JEE Main 2023 · 11 April · morning · Single correct · +4 / −1
Let f:[2,4] → ℝ be a differentiable function such that (x log ₑ x) f′(x)+(log ₑ x) f(x)+f(x) ≥ 1, x ∈[2,4] with f(2)=1/2 and f(4)=1/4.
Consider the following two statements :
(A) : f(x) ≤ 1, for all x ∈[2,4]
(B) : f(x) ≥ 1/8, for all x ∈[2,4]
Then,
- (a) Neither statement (A) nor statement (B) is true
- (b) Only statement (A) is true
- (c) Only statement (B) is true
- (d) Both the statements (A) and (B) are true
Answer and solution
Answer (d) Both the statements (A) and (B) are true
Solution
Maths sir · 5 steps
The left side is the derivative of x ln x f(x), so trap that product from both ends.
- Spot the product. Let g(x) = x ln x f(x). Since (x ln x)' = ln x + 1, we get g'(x) = x ln x f'(x) + (ln x + 1)f(x), which is exactly the given left side. So g'(x) ≥ 1, with g(2) = 2 ln 2 × 1/2 = ln 2 and g(4) = 4 ln 4 × 1/4 = 2 ln 2.
- Apply MVT. On [2, x], the Mean Value Theorem gives g(x) - g(2) = g'(c)(x - 2) ≥ x - 2. On [x, 4], it gives g(4) - g(x) ≥ 4 - x. Together:
x - 2 + ln 2 ≤ x ln x f(x) ≤ 2 ln 2 - (4 - x)
- Lower bound. On [2,4], x - 2 ≥ 0, so x ln x f(x) ≥ ln 2. Also x ln x ≤ 4 ln 4 = 8 ln 2. Dividing gives f(x) ≥ (ln 2)/(8 ln 2) = 1/8, so (B) holds.
- Upper bound. On [2,4], 4 - x ≥ 0, so x ln x f(x) ≤ 2 ln 2. Also x ln x ≥ 2 ln 2 > 0. Dividing gives f(x) ≤ (2 ln 2)/(x ln x) ≤ 1, so (A) holds.
- So. Both statements are true, so the answer is (d).
The idea to keep. When a messy combination of f and f' appears, look for a product whose derivative it is. Then apply MVT from each end point to trap the product between two bounds.
Answer from the NTA final answer key. Solution by PrepWiser.