Question 1515 Marks
Differentiate $(x^2– 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
by logarithmic differentiation.
Answer$y = (x^2 - 5x + 8) (x^3 + 7x + 9)$
$\Rightarrow\ \log\text{y}=\log(\text{x}^2-5\text{x}+8)+\log(\text{x}^3+7\text{x}+9)$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\log(\text{x}^2-5\text{x}+8)+\frac{\text{d}}{\text{dx}}\log(\text{x}^3+7\text{x}+9)$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2-5\text{x}+8}\frac{\text{d}}{\text{dx}}(\text{x}^2-5\text{x}+8)+\frac{1}{\text{x}^3+7\text{x}+9}\frac{\text{d}}{\text{dx}}(\text{x}^3+7\text{x}+9)$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2-5\text{x}+8}(2\text{x}-5)+\frac{1}{\text{x}^3+7\text{x}+9}(3\text{x}^2+7)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{2\text{x}-5}{\text{x}^2-5\text{x}+8}+\frac{3\text{x}^2+7}{\text{x}^3-7\text{x}+9}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{(2\text{x}-5)(\text{x}^3+7\text{x}+9)+(3\text{x}^2+7)(\text{x}^2-5\text{x}+8)}{(\text{x}^2-5\text{x}+8)(\text{x}^3+7\text{x}+9)}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{2\text{x}^4+14\text{x}^2+18\text{x}-5\text{x}^3-35\text{x}-45+3\text{x}^4-15\text{x}^3+24\text{x}^2+7\text{x}^2-35\text{x}+56}{(\text{x}^2-5\text{x}+8)(\text{x}^3+7\text{x}+9)}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{5\text{x}^4-20\text{x}^3+45\text{x}^2-52\text{x}+11}{(\text{x}^2-5\text{x}+8)(\text{x}^3+7\text{x}+9)}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\text{x}^2-5\text{x}+8)(\text{x}^3+7\text{x}+9)\Big[\frac{5\text{x}^4-20\text{x}^3+45\text{x}^2-52\text{x}+11}{(\text{x}^2-5\text{x}+8)(\text{x}^3+7\text{x}+9)}\Big]\ \text{[From eq.(i)}]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=5\text{x}^4-20\text{x}^3+45\text{x}^2-52\text{x}+11\ \dots\text{(iv)}$
From eq. (ii), (iii) and (iv), we can say that value of $\frac{\text{dy}}{\text{dx}}$ is same obtained by three different methods
View full question & answer→Question 1525 Marks
Discuss the continuity of the function f, where f is defined by:
$\text{f(x)}= \begin{cases}-2,\ \text{if}\ \text{x}\leq-1 \\\text{2x},\text{if}\ -1<\text{x}\leq1\\2,\text{if}\ \text{x}>1\end{cases}$
Answer$\text{f(x)}= \begin{cases}-2,\ \text{if}\ \text{x}\leq-1 \\\text{2x},\text{if}\ -1<\text{x}\leq1\\2,\text{if}\ \text{x}>1\end{cases}$The function f is defined at all points of the real line.
When x < -1, we have f(x) = -2, which it is constant and so is continuous.
At x = -1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}(-{2}) = -2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}(\text{2x}) = 2(-1)= -2$
Also f(-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 x = -1
in the interval -1 < x < 1, we have f(x) = 2 x, which being a linear polynomial, is continuous.
At x = -1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{-}}(\text{2x}) = 2(1)= 2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{-1}^{+}}({2}) = 2$
Also f(1) = 2(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 x = 1
When x > 1, we have f(x) = 2, which is constant and so it is continuous.
View full question & answer→Question 1535 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}$
AnswerHere,
$\text{y}=\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x})$
$\log\text{y}=\log\sin\text{x}+\log\sin2\text{x}+\log\sin3\text{x}+\log\sin4\text{x}$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\frac{\text{d}}{\text{dx}}\log\sin3\text{x}+\frac{\text{d}}{\text{dx}}\log\sin4\text{x}$
$=\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\frac{1}{\sin3\text{x}}\frac{\text{d}}{\text{dx}}(\sin3\text{x})+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4 \text{x})$
$=\frac{1}{\sin\text{x}}(\cos\text{x})+\frac{1}{\sin2\text{x}}(\cos2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x}) \\ +\frac{1}{\sin3\text{x}}(\cos3\text{x})\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\big[\cot\text{x}+\cot2\text{x}(2)+\cot3\text{x}(3)+\cot4\text{x}(4)\big]$
$\frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}) \\ \big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
[Using equation (i)]
View full question & answer→Question 1545 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
AnswerLet $\text{x}-1=\text{y}$
$\Rightarrow\text{x}=\text{y}+1$
Thus,
$\lim_\limits{\text{x}\rightarrow 1}(\text{x}-1)\tan\frac{\pi\text{x}}{2}=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\frac{\pi(\text{y}+1)}{2}$
$=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\Big(\frac{\pi\text{y}}{2}+\frac{\pi}{2}\Big)$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\cot\frac{\pi\text{y}}{2}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\sin\frac{\pi\text{y}}{2}}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{2}{\pi}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\frac{2}{\pi}\lim_\limits{\text{y}\rightarrow0}\cos\frac{\pi\text{y}}{2}$
$=-\frac{2}{\pi}$
Since the function is continuous, LHL = RHL.
Thus, $\text{k}=-\frac{2}{\pi}$
View full question & answer→Question 1555 Marks
Find the value of a and b so that the function f given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$ is continuous x = 3 and x = 5.
AnswerGiven,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}(1)=1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3+\text{h})+\text{b}=3\text{a}+\text{b}$
$(\text{LHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{a}(5-\text{h})+\text{b})=5\text{a}+\text{b}$
$(\text{RHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}7=7$
If f(x) is continuous at x = 3 and 5, then
$\therefore\ \lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow1=3\text{a}+\text{b}\ .... (\text{i})$ and $5\text{a}+\text{b}=7\ .... (\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=3$ and $\text{b}=-8$
View full question & answer→Question 1565 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 - 1)(x - 2)(x - 3) on [0, 4]
Answerf(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
We know that, polynomial function is continuous and differentiable everywhere. So, f(x) is continuous in [0, 4] and differentiable in (0, 4). So Lagrange's mean value theorem is applicable. Thus, there exist a point $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow(\text{c}-1)(\text{c}-2)(\text{c}-3)+(\text{c}-1)(\text{c}-3)\\=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4-1}$
$\Rightarrow\text{c}^2-3\text{c}+2+\text{c}^2+5\text{c}+6+\text{c}^2-4\text{c}+3=\frac{6+6}{4}$
$\Rightarrow3\text{c}^2-12\text{c}+11=3$
$\Rightarrow3\text{c}^2-12\text{c}+8=0$
$\Rightarrow\text{c}=\frac{-(-12)\pm\sqrt{144-4\times3\times8}}{6}$
$\Rightarrow\text{c}=\frac{12\pm\sqrt{48}}{6}$
$\Rightarrow\text{c}=2\pm\frac{2\sqrt3}{3}\in(0,4)$
$\Rightarrow\text{c}=2\pm\frac{2}{\sqrt3}\in(0,4)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1575 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
AnswerGiven: $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}\ \Rightarrow\ \log(\cos\text{x})^\text{y}=\log(\cos\text{y})^\text{x}$
$\Rightarrow\ \text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}\ \Rightarrow\ \frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{y})$
$\Rightarrow\ \text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{y}+\log\cos\text{y}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\frac{\text{d}}{\text{dx}}\cos\text{y}+\log\cos\text{y}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\Big(-\sin\text{y}\frac{\text{dy}}{\text{dx}}\Big)+\log\cos\text{y}$
$\Rightarrow\ -\text{y}\tan\text{x}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=-\text{x}\tan\text{y}.\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\ \text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}(\text{x}\tan\text{y}+\log\cos\text{x})=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\tan\text{x}+\log\cos\text{y}}{\text{x}\tan\text{y}+\log\cos\text{x}}$
View full question & answer→Question 1585 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
Since $\sin\text{x}$ and $\cos\text{x}$ are everywhere continuous and differentiable $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big).$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditionss of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
$\Rightarrow\text{f}'(\text{x})=4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\sin^3\text{x}\cos\text{x}-\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\tan^3\text{x}-\tan\text{x}=0$
$\Rightarrow\tan\text{x}(\tan^2\text{x}-1)=0$
$\Rightarrow\tan\text{x}=0,\tan^2\text{x}=1$
$\Rightarrow\tan\text{x}=0,\tan\text{x}=\pm1$
$\Rightarrow\text{x}=0,\text{x}=\frac{\pi}{4},\frac{3\pi}{4}$
Since $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c)=0
Hence, Rolle's theorem is verified.
View full question & answer→Question 1595 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sec^{-1}\bigg(\frac{1}{2\text{x}^{2}-1}\bigg), 0<\text{x}<\frac{1}{\sqrt{2}}$
AnswerThe given relationship is,$\text{y}=\sec^{-1}\bigg(\frac{1}{2\text{x}^{2}-1}\bigg)$
$\text{y}=\sec^{-1}\bigg(\frac{1}{2\text{x}^{2}-1}\bigg)$
$\Rightarrow\sec\text{y}=\frac{1}{2\text{x}^{2}-1}$
$\Rightarrow\cos\text{y}=2\text{x}^{2}-1$
$\Rightarrow2\text{x}^{2}=1+\cos\text{y}$
$\Rightarrow2\text{x}^{2}=2\cos^{2}\frac{\text{y}}{2}$
$\Rightarrow\text{x}=\cos\frac{\text{y}}{2}$
Differentiating this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\cos\frac{\text{y}}{2}\bigg)$
$\Rightarrow1=-\sin\frac{\text{y}}{2}.\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{y}}{2}\bigg)$
$\Rightarrow \frac{-1}{\sin\frac{\text{y}}{2}}=\frac{1}{2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{\sin\frac{\text{y}}{2}}=\frac{-2}{\sqrt{1-\cos^{2}\frac{\text{y}}{2}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{\sqrt{1-\text{x}^{2}}}$
View full question & answer→Question 1605 Marks
If $\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}$
AnswerHere,
$\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}}$
$\text{y}=(\cos\text{x})^\text{y}$
Taking log on both the sides,
$\log\text{y}=\log(\cos\text{x})^\text{y}$
$\log\text{y}=\text{y}\log(\cos\text{x}),\big\{\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big\}$
Differentiating it with resepect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\log(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{1}{\cos\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\cos\text{x}\Big)=\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\log\cos\text{x}}{\text{y}}\Big)=-\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{(1-\log\cos\text{x})}$
View full question & answer→Question 1615 Marks
If $\text{f(x)}=\sqrt{\text{x}^2+9},$ Write the value of $\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}.$
AnswerGiven: $\text{f(x)}=\text{x}^2+9$
Now,
$\text{f}(4)=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
So,
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}$
On rationalising the numeratore, we get
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}$
$=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}\times\frac{\sqrt{\text{x}^2+9}+5}{\sqrt{\text{x}^2+9}+5}$
$=\frac{\text{x}^2+9-25}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}^2-16}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
Taking limit x → 4, we have
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
$=\frac{8}{10}$
$=\frac{4}{5}$
View full question & answer→Question 1625 Marks
Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Differential at x = 0, if m > 1
AnswerLet m = 2, then the function $\text{f(x)}=\begin{Bmatrix}\text{x}^2\sin(\frac{1}{\text{x}}) &\text{x}\neq0 \\0 & \text{x}=0 \end{Bmatrix}$
Differentiability at x = 0:
$\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f(0)}}{\text{x}-0}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}}=\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)=0.$
$\big[\therefore\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)\text{x}=0,$ as
$\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}}-0\end{vmatrix}=\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}=\begin{vmatrix}\text{x} \end{vmatrix}\begin{vmatrix}\sin\frac{1}{\text{x}} \end{vmatrix} \leq\begin{vmatrix}\text{x}\end{vmatrix}$
$\because\ |\sin\theta|\leq1\ \text{for all }\theta$
Hence, $\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}<0\ \text{when}\ |\text{x-0}|<\in|\text{x}-0|<\in\big]$
Therefore, f'(x) = 0, which means f is differentiable at x = 0.
Hence the given function is differentiable at x = 0.
View full question & answer→Question 1635 Marks
If $\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that
AnswerHere,
$\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{a}+\text{a}\cos\theta}{-\text{a}\sin\theta}=\frac{1+\cos\theta}{-\sin\theta}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{d}\theta}\Big\{\frac{\text{dy}}{\text{dx}}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{-\sin\theta\cos\theta-\cos^2\theta}{\sin^2\theta}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{1+\cos\theta}{\sin^2\theta}\times\frac{-1}{\sin^2\theta}$
$\frac{-(1+\cos\theta)}{\sin^3\theta}$
At $\theta=\frac{\pi}{2}:\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-(1+\cos\frac{\pi}{2})}{\text{a}(\sin\frac{\pi}{2})^3}=\frac{-1}{\text{a}}$
View full question & answer→Question 1645 Marks
Verify mean value theorem for the function:
$\text{f(x)}=\sin\text{x}-\sin2\text{x in }[0,\pi].$
AnswerWe have, $\text{f(x)}=\sin\text{x}-\sin2\text{x in }[0,\pi]$
- Since, we know that sine functions are continuous functions hence
$\text{f(x)}=\sin\text{x}-\sin2\text{x}$ is a continuous function in $[0,\pi].$
- $\text{f}'(\text{x})=\cos\text{x}-\cos2\text{x}.2=\cos\text{x}-2\cos2\text{x},$ which exists in $(0,\pi)$
So, f(x) is differentiable in $(0,\pi).$ Continuous of mean value theorem are satisfied.
Hence, $\exists\text{ c}\in(0,\pi)$ such that, $\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}$
$\Rightarrow\ \cos\text{c}-2\cos2\text{c}=\frac{\sin\pi-\sin2\pi-\sin0+\sin2.0}{\pi-0}$
$\Rightarrow\ 2\cos2\text{c}-\cos\text{c}=\frac{0}{\pi}$
$\Rightarrow\ 2.(2\cos^2\text{c}-1)-\cos\text{c}=0$
$\Rightarrow\ 4\cos^2\text{c}-2-\cos\text{c}=0$
$\Rightarrow\ 4\cos^2\text{c}-\cos\text{c}-2=0$
$\Rightarrow\ \cos\text{c}=\frac{1\pm\sqrt{1+32}}{8}=\frac{1\pm\sqrt{33}}{8}$
$\therefore\ \text{c}=\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)$
Also, $\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)\in(0,\pi)$
Hence, mean value theorem has been verified. View full question & answer→Question 1655 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x(x - 1)^2 $on $[0, 1]$
AnswerGiven $f(x) = x(x - 1)^2$
$\Rightarrow f(x) = x(x^2 - 2x + 1)$
$\therefore$ $f(x) = (x^3- 2x^2+ x)$
We know that a polynominal function is everywhere derivable and hence continuous.
So, being a polynomial function, f(x) is continuous and derivable on $[0, 1]$
Also,
$f(0) = f(1) = 0$
Thus, all the continuous of Solids theorem are satisfied.
Now, we have to show that there exists $\text{c}\in(0,1)$ such that $f'(c) = 0.$
We have
$f(x) = x^3 -^2x^2 +^x$
$\Rightarrow f'(x) = 3x^2 - 4x + 1$
$\therefore$ $f'(x) = 0 \Rightarrow 3x^2- 4x + 1 = 0$
$\Rightarrow 3x^2- 3x - x + 1 = 0$
$\Rightarrow 3x(x - 1) - 1(x - 1) = 0$
$\Rightarrow (x - 1)(3x - 1) = 0$
$\Rightarrow\text{x}=1,\frac{1}{3}$
Thus, $\text{c}=\frac{1}{3}\in(0,1)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
View full question & answer→Question 1665 Marks
If $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^\text{x}}}\times\text{x}^{\text{e}^{\text{x}}}\Big\{\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{e}^{\text{e}^\text{x}}\Big\{\frac{1}{\text{x}}+\text{e}^\text{x}\times\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^{\text{e}}}\times\text{x}^{\text{e}-1}\Big\{\text{x}+\text{e}\log\text{x}\Big\}$
AnswerWe have, $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\text{y}=\text{u}+\text{v}+\text{w}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}\ .....(\text{i})$
Where $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}},\text{v}=\text{x}^{\text{e}^{\text{e}^{\text{x}}}}\text{ and w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
Now, $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\ .....(\text{ii})$
Taking log on both sides,
$\log\text{u}=\log\text{e}^{\text{x}^{\text{e}^{\text{x}}}}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\log\text{e}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\ .....(\text{iii})$
Taking $\log$ on both sides,
$\log\log\text{u}=\log\text{x}^{\text{e}^\text{x}}$
$\Rightarrow\log\log\text{u}=\text{e}^\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\log\text{u}}\frac{\text{d}}{\text{dx}}(\log\text{u})=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{1}{\log\text{u}}\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\log\text{u}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{A})$
[Using equation (ii) and (iii)]
Now, $\text{v}=\text{x}^{\text{e}^{\text{e}^\text{x}}}\ .....(\text{iv})$
Taking log on both sides,
$\log\text{v}=\log\text{x}^{\text{e}^{\text{e}^\text{x}}}$
$\Rightarrow\log\text{v}=\text{e}^{\text{e}^\text{x}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{e}^\text{x}}\big)$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\text{e}^\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{B})$
[Using equation (4)]
Now, $\text{w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}\ .....(\text{v})$
Taking log on sides,
$\log\text{w}=\log\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^\text{e}}\log\text{e}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^{\text{e}}}\ .....(\text{vi})$
Taking log on both sides,
$\log\log\text{w}=\log\text{x}^{\text{x}^{\text{e}}}$
$\Rightarrow\log\log\text{w}=\text{x}^{\text{e}}\log\text{x}$
$\Rightarrow\frac{1}{\log\text{w}}\frac{\text{d}}{\text{dx}} (\log\text{w})=\text{x}^\text{e}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^\text{e})$
$\Rightarrow\frac{1}{\log\text{w}}\big(\frac{1}{\text{w}}\big)\frac{\text{dw}}{\text{dx}}=\text{x}^{\text{e}}\big(\frac{1}{\text{x}}\big)\log\text{xex}^{\text{e}-1}$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{w}\log\text{w}\big[\text{x}^{\text{e}-1}+\text{e}\log\text{xx}^{\text{e}-1}\big]$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})\ .....(\text{C})$
Using equation (A), (B) and (C) in equation (i), we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})$
View full question & answer→Question 1675 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-4}{\text{|x}-4|}+\text{a}, &\text{if x} <4\\\text{a}+\text{b},&\text{if x}=4\\\frac{\text{x}-4}{\text{|x}-4|}+\text{b}, & \text{if x}>4\end{cases}$ is continuous at x = 4. Find a, b.
AnswerGiven,
f(x) is continuous at x = 4 & f(4) = a + b
For f(x) to be continuous at $x = 4, f(4)^- = f(4)^+= f(4)$
$\text{L.H.L}=\text{f(4)}^-=\lim\limits_{\text{x} \rightarrow 4}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h})-4}{|(4-\text{h})-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h}-4)}{|(4-\text{h}-4)|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{|(-\text{h})|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{\text{h}}+\text{a}$
$\Rightarrow\text{a}-1$
$\text{L.H.L}=\text{f(4)}^+=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}-4}{|\text{x}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4+\text{h})-4}{|(4+\text{h})-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{4+\text{h}-4}{|4+\text{h}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{|\text{h}|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{1}{|1|}$
$\Rightarrow1$
Since, f(x) is is continuous at x = 4 & f(4) = a + b
$\text{f(4)}^-=\text{f(4)}^+=\text{f(4)}$
$\therefore\ \text{a}-1=\text{a}+\text{b}=1$
$\Rightarrow\text{a}-1=1$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{a}+\text{b}=1$
$\Rightarrow\text{b}=1-2$
$\Rightarrow\text{b}=-1$
View full question & answer→Question 1685 Marks
If $\text{y}=\sin\Big[2\tan^{-1}\Big\{\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}\Big],$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\sin\Big[2\tan^{-1}\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big]\Big]$
Put, $\text{x}=\cos2\theta, \text{So},$
$\text{y}=\sin\Big[2\tan^{-1}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\Big]$
$=\sin\Big[2\tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\Big]$
$=\sin\big[2\tan^{-1}\sqrt{\tan^2\theta}\big]$
$=\sin\big[2\tan^{-1}(\tan\theta)\big]$
$=\sin{2\theta}$
$=\sin\Big[2\times\frac{1}{2}\cos^{-1}\text{x}\Big]\ \big[\text{Since, x}=\cos2\theta\big]$
$=\sin\big(\sin^{-1}\sqrt{1-\text{x}^2}\big)$
$\text{y}=\sqrt{1-\text{x}^2}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1695 Marks
If $\text{x}^\text{m}.\text{y}^\text{n}=(\text{x}+\text{y})^{\text{m}+\text{n}},$ prove that:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerWe have,
$\text{x}^\text{m}.\text{y}^\text{n}=(\text{x}+\text{y})^{\text{m}+\text{n}}\ \ \dots(\text{i})$
Differentiating Eq. (i) w.r.t. x, we get
$\frac{\text{d}}{\text{dx}}(\text{x}^\text{m}.\text{y}^\text{n})=\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})^{\text{m}+\text{n}}$
$\Rightarrow\ \text{x}^\text{m}.\frac{\text{d}}{\text{dy}}\text{y}^\text{n}.\frac{\text{dy}}{\text{dx}}+\text{y}^\text{n}.\frac{\text{d}}{\text{dx}}\text{x}^\text{m}$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\Rightarrow\ \text{x}^\text{m}.\text{ny}^{\text{n}-1}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{n}.\text{mx}^{\text{m}-1}$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\text{x}^\text{m}.\text{ny}^{\text{n}-1}-(\text{m}+\text{n}).(\text{x}+\text{y})^{\text{m}+\text{n}-1}\big]$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}-\text{y}^\text{n}\text{mx}^{\text{m}-1}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\text{nx}^\text{m}\text{y}^{\text{n}-1}-(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\big]$ $=(\text{m}+\text{n}(\text{x}+\text{y})^{\text{m}+\text{n}-1}-\frac{\text{y}^{\text{n}-1}.\text{y}.\text{mx}^{\text{m}-1}}{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}}{(\text{x}-\text{y})}-\frac{\text{y}^{\text{n}-1}.\text{y}.\text{mx}^\text{m}}{\text{x}}}{\frac{\text{nx}^\text{m}\text{y}^\text{n}}{\text{y}}-(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}\frac{1}{(\text{x}+\text{y})}}$
$=\frac{\frac{\text{x}(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}-(\text{x}+\text{y}).\text{y}.^{\text{n}-1}\text{y}.\text{mx}^\text{m}}{(\text{x}+\text{y}).\text{x}}}{\frac{(\text{x}+\text{y})\text{nx}^\text{m}\text{y}^\text{n}-\text{y}(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}}{(\text{x}+\text{y}).\text{y}}}$
$=\frac{\frac{\text{x}(\text{m}+\text{n})\text{x}^\text{m}.\text{y}^\text{n}-\text{m}(\text{x}+\text{y})\text{y}^\text{n}\text{x}^\text{m}}{(\text{x}+\text{y}).\text{x}}}{\frac{(\text{x}+\text{y})\text{nx}^\text{m}.\text{y}^\text{n}-\text{y}(\text{m}+\text{n}).\text{x}^\text{m}.\text{y}^\text{n}}{(\text{x}+\text{y}).\text{y}}}$ $\big[\because(\text{x}+\text{y})^{\text{m}+\text{n}}=\text{x}^\text{m}.\text{y}^\text{n}\big]$
$=\frac{\text{x}^\text{m}\text{y}^\text{n}[\text{mx}+\text{nx}-\text{mx}-\text{my}].(\text{x}+\text{y})\text{y}}{\text{x}^\text{m}\text{y}^\text{n}[\text{nx}+\text{ny}-\text{my}-\text{ny}].(\text{x}+\text{y})\text{x}}$
$=\frac{\text{y}}{\text{x}}\ \ \dots(\text{i})$
Hence proved.
View full question & answer→Question 1705 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.
$f(x) = x^2- 1$ on $[2, 3]$
AnswerWe have
$f(x) = x^2- 1$
Since a polynomial function is everywhere continuous and differentiable, $f(x)$ is continuous on $2, 3$ and differentiable on $2, 3$.
Thus, both conditions of Lagrange's mean value theorem is satisfied.
So, there must exist at least one real number $\text{c}\in2,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
Now,
$f(x) = x^2 - 1$
$\Rightarrow f'(x) = 2x,$
$\Rightarrow f(3) = (3)^2 - 1 = 8$
$\Rightarrow f(2) = (2)^2 - 1 = 3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
$\Rightarrow2\text{x}=\frac{8-3}{1}$
$\Rightarrow\text{x}=\frac{5}{2}$
Thus,
$\text{c}=\frac{5}{2}\in(2,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
View full question & answer→Question 1715 Marks
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
AnswerConsider $\text{y}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2+2)^3\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\sin\text{x})-\text{e}^\text{x}\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2+2)^3}{\big[(\text{x}^2+2)^3\big]^2}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-\text{e}^\text{x}\sin\text{x}3(\text{x}^2+2)^2(2\text{x})}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^3[\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x}]-6\text{xe}^\text{x}\sin\text{x}(\text{x}^2+2)^2}{(\text{x}^2+2)^6}$
$=\frac{(\text{x}^2+2)^2\big[(\text{x}^2+2)(\text{e}^\text{x}\cos\text{x}+\sin\text{xe}^\text{x})-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^6}$
$=\frac{\text{x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^2\sin\text{xe}^\text{x}+2\text{e}^\text{x}\cos\text{x}+2\sin\text{xe}^\text{x}-6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
$=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}+\frac{\text{e}^\text{x}\cos\text{x}}{(\text{x}^2+2)^3}-\frac{6\text{xe}^\text{x}\sin\text{x}}{(\text{x}^2+2)^4}$
View full question & answer→Question 1725 Marks
Differentiate the following functions with respect to x:
$\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
AnswerLet $\text{y}=\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
$\Rightarrow\text{y}=\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}-1}\times\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}\Big)$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\times\Bigg\{\frac{\big(\text{a}^2+\text{x}^2\big)\frac{\text{a}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)-\big(\text{a}^2-\text{x}^2\big)\frac{\text{d}}{\text{dx}}(\text{a}^2+\text{x}^2)}{\big(\text{a}^2+\text{x}^2\big)}\Bigg\}$
[Using chain rule]
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg\{\frac{-2\text{x}\big(\text{a}^2+\text{x}^2\big)-2\text{x}\big(\text{a}^2-\text{x}^2\big)}{\big(\text{a}^2+\text{a}^2\big)^2}\Bigg\}$
$=\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\bigg\{\frac{-2\text{xa}^2-2\text{x}^3-2\text{xa}^2+2\text{x}^3}{\big(\text{a}^2+\text{a}^2\big)^2}\bigg\}$
$ =\frac{1}{2}\Big({\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\Big)^{\frac{1}{2}}\Bigg(\frac{-4\text{xa}^2}{\big(\text{a}^2+\text{x}^2\big)^3}\Bigg)$
$=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\bigg)=\frac{-2\text{xa}^2}{\sqrt{\text{a}^2-\text{x}^2}\big(\text{a}^2+\text{x}^2\big)^{\frac{3}{2}}}$
View full question & answer→Question 1735 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.
$f(x) = 2x - x^2$ on $[0, 1]$
AnswerWe have,$f(x) = 2x - x^2$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on $0, 1$ and differentiable on $0,1$
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in0,1$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$
Now,
$f(x) = 2x - x^2$
$\Rightarrow f'(x) = 2 - 2x,$
$\Rightarrow f(1) = 2 - 1$
$\Rightarrow f(1) = 1,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow2-2\text{x}=\frac{1-0}{1}$
$\Rightarrow-2\text{x}=1-2$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,0)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1745 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\theta+\sin\theta)$ and $\text{y}=\text{a}(1-\cos\theta)$
AnswerHere, $\text{x}=\text{a}(\theta+\sin\theta)$Differentiating it with respect to $\theta$
$\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{dy}}{\text{d}\theta}=\text{a}(\theta+\sin\theta)$
and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\sin\theta...(\text{ii}) $
Using equation (i) and (ii),
$=\frac{\text{a}\sin\theta}{\text{a}(1-\cos\theta)} $
$=\frac{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\sin^{2}\theta}{2}}, \begin{Bmatrix} \text{Since, }1-\cos\theta=\frac{2\sin^{2\theta}}{2}\\\frac{2\sin\theta}{2}\frac{\cos\theta}{2}=\sin\theta \end{Bmatrix}$
$=\frac{\text{dy}}{\text{dx}}=\frac{\tan}{2}$
View full question & answer→Question 1755 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg)$
AnswerThe given relationship is $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\Rightarrow \sin\text{y}=\frac{2\text{x}}{1+\text{x}^{2}}$
Differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big) ...(\text{i})$
The function$\frac{2\text{x}}{1+\text{x}^{2}}$, is of the from of $\frac{\text{u}}{\text{v}}.$
Therefore, by quotient rule, we obtain
$\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)=\frac{(1+\text{x}^{2}).\frac{\text{d}}{\text{dx}}(2\text{x})-2\text{x}.\frac{\text{d}}{\text{dx}}(1 +\text{x}^{2})}{(1 +\text{x}^{2})^2}$
$=\frac{(1+\text{x}^{2}).2-2\text{x}.[0+2\text{x}]}{(1 +\text{x}^{2})^2}$$=\frac{2 + 2\text{x}^{2}-4\text{x}^{2}}{(1 +\text{x}^{2})^2}=\frac{2(1-\text{x}^{2})}{(1+\text{x}^{2})^2} ...(\text{ii})$
Also, $ \sin\text{y}=\frac{2\text{x}}{1+\text{x}^{2}}$
$\Rightarrow\cos\text{y}=\sqrt{1-\sin^{2}\text{y}}=\sqrt{1-\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)^2}=\sqrt{\frac{(1+ \text{x}^{2})^2-4\text{x}^{2}}{(1 +\text{x}^{2})^{2}}}$
$=\sqrt{\frac{1+\text{x}^4+2\text{x}^2-4\text{x}^2}{(1+\text{x}^2)^2}}=\sqrt{\frac{(1-\text{x}^2)^2}{(1+\text{x}^2)^2}}$
$=\sqrt{\frac{(1- \text{x}^{2})^{2}}{(1 +\text{x}^{2})^{2}}}=\frac{1-\text{x}^{2}}{1+\text{x}^{2}} ...\text{(iii)}$
Form(1), (2), and (3), we obtain
$\frac{1-\text{x}^{2}}{1+\text{x}^{2}}\times\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^{2})}{(1 +\text{x}^{2})^{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^{2}}$
View full question & answer→Question 1765 Marks
If $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where -1 < x < 1 then write the value of $\frac{\text{dy}}{\text{dx}}.$
Answer$\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1+\text{x}^3}\Big)$
We know, $\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
Using the chain rule of differentiation,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)^2}\cdot\frac{(1+\text{x}^2)\cdot(2\text{x})'-(1+\text{x}^2)'\cdot(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{(1+\text{x}^2)^2}{(1+\text{x}^2)^2+(2\text{x})^2}\cdot\frac{2(1+\text{x}^2)-(2\text{x})(2\text{x})}{(1+\text{x}^2)^2}$
$=\frac{2(1-\text{x}^2)}{(1+\text{x}^2)^2+(2\text{x})^2}$
Using Chain Rule of Differentiation,
$\frac{\text{du}}{\text{dv}}=\frac{\text{du}}{\text{dx}}\cdot\frac{\text{dx}}{\text{dv}}$
$=\frac{2}{1+\text{x}^2}\cdot\frac{(1+\text{x}^2)^2+(2\text{x})^2}{2(1-\text{x})^2}$
$=\frac{(1+\text{x}^2)^2+(2\text{x})^2}{(1+\text{x}^2)(1-\text{x}^2)}$
Dividing numerator and denominator by $(1 + x^2)^2$,
$\frac{\text{du}}{\text{dv}}=\frac{1+\big(\frac{2\text{x}}{1+\text{x}^2}\big)}{\frac{1-\text{x}^2}{1+\text{x}^2}}$
$=\frac{1+\sin^2\text{u}}{\cos\text{u}}$
$=\sec\text{u}(1+\tan\text{u})$
View full question & answer→Question 1775 Marks
Prove that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{|\text{x|+2}\text{x}^2}, &\text{ x}\neq0\\\text{k}, &\text{ x}=0\end{cases}$
remains discontinuous at x = 0, regardless the choice of k.
AnswerThe given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{\text{x}+2\text{x}^2}, & \text{x} > 0\\\frac{-\text{x}}{\text{x}-2\text{x}^2}, &\text{x} <0\\ \text{k},&\text{x}=0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{1}{2\text{x}+1}, &\text{x} > 0\\\frac{1}{2\text{x}-1}, & \text{x} <0\\\text{k},&\text{x} =0\end{cases}$
We observe
$\text{(LHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{-2\text{h}-1}=-1$
$\text{(RHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{2\text{h}+1}=1$
So,$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$ Such that
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\&\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
are independent of k.
Thus, f(x) is discontinuous at x = 0,
regardless of the choice of k.
View full question & answer→Question 1785 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\frac{1}{\text{x}}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{x}^{\frac{1}{\text{x}}}$
$\Rightarrow\log\text{y}=\frac{1}{\text{x}}\log\text{x}\ \big[\because\log\text{a}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
[Using product rule]
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\times\frac{1}{\text{x}}(\log\text{x})\times\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}^2}-\frac{\log\text{x}}{\text{x}^2}$
$\Rightarrow \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1-\log\text{x})}{\text{x}^2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{x}^\frac{1}{\text{x}}\Big[\frac{1-\log\text{x}}{\text{x}}\Big]$
View full question & answer→Question 1795 Marks
Differentiate the following functions with respect to x:
$\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\},-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\cos^{-1}\Big\{\frac{\cos\text{x}+\sin\text{x}}{\sqrt{2}}\Big\}$
$\text{y}=\cos^{-1}\Big\{\cos\text{x}\Big(\frac{1}{\sqrt{2}}\Big)+\sin\text{x}\Big(\frac{1}{\sqrt{2}}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\text{x}\cos\Big(\frac{\pi}{4}\Big)+\sin\text{x}\sin\text{x}\Big(\frac{\pi}{4}\Big)\Big\}$
$\text{y}=\cos^{-1}\Big[\cos\Big(\text{x}-\frac{\pi}{4}\Big)\Big]$
Here, $-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow\ \Big(-\frac{\pi}{4}-\frac{\pi}{4}\Big)<\Big(\text{x}-\frac{\pi}{4}\Big)<\Big(\frac{\pi}{4}-\frac{\pi}{4}\Big)$
$\Rightarrow-\frac{\pi}{2}<\Big(\text{x}-\frac{\pi}{4}\Big)<0$
So, from equation (i),
$\text{y}=-\Big(\text{x}-\frac{\pi}{4}\Big)$
$\Big[\text{Since}, \cos^{-1}(\cos\theta)=-\theta,\text{ if }\theta\in [-\pi, 0]\Big]$
$\text{y}=-\text{x}+\frac{\pi}{4}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 1805 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
AnswerWe have, $\text{x}=\text{a}(\cos\theta+\theta\sin\theta)$ and $\text{y}=\text{a}(\sin\theta-\theta\sin\theta-\theta\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\cos\theta+\frac{\text{d}}{\text{d}\theta}(\theta\sin\theta)\Big] $ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\frac{\text{d}}{\text{d}\theta}\sin\theta-\frac{\text{d}}{\text{d}\theta}(\theta\cos\theta)\Big] $
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big[-\sin\theta+\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)+\sin\theta\frac{\text{d}}{\text{d}\theta}(\theta)\Big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\Big[\cos\theta-\left\{\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)+\cos\theta\frac{\text{d}}{\text{d}\theta}(\theta)\right\}\Big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\big[-\sin\theta+\theta\cos\theta\big]$ and
$\frac{\text{dy}}{\text{d}\theta}=\text{a}\big[\cos\theta+\theta\sin\theta-\cos\theta\big]$
$ \Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\theta\cos\theta$ and $\frac{\text{dy}}{\text{d}\theta}=\text{a}\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{\text{a}\theta\sin\theta}{\text{a}\theta\cos\theta}=\tan\theta$
View full question & answer→Question 1815 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$
AnswerHere$\text{f}(\text{x})=\cos2{\text{x}}\text{ on }[0,\pi]$
We know that, cosine function is continuous and differentiable every where, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=\cos0=1$
$\text{f}(\pi)=\cos(2\pi)=1$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0.
Now,
$\text{f}(\text{x})=\cos2\text{x}$
$\text{f}'(\text{x})=-2\sin2\text{x}$
So, $\text{f}'(\text{c})=0$
$\Rightarrow-2\sin2\text{c}=0$
$\Rightarrow\sin2\text{c}=0$
$\Rightarrow2\text{c}=0$ or $2\text{c}=\pi$
$\Rightarrow\text{c}=0$ or $\text{c}=\frac{\pi}{2}\in(0,\pi)$
Hence, Rolle's theorem is verified.
View full question & answer→Question 1825 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x + y)^2 = 2axy$
AnswerWe Have, $(x + y)^2 = 2axy$
Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{x}+\text{y}\big)^2=\frac{\text{d}}{\text{dx}}\big(2\text{axy}\big)$
$\Rightarrow2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$\Rightarrow2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=2\text{a}\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]$
$\Rightarrow2(\text{x}+\text{y})+2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=2\text{a}\text{x}\frac{\text{dy}}{\text{dx}}+2\text{ay}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[2(\text{x}+\text{y})-2\text{a}\text{x}\big]=2\text{ay}-2(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2[\text{ay}-\text{x}-\text{y}]}{2[\text{x}+\text{y}-\text{a}\text{x}]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{ay}-\text{x}-\text{y}}{\text{x}+\text{y}-\text{a}\text{x}}\Big)$
View full question & answer→Question 1835 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
AnswerWe have, $4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big(4\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(3\text{y}\big)=\frac{\text{d}}{\text{dx}}\big\{\log\big(4\text{x}-3\text{y}\big)\big\}$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\frac{\text{d}}{\text{dx}}\big(4\text{x}-3\text{y}\big)$
$\Rightarrow4+3\frac{\text{dy}}{\text{dx}}=\frac{1}{\big(4\text{x}-3\text{y}\big)}\Big(4-3\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}+\frac{3}{\big(4\text{x}-3\text{y}\big)}\frac{\text{dy}}{\text{dx}}=\frac{4}{\big(4\text{x}-3\text{y}\big)}-4$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{1+\frac{1}{(4\text{x}-3\text{y})}\Big\}=4\Big\{\frac{1}{(4\text{x}-3\text{y})}-1\Big\}$
$\Rightarrow3\frac{\text{dy}}{\text{dx}}\Big\{\frac{4\text{x}-3\text{y}+1}{(4\text{x}+3\text{y})}\Big\}=4\Big\{\frac{1-4\text{x}-3\text{y}}{4\text{x}-3\text{y}}\Big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big\{\frac{1-4\text{x}+3\text{y}}{(4\text{x}-3\text{y})}\Big\}\Big(\frac{4\text{x}-3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{3}\Big(\frac{1-4\text{x}+3\text{y}}{4\text{x}-3\text{y}+1}\Big)$
View full question & answer→Question 1845 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
AnswerWe have, $\text{y}=\text{x}^{\text{x}}+(\sin\text{x})^\text{x}$
$\Rightarrow\text{y}=\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log(\sin\text{x})^\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})$
$=\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{x}^{\text{x}}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big] \\ +(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\Big[\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\log\sin\text{x}\Big]$
$=\text{x}^{\text{x}}\big[1+\log\text{x}\big]+(\sin\text{x})^\text{x}\big[\text{x}\cot\text{x}+\log\sin\text{x}\big]$
View full question & answer→Question 1855 Marks
If $\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, find k.
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\sin^2\text{x}-\sin^2\text{x}-1}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2\sin^2\text{x}}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\big({\sqrt{\text{x}^2+1}-1}\big)\big({\sqrt{\text{x}^2+1}+1}\big)}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\frac{(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^2\lim_\limits{\text{x}\rightarrow 0}\Big({\sqrt{\text{x}^2+1}-1}\Big)$
$\Rightarrow-2\times-1\times(1+1)=\text{k}$
$\Rightarrow\text{k}=-4$
View full question & answer→Question 1865 Marks
Differentiate the following functions with respect to x:
$\log(\text{cosec x}-\cot\text{x})$
AnswerConsider $\text{y}=\log(\text{cosec x}-\cot\text{x})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\big(\text{cosec x}-\cot\text{x}\big)$
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
[Using chain rule]
$=\frac{1}{(\text{cosec x}-\cot\text{x})}\times\big(-\text{cosec x}\cot\text{x}+\text{cosec}^2\text{x}\big)$
$=\frac{\text{cosec x}(\text{cosec x}-\cos\text{x})}{(\text{cosec x}-\cot\text{x})}$
$=\text{cosec x}$
Hence, the solution is, $\frac{\text{d}}{\text{dx}}\big(\log(\text{cosec x}-\cot\text{x})\big)=\text{cosec x}$
View full question & answer→Question 1875 Marks
Differentiate the following functions with respect to x:
$\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
AnswerLet $\text{y}=\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3\big]$
$=\frac{\text{d}}{\text{dx}}(\text{x}\sin2\text{x})+\frac{\text{d}}{\text{dx}}(5^\text{x})+\frac{\text{d}}{\text{dx}}(\text{k}^\text{k})+\frac{\text{d}}{\text{dx}}\big(\tan^6\text{x}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\sin2\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +5^\text{x}\log5+0+6\tan^5\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
[Using product rule and chain rule]
$=\Big[\text{x}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\sin2\text{x}\Big]+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
$=2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{x}\sin2\text{x}+5^\text{x}+\text{k}^\text{k}+(\tan^2\text{x})^3\big) \\ =2\text{x}\cos2\text{x}+\sin2\text{x}+5^\text{x}\log5+6\tan^5\text{x}\sec^2\text{x}$
View full question & answer→Question 1885 Marks
If $\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)},$ find $\frac{\text{dy}}{\text{dx}}$
Answer$\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)}$
Using the theorem,
If $\text{y}=1+\frac{\text{ax}^2}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{bx}}{(\text{x}-\text{b})(\text{x}-\text{c})}+\frac{\text{c}}{(\text{x}-\text{x})}$ then,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big\{\frac{\text{a}}{\text{a}-\text{x}}+\frac{\text{b}}{\text{b}-\text{x}}+\frac{\text{c}}{\text{c}-\text{x}}\Big\}$
Here, we have $\frac{1}{\text{x}}$ instead of x,
So, using above theorem we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\alpha}{\big(\frac{1}{\text{a}}-\alpha\big)}+\frac{\beta}{\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\gamma}{\big(\frac{1}{\text{x}}-\gamma\big)}$
View full question & answer→Question 1895 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.
$f(x) = x(x + 4)^2$ on $[0,4]$
AnswerHere,$f(x) = x(x + 4)^2$
$\Rightarrow f(x) = x(x^2+ 16 + 8x)$
$\Rightarrow f(x) = x^3 + 8x^2+ 16x$
Since f(x) is a polynomial function which is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [0, 4] and derivable on (0, 4)
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now,$ f(x) = x^3 + 8x^2+ 16x$
$\Rightarrow f(x) = 3x^2+ 16x + 16,$
$\Rightarrow f(4) = 64+ 128 + 64 = 256,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1905 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$
at x = 1
AnswerWe have, $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$ at x = 1.
At x = 1, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow1^-}\frac{\text{x}^2}{2}=\lim\limits_{\text{h}\rightarrow0}\frac{(1-\text{h})^2}{2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1+\text{h}^2-2\text{h}}{2}=\frac{1}{2}$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow1^+}\Big(2\text{x}^2-3\text{x}+\frac{3}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big[2(1+\text{h})^2-3(1+\text{h})+\frac{3}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big(2+2\text{h}^2+4\text{h}-3-3\text{h}+\frac{3}{2}\Big)$ $=-1+\frac{3}{2}=\frac{1}{2}$
And $\text{f}(1)=\frac{1^2}{2}=\frac{1}{2}$
$\therefore$ L.H.L = R.H.L = f(1)
Hence, f(x) is continuous at x = 1.
View full question & answer→Question 1915 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
AnswerLet $\text{y}=\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})$
$=\text{e}^{\text{ax}}\frac{\text{d}}{\text{dx}}\big\{\sec\text{x}\tan2\text{x}\big\}+\sec\text{x}\tan2\text{x}\frac{\text{d}}{\text{dx}}\big\{\text{e}^{\text{ax}}\big\}$
$=\text{e}^{\text{ax}}\big[\text{sec}\text{x}\tan\text{x}\tan2\text{x}+2\sec^2 2\text{x}\sec\text{x}\big]+\text{ae}^{\text{ax}}\sec\text{a}\tan^{2\text{x}}$
$=\text{ae}^{\text{ax}}\sec\text{x}\tan2\text{x}+\text{e}^{\text{ax }}\sec\text{x}\tan\text{x}\tan2\text{x}+2\text{e}^{\text{ax}}\sec\text{x}\sec^2 2\text{x}$
$=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
So,
$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}\sec\text{x}\tan2\text{x})=\text{e}^{\text{ax}}\sec\text{x}\big\{\text{a}\tan2\text{x}+\tan\text{x}\tan2\text{x}+2\sec^22\text{x}\big\}$
View full question & answer→Question 1925 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1-\text{x}}{2}}\Big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\Big\{\sqrt{\frac{1+\cos2\theta}{2}}\Big\}$
$=\sin^{-1}\Big\{\sqrt{\frac{2\sin^2\theta}{2}}\Big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
$\Rightarrow\ 0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\frac{1}{2}\cos^{-1}\text{x}\ \big[\text{Since x}=\cos2\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1935 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\sin{\text{x}}}$
AnswerLet $\text{y}=\text{x}^{\sin{\text{x}}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin{\text{x}}}$
$\log\text{y}=\sin\text{x}\log\text{x}\ \big[\text{Since,}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}$
[Using product rule]
$\frac{1}{\text{y}}\frac{\text{dt}}{\text{dx}}=\sin\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(\cos\text{x})$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
Put the value of y,
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big[\frac{\sin\text{x}}{\text{x}}+(\log\text{x})(\cos\text{x})\Big]$
View full question & answer→Question 1945 Marks
If $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$ find the value of $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
AnswerWe have,
$\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\cos\text{t}+\text{t}\sin\text{t})]=-\text{a}\sin\text{t}+\text{a}\sin\text{t}+\text{at}\cos\text{t}=\text{at}\cos\text{at}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\sin\text{t}-\text{t}\cos\text{t})]=\text{a}\cos\text{t}-\text{a}\cos\text{t}+\text{at}\sin\text{t}=\text{at}\sin\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})$
$=\frac{\text{d}}{\text{dt}}(\tan\text{t})\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{1}{\text{at}\cos\text{t}}$
$=\frac{1}{\text{at}\cos^3\text{t}}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{1}{\text{a}\Big(\frac{\pi}{4}\Big)\cos^3\Big(\frac{\pi}{4}\Big)}=\frac{8\sqrt{2}}{\text{a}\pi}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{4},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{2}}{\text{a}\pi}$
View full question & answer→Question 1955 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\text{x}^{10}-1,&\text{if }\text{ x}\leq1\\\text{x}^2,&\text{if }\text{ x}>1\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\text{x}^{10}-1,&\text{if }\text{ x}\leq1\\\text{x}^2,&\text{if }\text{ x}>1\end{cases}$ The given function f is defind at all the points of the relline. Let c be a point on the real line.
Case I:
If c < 1, then $\text{f(c)}=\text{c}^{10}-1$ and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\text{x}^{10}-1)=\text{c}^{10}-1$ $\therefore\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x such, that x < 1
Case II:
If c = 1, then the left hand limit of at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(\text{x}^{10}-1)=1^{10}-1=1-1=0$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(\text{x}^2)=1^1=1$ It is observed that the left and right hand limit of at x = 1 do not coincide. Therefore, f is not continuouse at x = 1
Case III:
If c > 1, then $f(c) = c^2$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\text{x}^2)=\text{c}^2$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, if can be concluded that x = 1 is the only point of discontinuity of y.
View full question & answer→Question 1965 Marks
Differentiate $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if 0 < x < 1.
AnswerLet $\text{u}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Put $\text{x}=\tan\theta,\text{so}$
$\text{u}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{u}=\tan^{-1}(\tan2\theta)\ .....(\text{i})$
Let $\text{v}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\cos^{-1}\Big(\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big)$
$\text{v}=\cos^{-1}(\cos2\theta)\ .....(\text{ii})$
Hrer, 0 < x < 1
$\Rightarrow0< \tan\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=2\tan^{-1}\text{x }[\text{Since,x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta \big[\text{Since,} \cos^{-1}(\cos\theta)=\theta,\text{if }\theta\in[0,\pi]\big]$
$\text{v}=2\tan^{-1}\text{x }[\text{Since},\text{x}=\tan\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{1+\text{x}^2}\ .....(\text{iv})$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{}du}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{1+\text{x}^2}\times\frac{1+\text{x}^2}{2}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 1975 Marks
If $\text{y}=\text{cosec}^{-1}\text{x},\text{x}>1$ prove that $\text{x}(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0.$
AnswerHere,
$\text{y}=\text{cosec}^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}\sqrt{{}\text{x}^2-1}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sqrt{\text{x}^2-1}+\frac{\text{x}^2}{\sqrt{\text{x}^2-1}}}{\text{x}^2(\text{x}^2-1)}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{x}^2-1+\text{x}^2}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}^2-1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{(\text{x}^2-1)\sqrt{\text{x}^2-1}}-\frac{1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{\sqrt{\text{x}^2-1}}-\frac{1}{\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0$
Hence proved
View full question & answer→Question 1985 Marks
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, -\frac{1}{2}<\text{x}<0,$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}$
Put $2\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\sqrt{1-\cos^2\theta}$
$=\cos^2(\cos\theta)+2\cos^{-1}(\sin\theta)$
$\text{y}=\cos^{-1}(\cos\theta)+2\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\theta\Big)\Big)\ .....(\text{i})$
Now, $-\frac{1}{2}<\text{x}<0$
$\Rightarrow -1<2\text{x}<0$
$\Rightarrow -1<\cos\theta<0$
$\Rightarrow \frac{\pi}{2}<\theta<\pi$
And
$\Rightarrow -\frac{\pi}{2}>-\theta>-\pi$
$\Rightarrow \Big(\frac{\pi}{2}-\frac{\pi}{2}\Big)>\Big(\frac{\pi}{2}-\theta\Big)>\Big(\frac{\pi}{2}-\pi\Big)$
$\Rightarrow 0>\Big(\frac{\pi}{2}-\theta\Big)>-\frac{\pi}{2}$
So, from equation (i),
$\text{y}=\theta+2\Big[-\Big(\frac{\pi}{2}-\theta\Big)\Big]$
$\begin{bmatrix} \text{Since}, \cos^{-1}\cos(\theta)=\theta, \text{if }\theta\in[0,\pi] \\ \cos^{-1}\cos(\theta)=-\theta, \text{if }\theta\in[-\pi,0] \end{bmatrix}$
$\text{y}=\theta-2\times\frac{\pi}{2}+2\theta$
$\text{y}=-\pi+3\theta$
$\text{y}=-\pi+3\cos^{-1}(2\text{x})\ \big[\text{Since}, 2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=0+3\Big(\frac{1}{\sqrt{1-(2\text{x})^2}}\Big)\frac{\text{d}}{\text{dx}}(2\text{x})$
$=\frac{-3}{\sqrt{1-4\text{x}^2}}(2)$
$\frac{\text{dy}}{\text{dx}}=-\frac{6}{\sqrt{1-4\text{x}^2}}$
View full question & answer→Question 1995 Marks
Verify Rolle's theorem for the following function on the indicated intervals
$f(x) = x^2 -4x + 3$ on $[1, 3]$
AnswerThe given function is $f(x) = x^2 -4x + 3$
f, being a pollynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is $2x - 4.$
$f(1) = 1^2 - 4 \times 1 + 3 = 0$
$f(4) = 4^2 - 4 \times 4 + 3 = 3$
$\therefore\ \frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}=\frac{\text{f}(4)-\text{f}(1)}{4-1}=\frac{3-(0)}{3}=\frac{3}{3}=1$
Mean Value Theorem states that there is a point $\text{c}\in(1,4)$ such that $f'(c) = 1$
$f'(c) = 1$
$\Rightarrow 2c - 4 = 1$
$\Rightarrow\text{c}=\frac{5}{2},$ where $\text{c}=\frac{5}{2}\in(1,4)$
Hence, Mean Value Theorem is verified for the given function.
View full question & answer→Question 2005 Marks
If the value of c prescribed bye Lagrange's mean value theorem for the function
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$ defined on $[2, 3]$.
AnswerHere,
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$ defined on [2, 3].
We have to find c prescribed by Lagrange's mean value theorem, so
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow\frac{2\text{c}}{2\sqrt{\text{c}^2-4}}=\frac{(\sqrt{9-4})-(\sqrt{4-4})}{3-2}$
$\Rightarrow\frac{\text{c}}{\sqrt{\text{c}^2-4}}=\frac{\sqrt5-0}{1}$
$\Rightarrow\frac{\text{c}}{\sqrt{\text{c}^2-4}}=\sqrt5$
Squaring both sides,
$\Rightarrow c^2 = (c^2- 4)5$
$\Rightarrow 5c^2 - c^2 = 20$
$\Rightarrow 4c^2 = 20$
$\Rightarrow c^2 = 5$
$\Rightarrow\text{c}=\pm\sqrt5$
but $\text{c}=\sqrt5\text{ as }\sqrt5\in(2,3).$
View full question & answer→