Question

If
\[ \frac{d}{dx}\left[a\tan^{-1}x+b\log\left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1} \] then \(a-2b\) is equal to

• A. \(1\)
• B. \(-1\)
• C. \(0\)
• D. \(2\)

Hint:

Convert rational expressions into partial fractions and compare coefficients with derivative form to find constants quickly.

Answer 1

The Correct Option is B

Step 1: Differentiate given expression.
\[ \frac{d}{dx}\left(a\tan^{-1}x\right)=\frac{a}{1+x^2} \]
Now:
\[ \frac{d}{dx}\left[b\log\left(\frac{x-1}{x+1}\right)\right] =b\cdot\frac{d}{dx}\left[\log(x-1)-\log(x+1)\right] \]
\[ =b\left(\frac{1}{x-1}-\frac{1}{x+1}\right) =b\left(\frac{(x+1)-(x-1)}{x^2-1}\right) =\frac{2b}{x^2-1} \]
So total derivative:
\[ \frac{a}{1+x^2}+\frac{2b}{x^2-1} \]
Step 2: Write RHS using partial fractions.
\[ \frac{1}{x^4-1}=\frac{1}{(x^2-1)(x^2+1)} \]
Assume:
\[ \frac{1}{(x^2-1)(x^2+1)}=\frac{A}{x^2+1}+\frac{B}{x^2-1} \]
Multiply both sides by \((x^2-1)(x^2+1)\):
\[ 1=A(x^2-1)+B(x^2+1) \]
\[ 1=(A+B)x^2+(-A+B) \]
Compare coefficients:
\[ A+B=0 \Rightarrow B=-A \]
\[ -B+A? \Rightarrow -A+B=1 \]
Substitute \(B=-A\):
\[ -A-A=1 \Rightarrow -2A=1 \Rightarrow A=-\frac{1}{2} \]
\[ B=\frac{1}{2} \]
Thus:
\[ \frac{1}{x^4-1}=-\frac{1}{2(x^2+1)}+\frac{1}{2(x^2-1)} \]
Step 3: Compare with derivative.
\[ \frac{a}{x^2+1}+\frac{2b}{x^2-1}=-\frac{1}{2(x^2+1)}+\frac{1}{2(x^2-1)} \]
So:
\[ a=-\frac{1}{2},\quad 2b=\frac{1}{2}\Rightarrow b=\frac{1}{4} \]
Step 4: Compute \(a-2b\).
\[ a-2b=-\frac{1}{2}-2\cdot\frac{1}{4} =-\frac{1}{2}-\frac{1}{2}=-1 \]
Final Answer:
\[ \boxed{-1} \]