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)

Given, $\mathrm{F}^{\prime}(x)=f^{\prime}(x)\quad \text{.... (i)}$

And $\quad \mathrm{F}(x)=\int_0^{x^2} f(\sqrt{t}) d t$

Differentiate the equation (ii) w.r.t. $x$

$\begin{aligned} & \Rightarrow \quad \mathrm{F}^{\prime}(x)=\frac{d\left(x^2\right)}{d x} \cdot f\left(\sqrt{x^2}\right)-\frac{d(0)}{d x} \cdot f(\sqrt{0}) \\ & \Rightarrow \quad f^{\prime}(x)=2 x \cdot f(x)-0 \\ & \Rightarrow \quad \frac{d f(x)}{f(x)}=2 x d x \\ & \Rightarrow \int \frac{d f(x)}{f(x)}=\int 2 x d x \\ & \Rightarrow \ln f(x)=2 \cdot \frac{x^2}{2}+\mathrm{C} \\ & \Rightarrow \ln f(x)=x^2+\mathrm{C} \quad \text{... (iii)} \end{aligned}$

Given, $f(0)=1$

$\begin{array}{llrl} & \therefore & \ln (1) & =0^2+C \\ & \Rightarrow & 0 & =0+C \\ \Rightarrow & & C=0 \end{array}$

Put $c=0$ in the equation (iii)

$\begin{array}{ll} \Rightarrow & \ln f(x)=x^2 \\ \Rightarrow \quad f(x) & =e^{x^2} \\ \Rightarrow \quad f(\sqrt{t}) & =e^t \end{array}$

Put $f(\sqrt{t})=e^t$ in the equation (ii)

$\begin{array}{ll} \Rightarrow & \mathrm{F}(x)=\int_0^{x^2} e^t \cdot d t \\ \Rightarrow & \mathrm{F}(x)=\left[e^t\right]_0^{x^2} \\ \Rightarrow & \mathrm{F}(x)=e^{x^2}-1 \end{array}$

Put $\quad x=2$

$\Rightarrow \mathrm{F}(2)=e^4-1$

Hint:

(i) If $y=\int_{\phi(x)}^{\psi(x)} f(t) d t$, then $\frac{d y}{d x}=\frac{d \psi(x)}{d x} \cdot f(\psi(x))-\frac{d \phi(x)}{d x} \cdot f(\phi(x))$

(ii) $\int \frac{f^{\prime}(x) d x}{f(x)}=\ln (f(x))+c$

Given that,

$g(x) = \log f(x)$

$ \Rightarrow g(x + 1) = \log f(x + 1)$

$ \Rightarrow g(x + 1) = \log xf(x)$

[$\because$ $f(x + 1) = xf(x)$]

$g(x + 1) = \log x + \log f(n)$

$ \Rightarrow g(x + 1) - g(x) = \log (x)$

$g'(x + 1) - g'(x) = {1 \over x}$

$ \Rightarrow g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$

Putting, $x = x - {1 \over 2}$, we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) = - {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$ .........., N we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) - = {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$, .........., N we get

$g''\left( {{3 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ .... (i)

$g''\left( {{5 \over 2}} \right) - g''\left( {{3 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ ..... (ii)

$g''\left( {{7 \over 2}} \right) - g''\left( {{5 \over 2}} \right) = {{ - {2^2}} \over {{5^2}}}$ ..... (iii)

$g''\left( {N + {1 \over 2}} \right) - g''\left( {N - {1 \over 2}} \right) = {{ - {2^2}} \over {{{(2N - 1)}^2}}}$ ..... (iv)

Adding all the above equations, we get

$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = 4$

$\left[ {1 + {1 \over {{3^2}}} + {1 \over {{5^2}}}\, + \,...\, + \,{1 \over {{{(2N - 1)}^2}}}} \right]$

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

for differentiation, we can write f(x) as

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

Now, on differentiation

$f'(x) = {{2a({x^2} - 1)} \over {{{({x^2} + ax + 1)}^2}}}$ ..... (i)

$f'(1) = 0 = f'( - 1)$

Then again differentiation (1)

${({x^2} + ax + 1)^2}\,.\,2x - ({x^2} - 1)$

$f''(x) = 2a.\,{{2({x^2} + ax + 1)(2x + a)} \over {{{({x^2} + ax + 1)}^4}}}$

$f''( - 1) = {{ - 4a} \over {{{(2 - a)}^2}}}$

$f''(1) = {{ - 4a} \over {{{(2 - a)}^2}}}$

Combining both

$f''(1){(2 + a)^2} + f''( - 1){(2 - a)^2} = 0$

$\mathop {\lim }\limits_{x \to 0} {{g(x)\cos x - g(0)} \over {\sin x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{g'(x)\cos x - g(x)\sin x} \over {\cos x}}$ (Applying L-Hospital rule)

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

Statement - 1 is True.

$f'(x) = g(x)\cos x + g'(x)\sin x$

$f'(0) = g(0)$

Statement - 2 is True.

We have

${y^3} - 3y + x = 0$

Differentiate both sides we get

$3{y^2}.y' - 3y' + 1 = 0$ ..... (i)

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2 $ then

$y'( - 10\sqrt 2 ) = {{ - 1} \over {21}}$

Differentiate eq. (i), we get

$3{y^2}y'' + 6y{(y')^2} - 3y'' = 0$

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2 ,y' = {{ - 1} \over {21}}$ then

$y''( - 10\sqrt 2 ) = {{ - 4\sqrt 2 } \over {{7^3}\,.\,{3^2}}}$

Let’s start by finding the relationship between derivatives.

We know that $\frac{dy}{dx} = \left( \frac{dx}{dy} \right)^{-1}$. This just means that the derivative going from $y$ to $x$ is the reciprocal of going from $x$ to $y$.

To find the second derivative, we get the derivative of the first derivative. So, $\frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dx}\left( \frac{dx}{dy} \right)^{-1}$.

Instead of $x$, let’s write the second derivative in terms of $y$. We use the chain rule for derivatives: $\frac{d}{dx} = \frac{d}{dy} \cdot \frac{dy}{dx}$.

So, $\frac{d^2y}{dx^2} = \frac{d}{dy}\left( \frac{dx}{dy} \right)^{-1} \cdot \frac{dy}{dx}$.

Taking the derivative of $\left( \frac{dx}{dy} \right)^{-1}$ with respect to $y$ gives us $-\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d^2x}{dy^2}$.

Including the extra $\frac{dy}{dx}$ from the chain rule, we get: $\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \frac{d^2x}{dy^2} \cdot \frac{dy}{dx}$.

Now, solve this for $\frac{d^2x}{dy^2}$:

$\frac{d^2x}{dy^2} = \frac{ - d^2y }{ dx^2 } \left( \frac{dy}{dx} \right)^{-3}$.

The statement "The derivative of an even function is always an odd function" is Option A: TRUE.

To understand why this is true, let's first define even and odd functions:

An even function is a function $ f(x) $ that satisfies the condition:

$ f(-x) = f(x) $

An odd function is a function $ g(x) $ that satisfies the condition:

$ g(-x) = -g(x) $

Now, let's consider the derivative of an even function. Let $ f(x) $ be an even function, meaning:

$ f(-x) = f(x) $

We want to find the derivative $ f'(x) $ and determine if it is odd. Using the definition of the derivative, we have:

$ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} $

Since $ f(x) $ is even, we have:

$ f(x) = f(-x) $

Taking the derivative of both sides with respect to $ x $, we get:

$ f'(-x) \cdot (-1) = f'(x) $

or

$ f'(-x) = -f'(x) $

This shows that the derivative $ f'(x) $ satisfies the condition for being an odd function. Therefore, the derivative of an even function is indeed always an odd function.