-\text{x}+1\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^3-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+\text{x}+1)(\text{x}^2+\text{x}+1)}$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+1)^2-(\text{x})^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+1+2\text{x}^2-\text{x}^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\Big\}=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
View full question & answer→Question 3925 Marks
If $\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
Answer$\text{x} =\text{a} (\theta -\sin\theta)\dots\text{ eq. } 1$
$\text{y}=\text{a}(1+\cos\theta)\dots\text{ eq. 2}$
To 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, lets first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and differentiate it again.
$\frac{\text{dx}\text{}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(\theta-\sin\theta)=\text{a}(1-\cos\theta)\dots\ \text{eq. 3}$
Similarly,
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}(1+\cos\theta)=-\text{a}\sin\theta\dots\text{ eq. }4$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}=\frac{-\sin\theta}{(1-\cos\theta)}\dots\ \text{eq. }5$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=-\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\theta}{1-\cos\theta}\Big)$
Using product rule and chain rule of differentiation together:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{-\frac{1}{1-\cos\theta}\frac{\text{d}}{\text{d}\theta}\sin\theta-\sin\theta\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}\Big\}\frac{\text{d}\theta}{\text{dx}}$
Apply chain rule to determine $\frac{\text{d}}{\text{d}\theta}\frac{1}{(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}}=\Big\{\frac{-\cos\theta}{1-\cos\theta}+\frac{\sin^2\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}$ [using eq.3]
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta(1-\cos\theta)+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\bigg\{\frac{-\cos\theta+\cos^2\theta+\sin^2\theta}{(1-\cos\theta)^2}\bigg\}\frac{1}{\text{a}(1-\cos\theta)}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big\{\frac{1-\cos\theta}{(1-\cos\theta)^2}\Big\}\frac{1}{\text{a}(1-\cos\theta)}[\because\cos^2\theta+\sin^2\theta=1]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a(1}-\cos\theta)^2}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{a}\Big(2\sin^2\frac{\theta}{2}\Big)^2}\big[\because-\cos\theta=2\sin^2\frac{\theta}{2}\big]$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{4a}}\text{cosec}^4\frac{\theta}{2}$
View full question & answer→Question 3935 Marks
Verify Rolle's theorem for the following function on the indicated intervals$f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
AnswerHere,$f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
f(x) is continuous is [-1, 2] and differentiable in (-1, 2) as it is a polynomial functions.
Now,
$f(-1) = (1-1)(-1-2) = 0$
$f(2) = (4-1)(2-2) = 0$
$\Rightarrow f(-1) = f(2)$
So, Rolle's theorem is applicable on $f(x) is [-1, 2]$ therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that f'(c) = 0
Now,
$f(x) = (x^2- 1)(x - 2)$
$f'(x) = 2x(x - 2) + (x^2 - 1)$
$= 2x^2 - 4 + x^2 - 1$
$f'(x) = 3x^2 - 5$
Now,
$f'(c) = 0$
$\Rightarrow 3x^2 - 5 = 0$
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
View full question & answer→Question 3945 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\},-\frac{3\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\bigg\{\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\bigg\}$
$=\sin^{-1}\Big\{\sin{\text{x}}\cos\frac{\pi}{4}+\cos\text{x}\times\sin\frac{\pi}{4}\Big\}$
$\text{y}=\sin^{-1}\Big\{\sin\Big(\text{x}+\frac{\pi}{4}\Big)\Big\}$
Here, $\frac{-3\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\Big(\frac{-3\pi}{4}+\frac{\pi}{4}\Big)$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta, \text{ if }\theta\in\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=1+0$
$\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 3955 Marks
If $x^{16}y^9 = (x + y)^{17}$, prove that $\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
AnswerHere,$x^{16}y^9 = (x + y)^{20}$
Taking log on both the siede,
$\log(\text{x}^{16}\times\text{y}^{19})=\log(\text{x}^2+\text{y})^{17}$
$\big[\text{since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$16\log\text{x}+9\log\text{y}=17\log(\text{x}^2+\text{y})$
Differentiating it with respect to x using chain rule
$16\frac{\text{d}}{\text{dx}}(\log\text{x})+9\frac{\text{d}}{\text{dx}}(\log\text{y})=17\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=17\frac{1}{(\text{x}^2+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y})$
$\frac{16}{\text{x}}+\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{17}{\text{x}^2+\text{y}}\Big[2\text{x}+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{9}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{17}{(\text{x}^2+\text{y})}\frac{\text{dy}}{\text{dx}}=\Big(\frac{34\text{x}}{\text{x}^2+\text{y}}\Big)-\frac{16}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9}{\text{y}}-\frac{17}{(\text{x}^2+\text{y})}\Big]=\frac{34\text{x}^2-16\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{9\text{x}^2+9\text{y}-17\text{y}}{\text{y}(\text{x}^2+\text{y})}\Big]=\frac{18\text{x}^2-16\text{y}}{\text{x}(\text{x}^2+\text{y})}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\frac{2(9\text{x}^2-8\text{y})}{9\text{x}^2-8\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}$
View full question & answer→Question 3965 Marks
If $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{x}=10(\text{t}-\sin\text{t}),\text{y}=12(1-\cos\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=10(1-\cos\text{t})\ ...(\text{i})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=12(\sin\text{t})\ ...(\text{ii})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{12(\sin\text{t})}{10(1-\cos\text{t})}$ From equation (i) and (ii)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{12\sin\frac{\text{t}}{2}\cdot\cos\frac{\text{t}}{2}}{10\sin^2\frac{\text{t}}{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6}{5}\cot\frac{\text{t}}{2}$
View full question & answer→Question 3975 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$
AnswerHere,
$\text{y}=(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$=\text{e}^{\log(\sin\text{x})^\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\text{y}=\text{e}^{\text{x}\log\sin\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\big[\text{Since},\log_\text{e}^\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentitating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\sin^{-1}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})+\frac{1}{\sqrt{1-\big(\sqrt{\text{x}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\Big]$
$=(\sin\text{x})^\text{x}\Big[\text{x}\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(1)\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$=(\sin\text{x})^\text{x}\Big[\frac{\text{x}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\text{x}\cot\text{x}+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
View full question & answer→Question 3985 Marks
Differentiate the following functions with respect to x:
$\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
AnswerLet, $\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
$=\frac{1}{\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)$
[Using chain rule]
$=\Big(\frac{1+\cos\text{x}}{\sin\text{x}}\Big)\Bigg[\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})-\sin\text{x}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}\Bigg]$
[Using quotient rule]
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\bigg[\frac{(1+\cos\text{x})(\cos\text{x})-\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}\bigg]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]$
$=\frac{(1+\cos\text{x})}{\sin\text{x}}\Big[\frac{(1+\cos\text{x})}{(1+\cos\text{x})^2}\Big]$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec x}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\Big(\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\Big)=\text{cosec x}$
View full question & answer→Question 3995 Marks
Differentiate the following functions from first principles:
$\log\text{cosec x}$
AnswerLet $\text{f(x)}=\log\text{cosec x}$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\log\text{cosec x}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\text{cosec}(\text{x}+\text{h})-\log\text{cosec x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{\frac{\text{cosec}(\text{x}+\text{h})}{\text{cosec x}}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}}{\sin(\text{x}+\text{h})}-1\Big)\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\begin{Bmatrix}\frac{\log\Big\{1+\Big(\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big)\Big\}}{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}} \end{Bmatrix}\frac{\Big\{\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}\Big\}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{\text{x}+\text{x}+\text{h}}{2}\Big)\sin\Big(\frac{\text{x}-\text{x}-\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})\text{h}}$
$\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}=1\text{ and }\sin\text{A}-\sin\text{B} \\ =2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{\sin(\text{x}+\text{h})(-2)}\bigg\{\frac{\sin\big(-\frac{\text{h}}{2}\big)}{-\frac{\text{h}}{2}}\bigg\}$
$=-\cot\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}(\log\text{cosec x})=-\cot\text{x}$
View full question & answer→Question 4005 Marks
Examine the differentiability of f, where f is defined by:
$\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x}=0\end{cases}$
at x = 0.
AnswerWe have, $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x}=0\end{cases}$ at x = 0.
For differentiability at x = 0,
$\text{Lf}'(0)=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}^2\sin\frac{1}{\text{x}}-0}{\text{x}-0}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(0-\text{h})^2\sin\Big(\frac{1}{0-\text{h}}\Big)}{0-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2\sin\Big(\frac{-1}{\text{h}}\Big)}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}+\text{h}\sin\Big(\frac{1}{\text{h}}\Big)\ [\because\sin(-\theta)=-\sin\theta]$
= 0 × [an oscillating number between -1 and 1] = 0
$\text{Rf}'(0)=\lim\limits_{\text{x}\rightarrow0^+}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^+}\frac{\text{x}^2\sin\frac{1}{\text{x}}-0}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(0+\text{h})^2\sin\Big(\frac{1}{0+\text{h}}\Big)}{0+\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2\sin\big(\frac{1}{\text{h}}\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\sin\Big(\frac{1}{\text{h}}\Big)$
= 0 × [an oscillating number between -1 and 1] = 0
$\because$ Lf'(0) = Rf'(0)
So, f(x) is differentiable at x = 0.
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