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Which Transformations of an Increasing Function Stay Increasing | JEE

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Question
Let f(x)f(x) be a derivable function which is increasing for all x∈Rx \in \mathbb{R} (having no critical point), then
Af(3−4x)f(3-4x) is an increasing function for all xx
Bf(3−4x)f(3-4x) is a decreasing function for all xxcorrect
Cf(x2−x)f\left(x^{2}-x\right) increases for x>12x > \dfrac12correct
D(f(x))3\left(f(x)\right)^{3} is an increasing function for all xxcorrect
Solution
Step 1: Turn the hypothesis into an inequality. "Increasing with no critical point" means
f′(x)>0for every x∈R.f'(x) > 0 \quad \text{for every } x \in \mathbb{R} .
 Step 2: Differentiate the function in (1) and (B) by the chain rule.
ddxf(3−4x)=f′(3−4x)⋅(−4).\frac{d}{dx}f(3-4x) = f'(3-4x)\cdot(-4) .
 Step 3: Read off the sign. The factor f′(3−4x)f'(3-4x) is positive by Step 1 and −4-4 is negative, so the derivative is negative everywhere:
f(3−4x) is decreasing for all x.f(3-4x) \ \text{is decreasing for all } x .
 So (B) is true and (1) is false. Step 4: Differentiate the function in (3).
ddxf(x2−x)=f′(x2−x)⋅(2x−1).\frac{d}{dx}f\left(x^{2}-x\right) = f'\left(x^{2}-x\right)\cdot(2x-1) .
 Step 5: Again f′(x2−x)>0f'\left(x^{2}-x\right) > 0, so the sign is that of 2x−12x-1:
2x−1>0  ⟺  x>12.2x-1 > 0 \iff x > \frac12 .
 So (3) is true. Step 6: Differentiate the function in (4).
ddx(f(x))3=3(f(x))2f′(x).\frac{d}{dx}\left(f(x)\right)^{3} = 3\left(f(x)\right)^{2}f'(x) .
 Step 7: Both factors are non-negative: (f(x))2≥0\left(f(x)\right)^{2} \ge 0 always and f′(x)>0f'(x) > 0. The derivative can be zero only where f(x)=0f(x) = 0, and since ff is strictly increasing that happens at most once - a single point, which does not stop the function increasing. So (4) is true. Answer: (2), (3) and (4).
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