If the function
f(x) = { 1/xlog ₑ( (1 + x/a)/(1 - x/b) ), ,, x < 0; k, ,, x = 0; (cos ²x - sin ²x - 1)/(√(x² + 1) - 1), ,, x > 0 is continuous
at x = 0, then 1/a + 1/b + 4/k is equal to :

Solution

Maths sir · 4 steps

Both one-sided limits must equal k, and the left one gives 1/a+1/b directly.

1. Left-hand limit. Split the log: 1/x[ln(1+x/a) - ln(1-x/b)]. Since ln(1+t)/t → 1 as t → 0, the limit is 1/a + 1/b.
2. Right-hand limit. The numerator is cos 2x - 1 = -2 sin² x. Rationalise the denominator: √(x²+1)-1 = x²/(√(x²+1)+1).

   lim(x → 0⁺) (-2 sin² x(√(x²+1)+1))/x² = -2 × 1 × 2 = -4

3. Match with k. Continuity at x = 0 needs both limits equal to f(0) = k. So k = -4 and 1/a + 1/b = -4.
4. So. 1/a + 1/b + 4/k = -4 + 4/-4 = -4 - 1 = -5, which is option (a).

The idea to keep. For continuity, set the left limit, the right limit and f(0) all equal. Use standard limits like ln(1+t)/t → 1 and (sin x)/x → 1, and rationalise any root minus 1.

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