JEE Main 2021 · 18 March · evening · Single correct · +4 / −1
Let f : R → R be a function defined as
f(x) = { (sin (a + 1)x + sin 2x)/2x,if x < 0; b, if x = 0; (√(x + bx³) - √x)/bx^(5/2), if x > 0
If f is continuous at x = 0, then the value of a + b is equal to :
- (a) -3
- (b) -2
- (c) - 5/2
- (d) - 3/2
Answer and solution
Answer (d) - 3/2
Solution
Maths sir · 4 steps
Find both one-sided limits, set each equal to b, and the two equations give a and b.
- Left-hand limit. Split the fraction as (sin (a+1)x)/2x + (sin 2x)/2x. Using (sin kx)/x → k, the first part tends to (a+1)/2 and the second to 1. So the left-hand limit is (a+3)/2.
- Right-hand limit. Take √x common from the numerator, which leaves (√(1+bx²)-1)/bx². Multiply top and bottom by √(1+bx²)+1, and the bx² cancels.
lim(x → 0⁺) (√(1+bx²)-1)/bx² = lim(x → 0⁺) 1/(√(1+bx²)+1) = 1/2
- Apply continuity. Continuity at 0 needs both limits to equal f(0) = b. The right-hand limit gives b = 1/2. Then (a+3)/2 = 1/2 gives a = -2.
- So. a + b = -2 + 1/2 = -3/2, which is option (d).
The idea to keep. For continuity at a point, left limit = right limit = function value. When a square-root difference tends to 0/0, multiply by its conjugate to cancel the factor that is going to zero.
Answer from the NTA final answer key. Solution by PrepWiser.