If the function f(x) = { (log ₑ(1 - x + x²) + log ₑ(1 + x + x²))/(sec x - cos x), ,, x ∈ ( (- π)/2,π/2 ) - { 0}; k, ,, x = 0 is continuous at x = 0, then k is equal to:

Solution

Maths sir · 4 steps

Join the two logs into one log and rewrite sec x - cos x; the 0/0 form then falls apart.

1. Combine the logs. The numerator is ln[(1+x²-x)(1+x²+x)] = ln[(1+x²)² - x²] = ln(1+x²+x⁴).
2. Rewrite the denominator. sec x - cos x = (1-cos² x)/(cos x) = (sin² x)/(cos x). At x = 0 both the top and the bottom are 0, so you have to find the limit.

   lim(x→ 0) f(x) = lim(x→ 0) ln(1+x²+x⁴)/(x²+x⁴) × x²(1+x²)/(sin² x) × cos x

3. Use standard limits. As x → 0, ln(1+t)/t → 1 with t = x²+x⁴, and x²/(sin² x) → 1. The remaining factors 1+x² and cos x both tend to 1, so the limit is 1 × 1 × 1 × 1 = 1.
4. So. For continuity, k = lim(x→ 0) f(x) = 1. The answer is (a).

The idea to keep. In a 0/0 limit, first simplify: merge logs, change sec and tan into sin and cos, then pair each piece with a standard limit like ln(1+t)/t or (sin x)/x.

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