Let f and g be the twice differentiable functions on ℝ such that
f''(x)=g''(x)+6x
f'(1)=4g'(1)-3=9
f(2)=3g(2)=12.
Then which of the following is NOT true?

Solution

Maths sir · 3 steps

Do not chase f and g separately: put h = f − g, integrate 6x twice, and test every option on h.

1. One function only. Let h(x) = f(x) - g(x). Then h''(x) = 6x. The conditions give f'(1) = 9, g'(1) = 3, f(2) = 12 and g(2) = 4, so h'(1) = 6 and h(2) = 8.
2. Integrate twice. h'(x) = 3x² + c, and h'(1) = 6 gives c = 3. Then h(x) = x³ + 3x + d, and h(2) = 8 + 6 + d = 8 gives d = -6.

   h(x) = f(x) - g(x) = x³ + 3x - 6, h'(x) = 3x² + 3

   Option	Check on h	Verdict
   (a)	g(-2) - f(-2) = -h(-2) = -(-8 - 6 - 6) = 20	true
   (b)	h(1) = -2 < 0 and h(3/2) = 1.875 > 0, so h has a root in (1, 3/2)	true
   (c)	3x² + 3 < 6 gives x² < 1, that is -1 < x < 1	true
   (d)	h is increasing, h(-1) = -10 and h(2) = 8, so |h| gets close to 10 near x = -1	not true

3. So. Since h'(x) = 3x² + 3 > 0, h takes every value between -10 and 8 on (-1, 2). For x just above -1, |f(x) - g(x)| is close to 10, which is more than 8. The statement that is NOT true is (d).

The idea to keep. When conditions link f and g through derivatives, work with their difference. On an interval, check |h| at both ends. An increasing function is largest in size at whichever end is farther from zero.

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