JEE Main 2020 · 9 January · morning · Single correct · +4 / −1
If for all real triplets (a, b, c), ƒ(x) = a + bx + cx²; then ∫₀¹ f(x)dx is equal to :
- (a) 1/6{ f(0) + f(1) + 4f( 1/2 ) }
- (b) 2{ 3f(1) + 2f( 1/2 ) }
- (c) 1/3{ f(0) + f( 1/2 ) }
- (d) 1/2{ f(1) + 3f( 1/2 ) }
Answer and solution
Answer (a) 1/6{ f(0) + f(1) + 4f( 1/2 ) }
Solution
Maths sir · 4 steps
Integrate the quadratic, then expand option (a). Both give exactly a + b/2 + c/3.
- Integrate directly. Integrate term by term: ∫₀¹ (a + bx + cx²) dx = a + b/2 + c/3.
- Values of f. f(0) = a, f(1) = a + b + c and f(1/2) = a + b/2 + c/4.
f(0) + f(1) + 4f(1/2) = a + (a + b + c) + (4a + 2b + c) = 6a + 3b + 2c
- Divide by 6. 1/6(6a + 3b + 2c) = a + b/2 + c/3. This is the integral for every a, b, c. This is Simpson's rule, which is exact for quadratics.
- So. The answer is (a). Test the others with f(x) = 1, whose integral is 1: option (b) gives 10, (c) gives 2/3 and (d) gives 2. All three fail.
The idea to keep. When a formula must hold for every a, b, c, test it with the simplest case first. f(x) = 1 alone rules out most wrong options in one line.
Answer from the NTA final answer key. Solution by PrepWiser.