Application of Derivatives

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 $x \in\left(\frac{1}{4}, 1\right)$.

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.

Given that, $f(x) = \left\{ {\matrix{ {{{(2 + x)}^3},} & { - 3 < x \le - 1} \cr {{x^{2/3}},} & { - 1 < x < 2} \cr } } \right.$

$f'(x) = \left\{ {\matrix{ {3{{(2 + x)}^2},} & { - 3 < x < - 1} \cr {{2 \over 3}{x^{{{ - 1} \over 3}}},} & { - 1 < x < 2} \cr } } \right.$

Clearly, $f'(x)$ changes signs from positive to negative as x passes through x $=-1$

$f'(x) < 0$ for all $x\in(-3,-1)$

Also, $f'(x)>0$ for all $x\in$

$f'(x) < 0$ for all $x\in(-1,0)$

But $f'(0)$ does not exist

So, $f(x)$ attains a local minimum at x = 0

Hence, the total number of local maxima and local minima is 2.

We have,

$\mathop {\lim }\limits_{t \to x} {{{t^2}f(x) - {x^2}f(t)} \over {t - x}} = 1$

$ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2tf(x) - {x^2}f'(t)} \over {1 - 0}} = 1$

$ \Rightarrow 2xf(x) - {x^2}f'(x) = 1$

$ \Rightarrow {{{x^2}f'(x) - 2xf(x)} \over {{x^4}}} = {{ - 1} \over {{x^4}}}$

$ \Rightarrow {d \over {dx}}\left( {{{f(x)} \over {{x^2}}}} \right) = {{ - 1} \over {{x^4}}}$ ...... (i)

On integrating both sides of this equation, we get

${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + c$

Now, since

$f(1) = 1 \Rightarrow c = {2 \over 3}$

We get

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

Required equation of tangent is $y-2=0$.

As $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq\left(x_{1}-x_{2}\right)^{2} \forall x_{1}, x_{2} \in \mathrm{R}$

$\Rightarrow\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \leq\left|x_{1}-x_{2}\right|\left(x^{2}=|x|^{2}\right)$

$\therefore \quad \lim_\limits{x_{1} \rightarrow x_{2}}\left|\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}\right| \leq \lim_\limits{x_{1}-x_{2}}\left|x_{1}-x_{2}\right|$

$\Rightarrow\left|f^{\prime}\left(x_{1}\right)\right| \leq 0, \forall x_{1} \in \mathrm{R}$

$\therefore\left|f^{\prime}(x)\right| \leq 0$ But $\left|f^{\prime}(x)\right| \geq 0$

Hence, $f^{\prime}(x)=0$

$\Rightarrow f(x)$ is a constant function.

$\Rightarrow$ Equation of tangent at $(1,2)$ is

$\frac{y-2}{x-1}=f^{\prime}(x) \text { or } y-2=0 \quad\left(\because f^{\prime}(x)=0\right) $

i.e., $y=2$ or $y-2=0$ is required tangent.

Let the polynomial be $p(x) = a{x^3} + b{x^2} + cx + d$.

We know $p(-1) = 10$. So, $-a + b - c + d = 10$.

We also know $p(1) = -6$. So, $a + b + c + d = -6$.

It is given that $p(x)$ has a maximum at $x = -1$. For a maximum or minimum, the first derivative is zero: $p'(-1) = 0$.

First, find the derivative: $p'(x) = 3a{x^2} + 2bx + c$.

Plug in $x = -1$ in the derivative: $3a(-1)^2 + 2b(-1) + c = 0$, which means $3a - 2b + c = 0$.

It is given that $p'(x)$ has a minimum at $x = 1$. For a minimum or maximum of the derivative, the second derivative is zero: $p''(1) = 0$.

Find the second derivative: $p''(x) = 6a x + 2b$.

Plug in $x = 1$: $6a(1) + 2b = 0$, which gives $6a + 2b = 0$.

Now, we have four equations:
1) $-a + b - c + d = 10$
2) $a + b + c + d = -6$
3) $3a - 2b + c = 0$
4) $6a + 2b = 0$

Solving these equations, we find:
$a = 1$, $b = -3$, $c = -9$, $d = 5$

So the polynomial is $p(x) = x^3 - 3x^2 - 9x + 5$.

Find where $p'(x) = 0$ (for local maximum and minimum points).
$p'(x) = 3x^2 - 6x - 9$

Set this equal to zero: $3x^2 - 6x - 9 = 0$

Divide both sides by 3: $x^2 - 2x - 3 = 0$

Factorize: $(x + 1)(x - 3) = 0$

So $x = -1$ and $x = 3$.

At $x = -1$, we have a maximum and at $x = 3$, we have a minimum.

Calculate the points:
At $x = -1$, $p(-1) = 10$, so the point is $(-1, 10)$.
At $x = 3$, $p(3) = (3)^3 - 3(3)^2 - 9 \times 3 + 5 = 27 - 27 - 27 + 5 = -22$, so the point is $(3, -22)$.

The distance between these two points is:
$\sqrt{(3 - (-1))^2 + ((-22) - 10)^2}$

$= \sqrt{(4)^2 + (-32)^2}$

$= \sqrt{16 + 1024} = \sqrt{1040} = 4\sqrt{65}$

$ \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. } $