Which of the following statements are correct ?A. The function f(x)=a…

The correct answer is Option (2) → A, B and D only

A function increases where f′(x) > 0 and decreases where f′(x) < 0.

For \( f(x)=a^x \):

    \( \displaystyle f'(x)=a^x\log_e a \)

For \( f(x)=x^2-2x \):

    \( \displaystyle f'(x)=2x-2=2(x-1) \)

For \( f(x)=x(x-3)^2 \):

    \( \displaystyle f'(x)=(x-3)^2+2x(x-3) \)

    \( \displaystyle =(x-3)(3x-3) \)

    \( \displaystyle =3(x-1)(x-3) \)

For \( f(x)=e^{-x} \):

    \( \displaystyle f'(x)=-e^{-x} \)

So \( a^x\log_e a\gt 0 \) for a > 1, \( 2(x-1)\lt 0 \) for x < 1, \( 3(x-1)(x-3)\lt 0 \) for 1 < x < 3, and \( -e^{-x}\lt 0 \) for every x.

Checking the statements

(A) The function \( f(x)=a^x \) is increasing on \( (-\infty,\infty) \) if \( a\gt 1 \). — Correct. \( a^x\gt 0 \) and \( \log_e a\gt 0 \), so f′(x) > 0.

(B) \( f(x)=x^2-2x \) is decreasing in \( (-\infty,1) \). — Correct. f′(x) = 2(x − 1) < 0 for x < 1.

(C) \( f(x)=x(x-3)^2 \) increases for \( 1\le x\le 3 \). — Incorrect. f′(x) = 3(x − 1)(x − 3) < 0 there, so f decreases.

(D) \( f(x)=e^{-x} \) decreases in \( [0,\infty) \). — Correct. f′(x) = −e^−x < 0 for every x.