Application of Derivatives

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

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

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 $x=2$.

The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \, \text{cm/s}^2$, can be described by the equation of motion :

$D_l = \frac{1}{2} a t^2$

where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.

The insect, moving at a constant speed of $v = 20 \, \text{cm/s}$, has a motion that can be simply described as:

$D_i = v t$

where $D_i$ is the distance the insect travels, $v$ is its velocity, and $t$ is the time.

Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm. Equating these two distances gives us :

$\frac{1}{2} a t^2 = v t + 21$

Substituting the given values into this equation gives :

$t^2 = \frac{(v t + 21) \cdot 2}{a}$

Substituting $a = 2 \, \text{cm/s}^2$ and $v = 20 \, \text{cm/s}$, we simplify to :

$t^2 = 20t + 21$

This simplifies to a quadratic equation :

$t^2 - 20t - 21 = 0$

Solving this quadratic equation for $t$ gives the solutions $t = 21 \, \text{s}$ and $t = -1 \, \text{s}$. Since time cannot be negative, we discard the -1 s solution. Therefore, the lizard will catch the insect after 21 seconds.

So, the correct answer is Option C : 21 s.

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:
$\sqrt{(3 - (-1))^2 + ((-22) - 10)^2}$

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

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