Question 1015 Marks
Find all points of discontinuity of f, where f is defined by:
$\text {f(x)}=\begin{cases}\frac{\left|\text x\right|}{\text x},\ \ \text {if x}\neq0\\0,\ \ \ \ \text{if x} = 0\end{cases}$
AnswerHere $\text {f(x)}=\begin{cases}\frac{\left|\text x\right|}{\text x},\ \ \text {if x}\neq0\\0,\ \ \ \ \text{if x} = 0\end{cases}$Function f is defined for all points of the real line.
Let c be any real number.
Three cases arise:
Case I: c < 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{|\text{x}|}{\text{x}} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\big(\frac{\text{-x}}{\text{x}}\big) =^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(-1) = -1 $
Also $\text{f(c)} = \frac{|c|} {3} = \frac{-\text{c}}{\text{c}} = -1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all point x < 0
Case II: c > 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\frac{|\text{x}|}{\text{x}} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\big(\frac{\text{x}}{\text{x}}\big) =^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(1) = 1 $
Also $\text{f(c)} = \frac{|c|} {\text c} = \frac{\text{c}}{\text{c}} = 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x > 0
Case III: c = 0
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\frac{|\text{x}|}{\text{x}} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\big(\frac{\text{-x}}{\text{x}}\big) =^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}(-1) = -1 $
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\frac{|\text{x}|}{\text{x}} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\big(\frac{\text{x}}{\text{x}}\big) =^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}(1) = 1 $
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = 0.
View full question & answer→Question 1025 Marks
Find the points on the curve $y = x^3 - 3x$, where the tangent to the curve is parallel to the chord joining $(1, -2)$ and $(2, 2).$
AnswerHere,$y = x^3 − 3x$
y is polynomial function, so it is continuous and differentiable,
so Lagrange's mean value theorem is applicable thus there exists a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow3\text{c}^2-3=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow3\text{c}^2-3=\frac{2+2}{1}$
$\Rightarrow3\text{c}^2=7$
$\Rightarrow\text{c}=\pm\sqrt{\frac{7}{3}}$
$\Rightarrow\text{y}=\pm\frac{2}{3}\sqrt{\frac{7}{3}}$
So, $(\text{c},\text{y})=\Big(\pm\sqrt{\frac{7}{3}},\pm\frac{2}{3}\sqrt{\frac{7}{3}}\Big)$ is the required point.
View full question & answer→Question 1035 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 1045 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 1055 Marks
Differentiate $\text{w.r.t. x}$ the function in Exercise :
$\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a},$ for some fixed $a > 0$ and $x > 0$
AnswerLet $\text{y}=\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a}$
Also, let $x^x = u, x^a = v, a^x = w,$ and $a^a = s$
$\therefore y = u + v + w + s$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}+\frac{\text{ds}}{\text{dx}}$
$U = x^x$
$\Rightarrow\ \log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\ \log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to $x,$ we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x)}+\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\log\text{x}.1+\text{x}.\frac{1}{\text{x}}\Big]$
$V = x^a$
$\therefore\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{a})$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{ax}^{\text{a}-1}\ \dots(3)$
$W = a^x$
$\Rightarrow\ \log\text{w}=\log\text{a}^\text{x}$
$\Rightarrow\ \log\text{w}=\text{x}\log\text{a}$
Differentiating both sides with respect to $x,$ we obtain
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{w}\log\text{a}$
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{a}^\text{x}\log\text{a}\ \dots(4)$
$S = a^a$
Since a is constant$, a^a$ is also a constant.
$\therefore\ \frac{\text{ds}}{\text{dx}}=0\ \dots(5)$
From $(1), (2), (3), (4),$ and $(5),$ we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})+\text{ax}^{\text{a}-1}+\text{a}^\text{x}\log\text{a}+0$
View full question & answer→Question 1065 Marks
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases: $y^3 - 3xy^2 = x^3 + 3x^2y$
AnswerHere, $y^3 - 3xy^2 = x^3 + 3x^2y$ Differentiating with respect to $x,$
$\Rightarrow\frac{\text{d}}{\text{dy}}(\text{y}^3)-\frac{\text{d}}{\text{dx}}(3\text{xy}^2)=\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{\text{d}}{\text{dx}}(3\text{x}^2\text{y})$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}\frac{\text{d}}{\text{dx}}\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(\text{y})+\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^2)\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-3\Big[\text{x}(2\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big]=3\text{x}^2+3\Big[\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{y}(2\text{x})\Big]$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{y}^2+3\text{x}^2+3\text{x}^2\frac{\text{dy}}{\text{dx}}+6\text{xy}$
$\Rightarrow 3\text{y}^2\frac{\text{dy}}{\text{dx}}-6\text{xy}\frac{\text{dy}}{\text{dx}}-3\text{x}^2\frac{\text{dy}}{\text{dx}}=3\text{x}^2+6\text{xy}+3\text{y}^2$
$=3\frac{\text{dy}}{\text{dx}}(\text{y}^2-2\text{xy}-\text{x}^2)=3(\text{x}^2+2\text{xy}+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3(\text{x}+\text{y})^2}{3(\text{y}^2-2\text{xy}-\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+\text{y})^2}{\text{y}^2-2\text{xy}-\text{x}^2)}$
View full question & answer→Question 1075 Marks
Differentiate $\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$ with respect to $\sqrt{1-\text{x}^2},$ if -1 < x < 1.
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Put $\text{x}=\tan\theta$
$\Rightarrow\theta=\tan^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}-\theta\Big)\Big]\ .....(\text{i})$
Here,
$-1<\text{x}<1$
$\Rightarrow-1<\tan\theta<1$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
$\Rightarrow\frac{\pi}{4}>-\theta>\frac{\pi}{4}$
$\Rightarrow-\frac{\pi}{4}<-\theta<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}-\theta<\frac{\pi}{2}$
So, from equation (i)
$\text{u}=\frac{\pi}{4}-\theta$
$\Big[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\tan^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}}=0-\Big(\frac{1}{1+\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=-\frac{1}{1+\text{x}^2}\ .....(\text{ii})$
And
Let, $\text{v}=\sqrt{1-\text{x}^2}$
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\times\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
Dividing equation (ii) by (iii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=-\frac{1}{1+\text{x}^2}\times\frac{\sqrt{1-\text{x} ^2}}{-\text{x}}$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{\sqrt{1-\text{x}^2}}{\text{x}(1+\text{x}^2)}$
View full question & answer→Question 1085 Marks
Find the second order derivatives of the following functions:$\log(\sin\text{x})$
AnswerWe have,
$\text{y}=\text{e}^\text{e}\sin(5\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\sin5\text{x}+\text{e}^\text{x}\cos5\text{x}\times5$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^\text{x}\sin5\text{x}+\text{e}^\text{x}\cos5\text{x}\times5+5\text{e}^\text{x}\cos5\text{x}$
$=-24\text{e}^\text{x}\sin5\text{x}+10\text{e}^\text{x}\cos5\text{x}$
$=2\text{e}^\text{x}(5\cos5\text{x}-12\sin5\text{x})$
View full question & answer→Question 1095 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 1105 Marks
Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\sin2\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin2\text{x}$ Since, $\sin2\text{x}$ is everywhere continuous and differentiable. Therefore $\sin2\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big],$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$ Also, $\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=0$ Thus, f(x) satisfies all the conditions 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
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
View full question & answer→Question 1115 Marks
If $\text{f}:[-5,5]\rightarrow\text{R}$ is differentiable and if f'(x) does not vanish anywhere, then prove that $\text{f}(-5)\pm\text{f}(5).$
AnswerIt is given that $\text{f}:[-5,5]\rightarrow\text{R}$ is a differentiable function. Since every differentiable function is continuous function, we obtain
- f is continuous on [-5, 5].
- f is differentiable on (-5, 5).
Therefore, by the Mean Value Theorem, there exists $\text{c}\in(-5,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-5)}{5-(-5)}$
$\Rightarrow10\text{f}'(\text{c})=\text{f}(5)-f(-5)$
It is also given that f'(x) does not vanish anywhere.
$\therefore\ \text{f}'(\text{c})\neq0$
$\Rightarrow10\text{f}'(\text{c})\neq0$
$\Rightarrow\text{f}(5)-\text{f}(-5)\neq0$
$\Rightarrow\text{f}(5)\neq\text{f}(-5)$
Hence, proved. View full question & answer→Question 1125 Marks
If $\text{y}=\text{x}^3\log\text{x},$ Prove that $\frac{\text{d}^4\text{y}}{\text{dx}^4}=\frac{6}{\text{x}}$
Answerhere,
$\text{y}=\text{x}^3\log\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2\log{x}+\text{x}^3\times\frac{1}{\text{x}}$
$=3\text{x}^2\log{\text{x}}+\text{x}^2$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{x}\log\text{x}+3\text{x}^2\times\frac{1}{\text{x}}+2\text{x}$
$=6\text{x}\log\text{x}+5\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\log\text{x}+6\text{x}\times\frac{1}{\text{x}}+5=6\log\text{x}+11$
Differentiating w.r.t.x, we get
View full question & answer→Question 1135 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x(x^{ }- 4)^2$ on the interval $[0, 4]$
AnswerGiven function is $f(x) = x(x^{ }- 4)^2$. Which can be rewritten as $f(x) = x^3 - 8x^2 + 16x.$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function $f(x)$ is continuous and derivable on $[0, 4].$
Also,
$f(0) = f(4) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,4]$ such that $f'(c) = 0.$
We have
$f(x) = x^3- 8x^2 + 16x$
$\Rightarrow f'(x) = 3x^{2 }- 16x + 16$
$\therefore f'(x) = 0$
$\Rightarrow 3x^{2 }- 16x + 16 = 0$
$\Rightarrow 3x^{2 }- 12x - 4x + 16 = 0$
$\Rightarrow 3x(x - 4) - 4(x - 4) = 0$
$\Rightarrow (x - 4)(3x - 4)$
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.
View full question & answer→Question 1145 Marks
Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\cos2\Big(\text{x}-\frac{\pi}{4}\Big)$ $=\cos\Big(2\text{x}-\frac{\pi}{2}\Big)=\sin2\text{x}.$ Thus, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
View full question & answer→Question 1155 Marks
If $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}},$ then find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big), -\frac{1}{3\sqrt{2}}<\text{x}<\frac{1}{3\sqrt{2}}$
So, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\big]$
$=\frac{\text{d}}{\text{dx}}\Big[\sin^{-1}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)\Big]$
$=\frac{1}{\sqrt{1-\big(6\text{x}\sqrt{1-9\text{x}^2}\big)^2}}\times\frac{\text{d}}{\text{dx}}\big(6\text{x}\sqrt{1-9\text{x}^2}\big)$
$=\frac{1}{\sqrt{1-[36\text{x}^2(1-9\text{x}^2)]}}\times\Big(6\text{x}\frac{\text{d}}{\text{dx}}\sqrt{1-9\text{x}^2}+\sqrt{1-9\text{x}^2}\frac{\text{d}}{\text{dx}}(6\text{x})\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-9\text{x}^2)+\sqrt{1-9\text{x}^2}(6)\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(6\text{x}\times\frac{1}{2\sqrt{1-9\text{x}^2}}(-18\text{x}^2)+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2}{\sqrt{1-9\text{x}^2}}+6\sqrt{1-9\text{x}^2}\Big)$
$=\frac{1}{\sqrt{1-36\text{x}^2-324\text{x}^4}}\times\Big(\frac{-54\text{x}^2+6\sqrt{1-9\text{x}^2}}{\sqrt{1-9\text{x}^2}}\Big)$
$=\frac{-54\text{x}^2+6-54\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
$=\frac{6-108\text{x}}{\sqrt{1-9\text{x}^2}\sqrt{1-36\text{x}^2-324\text{x}^4}}$
View full question & answer→Question 1165 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 1175 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 1185 Marks
If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
AnswerWe have, $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\sin\text{x})^\text{y}=\log(\cos\text{y})^{\text{x}}$
$\Rightarrow\text{y}\log(\sin\text{x})=\text{x}\log(\cos\text{y})$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\big[\text{y}\log\sin\text{x}\big]=\frac{\text{d}}{\text{dx}}\big[\text{x}\log\cos\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\text{x}\frac{\text{dy}}{\text{dx}}(\log\cos\text{y})+\log\cos\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\text{y}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}\frac{\text{x}}{\text{dx}}(\cos\text{y})+\log\cos\text{y}(1)$
$\Rightarrow\frac{\text{y}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}(-\sin\text{y})\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow \text{y}\cot\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\sin\text{x}+\text{x}\tan\text{y}) \\ =\log\cos\text{y}-\text{y}\cot\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$
View full question & answer→Question 1195 Marks
If $y^x = e^{y-x},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
AnswerHere,
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to $x$ using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
View full question & answer→Question 1205 Marks
If $x^x + y^x = 1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{y}+\text{x}\log\text{y})}{\text{x}(\text{y}\log\text{x}+\text{x})}$
AnswerHere,
$x^x + y^x = 1$
Taking on bith sides,
$\log(\text{x}^\text{y}\times\text{y}^\text{x})=\log(1)$
$\text{y}=\log\text{x}+\text{x}\log\text{y}=\log1$
$\big[\text{Since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\log1)$
$\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}\text{(x)}\Big]=0$
$\Big[\text{y}\Big(\frac{1}{\text{x}}\Big)\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\Big(\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)+\log\text{y}(1)\Big]=0$
$\frac{\text{y}}{\text{x}}+\log\text{x}\log\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}=0$
$\frac{\text{dy}}{\text{dx}}\Big(\log\text{x}+\frac{\text{x}}{\text{y}}\Big)=-\Big[\log\text{y}+\frac{\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{y}\log\text{x}+\text{x}}{\text{y}}\Big]=-\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{y}\log\text{x}+\text{x}}\Big]$
View full question & answer→Question 1215 Marks
If $(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\tan\text{y}-\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}$
AnswerHere,
$(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\cos\text{x})^{\text{y}}=\log(\tan\text{y})^{\text{x}}$
$\text{y}\log(\cos\text{x})=\text{x}\log(\tan\text{y})$
$\big[\text{Since}, \log\text{e}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{y})$
$\Big(\text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{y}+\log\tan\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big)$
$\Big(\text{y}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{1}{\tan\text{y}}\frac{\text{d}}{\text{dx}}(\tan\text{y})+\log\tan\text{y}(1)\Big)$
$\Big(\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big)\\ =\Big(\frac{\text{x}}{\tan\text{y}}(\sec^2\text{y})\Big)\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}-\text{y}\tan\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}} \\ =\Big(\sec\text{y cosec y}\times\text{y}\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}\Big)$
$\frac{\text{dy}}{\text{dx}}\big[\log\cos\text{x}-\text{x}\sec\text{y cosec y}\big] \\ =\log\tan\text{y}+\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{\log\tan\text{x}+\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}\Big]$
View full question & answer→Question 1225 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 1235 Marks
If $e^y= y^x$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{\log\text{y}-1}$
AnswerWe have, $e^y = y^x$
Taking log on both sides,
$\log\text{e}^{\text{y}}=\log\text{y}^\text{x}$
$\Rightarrow\text{y}\log\text{e}=\text{x}\log\text{y}$
$\Rightarrow\text{y}=\text{x}\log\text{y}\ .....(\text{i})$
Differentiating with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(1-\frac{\text{x}}{\text{y}}\Big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\frac{\text{y}-\text{x}}{\text{y}}\big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{y}\log\text{y}}{\Big(\text{y}-\frac{\text{y}}{\log\text{y}}\Big)}$
$[$Using equation $(i)]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}(\log\text{y})}{\text{y}\log\text{y}-\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\log\text{y})^2}{\text{y}(\log\text{y}-1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{(\log\text{y}-1)}$
View full question & answer→Question 1245 Marks
Differentiate $\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$ with respect to $\sqrt{1-\text{a}^2\text{x}^2},$ if $-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$.
AnswerLet $\text{u}=\sin^{-1}\Big(2\text{ax}\sqrt{1-\text{a}^2\text{x}^2}\Big)$
Put $\text{ax} =\sin\theta\Rightarrow\theta=\sin^{-1}(\text{ax})$
$\therefore\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^{2}\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sqrt{1-\text{a}^2\text{x}^2}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{a}^2\text{x}^2}}\times\frac{\text{d}}{\text{dx}}\big(1-\text{a}^2\text{x}^2\big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\Big(\frac{0-2\text{a}^2\text{x}}{2\sqrt{1-\text{a}^2\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-\text{a}^2\text{x}}{\sqrt{1-\text{a}^2\text{x}^2}}\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{ax}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}=2\theta\Big[\text{Since},\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=2\times\frac{1}{\sqrt{1-(\text{ax})^2}}\frac{\text{d}}{\text{dx}}(\text{ax})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{1-\text{a}^2\text{x}^2}(\text{a})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2\text{a}}{1-\text{a}^2\text{x}^2}\ .....(\text{iii})$
Dividing equation (iii) by (ii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{2\text{a}}{\sqrt{1-\text{a}^2\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{a}^2\text{x}^2}}{\text{-a}^2\text{x}}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=-\frac{2}{\text{ax}}$
View full question & answer→Question 1255 Marks
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x^2+ 5 x + 6$ on the interval $[-3, -2]$
AnswerHere$, f(x) = x^2+ 5 x + 6$ on $[-3, -2]$
$f(x)$ is continuous is $[-3, -2]$ and $f(x)$ is differentiable is $(-3, -2)$ since it is a polynomial function.
Now,
$f(x) = x^2+ 5x + 6$
$f(-3) = (-3)^2+5(-3) + 6$
$= 9 - 15 + 6$
$f(-3) = 0 ....(i)$
$f(-2) = (-2)^2+ 5(-2) + 6$
$= 4 - 10 + 6$
$f(-2) = 20 ....(ii)$
From equation $(i)$ and $(ii),$
$f(-3) = f(-2)$
So, Rolle's theorem is applicable is $[-3, -2],$ we have to show that
$f\ '(c) = 0$ as $\text{c}\in (-3,-2)$
Now,
$f(x) = x^2 + 5x + 6$
$f\ '(x) = 2x + 5$
$\Rightarrow f'(c) = 0$
$2c + 5 = 0$
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.
View full question & answer→Question 1265 Marks
If $\text{x}=3\cot-2\cos^3\text{t},\text{y}=3\sin\text{t}-2\sin^3\text{t}$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
AnswerGiven,
$\text{x}=3\cot-2\cos^3\text{t},$
$\text{y}=3\sin\text{t}-2\sin^3\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}-6\cos^2\text{t}(-\sin\text{t})$
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
And $\text{y}=3\sin\text{t}-2\sin^2\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dy}}{\text{dt}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot-2\sin^2\text{t}\cos\text{t}}{-\sin\text{t}+2\cos^2\text{t}\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot[1-2\sin^2\text{t]}}{\sin\text{t}[2\cos^2\text{t}-1]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cot\text{t}$
Differentiating both w.r.t. x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}(\cot\text{x})}{\text{dx}}=-\text{cosec}^2\text{x}$
View full question & answer→Question 1275 Marks
Finde the value of a and b, if the function f(x) defined by $\text{f(x)}\begin{cases}\text{x}^2+3\text{x}+\text{a}, &\text{x}\leq1\\\text{bx}+2, & \text{x}>1\end{cases}$is differentiable at x = 1.
AnswerGiven that f(x) is differentiable at x = 1, Therefore, f(x) is countinuous at x = 1.
$\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\text{f(1)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}^2+3\text{x}+\text{a})=\lim\limits_{\text{x}\rightarrow1}(\text{bx}+2)=1+3+\text{a}$
$\Rightarrow1+3+\text{a}=\text{b}+2$
$\Rightarrow\text{a}-\text{b}+2=0\dots(1)$
Again, f(x) is differentiable at x = 1. So,
(LHL at x = 1) = (RHL at x = 1)
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(z)}-\text{f}(1)}{\text{z}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+3\text{x}+\text{a})-(4+\text{a})}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{bx}+2)-(4+\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2+3\text{x}-4}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{bx}-2-\text{a})}{\text{x}-1}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}(\text{x}-4)=\lim\limits_{\text{x}\rightarrow1}\frac{\text{b}(\text{x}-1)}{\text{x}-1}$
$\Rightarrow5=\text{b}$
Hence, a = 3 and b = 5.
View full question & answer→Question 1285 Marks
Discuss the continuity and differentiability of,$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
Answer$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
(LHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
(RHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{c}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
f(c) = 0
Since, LHL = f(x) = RHL at x = c
⇒ f(x) is continuous at x = c
(LHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}-\text{h})-\text{f(c)}}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(RHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}+\text{h})-\text{f(c)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\cos\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(LHL at x = c) = (RHL at x = c)
So,
f(x) is differentiable and continuous at x = c.
View full question & answer→Question 1295 Marks
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cot^2\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if 0 < x < 1.
AnswerLet $\text{u}=\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
Put $\text{x}=\cos\theta\Rightarrow\theta=\cos^{-1}\text{x},\text{ so}$
$\text{u}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
And,
Let $\text{v}=\cot^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
$=\cot^{-1}\Big(\frac{\cot\theta}{\sqrt{1-\cos^2\theta}}\Big)$
$=\cot^{-1}\Big(\frac{\cos\theta}{\sin\theta}\Big)$
$\text{v}=\cot^{-1}(\cot\theta)\ .....(\text{ii})$
Here, $0<\text{x}<1$
$\Rightarrow0<\cos\theta<1$
$\Rightarrow0<\theta<\frac{\pi}{2}$
So, from equation (i),
$\text{u}=\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{u}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\big[\text{Since,}\cot^{-1}(\cot\theta)=\theta,\text{if }\theta\in(0,\pi) \big]$
$\text{v}=\cos^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-1}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{-1}$
$\frac{\text{du}}{\text{dv}}=1$
View full question & answer→Question 1305 Marks
If $x^x + y^x = 1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\Big\{\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\times\log\text{y}}{\text{x}\times\text{y}^{\text{x}-1}}\Big\}$
AnswerHere,
$x^x + y^x = 1$
$\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{y}^\text{x}}=1$
$\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=1$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a}.\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to $x$ using product rule and chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}(1)$
$\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=0$
$\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$+\text{e}^{\log\text{y}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=0$
$\text{x}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]+\text{y}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big]=0$
$\text{x}^\text{x}[1+\log\text{x}]+\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=0$
$\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\big(\text{xy}^{\text{x}-1}\big)\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\frac{\text{dy}}{\text{dx}}=-\Big[\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}}{\text{xy}^{\text{x}-1}}\Big]$
View full question & answer→Question 1315 Marks
If $\text{y}\sin(\text{x}^\text{x}),$ prove that $\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
AnswerLet $\text{y}=\sin(\text{x}^\text{x})\ .....(\text{i})$
Also, Let $\text{u}=\text{x}^\text{x}\ .....(\text{ii})$
Taking log on both sides,
$\Rightarrow\log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}(1)$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iii})$
[Using equation (ii)]
Now, using equation (ii) in equation (i),
$\text{y}=\sin\text{u}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin\text{u})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cos\text{u}\frac{\text{du}}{\text{dx}}$
Using equation (ii) and (iii),
$\frac{\text{dy}}{\text{dx}}=\cos(\text{x}^\text{x})\times\text{x}^\text{x}(1+\log\text{x})$
View full question & answer→Question 1325 Marks
Find the derivative of the function given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f\ ′(1).$
AnswerGiven: $\text{f(x)}=(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\ \dots\text{(i)}$
$\Rightarrow\ \log\text{f(x)}=\log\big\{(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\big\}$
$\Rightarrow\ \log\text{f(x)}=\log(1+\text{x})+\log(1+\text{x}^2)+\log(1+\text{x}^4)+\log(1+\text{x}^8)$
$\Rightarrow\ \frac{\text{1}}{\text{f(x)}}\frac{\text{d}}{\text{dx}}\text{f(x)}=\frac{\text{1}}{1+\text{x}}\frac{\text{d}}{\text{dx}}(1+\text{x})+\frac{\text{1}}{\text{1+x}^2}\frac{\text{d}}{\text{dx}}(1+\text{x}^2)+\frac{\text{1}}{\text{1+x}^4}\frac{\text{d}}{\text{dx}}(1+\text{x}^4)+\frac{\text{1}}{\text{1+x}^8}\frac{\text{d}}{\text{dx}}(1+\text{x}^8)$
$\Rightarrow\ \frac{1}{\text{f(x)}}\text{f}'\text{(x)}=\frac{1}{1+\text{x}}.1+\frac{1}{1+\text{x}^2}.2\text{x}+\frac{1}{1+\text{x}^4}.4\text{x}^3+\frac{1}{1+\text{x}^8}8\text{x}^7$
$\Rightarrow\ \text{f}'\text{(x)}=\text{f}\text{(x)}\Big[\frac{1}{1+\text{x}}+\frac{2\text{x}}{1+\text{x}^2}+\frac{4\text{x}^3}{1+\text{x}^4}+\frac{8\text{x}^7}{1+\text{x}^8}\Big]$
Putting the value of $f(x)$ from eq. $(i),$
$\text{f}'\text{(x)}=(1+\text{x})(1+\text{x}^2)(1+\text{x}^4)(1+\text{x}^8)\Big[\frac{1}{1+\text{x}}+\frac{2\text{x}}{1+\text{x}^2}+\frac{4\text{x}^3}{1+\text{x}^4}+\frac{8\text{x}^7}{1+\text{x}^8}\Big]$
$\Rightarrow\ \text{f}'\text{(x)}=(1+1)(1+1^2)(1+1^4)(1+1^8)\Big[\frac{1}{1+1}+\frac{2\times1}{1+1^2}+\frac{4\times1^3}{1+1^4}+\frac{8\times1^7}{1+1^8}\Big]$
$\Rightarrow\ \text{f}'(1)=(2)(2)(2)(2)\Big[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\Big]$
$\Rightarrow\ \text{f}'(1)=16\Big[\frac{15}{2}\Big]=8\times15=120$
View full question & answer→Question 1335 Marks
Discuss the continuity and differentiability of $\text{f(x)}=\text{e}^{|\text{x}|}.$
AnswerGiven:
$\text{f(x)}=\text{e}^{|\text{x}|}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{e}^\text{x},&\text{x}\geq0\\\text{e}^{-\text{x}},&\text{x}<0\end{cases}$
f is Continuity:
(LHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-(0-\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{-\text{h}}$
$=1$
(RHL at x = 0)
$\lim_\limits{\text{x}\rightarrow0{^{+}}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{(0+\text{h})}$
$=1$
and f(0)
$=\text{e}^0=1$
Thus, $\lim_\limits{\text{x}\rightarrow0^{-}}-\text{f(x)}=\lim_\limits{\text{h}\rightarrow0^{+}}-\text{f(x)}=\text{f(0)}$
Hence, function is continuous at x = 0.
Differentiability at x = 0.
(LHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{-\text{h}}=-1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
(RHL at x = 0)
$=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{-(0-\text{h})}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
LHL at (x = 0) $\neq$ RHL at (x = 0)
Hence the function is not differentiable at x = 0.
View full question & answer→Question 1345 Marks
Differentiate the following functions from first principles : $e^{ax+b}.$
AnswerLet $f(x) = e^{ax+b}\Rightarrow f(x + h) = e^{a(x+h)+b}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}(\text{x}+\text{h})+\text{b}}-\text{e}^{(\text{ax}+\text{b})}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}+\text{b}}\text{e}^{\text{ah}}-\text{e}^{\text{ax}+\text{b}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\text{ax}+\text{b}}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}\times\text{a}$
$=\text{ae}^{\text{ax}+\text{b}} \lim\limits_{\text{h}\rightarrow0}\left\{\frac{(\text{e}^{\text{ah}}-1)}{\text{ah}}\right\}$
$=\text{ae}^{\text{ax}+\text{b}}$
So, $\frac{\text{d}}{\text{dx}}(\text{e}^{\text{ax}}+\text{b})=\text{ae}^{\text{ax}+\text{b}}$
View full question & answer→Question 1355 Marks
Differentiate $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$ with respect to $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)$
Put $\text{x}=\sin\theta$
$\Rightarrow\theta=\sin^{-1}\text{x}$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{u}=\tan^{-1}(\tan\theta)\ .....(\text{i})$
And
Let, $\text{v}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
$\text{v}=\sin^{-1}(2\sin\theta\sqrt{1-\sin^{2}\theta})$
$\text{v}-\sin^{-1}(2\sin\theta\cos\theta)$
$\text{v}=\sin^{-1}(\sin2\theta)\ .....(\text{ii})$
Here,
$-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i)
$\text{u}=\theta\Big[\text{since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\Big]$
$\Rightarrow\text{u}=\sin^{-1}\text{x}$
Differentiatiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=2\theta\Big[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\text{v}=2\sin^{-1}\text{x}$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iv})$
Dividing equation (iii) by (iv),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}}{2}\Big)$
$\therefore\frac{\text{du}}{\text{dv}}=\frac{1}{2}$
View full question & answer→Question 1365 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}-3|,&\text{if }\text{ x}\geq1\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},&\text{if }\text{ x}<1\end{cases}$
AnswerWhen $x >,$ then
$f(x) = |x - 3|$
Since modulus function is a continuous function, $f(x)$ is continuous for each$ x > 1$
When $x < 1, $then
$\text{f(x)}=\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$
Since, $x^2 3x$ are continuous being polynomial functions, $x^2 3x$ will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polymomial function.
$\Rightarrow\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}$ is continuous for eqch $x < 1$
$\Rightarrow f(x)$ is continuous for each $x < 1$
At $x = 1$, we have
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big[\frac{(1-\text{h}^2)}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}\Big]=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|1+\text{h}-3|\big]=|-2|=2$
Also, $\text{f}(1)=|1-3|=|-2|=2$
Thus, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=(1)$
Hence, $f(x)$ is continuous at $x = 1$
Thus, the given function is now where discontinuous.
View full question & answer→Question 1375 Marks
Differentiate the following functions with respect to x:
$\text{e}^{\sqrt{\cot\text{x}}}$
AnswerLet, $\text{y}=\text{e}^\sqrt{{\cot\text{x}}}$
$\Rightarrow\ \text{y}=\text{e}^{(\cot\text{x})^\frac{1}{2}}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{(\cot\text{x})^\frac{1}{2}}\Big)$
$=\text{e}^{(\cot\text{x})^\frac{1}{2}}\frac{\text{d}}{\text{dx}}(\cot\text{x})^\frac{1}{2}$
[Using chain rule]
$=\text{e}^\sqrt{\cot\text{x}}\times\frac{1}{2}(\cot\text{x})^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}(\cot\text{x})$
$=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
View full question & answer→Question 1385 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:
$\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
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,
$\text{x}\in\Big(\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\Rightarrow2\text{x}\in\Big(-1,-\frac{1}{\sqrt{2}},\Big)$
$\Rightarrow\theta\in\Big(\frac{3\pi}{4},\pi\Big)$
So, from equation (i),
$\text{u}=\pi-2\theta$
$\Big[\text{Since},\sin^{-1}(\sin\theta)=\pi-\theta,\text{ if }\theta\in\Big[-\frac{\pi}{2},\frac{3\pi}{2}\Big]\Big]$
$\Rightarrow\text{u}=\pi-2\cos^{-1}(2\text{x})\big[\text{Since},2\text{x}=\cos\theta\big]$
Differentiating it with respect to x using chain rule,
$\frac{\text{du}}{\text{dx}}=0-2\bigg(\frac{-1}{\sqrt{1-(2\text{x})^2}}\bigg)\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-4\text{x}^2}}(2)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\ .....(\text{vi})$
From equation (iv)
$\frac{\text{dv}}{\text{dx}}=\frac{4}{\sqrt{1-4\text{x}^2}}$
But, $\text{x}\in\Big(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\Big)$
$\frac{\text{dv}}{\text{dx}}=\frac{-4(-\text{x})}{\sqrt{1-4(-\text{x})^2}}$
$\frac{\text{dv}}{\text{dx}}=\frac{4\text{x}}{\sqrt{1-4\text{x}^2}}\ .....(\text{vii})$
Dividing equation (vi) by (vii),
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{-4}{\sqrt{1-4\text{x}^2}}\times\frac{\sqrt{1-4\text{x}^2}}{4\text{x}}$
$\frac{\text{du}}{\text{dv}}=-\frac{1}{\text{x}}$
View full question & answer→Question 1395 Marks
At what points on the following curves, is the tangent parallel to $x-$axis? $y = 12(x + 1)(x - 2)$ on $[-1, 2]$
AnswerLet$ f(x) = 12(x + 1)(x - 2) ...(1)$
$\Rightarrow f(x) = 12(x^2 - x - 2)$
$\Rightarrow f(x) = 12x^2 - 12x - 24$
Since $f(x)$ is a polynomial function, $f(x)$ is continuous on $[-1, 2]$ and differentiable on $(1, 2).$
Also,
$f(2) = f(-1) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point $\text{c}\in(-1,2)$ for which $f'(c) = 0.$
But $\text{f}'(\text{c})=0$
$\Rightarrow24\text{c}-12=0$
$\Rightarrow\text{c}=\frac{1}{2}$
$\therefore\ \text{f}(\text{c})=\text{f}\Big(\frac{1}{2}\Big)=-12\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)=-27$
By the geometrical interepretetion of Rolle's theorem, $\Big(\frac{1}{2}\Big),-27$ is
the point on $y = 12(x + 1)(x - 2)$ where the tangent is parallel to the $x-$axis.
View full question & answer→Question 1405 Marks
If $\text{x}=\cot\text{t and y}=\sin\text{t},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{3}}\text{ at t}=\frac{2\pi}{3}$
AnswerWe have, $\text{x}=\cos\text{t}$ and $\text{y}=\sin\text{t}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos\text{t}) $ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t}) $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\sin\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\cos\text{t}$
$\therefore\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\cos\text{t}}{-\sin\text{t}}=-\cot{\text{t}}$
Now, $\big(\frac{\text{dy}}{\text{dx}}\big)_{\text{t}=\frac{2\pi}{3}}=-\cot\big(\frac{2\pi}{3}\big)=\frac{1}{\sqrt{3}}$
View full question & answer→Question 1415 Marks
If u, v and w are functions of x, then show that
$\frac{\text{d}}{\text{dx}}(\text{u. v. w)}=\frac{\text{du}}{\text{dx}}\text{ v. w}+\text{u }\frac{\text{dv}}{\text{dx}}\text{ w}+\text{u.v}\frac{\text{dw}}{\text{dx}}$
in two ways - first by repeated application of product rule, second by logarithmic
differentiation.
AnswerGiven: u, v and w are functions of x. To prove: $\frac{\text{d}}{\text{dx}}(\text{u. v. w)}=\frac{\text{du}}{\text{dx}}\text{ y.w}+\text{u }\frac{\text{dv}}{\text{dx}}\text{ w}+\text{u.v}\frac{\text{dw}}{\text{dx}}$
- By repeated application of product rule:
$\text{L.H.S } \frac{\text{d}}{\text{dx}}(\text{u.v.w)}=\frac{\text{d}}{\text{dx}}[(\text{uv).w}]=\text{uv}\frac{\text{d}}{\text{dx}}\text{w}+\text{w}\frac{\text{d}}{\text{dx}}\text{(uv)}$
$=\text{uv}\frac{\text{dw}}{\text{dx}}+\text{w}\Big[\text{u}\frac{\text{d}}{\text{dx}}\text{v}+\text{v}\frac{\text{d}}{\text{dx}}\text{u}\Big]=\text{uv}\frac{\text{dw}}{\text{dx}}+\text{uw}\frac{\text{dv}}{\text{dx}}+\text{vw}\frac{\text{du}}{\text{dx}}$
$=\frac{\text{du}}{\text{dx}}\text{.v.w}+\text{u}\frac{\text{dv}}{\text{dx}}\text{.w}+\text{u.v.}\frac{\text{dw}}{\text{dx}}=\text{R.H.S}\ \ \ \text{Hence proved}.$
- By Logarithmic differentiation:
Let y = uvw $\ \Rightarrow\ \log\text{y}=\log(\text{u.v.w)}$
$\Rightarrow\ \log\text{y}=\log\text{u}+\log\text{v}+\log\text{w}\ \Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\log\text{u}+\frac{\text{d}}{\text{dx}}\log\text{v}+\frac{\text{d}}{\text{dx}}\log\text{w}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\Big]$
Putting y = uvw, we get
$\frac{\text{d}}{\text{dx}}(\text{u.v.w)}=\text{uvw}\Big[\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}+\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}+\frac{1}{\text{w}}\frac{\text{dw}}{\text{dx}}\Big]$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\text{(u.v.w)}=\frac{\text{du}}{\text{dx}}\text{.v.w}+\text{u.}\frac{\text{dv}}{\text{dx}}\text{.w}+\text{u.v.}\frac{\text{dw}}{\text{dx}}\ \ \ \text{Hence proved}.$ View full question & answer→Question 1425 Marks
Find the points of discontinuity, if any of the following function: $\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{x}^4-16}{\text{x}-2},&\text{if }\text{ x}\neq2\\16,&\text{if }\text{ x}=2\end{cases}$
When $\text{x}\neq2,$ then
$\text{f(x)}=\frac{\text{x}^4-16}{\text{x}-2}$
$=\frac{\text{x}^4-2^4}{\text{x}-2}$
$=\frac{(\text{x}^2-4)(\text{x}-2)(\text{x}+2)}{\text{x}-2}$
$=(\text{x}^2+4)(\text{x}+2)$
We know that a polynomial function is everywhere continuous.
Therefore, the functions $(x^2 + 4)$ and $(x + 2)$ are everywhere continuous.
So, the product function $x^2 + 4x + 2$ is everywhere continuous.
Thus, $f(x)$ is continuous at every $\text{x}\neq2$
At $x = 2,$ we have
$(\text{LHL at x}=2)=\lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2-\text{h})^2+4\big](2-\text{h}+2)=8(4)=32$
$(\text{RHL at x}= 2)=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[(2+\text{h})^2+4\big](2+\text{h}+2)=8(4)=32$
Also, $f(2) = 16$
$\therefore\ \lim_\limits{\text{x}\rightarrow2^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^+}\text{f(x)}\neq\text{f}(2)$
Thus, $f(x)$ is discontinuous at $x = 2$
Hence, the only point of discontinuity for $f(x)$ is $x = 2$
View full question & answer→Question 1435 Marks
Differentiate the functions given in Exercise:
$\text{x}^\text{x}-2^{\sin\text{x}}$
AnswerLet $\text{y}=\text{x}^\text{x}-2^{\sin\text{x}}$
Putting $\text{u}=\text{x}^\text{x}\text{ and v }=2^{\sin\text{x}}$
$\Rightarrow\ \text{y}=\text{u}-\text{v}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}-\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now, $u = x^x$ $\ \Rightarrow\ \log\text{u}=\log\text{x}^\text{x}=\text{x}\log\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\text{x}}+\log\text{x}.1$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1+\log\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}(1+\log\text{x})$ $=\text{x}^\text{x}(1+\log\text{x})\ \dots\text{(ii)}$
Again, $\text{v}=2^{\sin\text{x}}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}2^{\sin\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=2^{\sin \text{x}}\log2\frac{\text{d}}{\text{dx}}\sin\text{x}=\ \Big[\because\frac{\text{d}}{\text{dx}}\text{a}^{\text{f(x)}}=\text{a}^{\text{f(x)}}\log\text{a}\frac{\text{d}}{\text{dx}}\text{f(x)}\Big]$
$\frac{\text{dv}}{\text{dx}}=2^{\sin\text{x}}(\log2).\cos\text{x}=\cos\text{x}.2^{\sin\text{x}}\log2\ \dots\text{(iii)}$
Putting the values from eq. $(ii)$ and $(iii)$ in eq. $(i),$
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})-\cos\text{x}.2^{\sin\text{x}}\log2$
View full question & answer→Question 1445 Marks
If $\text{y}\log(1+\cos\text{x}),$ prove that $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^\text{y}}{\text{dx}^2}.\frac{\text{d}\text{y}}{\text{dx}}=0$
Answer$\text{y}\log(1+\cos\text{x}),$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\cos\text{x}}\times-\sin\text{x}=\frac{-\sin\text{x}}{1+\cos\text{x}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{(1+\cos\text{x})\cos\text{x}-\sin(-\sin\text{x})}{(1+\cos\text{x})^2}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big[\frac{\cos\text{x}+\cos^2\text{x}+\sin^2\text{x}}{(1+\cos\text{x})^2}\Big]=-\Big[\frac{1+\cos\text{x}}{(1+\cos\text{x})^2}\Big]=\frac{-1}{1+\cos\text{x}}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^3}=-\Big(\frac{+1}{(1+\cos\text{x})^2}\times+\sin\text{x}\Big)\\=-\Big(\frac{-\sin\text{x}}{1+\cos\text{x}}\Big)\times\Big(\frac{-1}{1+\cos\text{x}}\Big)=-\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}.\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 1455 Marks
Determine if f defined by:
$\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
AnswerIt is given that $\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
We know that f is defined at all point of the real line.
Let k be a real number.
Case I: $\text{k} \neq 0,$
Then $\text{f(k)} =\text{k}^{2} \sin\frac{1}{\text{k}}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = \text{k}^{2}\sin \frac{\text{1}}{\text{k}}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$
Thus, f is continuous at all points x that is $\text{x}\neq0.$
Case II: k = 0
Then f(k) = f(0) = 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big)$
We know that $-1 \leq\sin\frac{1}{\text {x}}\leq1, \text{x}\neq0$
$\rightarrow \text{x}^{2}\leq \text{x}^{2}\sin\frac{1}{\text {x}}\leq0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = 0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = 0$
similarly,
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big) = 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = \text{f(0)}= ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)}$
Therefore , f is continuous at x = 0.
Therefore, f is has no point of discontinuity.
View full question & answer→Question 1465 Marks
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
AnswerHere, $\text{y}=\sqrt{\text{x}^2+\text{a}^2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+\text{a}^2}\big)$
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{a}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\times(2\text{x})$
$=\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\ \Big[\text{Since},\sqrt{\text{x}^2+\text{a}^2}=\text{y}\Big]$
$\Rightarrow \text{y}\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
View full question & answer→Question 1475 Marks
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
AnswerHere,
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
View full question & answer→Question 1485 Marks
If $\text{y}=\text{e}^{\text{x}}+\text{e}^{-\text{x}},$ prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
AnswerDifferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)$
$=\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}+\frac{\text{d}}{\text{dx}}{\text{e}}^{-\text{x}}$
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\big(-\text{x}\big)$
[Using chain rule]
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}(-1)$
$=\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)$
$=\sqrt{\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)^2-4\text{e}^{\text{x}}\times\text{e}^{-\text{x}}}$
$\Big[\text{Since},(\text{a}-\text{b}=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}\Big]$
$=\sqrt{\text{y}^2-4}$
$\big[\text{Since e}^\text{x}+\text{e}^{-\text{x}}=\text{y}\big]$
Hence, the solution is, $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$
View full question & answer→Question 1495 Marks
Prove that the function $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$ is everywhere continuous.
AnswerWhen x < 0, we have
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ as well as the identity function x are everywhere continuous.
So, the quotient function $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous at each x < 0
When x > 0, we have f(x) = x + 1, which is a polynomial function.
Therefore, f(x) is continuous at each x > 0
Now, Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{x}<0\\\text{x}+1,&\text{x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(\text{h})}{\text{h}}\Big)=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}(\text{h}+1)=1$
Also,
$\text{f}(0)=0+1=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0
Hence, f(x) is everywhere continuous.
View full question & answer→Question 1505 Marks
If $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
AnswerWe have, $\sec\Big(\frac{\text{x}+\text{y}}{\text{x}-\text{y}}\Big)=\text{a}$
$\Rightarrow\frac{\text{x}+\text{y}}{\text{x}-\text{y}}=\sec^{-1}({\text{a}})$
Differentiate with respect to x, we get,
$\Rightarrow\bigg[\frac{(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})-(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})}{(\text{x}-\text{y}}\bigg]=0$
$\Rightarrow(\text{x}-\text{y})\Big(1+\frac{\text{d}}{\text{dx}}\Big)-(\text{x}+\text{y})\Big(1-\frac{\text{d}}{\text{dx}}\Big)=0$
$\Rightarrow(\text{x}-\text{y})+(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}-(\text{x}+\text{y})+(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[\text{x}-\text{y}+\text{x}+\text{y}]=\text{x}+\text{y}-\text{x}+\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{x})=2\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
View full question & answer→