Let a function f: ℝ → ℝ be defined as :
f(x)= ∫₀ˣ(5-|t-3|) d t, x>4 ; x²+b x , x ≤ 4
where b ∈ ℝ. If f is continuous at x=4, then which of the following statements is NOT true?

Solution

Maths sir · 4 steps

Use continuity to find b, then read the sign of f′ on each piece; option (c) has the intervals backwards.

1. Open the integral. For t < 3 the integrand is 2 + t, and for t > 3 it is 8 - t. Splitting at t = 3 gives 21/2 + 8(x-3) - (x² - 9)/2 for x > 4, which simplifies to the form below.

   f(x) = -x²/2 + 8x - 9 (x > 4), f(x) = x² + bx (x ≤ 4)

2. Continuity gives b. The right piece tends to -8 + 32 - 9 = 15 as x → 4⁺. The left piece gives 16 + 4b. Setting 16 + 4b = 15 gives b = -1/4.
3. Derivatives. For x < 4, f'(x) = 2x - 1/4, which is negative before 1/8 and positive after it. For x > 4, f'(x) = 8 - x, which is positive on (4, 8) and negative after 8.

   Option	Check	Statement is
   (a)	Left derivative at 4 is 31/4, right derivative is 4	true
   (b)	f'(3) + f'(5) = 23/4 + 3 = 35/4	true
   (c)	f' < 0 on (-∞, 1/8) and on (8, ∞), so f decreases there	NOT true
   (d)	f' changes from negative to positive at 1/8	true

4. So. f increases on (1/8, 8) and decreases on (-∞, 1/8) ∪ (8, ∞). The statement that is NOT true is (c).

The idea to keep. For an integral with |t - a| inside, split it at t = a before you integrate. For monotonicity, find where f' changes sign on each piece separately, and check the joining point on its own.

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