Question

Let $f\left(x\right)=\int \frac{x}{2}{h}^{\prime }\left(x\right)$, where $h\left(x\right)=\frac{x^2-1}{x^2}$. If $f\left(2\right)=\frac{1}{2}$ and $g_r\left(x\right)=\underset{rtimes}{\underset{}{f\left(f\right(f(....f(x)....)\left)\right)}}$ i.e., $g_1\left(x\right)=f\left(x\right),g_2\left(x\right)=f\left(f\right(x\left)\right)$ and so on, then identify the CORRECT statement(s).

• A. $\frac{d}{dx}\left(g_{(3n-2)}\right(x\left)\right)=1$ (whenever exists) for $n\in N$
• B. $\frac{d}{dx}\left(g_{\left(3n\right)}\right(x\left)\right)=1$ (whenever exists) for $n\in N$
• C. $\underset{x\to 1}{lim}\frac{\sum _{r=1}^{100}\left(g_{3r}\right(x){)}^r+x-101}{x-1}=5050$
• D. Equation of the tangent to the curve $y=g_{80}\left(x\right)$ when $x=2$ is $x-y-3=0$
  

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{d}{dx}\left(g_{\left(3n\right)}\right(x\left)\right)=1$ (whenever exists) for $n\in N$
D Equation of the tangent to the curve $y=g_{80}\left(x\right)$ when $x=2$ is $x-y-3=0$


$f\left(x\right)=\int \frac{x}{2}{h}^{\prime }\left(x\right)=\int \frac{x}{2}\frac{d}{dx}(1-\frac{1}{x^2})=\int \frac{x}{2}\cdot \frac{2}{x^3}dx=\int \frac{1}{x^2}dx=-\frac{1}{x}+C$
Now, $f\left(2\right)=-\frac{1}{2}+C=\frac{1}{2}$
$\Rightarrow C=1$
$\therefore f\left(x\right)=\frac{x-1}{x}$

$g_1\left(x\right)=f\left(x\right)=\frac{x-1}{x}$
$g_2\left(x\right)=f\left(f\right(x\left)\right)=\frac{\frac{x-1}{x}-1}{\frac{x-1}{x}}=-\frac{1}{x-1}=\frac{1}{1-x}$

$g_3\left(x\right)=f\left(f\right(f\left(x\right)\left)\right)=\frac{1}{1-\left(\frac{x-1}{x}\right)}=x$
Now, $g_4\left(x\right)=g_1\left(x\right)=\frac{x-1}{x}$
$g_5\left(x\right)=g_2\left(x\right)=\frac{1}{1-x}$
$g_6\left(x\right)=g_3\left(x\right)=x$ and so on.
So, it is observed that
$g_{(3n-2)}=\frac{x-1}{x}$,
$g_{(3n-1)}=\frac{1}{1-x}$,
and $g_{\left(3n\right)}=x$

So, $\frac{d}{dx}\left(g_{(3n-2)}\right(x\left)\right)=\frac{d}{dx}(1-\frac{1}{x})=\frac{1}{x^2}$
$\frac{d}{dx}\left(g_{\left(3n\right)}\right(x\left)\right)=\frac{d}{dx}\left(x\right)=1$

$\underset{x\to 1}{lim}\frac{\sum _{r=1}^{100}\left(g_{3r}\right(x){)}^r+x-101}{x-1}$
$=\underset{x\to 1}{lim}\frac{x+x^2+\cdots +x^{100}+x-101}{x-1}$
Applying L'Hopitals' rule, we get
$\underset{x\to 1}{lim}\frac{1+2x+\cdots +100x^{99}+1}{1}=1+2+3+4+\cdots +100+1$
$=\frac{100\times 101}{2}+1=5051$

$y=g_{80}\left(x\right)=\frac{1}{1-x}$
At $x=2$, $y=-1$
$y^{\prime }=\frac{1}{(1-x{)}^2}$
$y^{\prime }{|}_{x=2}=1$
Equation of tangent at $(2,-1)$ is
$y+1=1(x-2)$
$\Rightarrow x-y=3$