Application of Derivatives

$ \begin{aligned} & \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=3 \\ & \text { since } 3>\frac{7}{3} \Rightarrow 3>f(0) \end{aligned} $

$\Rightarrow \mathrm{x}=0$ is local minima ….option (B) is correct

Now,

$ \begin{aligned} & f(x)=\frac{6 x+\sin x}{2 x+\sin x}=1+\frac{4 x}{2 x+\sin x} \\ & f^{\prime}(x)=\frac{4[(2 x+\sin x) \cdot 1-x(2+\cos x)]}{(2 x+\sin x)^2}=\frac{4(\sin x-x \cos x)}{(2 x+\sin x)^2} \\ & =\frac{4 \cos x(\tan x-x)}{(2 x+\sin x)^2} \end{aligned} $

$ \text { Options (C) & (D) are also correct. } $

Given,

$\alpha = \sum\limits_{k = 1}^\alpha {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)} $

$ = {\sin ^2}\left( {{\pi \over 6}} \right) + {\sin ^4}\left( {{\pi \over 6}} \right) + {\sin ^6}\left( {{\pi \over 6}} \right)\, + \,......$

This is a infinite G.P.

$\therefore$ $\alpha = {{{{\sin }^2}\left( {{\pi \over 6}} \right)} \over {1 - {{\sin }^2}\left( {{\pi \over 6}} \right)}}$

$ = {{{1 \over 4}} \over {1 - {1 \over 4}}}$

$ = {1 \over 4} \times {4 \over 3}$

$ = {1 \over 3}$

$\therefore$ $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

Given $x \in [0,1]$

We know,

$AM \ge GM$

For ${2^{{x \over 3}}}$ and ${2^{{1 \over 3}(1 - x)}}$ both are positive in $x \in [0,1]$

${{{x^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{x \over 3}}}\,.\,{2^{{1 \over 3}(1 - x)}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {{{2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{1 \over 3}}}} \right)^{{1 \over 2}}}$

$ \Rightarrow {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}} \ge 2\,.\,{2^{{1 \over 6}}}$

$ \Rightarrow g(x) \ge {2^{{7 \over 6}}}$

We know, $AM = GM$ when terms are equal.

$\therefore$ $g(x) = {2^{{7 \over 6}}}$ when

${2^{{x \over 3}}} = {2^{{1 \over 3}(1 - x)}}$

$ \Rightarrow {x \over 3} = {{1 - x} \over 3}$

$ \Rightarrow 2x = 1$

$ \Rightarrow x = {1 \over 2}$

$\therefore$ At $x = {1 \over 2}$, ${\left[ {g(x)} \right]_{\min }} = {2^{{7 \over 6}}}$

$\therefore$ Option (A) is correct.

And option (D) is wrong as at only at a single point $x = {1 \over 2},\,g(x)$ is minimum.

Now, $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

$ = {2^{{x \over 3}}} + {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}$

$ = {2^{{x \over 3}}} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}$

$\therefore$ $g'(x) = {2^{{x \over 3}}}\,.\,\log 2\,.\,{1 \over 3} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}\,.\,\log 2\,.\,\left( { - {1 \over 3}} \right)$

$ = {1 \over 3}\log 2\left[ {{2^{{x \over 3}}} - {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

$ = {1 \over 3}\log 2\left[ {{{{2^{{{2x} \over 3}}} + {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

We already found that at $x = {1 \over 2}$ $g(x)$ is minimum.

$\therefore$ $g'(x) = 0$ at $x = {1 \over 2}$

Now, $g'(x) > 0$ when

${1 \over 3}\log \left[ {{{{2^{{{2x} \over 3}}} - {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right] > 0$

$ \Rightarrow {2^{{{2x} \over 3}}} > {2^{{1 \over 3}}}$

$ \Rightarrow {{2x} \over 3} > {1 \over 3}$

$ \Rightarrow x > {1 \over 2}$

Similarly, $g'(x) < 0$ when $x < {1 \over 2}$

If we put it in number line we get this

We know $g'(x)$ represent slope of curve $g(x)$ and it is negative when $x < {1 \over 2}$ and positive when $x > {1 \over 2}$ and zero when $x = {1 \over 2}$.

$\therefore$ Graph of $g(x)$ is

From graph you can see value of $g(x)$ is maximum either at $x = 0$ or $x = 1$ in the range $x \in [0,1]$.

$\therefore$ $g(0) = {2^0} + {2^{{1 \over 3}}} = 1 + {2^{{1 \over 3}}}$

$g(1) = {2^{{1 \over 3}}} + {2^0} = 1 + {2^{{1 \over 3}}}$

$\therefore$ We get maximum value at $x = 0$ and $x = 1$ both.

$\therefore$ B and C options are correct.

Given, f : R $ \to $ and

f(x) = (x $-$ 1)(x $-$ 2)(x $-$ 5)

Since, $F(x) = \int_0^x {f(t)dt} $, x > 0

So, $F'(x) = f(x) = (x - 1)(x - 2)(x - 5)$

According to wavy curve method



F'(x) changes, it's sign from negative to positive at x = 1 and 5, so F(x) has minima at x = 1 and 5 and as F'(x) changes, it's sign from positive to negative at x = 2, so F(x) has maxima at x = 2.

$ \because $ $F'(2) = \int\limits_0^2 {f(t)dt = \int\limits_0^2 {({t^3} - 8{t^2}} } + 17t - 10)dt$

$ = \left[ {{{{t^4}} \over 4} - 8{{{t^3}} \over 3} + 17{{{t^2}} \over 2} - 10t} \right]_0^2$

$ = 4 - {{64} \over 3} + 34 - 20 = 38 - {{124} \over 3} = - {{10} \over 3}$

$ \because $ AT the point of maxima x = 2, the functional value F(2), = $ - {{10} \over 3}$, is negative for the interval, x $ \in $(0, 5), so F(x) $ \ne $ 0 for any value of x $ \in $(0, 5),

Hence, options (a), (b) and (d) are correct.

Given, $f(x) = {{\sin (\pi x)} \over {{x^2}}}$, x > 0

$ \Rightarrow f'(x) = {{{x^2}\pi cos(\pi x) - 2x\sin (\pi x)} \over {{x^4}}}$

$ = {{2x\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^4}}}$

$ = {{2\cos (\pi x)\left[ {{{x\pi } \over 2} - \tan (\pi x)} \right]} \over {{x^3}}}$

Since, for maxima and minima of f(x), f'(x) = 0

$\cos (\pi x) = 0$ or $\tan (\pi x) = {{\pi x} \over 2}$, (as x > 0)

$ \because $ $\cos (\pi x) \ne 0 \Rightarrow \tan (\pi x) = {{\pi x} \over 2}$



$ \because $ $f'(P_1^ - ) < 0$ and $f'(P_1^ + ) > 0$

$ \Rightarrow $ $x = {P_1} \in \left( {1,{3 \over 2}} \right)$

is point of local minimum.

$ \because $ $f'(P_2^ - ) > 0$ and $f'(P_2^ + ) < 0$

$ \Rightarrow x = {P_2} \in \left( {2,{5 \over 2}} \right)$

is point of local maximum.

From the graph, for points of maxima x₁, x₂, x₃ .... it is clear that

${5 \over 2} - {x_1} > {9 \over 2} - {x_2} > {{13} \over 2} - {x_3} > {{17} \over 2} - {x_4}$ .......

$ \Rightarrow {x_{n + 1}} - {x_n} > 2$, $\forall \,n$.

From the graph for points of minima y₁, y₂, y₃ ....., it is clear that

${3 \over 2} - {y_1} > {5 \over 2} - {x_1} > {7 \over 2} - {y_2} > {9 \over 2} - {x_2}$ ........

$|{x_n} - {y_n}|\, > 1$, $\forall \,n$ and x₁ > (y₁ + 1)

And ${x_1} \in \left( {2,\,{5 \over 2}} \right)$, ${x_2} \in \left( {4,\,{9 \over 2}} \right)$, ${x_3} \in \left( {6,\,{{13} \over 2}} \right)$ ..........

$ \Rightarrow $ ${x_n} \in \left( {2n,2n + \,{1 \over 2}} \right)$, $\forall \,n$.

Hence, options (a), (b) and (d) are correct.

$f'(x) > 2f(x)$

$ \Rightarrow {{dy} \over y} > 2dx$

$ \Rightarrow \int_1^{f(x)} {{{dy} \over y} > 2\int_0^x {dx} } $

$\ln (f(x)) > 2x$

$ \therefore $ $f(x) > {e^{2x}}$

Also, as $f'(x) > 2f(x)$

$ \therefore $ $f'(x) > 2{c^{2x}} > 0$

$f(x) = \left| {\matrix{ {\cos 2x} & {\cos 2x} & {\sin 2x} \cr { - \cos x} & {\cos x} & { - \sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right|$

$\cos 2x({\cos ^2}x + {\sin ^2}x) - \cos 2x( - {\cos ^2}x + {\sin ^2}x) + \sin 2x( - \sin 2x)$

$ = \cos 2x + \cos 4x$

$f'(x) = - 2\sin 2x - 4\sin 4x$

$ = - 2\sin 2x(1 + 4\cos 2x)$

At x = 0

f'(x) = 0 and f(x) = 2

Also, f'(x) = 0 sin2x = 0 or $\cos 2x = {{ - 1} \over 4}$

$ \Rightarrow x = {{n\pi } \over 2}$ or $\cos 2x = - {1 \over 4}$

Let f : R $\to$ (0, $\infty$) and g : R $\to$ R.

f(x) > 0 $\forall$x $\in$ R

It is given that f'(2) = 0, g(2) = 0, f''(2) $\ne$ 0 and g'(2) $\ne$ 0.

It is also given that

$\mathop {\lim }\limits_{x \to 2} {{f(x)g(x)} \over {f'(x)g'(x)}} = 1$ $\left( {{0 \over 0}} \right)$

Applying L' Hospital rule, we get

$\mathop {\lim }\limits_{x \to 2} {{f'(x)g(x) + g'(x)f(x)} \over {f''(x)g'(x) + f'(x)g''(x)}} = 1$

For finite limit, we get

${{f'(2)g(2) + g'(2)f(2)} \over {f''(2)g'(2) + f'(2)g''(2)}} = 1$

${{g'(2)f(2)} \over {f''(2)g'(2)}} = 1$

${{f(2)} \over {f''(2)}} = 1 \Rightarrow f''(2) = f(2) > 0$ and f'(2) = 0 which means that f(x) has local minima at x = 2.

Hence, option (A) is correct.

$f(2) - f''(2) = 0$

Therefore, we can say that $f(x) - f''(x) = 0$ has at least one solution in x $\in$ R.

Hence, option (D) is correct.

Let $F(x) = f(x) - 3g(x)$

$\therefore$ $F( - 1) = 3$, $F(0) = 3$ and $F(2) = 3$

So, $F'(x)$ will vanish at least twice in

$( - 1,0) \cup (0,2)$.

$\because$ $F''(x) > 0$ or $ < 0$, $\forall x \in ( - 1,0) \cup (0,2)$

Hence, $f'(x) - 3g'(x) = 0$ has exactly one solution in $( - 1,0)$ and one solution in $(0,2)$.

Here, $f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$

$ = \left\{ {\matrix{ { - 2x - (x + 2) + (x - 2),} & {when\,x \le - 2} \cr { - 2x + x + 2 + 3x + 2,} & {when\, - 2 < x \le - {2 \over 3}} \cr { - 4x,} & {when\, - {2 \over 3} < x \le 0} \cr {4x,} & {when\,0 < x \le 2} \cr {2x + 4,} & {when\,x > 2} \cr } } \right.$

$ = \left\{ {\matrix{ { - 2x - 4,} & {x \le - 2} \cr {2x + 4,} & { - 2 < x \le - 2/3} \cr { - 4x,} & { - {2 \over 3} < x \le 0} \cr {4x,} & {0 < x \le 2} \cr {2x + 4,} & {x > 2} \cr } } \right.$

Graph for y = f(x) is shown as

Let $a$ rectangular sheet with sides $15 a$ and $8 a$ and four squares of side length $b$ removing from each corners of the rectangular sheet

Given, the perimeter of rectangular sheet is constant.

$\therefore 15 a+8 a=$ Constant

$\Rightarrow a$ is constant

Let $V$ be the volume of open rectangular box

$\begin{aligned} & \Rightarrow \quad \mathrm{V}=(15 a-2 b) b(8 a-2 b) \\ & \Rightarrow \quad \mathrm{V}=4 b^3-46 a b^2+120 a^2 b \\ & \Rightarrow \quad \frac{d \mathrm{~V}}{d b}=12 b^2-92 a b+120 a^2=0 \\ & \Rightarrow \quad \frac{d \mathrm{~V}}{d b}=4(3 b-5 a)(b-6 a)=0 \\ & \Rightarrow \quad b=\frac{5 a}{3}, 6 a \\ \end{aligned}$

Now $\frac{d^2 v}{d b^2}=24 b-92 a<0$ at $b=\frac{5 a}{3}$

So, $b=\frac{5 a}{3}$ is the point of local maxima

Given, the area of removed squares is 100

$\begin{aligned} & \therefore & 4 b^2 & =100 \\ & \Rightarrow & b^2 & =25 \\ & \Rightarrow & \left(\frac{5 a}{3}\right)^2 & =25 \\ & \Rightarrow & a & =3 \end{aligned}$

Hence, the side length of the given rectangular sheet are $15 a=45$ and $8 a=24$.

Hints:

$\Rightarrow$ If $x=x_0$ is the point of local maxima of a function $y=f(x)$, then $f^{\prime}\left(x_0\right)=0$ and $f^{\prime \prime}\left(x_0\right)<0$

$\begin{aligned} & \text { Given, } f(x)=\int_0^x e^{t^2}(t-2)(t-3) d t, x \in(0, \infty) \\ & \Rightarrow \quad f^{\prime}(x)=e^{x^2}(x-2)(x-3) \end{aligned}$

Here, $f^{\prime}(x)$ changes its sign $+v e$ to $-v e$ about $x=2$ and $f^{\prime}(x)$ changes its sign $-v \mathrm{e}$ to $+v \mathrm{e}$ about $x=3$

Hence, $x=2$ is the point of local maxima and $x=3$ is the point of local minima

$\because f^{\prime}(x)<0$ for $x \in(2,3)$

$\therefore f(x)$ is decreasing on $x \in(2,3)$

$\because \quad f^{\prime}(x)$ is continuous and differentiable for all $x(0, \infty)$ and $f^{\prime}(2)=f^{\prime}(3)=0$

$\therefore$ According to Rolle's theorem, $f^{\prime \prime}(c)=0$ must have at least one root $\in(2,3)$

We have for

$f(x) = x\cos \left( {{1 \over x}} \right),x \ge 1$

$f'(x) = \cos \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) \to 1$ for $x \to \infty $

Also,

$f'(x) = \left( {{1 \over x}} \right) + {1 \over x}\sin \left( {{1 \over x}} \right) - {1 \over {{x^2}}}\sin \left( {{1 \over x}} \right) - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right)$

$ = - {1 \over {{x^3}}}\cos \left( {{1 \over x}} \right) < 0$ for $x \ge 1$

$ \Rightarrow f'(x)$ is decreasing for $[1,\infty )$

$ \Rightarrow f'(x + 2) < f'(x)$. Also,

$\mathop {\lim }\limits_{x \to \infty } f(x + 2) - f(x) = \mathop {\lim }\limits_{x \to \infty } \left[ {(x + 2)\cos {1 \over {x + 2}} - x\cos {1 \over x}} \right]=2$

Hence, $f(x + 2) - f(x) > 2\forall x \ge 1$.

We know that equation of tangent

$ (\mathrm{Y}-y)=\frac{d y}{d x}(\mathrm{X}-x) $

Intersection of above line with $x$-axis

i.e, $\quad \mathrm{Y}=0$

$ \begin{aligned} \Rightarrow & & (0-y) & =\frac{d y}{d x}(\mathrm{X}-x) \\ \Rightarrow & & \mathrm{X} \frac{d y}{d x} & =x \frac{d y}{d x}-y \\ \Rightarrow & & \mathrm{X} & =x-\frac{y}{\frac{d y}{d x}} \end{aligned} $

Similarly for intersection with $Y-$ axis $X=0$

$ \begin{array}{ll} \Rightarrow & \mathrm{Y}=y+\frac{d y}{d x}(0-x) \\ \Rightarrow & \mathrm{Y}=y-x \frac{d y}{d x} \\ \text { Here, } & \frac{\mathrm{BP}}{\mathrm{AP}}=\frac{3}{1} \\ \Rightarrow & \frac{-d y}{3 y}=\frac{d x}{x} \end{array} $

Integrating on both sides

$ \begin{aligned} \Rightarrow & & -\int \frac{d y}{3 y} & =\int \frac{d x}{x} \\ \Rightarrow & & \frac{-1}{3} \ln y & =\ln x+\ln c \\ \Rightarrow & & \ln y & =-3 \ln x-3 \ln c \\ \Rightarrow & & \ln y & =\ln \left(x^{-3}\right)+\ln c \\ \Rightarrow & & \ln y & =\ln \left(x^{-3} \cdot c\right) \end{aligned} $

(using properties of logarithm)

$ \Rightarrow \quad y=\frac{c}{x^3} $

$ \begin{aligned} &\begin{array}{lrl} \text { Since, } & f(1) & =1 \\ \Rightarrow & c & =1 \end{array}\\ &\text { Therefore, } y=\frac{1}{x^3} \end{aligned} $

Since, $f(x)$ has local maxima at $x=-1$

and $f^{\prime}(x)$ has local maxima at $x=0$

and we know if $f(x)$ has local maxima then $f^{\prime \prime}(x)<0$

also $f^{\prime}(x)$ has local maxima then $f^{\prime \prime}(x)=0$

So, Let us assume $f^{\prime \prime}(x)=\lambda x$

$ \begin{array}{r} (\because \lambda x=0 \text { at } x=0 \text { and } \\ \lambda x<0 \text { at } x=-1) \end{array} $

$ \begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\lambda \frac{x^2}{2}+c \\ & \text { As } \quad f^{\prime}(-1)=0 \\ & \Rightarrow \quad \frac{\lambda}{2}=-c \Rightarrow \quad \lambda=-2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again integrating $f^{\prime}(x)$, we have

$ f(x)=\frac{\lambda x^3}{6}+c x+d $

Since $\quad f(2)=18$ and $f(1)=-1$

$ \lambda\left(\frac{8}{6}\right)+2 c+d=18 \text { and } \frac{\lambda}{6}+c+d=-1 $

Use $\quad \lambda=-2 c$ in above equations.

$ \Rightarrow-2 c \times \frac{4}{3}+2 c+d=18 \text { and } \frac{-2 c}{6}+c+d=-1 $

$ \begin{aligned} &\begin{aligned} \Rightarrow & -2 c+3 d=54 \text { and } 4 c+6 d=-6 \text { or } \\ & 2 c+3 d=-3 \end{aligned}\\ &\text { Adding above equation. }\\ &\begin{aligned} 6 d & =51 \Rightarrow d=\frac{17}{2} \\ \text { and } c & =\frac{54-3 d}{-2}=\frac{-57}{4} \\ \lambda & =-2 c \\ & =\frac{57}{2} \\ \therefore \quad f(x) & =\frac{57}{2} \times \frac{x^3}{6}-\frac{57}{4} x+\frac{17}{2} \\ f(x) & =\frac{1}{4}\left(19 x^3-57 x+34\right) \\ f^{\prime}(x) & =\frac{1}{4}\left(57 x^2-57\right) \end{aligned} \end{aligned} $

for $\max$ or $\min \frac{57 x^2-57}{4}=0$

$ x^2-1=0 \quad \Rightarrow \quad x= \pm 1 $

$\therefore \quad$ Critical points are $x=-1$ and $x=1$

Now, $f^{\prime \prime}(x)=\frac{114 x}{4}$

$f^{\prime \prime}(-1)<0 \Rightarrow$ At $x=-1 f(x)$ is maximum

$f^{\prime \prime}(1)>0 \Rightarrow$ At $x=1 \quad f(x)$ is minimum

And $f(x)$ is increasing for $x>1$

Therefore $f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$

$ \begin{aligned} g(x) & =\int_0^x f(t) d t \quad x \in[1,3] \\ \therefore & g^{\prime}(x)=f(x)=\left\{\begin{array}{cc} e^x & 0 \leq x \leq 1 \\ 2-e^{x-1} & 1 < x \leq 2 \\ x-e & 2 < x \leq 3 \end{array}\right. \end{aligned} $

$ \begin{aligned} &\text { Now, calculating critical points }\\ &\begin{aligned} g^{\prime}(x) & =0 \\ \text { i.e, } \quad e^x & =0,2-e^{x-1}=0, \text { and } x-e=0 \end{aligned} \end{aligned} $

$ \Rightarrow \quad x=-\infty, 2=e^{x-1} \text { and } x=e $

$ \begin{aligned} \ln 2 & =(x-1) \ln e \\ x-1 & =\ln 2 \\ x & =1+\ln 2 \end{aligned} $

Therefore, critical points

$ \begin{aligned} x & =1+\ln 2 \text { and } x=e \\ g^{\prime \prime}(1+\ln 2) & =-e^{\ln 2}<0 \\ \Rightarrow \quad g(x) \text { is maximum at } x & =1+\ln 2 \\ g^{\prime \prime}(e) & =1>0 \\ \Rightarrow \quad g(x) \text { is minimum at } x & =e \end{aligned} $

Since, from graph $f(x)$ is discontinuous at $x=1$

Then we get local max at $x=1$ and

Local minima at $x=2$.