For a, b>0, let f(x)= (tan ((a+1) x)+b tan x)/x, x< 0 ; 3, x=0 ; (√(a x+b² x²)-√(a x))/(b √a x √x), x> 0 be a continuous function at x=0. Then b/a is equal to :

Solution

Maths sir · 4 steps

Pull √x out of the right-hand piece, rationalise, and the limit gives b/2a directly.

1. Left-hand limit. Use lim(x → 0) (tan kx)/x = k on each term. The left limit is (a+1) + b. Continuity needs it to equal f(0) = 3, so a + b = 2.
2. Simplify the right piece. For x > 0, write √(ax + b²x²) = √x √(a + b²x) and √ax = √x √a. Cancel √x from top and bottom, then multiply by the conjugate √(a + b²x) + √a.

   (√(a + b²x) - √a)/(b√a x) = b²x/(b√a x(√(a + b²x) + √a)) → b/(√a· 2√a) = b/2a

3. Match with f(0). The right limit must also equal 3, so b/2a = 3, which gives b = 6a. With a + b = 2 this means a = 2/7 and b = 12/7, both positive as required.
4. So. b/a = 6, which is option (d).

The idea to keep. For continuity at a point, set the left limit, the right limit and the function value equal. A difference of square roots over x almost always needs rationalising first.

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