Question

If the derivative of \(tan^{-1}(a+bx)\) with respect to x at \(x = 0\) is \(1\), then \(a^6-b^3 =\)

• A. \(3a^2b-1\)
• B. \(-3ab^2+1\)
• C. \(-1-3a^2b\)
• D. \(1+3ab^2\)

Hint:

Differentiate tan^-1(a + bx), set x = 0, and expand (1 + a^2)^3.

Answer 1

The Correct Option is C

Step 1: Differentiate:
\(\dfrac{d}{dx}\tan^{-1}(a+bx)=\dfrac{b}{1+(a+bx)^2}\).

Step 2: Apply the condition at x = 0:
\(\dfrac{b}{1+a^2}=1\), so \(b=1+a^2\).

Step 3: Compute a^6 - b^3:
\(b^3=(1+a^2)^3=1+3a^2+3a^4+a^6\). So \(a^6-b^3=-1-3a^2-3a^4\).

Step 4: Match with the options:
Option C: \(-1-3a^2b=-1-3a^2(1+a^2)=-1-3a^2-3a^4\). This matches.

Step 5: Why the other options are wrong.
Options A, B and D carry the wrong sign or wrong power pattern, for example D, \(1+3ab^2\), is positive and cannot be the negative expression found.

Final Answer:
a^6 - b^3 = -1 - 3a^2 b. \[ \boxed{\text{(C) }-1-3a^2b} \]