Application of Derivatives

The given function is

$f(x) = {x^2} - 3{x^2} + 5x + 7$

$f'(x) = 3{x^2} - 6x + 5$

The discriminant of the above quadratic equation is

$\Delta = 36 - 4(3)(5) = 36 - 60 < 0$

Therefore, $f'(x) > 0\,\forall x \in {R^ + }$

Also, $f'(x) > 0\,\forall x \in {R^ - }$

Therefore, the given function f is increasing in R.

We have

${{(y - {y_2})} \over {(x - {x_1})}} = f'({x_1})$

$ \Rightarrow y - {y_1} = f'({x_1})(x - {x_1})$

$\bullet$ When y = 0: ${{ - {y_1}} \over {f'({x_1})}} = x - {x_1}$

$ \Rightarrow x = {x_1} - {{{y_1}} \over {f'({x_1})}}$

Therefore, point A is $A\left( {{x_1} - {{{y_1}} \over {f'({x_1})}},0} \right)$

$\bullet$ When x = 0: $y - {y_1} = f'(x)\,.\,( - {x_1})$

$ \Rightarrow y = {y_1} - {x_1}f'({x_1})$

Therefore, point B is $B(0,{y_1} - {x_1}f'({x_1}))$

Point P divides AB in the ratio 1 : 3.

${x_1} = {{\left[ {3\left( {{x_1} - {{{y_1}} \over {f'({x_1})}}} \right)} \right]} \over 4}$

${y_1} = {{{y_1} - {x_1}f'({x_1})} \over 4}$

Therefore, $4{y_1} = {y_1} - {x_1}f'({x_1})$

$ \Rightarrow f'({x_1}) = {{ - 3{y_1}} \over {{x_1}}} \Rightarrow f'(x) = {{ - 3y} \over x}$

Now, ${{dy} \over {dx}} = {{ - 3y} \over x} \Rightarrow {{dy} \over y} = {{ - 3dx} \over x}$

On integrating, we get

$\ln y = - 3\ln x + C \Rightarrow y = k{x^{ - 3}}$

$y(1) = 1 \Rightarrow k = 1$

$y = {1 \over {{x^3}}}$

When we substitute the values from the given options, only option (c) satisfies the above equation.

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 $(0,2)$.

Given, $f(0)=f(1)=0$

$\begin{aligned} & \text { And } f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1] \\ & \Rightarrow e^{-x} f^{\prime \prime}(x)-e^{-x} f^{\prime}(x)-e^{-x} f^{\prime}(x)+e^{-x} f(x) \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)\right)^{\prime}-\left(e^{-x} f^{\prime}(x)\right)-\left(e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)-e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1 \end{aligned}$

Let $g(x)=e^{-x} f(x)$

Now, $g(0)=e^{-0} f(0)=0$ and $g(1)=e^{-1} f(1)=0$

Here, $g^{\prime \prime}(x)=\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1>0$

$\therefore$ Concavity of $g(x)$ is up and $g(0)=g(1)=0$

$\begin{aligned} & \Rightarrow g(x)=e^{-x} f(x)<0 \forall x \in(0,1) \\ & \Rightarrow f(x)<0 \end{aligned}$

Hints:

(i) $f^{\prime \prime}(x)>0$ implies the concavity of $f(x)$ is upward and $f^{\prime \prime}(x)<0$ implies the concavity of $f(x)$ is downward.

(ii) $f^{\prime \prime}(x)>0 \quad \forall x \in(a, b)$ and $f(a)=f(b)=0$ implies that the concavity of the function is upward and $f(x)$ lies below the $x$-axis in the interval $x \in(a, b)$

Given, $e^{-x} f(x)$ has point of minima at $x=\frac{1}{4}$ in the interval $x \in[0,1]$

Also given $f(0)=f(1)=0$

$\begin{aligned} & \text { And } f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1] \\ & \Rightarrow e^{-x} f^{\prime \prime}(x)-e^{-x} f^{\prime}(x)-e^{-x} f^{\prime}(x)+e^{-x} f(x) \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)\right)^{\prime}-\left(e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f^{\prime}(x)-e^{-x} f(x)\right)^{\prime} \geq 1 \\ & \Rightarrow\left(e^{-x} f(x)\right)^{\prime \prime} \geq 1>0 \end{aligned}$

Hence, the concavity of $e^{-x} f(x)$ is upward and given $x=\frac{1}{4}$ is the point of local minima of $e^{-x} f(x)$ in $x \in[0,1]$

$\therefore e^{-x} f(x)$ is decreasing in $x \in\left(0, \frac{1}{4}\right)$ and increasing in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow\left(e^{-x} f(x)\right)^{\prime}<0$ in $x \in\left(0, \frac{1}{4}\right)$ and $\left(e^{-x} f(x)\right)^{\prime}>0$ in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow e^{-x} f^{\prime}(x)-e^{-x} f(x)<0$ in $x \in$

$\therefore e^{-x} f(x)$ is decreasing in $x \in\left(0, \frac{1}{4}\right)$ and increasing in $x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow\left(e^{-x} f(x)\right)^{\prime}<0, x \in\left(0, \frac{1}{4}\right)$ and

$\left(e^{-x} f(x)\right)^{\prime}>0, x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow e^{-x} f^{\prime}(x)-e^{-x} f(x)<0, x \in\left(0, \frac{1}{4}\right)$ and

$e^{-x} f^{\prime}(x)-e^{-x} f(x)>0, \quad x \in\left(\frac{1}{4}, 1\right)$

$\Rightarrow f^{\prime}(x) < f(x), x \in\left(0, \frac{1}{4}\right)$ and

$f^{\prime}(x) > f(x), x \in\left(\frac{1}{4}, 1\right)$

Hints :

$\Rightarrow e^{-x} f(x)$ is minimum at $x=\frac{1}{4}$ in $x \in\left[0, \frac{1}{4}\right]$ implies $e^{-x} f(x)$ is decreasing in $x \in\left[0, \frac{1}{4}\right]$ and $e^{-x} f(x)$ is increasing in $x \in\left(\frac{1}{4}, 1\right)$.

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 $f^{\prime \prime}\left(x_0\right)<0$

Given,

$\begin{aligned} g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t \\ \text { and } f(x) & =(1-x)^2 \sin ^2 x+x^2 \\ \Rightarrow g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)\left((1-t)^2 \sin ^2 t+t^2\right) d t \end{aligned}$

On differentiating the above equation w.r.t. $x$

$\begin{aligned} & \Rightarrow g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \\ & \Rightarrow g^{\prime}(x)=\left(2-\frac{4}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \end{aligned}$

Here $(1-x)^2 \sin ^2 x+x^2>0$ for all $x \in \mathrm{R}$

Now, draw the graph of $y=2-\frac{4}{x+1}$ and $y=-\ln x$

For $x>0$

$\text { Add the graph of } y=-\ln x \text { and } y=2-\frac{4}{x+1}$

$\Rightarrow g^{\prime}(x)<0 \text { for } x \in(1, \infty)$

Hence, $g(x)$ is decreasing function for $x \in(1, \infty)$

For statement $P$

$\begin{aligned} & f(x)+2 x=2\left(1+x^2\right) \\ \Rightarrow & (1-x)^2 \sin ^2 x+x^2+2 x=2+2 x^2 \\ \Rightarrow & (1-x)^2 \sin ^2 x=x^2-2 x+1+1 \\ \Rightarrow & (1-x)^2 \sin ^2 x=(1-x)^2+1 \\ \Rightarrow & -(1-x)^2\left(1-\sin ^2 x\right)=1 \\ \Rightarrow & -\cos ^2 x=\frac{1}{(1-x)^2}>0 \\ \Rightarrow & \cos ^2 x<0 \end{aligned}$

Here, no value of $x$ that satisfy the above equation.

$\therefore$ Statement P is false

For Statement Q

$\begin{aligned} & 2 f(x)+1=2 x(1+x) \\ \Rightarrow \quad & 2(1-x)^2 \sin ^2 x+2 x^2+1=2 x+2 x^2 \\ \Rightarrow \quad & 2(x-1)^2 \sin ^2 x=2 x-1 \\ \Rightarrow \quad & \sin ^2 x=\frac{2 x-1}{2(x-1)^2} \\ \Rightarrow \quad & \sin ^2 x=\frac{1}{x-1}+\frac{1}{2(x-1)^2} \quad \text{... (i)} \end{aligned}$

$\begin{aligned} \text { Let } B(x) & =\frac{1}{x-1}+\frac{1}{2(x-1)^2} \\ \Rightarrow \quad B^{\prime}(x) & =\frac{-1}{(x-1)^2}-\frac{1}{(x-1)^3}=\frac{-x}{(x-1)^3} \text { and } \\ B^{\prime \prime}(x) & =\frac{2 x+1}{(x-1)^4} \end{aligned}$

Hence, $\mathrm{B}(x)$ is increasing function on $x \in(0,1)$ and decreasing on $x \in(-\infty, 0) \cup(1, \infty), \mathrm{B}^{\prime \prime}(x)=0$ at $x=-\frac{1}{2}$ and $\mathrm{B}^{\prime \prime}(x) \geq 0$ for $x \in \mathrm{R}-\left\{\frac{-1}{2}, 1\right\} \text {. }$

Draw the graph of the $y=\sin ^2 x$ and $y=\frac{1}{x+1}+\frac{1}{2(x-1)^2}$

The graph of $y=\sin ^2 x$ and $y=\frac{1}{x+1}+\frac{1}{2(x-1)^2}$ intersect at some $x$.

Hence, statement $Q$ is true.

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

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