Let f(x) = 2x + tan ⁻¹x and g(x) = log ₑ(√(1 + x²) + x),x ∈ [0,3]. Then

Solution

Maths sir · 3 steps

Differentiate both: f' stays above 2, g' never goes above 1, and every option follows from that.

1. Find the derivatives. f'(x) = 2 + 1/(1+x²). For g, the chain rule gives g'(x) = 1/(√(1+x²)+x)(x/√(1+x²) + 1) = 1/√(1+x²).
2. Compare their ranges. On [0,3] both derivatives decrease as x grows. So f' runs from 3 at x=0 down to 2 + 1/10 = 2.1 at x=3, and g' runs from 1 at x=0 down to 1/√10 ≈ 0.32 at x=3. Hence f'(x) > g'(x) everywhere on [0,3].

   Option	Check	Verdict
   (a)	f' ≥ 2.1 and g' ≤ 1, so f' < g' never happens	false
   (b)	f(0) = g(0) = 0 and f - g is increasing, so f(x) > g(x) for all x > 0	false
   (c)	min f' = 2.1 but 1 + max g' = 1 + 1 = 2	false
   (d)	max f = f(3) = 6 + tan⁻¹ 3 ≈ 7.25, while max g = g(3) = ln(3+√10) ≈ 1.82	true

3. So. Both functions are increasing, so their maxima sit at x = 3, and there f is far bigger than g. The answer is (d).

The idea to keep. To compare two functions on an interval, compare their derivatives first. If they start equal and one always grows faster, it stays ahead for the whole interval.

Answer from the NTA final answer key. Solution by PrepWiser.