Question 512 Marks
Verify Mean Value Theorem, if $f(x) = x^2 - 4x - 3$ in the interval $[a, b],$ where $a = 1$ and $b = 4$.
AnswerFunction is continuous in [1, 4] as it is a polynomial function and polynomial function is always continuous. f'(x) = 2x - 4, f'(x) exists in [1, 4], hence derivable. Conditions of MVT theorem are satisfied, hence there exists, at least one $\text{c}\in(1,\ 4)$ such that.
$\frac{\text{f}(4)-\text{(f)}(1)}{4-1}=\text{f}'\text{(c)}\ \Rightarrow\ \frac{-3-(-6)}{3}=2\text{c}-4$
$\Rightarrow\ 1=2\text{c}-4\ \Rightarrow\ \text{c}=\frac{5}{2}$
View full question & answer→Question 522 Marks
Differentiate the functions with respect to x.
$\sin(\text{x}^{2} + 5)$
Answer$\text{Let y} = \sin(\text{x}^{2} + 5)$
$\therefore \frac{\text{dy}}{\text{dx}} = \cos(\text{x}^{2} + 5)\frac{\text{d}}{\text{dx}}(\text{x}^{2} + 5)$$ = \cos (\text{x}^{2} + 5)(2\text{x} + 0) = 2\text{x}\cos(\text{x}^{2} +5)$
View full question & answer→Question 532 Marks
Find $\frac{\text{d}^2\text{y}}{\text{dx}^2},$ where $\text{y}=\log\Big(\frac{\text{x}^2}{\text{e}^2}\Big)$
AnswerHere
$\text{y}=\log\Big(\frac{\text{x}^2}{\text{e}^2}\Big)$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{\text{x}^2}{\text{e}^2}}\times\frac{2\text{x}}{\text{e}^2}=\frac{2}{\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2}{\text{x}^2}$
View full question & answer→Question 542 Marks
Differentiate the functions with respect to x.
$\cos(\sin \text{x})$
Answer$\text{Let y} = \cos(\sin\text{x})$
$\therefore \frac{\text{dy}}{\text{dx}} = -\sin(\sin\text{x})\frac{\text{d}}{\text{dx}}\sin\text{x} = -\sin(\sin\text{x})\cos\text{x}$
View full question & answer→Question 552 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}\cos\theta$ and $\text{y}=\text{b}\sin\theta$
View full question & answer→Question 562 Marks
If $\text{y}=\sin^{-1}\text{x}+\cos^{-1}\text{x},$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=\sin^{-1}\text{x}+\cos^{-1}\text{x}$
$\Rightarrow\text{y}=\frac{\pi}{2}$
$\Big[\because\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 572 Marks
Verify Rolle’s theorem for the function $\text{f(x)}=\text{x}^2+2\text{x}-8,\text{x}\in[-4,2].$
AnswerConsider $\text{f(x)}=\text{x}^2+2\text{x}-8,\text{x}\in[-4,\ 2]$
Function is continuous in [-4,2] as it is a polynomial function and polynomial function is always continuous.
$\text{f}'\text{(x)}=2\text{x}+2, \text{f}'\text{(x)}$ exists in [-4, 2], hence derivable.
$\text{f}(-4)=0\text{ and f }(2)=0$
$\therefore\ \text{f}(-4)=\text{f}(2)$
Conditions of Rolle’s theorem are satisfied, hence there exists, at least one $\text{c}\in(-4,2)$ such that $\text{f}'\text{(c)}=0$
$\Rightarrow\ 2\text{c}+2=0\ \Rightarrow\ \text{c}=-1$
View full question & answer→Question 582 Marks
If $\text{y}=\sin^{-1}(\sin\text{x}),-\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}.$ Then, wrrite tha value of $\frac{\text{dy}}{\text{dx}}\text{ for x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big).$
AnswerWe have, $\text{y}=\sin^{-1}(\sin\text{x})$
$\Rightarrow\text{y}=\text{x}$
$\Big[\because\sin^{-1}(\sin\text{x})=\text{x},\text{ if x}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→Question 592 Marks
Differentiate w.r.t. x the function in Exercise:
$(3\text{x}^2-9\text{x}+5)^9$
AnswerLet $\text{y}=(3\text{x}^2-9\text{x}+5)^9$
Using chain rule, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\text{x}^2-9\text{x}+5)^9$
$=9(3\text{x}^2-9\text{x}+5)^8.\frac{\text{d}}{\text{dx}}(3\text{x}^2-9\text{x}+5)$
$=9(3\text{x}^2-9\text{x}+5)^8.(6\text{x}-9)$
$=9(3\text{x}^2-9\text{x}+5)^8.3(2\text{x}-3)$
$=27(3\text{x}^2-9\text{x}+5)^8.3(2\text{x}-3)$
View full question & answer→Question 602 Marks
Differentiate the functions with respect to x.
$\cos(\sqrt{\text{x})}$
Answer$\text{Let y} =\cos(\sqrt{\text{x})}$
$\therefore \frac{\text{dy}}{\text{dx}}= -\sin\sqrt{\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}} = -\sin\sqrt{\text{x}}.\frac{1}{2}(\text{x})^{\frac{-1}{2}}=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
View full question & answer→Question 612 Marks
Differentiate $(x^2– 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
by expanding the product to obtain a single polynomial.
Answer$y = (x^2 - 5x + 8) (x^3 + 7x + 9)$
$\Rightarrow\ \text{y}=\text{x}^5+7\text{x}^3+9\text{x}^2-5\text{x}^4-35\text{x}^2-45\text{x}+8\text{x}^3+56\text{x}+72$
$\Rightarrow\ \text{y}=\text{x}^5-5\text{x}^4+15\text{x}^3-26\text{x}^2+11\text{x}+72$
$\frac{\text{dy}}{\text{dx}}=5\text{x}^4-20\text{x}^3+45\text{x}^2-52\text{x}+11\ \dots\text{(iii)}$
View full question & answer→Question 622 Marks
For what value of $\lambda$ is the function $\text{f(x)}=\begin{cases}\lambda(\text{x}^2-2\text{x}),&\text{if }\text{ x}\leq0\\4\text{x}+1,&\text{if }\text{ x}>0\end{cases}$ continuous at x = 0? What about continuity at $\text{x}=\pm1?$
View full question & answer→Question 632 Marks
If $\text{f(x)}=\begin{cases}\frac{\text{x}^2-16}{\text{x}-4},&\text{if }\text{ x}\neq4\\\text{k},&\text{if }\text{ x}=4\end{cases}$ is continuous at x = 4, find k.
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\text{x}^2-16}{\text{x}-4},&\text{if }\text{ x}\neq4\\\text{k},&\text{if }\text{ x}=4\end{cases}$
If f(x) is continuous at x = 4, then
$\lim\limits_{{\text{x}}\rightarrow4}\text{f(x})=\text{f(4)}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}\Big(\frac{\text{x}^2-16}{\text{x}-4}\Big)=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}\frac{(\text{x}+4)(\text{x}-4)}{(\text{x}-4)}=\text{k}$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow4}(\text{x}+4)=\text{k}$
$\Rightarrow\text{k}=8$
View full question & answer→Question 642 Marks
Find the second order derivatives of the function given in Exercise:
$\text{x}^3\log\text{x}$
AnswerLet $\text{y}=\text{x}^3\log\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{x}^3.\frac{1}{\text{x}}+\log\text{x}.3\text{x}^2=\text{x}^2+3\text{x}^2\log\text{x}=\text{x}^2(1+3\log\text{x})$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2.\frac{3}{\text{x}}+(1+3\log\text{x}).2\text{x}$
$=3\text{x}+2\text{x}(1+3\log\text{x})=\text{x}[3+2(1+3\log\text{x})]$
$=\text{x}[3+2+6\log\text{x}]=\text{x}(5+6\log\text{xz})$
View full question & answer→Question 652 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\cos\theta$
AnswerThe given equations are $\text{x}=\text{a}\cos\theta\text{ and y}=\text{b}\cos\theta$
Then, $\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{a}\cos\theta)=\text{a}(-\sin\theta)=-\text{a}\sin\theta$
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{b}\cos\theta)=\text{b}(-\sin\theta)=-\text{b}\sin\theta$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{d}\theta}\Big)}{\Big(\frac{\text{dx}}{\text{d}\theta}\Big)}=\frac{-\text{b}\sin\theta}{-\text{a}\sin\theta}=\frac{\text{b}}{\text{a}}$
View full question & answer→Question 662 Marks
If $\text{y}=\mid\log_\text{e}\text{x}\mid $ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
AnswerHere,
$\text{y}=\mid\log_\text{e}\text{x}\mid$
$=\begin{cases}-\log_\text{e}\text{x}&\text{if}&0<\text{x}< 1\\\log_\text{e}\text{x}&\text{if}&\text{x}>1\end{cases}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\begin{cases}\frac{-1}{\text{x}}&\text{if}&0<\text{x}<1\\\frac{1}{\text{x}}&\text{if}&\text{x}>1\end{cases}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\begin{cases}\frac{1}{\text{x}^2}&\text{if}&0<\text{x}<1\\\frac{-1}{\text{x}^2}&\text{if}&\text{x}>1 \end{cases}$
View full question & answer→Question 672 Marks
Write the points of non-differentiability of f(x) = |log |x||.
AnswerHere,
f(x) = |log |x||
f(x) will always positive and let two points x = 1 and x = -1
f(x) = 0
The function f(x) = |log |x|| is not differentiable at x = -1 and 1.
View full question & answer→Question 682 Marks
If $\text{f(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text{x}^2},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, find k.
AnswerSince f(x) is continuous at x = 0,
$\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})$
$\Rightarrow\text{k}=\lim\limits_{{\text{x}}\rightarrow0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{k}=\lim\limits_{{\text{x}}\rightarrow0}\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}$
$\Rightarrow\text{k}=\frac{1}{2}$
View full question & answer→Question 692 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=4\text{t, y}=\frac{4}{\text{t}}$
AnswerThe given equations are $\text{x}=4\text{t, and y}=\frac{4}{\text{t}}$
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(4\text{t)}=4$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big(\frac{4}{\text{t}}\Big)=4.\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\text{t}}\Big)=4.\Big(\frac{-1}{\text{t}^2}\Big)=\frac{-4}{\text{t}^2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\Big(\frac{-4}{\text{t}^2}\Big)}{4}=\frac{-1}{\text{t}^2}$
View full question & answer→Question 702 Marks
For what value of $\lambda$ is the function defined by
$\text{f(x)}= \begin{cases}\lambda(\text{x}^{2} - 2\text{x}), \text{if}\ \text{x} \leq0\\ \text{4x} + 1,\ \ \ \ \ \ \ \text{if}\ \text{x} > 0\end{cases}$
Answer$\text{f(x)}= \begin{cases}\lambda(\text{x}^{2} - 2\text{x}), \text{if}\ \text{x} \leq0\\ \text{4x} + 1,\ \ \ \ \ \ \ \text{if}\ \text{x} > 0\end{cases}$ At x = 0$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{-}}\lambda(\text{x}^{2} - \text{2x}) = \lambda(0 - 0) = 0$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}(4\text{x} + 1) = 4(0) + 1 = 0 + 1 = 1$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x }) $ At x = 1$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}}(4\text{x} + 1) = 4 + 1 = 5$
Also f(1) = 4 + 1 = 5 $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(1)}$ $\therefore$ f is continuous at x = 1 whatever value of $\lambda$ be.
View full question & answer→Question 712 Marks
Is every continuous function differentiable?
AnswerNo, function may be continuous at a point but may not be differentiable at that point .
For example: function f(x) = |x| is continuous at x = 0 but it is not differentiable at x = 0.
View full question & answer→Question 722 Marks
If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then write the value of $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
AnswerWe are given
$\text{x}=\text{f}(\text{t})$
$\text{Y}=\text{g}(\text{t})$
Then $\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\frac{\text{f}'(\text{t})\text{g}'\text({t})-\text{g}'(\text{t})\text{f}''\text({t})}{[\text{f}'(\text{t})]^3}$
View full question & answer→Question 732 Marks
If $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Big[\text{Since},\sin^{-1}\text{x}+\cos^{-1}=\frac{\pi}{2}\Big]$
So,
$\text{y}=\frac{\pi}{2}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→Question 742 Marks
Differentiate the functions with respect to x.
$\sin(\text{ax + b})$
Answer$\text{Let y} = \sin(\text{ax + b})$
$\therefore \frac{\text{dy}}{\text{dx}} = \cos(\text{ax + b})\frac{\text{d}}{\text{dx}}\text{(ax + b)}$
$= \cos(\text{ax + b)}\text{(a + 0)} = \text{a}\cos\text{(ax + b)}$
View full question & answer→Question 752 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\sin^{2}\text{x}+\cos^{2}\text{y}=1$
AnswerThe given relationship is $\sin^{2}\text{x}+\cos^{2}\text{y}=1$
differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{x}+\cos^{2}\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin^{2}\text{x})+\frac{\text{d}}{\text{dx}}(\cos^{2}\text{y})=0$
$\Rightarrow 2\sin\text{x}.\frac{\text{d}}{\text{dx}}(\sin\text{x})+2\cos\text{y}.\frac{\text{d}}{\text{dx}}(\cos\text{y})=0$
$\Rightarrow2\sin\text{x}\cos\text{x}+2\cos\text{y}(-\sin\text{y}).\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\sin2\text{x}-\sin2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\sin2\text{x}}{\sin2\text{y}}$
View full question & answer→Question 762 Marks
If $\text{x}=2\text{ at},\text{y}=\text{at}^2,$ where a is a constant, then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ at}\text{ x}=\frac{1}{2}.$
AnswerHere,
$\text{x}=2\text{at}\ \text{and}\ \text{y}=\text{at}^2$
Differentiating w.r.t.t, we get
$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$
Differentiating w.r.t.t, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$
Now $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$
View full question & answer→Question 772 Marks
If |x| < 1 and $\text{y}=1+\text{x}+\text{x}^2+\ .....\ \text{to}\ \infty,$ then find the value of $\frac{\text{dy}}{\text{dx}}.$
AnswerWe have, $\text{y}=1+\text{x}+\text{x}^2+\ .....\ \text{to}\ \infty,$
$\Rightarrow\text{y}=\frac{1}{1-\text{x}}$
[$\because$ It is a G.P with first term 1 and common ration x]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1}{1-\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}\frac{\text{d}}{\text{dx}}(1-\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}(-1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{(1-\text{x})^2}$
View full question & answer→Question 782 Marks
If $\frac{\pi}{2}\leq\text{x}\leq\frac{3\pi}{2}$ and $\text{y}=\sin^{-1}(\sin\text{x}),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere,
$\text{y}=\sin^{-1}(\sin\text{x}),\text{x}\in\Big[\frac{\pi}{2},\frac{3\pi}{1}\Big]$
$\Big[\text{Since},\sin^{-1}(\sin\text{x})=\text{x},\text{if x}\in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\pi-\text{x}$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\pi-\text{x})$
$0-1$
$\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 792 Marks
If $\text{x}=\text{t}^2$ and $\text{y}=\text{t}^3$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
AnswerHere,
$\text{x}=\text{t}^2\ \text{and}\ \text{y}=\text{t}^3$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}\ \text{and}\ \frac{\text{dy}}{\text{dt}^2}=3\text{t}^2$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{3}{2}\frac{\text{dt}}{\text{dx}}=\frac{3}{4\text{t}}$
View full question & answer→Question 802 Marks
Discuss the applicability of the Rolle's theorem for the following function on the indicated interval
$\text{f}(\text{x})=2\text{x}^2-5\text{x}+3\text{ on }[1,3]$
AnswerThe given function $\text{f}(\text{x})=2\text{x}^2-5\text{x}+3\text{ on }[1,3].$ The domain of f is given to be (1, 3). It is a polynomial function.Thus, it is everywhere derivable and hence continuous.
But
f(1) = 0 and f(3) = 6
$\Rightarrow\text{f}(3)\neq\text{f}(1)$ Hence, Rolle's theorem is not applicable for the given function.
View full question & answer→Question 812 Marks
Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}\text{x}+1,&\text{if}\ \text{x}\leq{5}\\3\text{x} - 5,& \text{if}\ \text{x}> 5\end{cases}$
$\text{at} \text{x} = 5$
AnswerHere $\text{f(x)}=\begin{cases}\text{k}\text{x}+1,&\text{if}\ \text{x}\leq{5}\\3\text{x} - 5,& \text{if}\ \text{x}> 5\end{cases}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}(\text{k}\text{x} + 1) = 5\text{k} + 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}({3}\text{x} - 5) = 15 - 5 = 10$
Since f is continuous at x = 1
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}\text{f(x)}$
$\therefore 5 \text{k} + 1 = 10 \Rightarrow 5\text{k} = 9 \Rightarrow\text{k} = \frac{9}{5}$
View full question & answer→Question 822 Marks
Differentiate the following w.r.t.x: $\frac{\text{e}^\text{x}}{\sin\text{x}}$
Answer$\text{Let}\ \text{y}=\frac{\text{e}^\text{x}}{\sin\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{x}.\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}})-\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}}(\sin\text{x})}{\sin^{2}\text{x}}$
$=\frac{\sin\text{x}.\text{e}^{\text{x}}-\text{e}^{\text{x}}.\cos\text{x}}{\sin^{2}\text{x}}=\text{e}^{\text{x}}\bigg(\frac{\sin\text{x}-\cos\text{x}}{\sin^{2}\text{x}}\bigg)$
View full question & answer→Question 832 Marks
Examine the continuity of the function $f(x) = 2x^2 – 1 at x = 3$.
AnswerHere $f(x) = 2x^2 - 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow{3}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow{3}}(2\text{x}^3-1) = 2(3)^2 - 1$
$= 2(9) - 1 = 18 - 1 = 17$
Now f is defined at $x = 3$
and $f(x) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$
$\therefore\ \ \text{Lt}\ \ \ \ \text{f(x)} = \text{f(3)} = 17\\ \ \ \ \ \text{x}\rightarrow3$
$\therefore$ f is continous at $x = 3$.
View full question & answer→Question 842 Marks
If $\text{f(x)}=\log_\text{e}(\log_\text{e}\text{x}),$ then write the value of f'(e).
AnswerWe have, $\text{f(x)}=\log_\text{e}(\log_\text{e}\text{x})$
Differentiating with respect to x,
$\text{f}'\text{(x)}=\frac{1}{\log_\text{e}\text{x}}\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\log_\text{e}\text{x}}\Big(\frac{1}{\text{x}}\Big)$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{\log_\text{e}\text{e}}\Big(\frac{1}{\text{e}}\Big) \big[\because\text{x}=\text{e}\big]$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{\text{x}}\big[\because\log_\text{e}=1\big]$
View full question & answer→Question 852 Marks
if f(1) = 4, f'(1) = 2, find the value of the derivative of $\log\Big(\text{f}\big(\text{e}^\text{x}\big)\Big)$ w.r.t x at the point x = 0.
AnswerWe have, f(1) = 4 and f'(1) = 2
Let $\text{y}=\log\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{f}\big(\text{e}^\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big\{\text{f}\big(\text{e}^\text{x}\big)\big\}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{f}(\text{e}^\text{x})}\times\text{f}'\big(\text{e}^\text{x}\big)\times\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\text{f}'\big(\text{e}^\text{x}\big)}{\text{f}\big(\text{e}^\text{x}\big)}$
Putting x = 0, we get,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^0\text{f}'\big(\text{e}^0\big)}{\text{f}\big(\text{e}^0\big)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1\text{f}'(1)}{\text{f}(1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{4}\big[\because\text{f}'(1)=2\text{ and f}(1)=4\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 862 Marks
Differentiate the following with respect to x:
$\cos^{-1}(\sin\text{ x})$
AnswerLet $\text{f(x)}=\cos^{-1}(\sin\text{x})$
We observe that this function is defined for all real numbers.
$\text{f(x)}=\cos^{-1}(\sin\text{x})$
$=\cos^{-1}\Big[\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big]=\frac{\pi}{2}-\text{x}$
Thus, $\text{f(x)}=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}-\text{x}\Big)=-1$
View full question & answer→Question 872 Marks
If x and y are connected parametrically by the equations given in Exercise without eliminating the parameter, Find $\frac{\text{dy}}{\text{dx}}.$
$\text{x}=\sin\text{t, y}=\cos2\text{t}$
AnswerThe given equations are $\text{x}=\sin\text{ t and y}=\cos=2\text{t}$
Then, $\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\sin\text{t)}=\cos\text{t}$
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos2\text{t})=-\sin2\text{t}.\frac{\text{d}}{\text{dt}}(2\text{t})=-2\sin2\text{t}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{-2\sin2\text{t}}{\cos\text{t}}=\frac{-2.2\sin\text{t}\cos\text{t}}{\cos\text{t}}=-4\sin\text{t}$
View full question & answer→Question 882 Marks
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{a}(1-\cos\theta)\text{ and y}=\text{a}(\theta+\sin\theta)\text{ at }\theta=\frac{\pi}{1}$
AnswerWe have, $\text{x}=\text{a}(1-\cos\theta)\text{ and y}=\text{a}(\theta+\sin\theta)$
$\therefore\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}[\text{a}(1-\cos\theta)]=\text{a}(\sin\theta)$
and
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}[\text{a}(\theta+\sin\theta)]=\text{a}(1+\cos\theta)$
$\therefore\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\theta=\frac{\pi}{2}}=\bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}\bigg]_{\theta=\frac{\pi}{2}} \\ =\Big[\frac{\text{a}(1+\cos\theta)}{\text{a}(\sin\theta)}\Big]_{\theta=\frac{\pi}{2}}=\frac{\text{a}(1+0)}{\text{a}}=1$
View full question & answer→Question 892 Marks
Find the second order derivatives of the function given in Exercise:
$\log(\log\text{x})$
AnswerLet $\text{y}=\log(\log\text{x})$ $\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}.\frac{1}{\text{x}}=\frac{1}{\text{x}\log\text{x}}$$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{(\text{x}\log\text{x}).\frac{\text{d}}{\text{dx}}(1)-1.\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})}{(\text{x}\log\text{x})^2}$
$=\frac{(\text{x}\log\text{x}).0-1.\Big(\text{x}.\frac{1}{\text{x}}+\log\text{x}.1\Big)}{(\text{x}\log\text{x})^2}$ $=\frac{0-(1+\log\text{x})}{(\text{x}\log\text{x})}$ $=-\frac{(1+\log\text{x})}{(\text{x}\log\text{x})^2}$
View full question & answer→Question 902 Marks
If f(x) = x + 1 then write the value of $\frac{\text{d}}{\text{dx}}\text{ fof }\text{(x)}.$
AnswerHere,
f(x) = x + 1
(fof)(x) = f(f(x))
= f(x + 1)
= (x + 1) + 1
(fof)(x) = x + 2
Differentiating it with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{fof})\text{(x)}=\frac{\text{d}}{\text{dx}}\text{(x)}+\frac{\text{d}}{\text{dx}}(2)$
$=1+0$
$\frac{\text{d}}{\text{dx}}\text{(fof)}\text{(x)}=1$
View full question & answer→Question 912 Marks
Find the second order derivatives of the function given in Exercise:
$\text{x }\cos\text{x}$
AnswerLet $\text{y}=\text{x}\cos\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{x}(-\sin\text{x})+\cos\text{x}.1=-\text{x}\sin\text{x}+\cos\text{x}$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{x}\cos\text{x}-\sin\text{x}.1-\sin\text{x}=-\text{x}\cos\text{x}-2\sin\text{x}$
View full question & answer→Question 922 Marks
Find $\frac{\text{dx}}{{\text{dy}}}$ in the following:
$\text{ax} + \text{by}^{2} = \cos\text{y}$
AnswerThe given relationship is $\text{ax} + \text{by}^{2} = \cos\text{y}$
Differentiating this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\text{ax}) + \frac{\text{d}}{\text{dx}}(\text{by}^{2}) =\frac{\text{d}}{\text{dx}}( \cos\text{y})$
$\Rightarrow \text{a + b}\frac{\text{d}}{\text{dx}}(\text{y}^{2}) =\frac{\text{d}}{\text{dx}}( \cos\text{y}) ...(\text{i})$
Using chain rule, we obtain $\frac{\text{d}}{\text{dx}}(\text{y}^{2}) = 2\text{y}\frac{\text{dy}}{\text{dx}}\ \text{and}\ \frac{\text{d}}{\text{dx}}( \cos\text{y}) = -\sin\text{y}\frac{\text{dy}}{\text{dx}} ...{\text{(ii)}}$
Form (i) and (ii), we obtain
$\text{a + b}\times 2\text{y}\frac{\text{dy}}{\text{dx}} = -\sin\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(2\text{by} + \sin\text{y})\frac{\text{dy}}{\text{dx}} = -\text{a}$
$\therefore\frac{\text{dy}}{\text{dx}} =\frac{\text{-a}}{2\text{by} + \sin\text{y}}$
View full question & answer→Question 932 Marks
If $\text{y}=\cot\text{x}$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
Answer$\text{y}=\cot\text{x}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{cosec}^2\text{x}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-[2\text{cosec}\text{ x}(-\text{cosec}^2\times\cot\text{x})]=-2\frac{\text{dy}}{\text{dx}}.\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
Hence proved.
View full question & answer→Question 942 Marks
If $\text{f(x)}=\begin{cases}\frac{\sin^{-1}\text{x}}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, write the value of k.
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin^{-1}\text{x}}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=\text{f}(0)$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=\text{k}$
$\Rightarrow\text{k}=1$ $\bigg[\because\ \lim\limits_{{\text{x}}\rightarrow0}\Big(\frac{\sin^{-1}\text{x}}{\text{x}}\Big)=1\bigg]$
View full question & answer→Question 952 Marks
If $\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$ then write $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y.
AnswerHere,
$\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$
Thus
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1+\frac{2\text{x}}{2!}-\frac{3\text{x}^2}{3!}+\frac{4\text{x}^3}{4!}...\infty$
$=-1+\text{x}-\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+...\infty$
$=\text{y}$
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If $\text{y}=\text{x}+\text{e}^\text{x},$ find $\frac{\text{d}^2\text{x}}{\text{dy}^2}.$
AnswerHere,
$\text{y}=\text{x}+\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{e}^\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{e}^\text{x}}{(1+\text{e}^\text{x})^2}$
$\frac{\text{dx}}{\text{dy}}=-\frac{-\text{e}^\text{x}}{(1+\text{e}^\text{x})^3}$
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