Differentiation

Let us first expand the given function:

$ g(x) = (x^2 - 1)^{10} $

Using the binomial theorem, we get:

$ g(x) = {^{10}C_0}x^{20} - {^{10}C_1}x^{18} + {^{10}C_2}x^{16} - \cdots + (-1)^{10}{^{10}C_{10}} $

Now, the function is defined as:

$ f(x) = \frac{d^{10}}{dx^{10}} \left( (x^2 - 1)^{10} \right) $

This means we need to take the 10th derivative of each term of $ g(x) $. When we differentiate $ x^{20}, x^{18}, \ldots $ ten times, their powers will reduce by 10 at each differentiation step. Hence:

$ f(x) = {^{10}C_0} (20 \cdot 19 \cdot \ldots \cdot 11)x^{10} - {^{10}C_1}(18 \cdot 17 \cdot \ldots \cdot 9)x^8 + {^{10}C_2}(16 \cdot 15 \cdot \ldots \cdot 7)x^6 - \cdots - {^{10}C_5}(10 \cdot 9 \cdot 8 \cdots 1) $

(A) The coefficient of $ x^8 $ is obtained from the second term:

$ \text{Coefficient of } x^8 = (-1)^{1} {^{10}C_1} \frac{18!}{8!} = (-10)\left( \frac{18!}{8!} \right) $

(B) Since $ g(x) = (x^2 - 1)^{10} $ is an even function, its even-order derivatives (like the 10th derivative) are also even. Therefore,

$ f(1) = f(-1) $

The sum can be written as:

$ \begin{aligned} f(1) + f(-1) &= 2\left[ \frac{20!}{10!} - {^{10}C_1}\frac{18!}{8!} + {^{10}C_2}\frac{16!}{6!} - {^{10}C_3}\frac{14!}{4!} + {^{10}C_4}\frac{12!}{2!} - {^{10}C_5}(10!) \right] \\ &= 2 \times 10! \left[ {^{10}C_0}{^{20}C_{10}} - {^{10}C_1}{^{18}C_{10}} + {^{10}C_2}{^{16}C_{10}} - {^{10}C_3}{^{14}C_{10}} + {^{10}C_4}{^{12}C_{10}} - {^{10}C_5}{^{10}C_{10}} \right] \end{aligned} $

This expression equals twice the coefficient of $ x^{10} $ in $ ((1 + x)^2 - 1)^{10} $. Hence:

$ f(1) + f(-1) = 10! \times 2^{11} $

(C) After differentiating $ (x^2 - 1)^{10} $ ten times, the highest power of $ x $ left is $ x^{10} $. Therefore,

$ \text{Degree of } f(x) = 10 $

(D) The constant term comes from the last term where the power of $ x $ becomes zero after the 10th derivative, which is:

$ \text{Constant term} = \frac{ - (10!)^2 }{ 5! \times 5! } $

$S=$ Set of all twice differentiable functions $\mathrm{f}: \mathrm{R} \rightarrow$ R

$ \frac{\mathrm{d}^2 \mathrm{f}}{\mathrm{dx}^2}>0 \text { in }(-1,1) $

Graph ' $\mathrm{f}$ ' is Concave upward.

Number of solutions of $f(x)=x \rightarrow x_f$

$\Rightarrow$ Graph of $y=f(x)$ can intersect graph of $y=x$ at atmost two points $\Rightarrow 0 \leq x_f \leq 2$

Option (A): It is given that $f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R}$ and $h: \mathbb{R} \rightarrow \mathbb{R}$ are differentiable functions.

Now, $\quad f(x)=x^3+3 x+2$

Differentiating w.r.t. to $x$, we get

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

Also, $ g(f(x))=x $

Now,

$ \begin{aligned} & g^{\prime}(f(x)) \cdot f^{\prime}(x)=1 \\\\ & f(x)=2 \Rightarrow x^3+3 x+2=2 \\\\ & \Rightarrow x^3+3 x=0 \Rightarrow x\left(x^2+3\right)=0 \\\\ & \Rightarrow x=0 \end{aligned} $ Now,

$ g^{\prime}(f(0))=\frac{1}{f^{\prime}(0)} \Rightarrow g^{\prime}(2)=\frac{1}{3} $

Hence, option (A) is incorrect.

Option (B): For all $x \in \mathbb{R}$ :

$ \begin{aligned} & h(g(g(f(x)))=x \\\\ & h(g(g(x)))=x \\\\ & \text { Now, } \quad x \rightarrow f(x) \Rightarrow h(g(g(f(x)))=f(x) \\\\ & \Rightarrow \quad h(g(x))=f(x) \\\\ & \Rightarrow \quad h^{\prime}(g(x)) \cdot g^{\prime}(x)=f^{\prime}(x)=3 x^2+3~~...(1) \end{aligned} $

For all $x \in \mathbb{R}: g(f(x))=x$

Now, $\quad x=1 \Rightarrow g(f(1))=1 \Rightarrow g(6)=1 \quad[\because f(1)=6]$

Substituting $x=6$ in Eq. (1), we get

$ h^{\prime}(g(6)) \cdot g^{\prime}(6)=3\left(6^2\right)+3=111 $

Therefore,

$ h^{\prime}(1)=\frac{111}{g^{\prime}(6)} \quad\left(g^{\prime}(6)=\frac{1}{f^{\prime}(1)}\right) $

That is, $h^{\prime}(1)=111 \cdot f^{\prime}(x)=111 \times(3+3)=666$

Hence, option (B) is correct.

Option (C): $h(g(g(x)))=x$.

For $g(g(x))=0$, we have

$ \begin{array}{ll} & g(x)=g^{-1}(0)=2 \\\\ \Rightarrow & x=g^{-1}(2)=f(2)=16 \\\\ \Rightarrow & h(0)=16 \end{array} $

Hence, option (C) is correct.

Option (D): Here, $g(g(x))=g(3)$ which implies that

$ g(x)=3 \Rightarrow x=g^{-1}(3)=f(3)=38 $

Hence, option (D) is incorrect.

Given,

F(1) = 0, F(3) = 4

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

and f(x) = xF(x)

Now, f'(x) = F(x) + xF'(x)

$\Rightarrow$ f'(1) = F(1) + 1 . F'(1)

$\Rightarrow$ f'(1) = 0 + F'(1) ....... (1)

As F'(x) < 0 for all x $\in$ (1, 3)

$\therefore$ F'(1) < 0

From (1), we get

f'(1) = F'(1) < 0

From Lagrange theorem

$F'(2) = {{F(3) - F(1)} \over {3 - 1}}$

$ = {{ - 4 - 0} \over 2}$

= $-$2

As f(x) = xF(x)

$\therefore$ f(3) = 3 . F(3) = 3 . ($-$4) = $-$12

and f(1) = 1 . F(1) = 0

$\therefore$ $f'(2) = {{f(3) - f(1)} \over {3 - 1}}$

$ = {{ - 12} \over 2} = - 6$

As, f'(x) = F(x) + x . F'(x)

$\therefore$ f'(2) = F(2) + 2 . F'(2)

$ \Rightarrow - 6 = {{f(2)} \over 2} + 2.\,( - 2)$

$ \Rightarrow {{f(2)} \over 2} = - 2$

$\Rightarrow$ f(2) = $-$4

$\therefore$ f(2) < 0

$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$ when x $\in$ (1, 3)

f(3) = $-$12

and f(1) = 0

$\therefore$ $f'(x) = {{ - 12} \over 2} = - 6$

$\therefore$ f'(x) $\ne$ 0 for any x $\in$(1, 3)