Question

Match List-I with List-II 

List-I	Function	List-II	\(n^{\text{th}}\) Derivative Form
A	\(e^{ax}\)	IV	\(a^n e^{ax}\)
B	\(\log(ax+b)\)	III	\(\dfrac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n}\)
C	\(\cos(ax+b)\)	II	\(a^n\cos\left(ax+b+\dfrac{n\pi}{2}\right)\)
D	\(e^{ax}\cos(bx+c)\)	I	\(r^n e^{ax}\cos(bx+c+n\theta)\),  where \(r=\sqrt{a^2+b^2}\), \(\theta=\tan^{-1}\left(\dfrac{b}{a}\right)\)

• A. A-IV, B-III, C-II, D-I
• B. A-III, B-I, C-IV, D-II
• C. A-II, B-I, C-III, D-IV
• D. A-III, B-IV, C-II, D-I

Hint:

For \(n^{th}\) derivatives, memorize standard forms of \(e^{ax}\), \(\log(ax+b)\), \(\cos(ax+b)\), and \(e^{ax}\cos(bx+c)\).

Answer 1

The Correct Option is A

Concept:
This question is based on standard formulae for \(n^{th}\) derivatives.

Step 1: \(e^{ax}\). \[ \frac{d^n}{dx^n}(e^{ax})=a^ne^{ax} \] So: \[ A\rightarrow IV \]

Step 2: \(\log(ax+b)\). \[ \frac{d^n}{dx^n}\log(ax+b)=\frac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n} \] So: \[ B\rightarrow III \]

Step 3: \(\cos(ax+b)\). \[ \frac{d^n}{dx^n}\cos(ax+b)=a^n\cos\left(ax+b+\frac{n\pi}{2}\right) \] So: \[ C\rightarrow II \]

Step 4: \(e^{ax}\cos(bx+c)\). \[ \frac{d^n}{dx^n}\left(e^{ax}\cos(bx+c)\right) = r^ne^{ax}\cos(bx+c+n\theta) \] where: \[ r=(a^2+b^2)^{1/2},\quad \theta=\tan^{-1}\frac{b}{a} \] So: \[ D\rightarrow I \] Therefore: \[ A-IV,\ B-III,\ C-II,\ D-I \] \[ \therefore \text{Correct Answer is (A)} \]