Question

Let $\ln x$ denote the logarithm of $x$ with respect to the base $e$. Let $S\subset R$ be the set of all points where the function $\ln (x^2–1)$ is well-defined. Then the number of functions $f:S\to R$ that are differentiable, satisfy $f^{\prime }\left(x\right)=\ln (x^2–1)$ for all $x\in S$ and $f\left(2\right)=0$, is

• A. 0.0
• B. 1
• C. 2
• D. infinite

Answer 1

The correct option is D infinite

Given:
$\ln (x^2-1)$ exists when $x^2-1>0$.

$\Rightarrow (x-1)(x+1)>0$

$\therefore S:x\in (-\infty ,-1)\cup (1,\infty )$

$f^{\prime }\left(x\right)=\ln (x^2-1)$

$\Rightarrow \int f^{\prime }\left(x\right)dx=\int \ln (x^2-1)dx$

using integration by parts, $\int f\left(x\right).g\left(x\right)dx=f\left(x\right).g^{\prime }\left(x\right)-\int \left(f^{\prime }\right(x)\int g(x\left)dx\right)dx+C$

$\Rightarrow f\left(x\right)=x.\ln (x^2-1)-\int x.\frac{2x}{x^2-1}dx$---(1)

finding $\int \frac{x^2}{x^2-1}dx$

$=\int \frac{x^2-1+1}{x^2-1}dx$

$=\int 1.dx+\int \frac{1}{x^2-1}dx$

$=x+\frac{1}{2}\ln |\frac{x-1}{x+1}|$ [Since, $\int \frac{1}{x^2-a^2}dx=\frac{1}{2}\ln |\frac{x-a}{x+a}|+C$]

From (1),

$f\left(x\right)=x.\ln (x^2-1)-\int x.\frac{2x}{x^2-1}dx$

$\Rightarrow f\left(x\right)=x.\ln (x^2-1)-2(x+\frac{1}{2}\ln |\frac{x-1}{x+1}|)$

$\Rightarrow f\left(x\right)=x\ln (x^2-1)-2x-\ln \left|\frac{x-1}{x+1}\right|+C$

$f\left(x\right)=x\ln (x^2-1)-2x-\ln \frac{x-1}{x+1}+C_1;\&x>1$ and

$x\ln (x^2-1)-2x-\ln \frac{1-x}{x+1}+C_2;\&x<-1$

given: $f\left(2\right)=0$