JEE Main 2020 · 9 January · morning · Single correct · +4 / −1
If f(x) = { (sin (a + 2)x + sin x)/x;, x < 0; b ;, x = 0; (( x + 3x² )^(1/3) - x^(1/3))/x^(4/3);, x > 0
is continuous at x = 0, then a + 2b is equal to :
- (a) 0
- (b) -1
- (c) -2
- (d) 1
Answer and solution
Answer (a) 0
Solution
Maths sir · 4 steps
The right-hand limit fixes b first. Then match the left-hand limit to it and a falls out.
- Right-hand limit. For x > 0, take x^(1/3) out of the top: (x + 3x²)^(1/3) - x^(1/3) = x^(1/3)[(1+3x)^(1/3) - 1]. Dividing by x^(4/3) leaves ((1+3x)^(1/3) - 1)/x.
lim(x → 0⁺) ((1+3x)^(1/3) - 1)/x = lim(x → 0⁺) 3 × ((1+3x)^(1/3) - 1)/3x = 3 × 1/3 = 1
- Value of b. Here we used lim(y → 0) ((1+y)ⁿ - 1)/y = n with y = 3x and n = 1/3. Continuity needs f(0) to equal this limit, so b = 1.
- Left-hand limit. For x < 0, split the fraction: (sin (a+2)x)/x + (sin x)/x → (a+2) + 1 = a + 3. Setting a + 3 = 1 gives a = -2.
- So. a + 2b = -2 + 2 × 1 = 0, which is option (a).
The idea to keep. For continuity at a point, the left limit, the right limit and the function value must all be equal. Find whichever side has no unknown first, then use it to fix the others.
Answer from the NTA final answer key. Solution by PrepWiser.