If the function
f(x)={2/x{sin (k₁+1) x+sin (k₂-1) x}, x<0; 4, x=0; 2/x log ₑ((2+k₁ x)/(2+k₂ x)), x>0
is continuous at x=0, then k₁²+k₂² is equal to :

Solution

Maths sir · 5 steps

Turn each side into a standard limit: the sine side gives k₁ + k₂, the log side gives k₁ - k₂.

1. Condition. Continuity at x = 0 means the left limit and the right limit both equal f(0) = 4.
2. Left limit. Use (sin ax)/x → a. The left limit is 2[(k₁ + 1) + (k₂ - 1)] = 2(k₁ + k₂). Setting it equal to 4 gives k₁ + k₂ = 2.
3. Right limit. Divide the top and bottom of the fraction inside the log by 2. The log becomes ln(1 + (k₁ x)/2) - ln(1 + (k₂ x)/2). Then use ln(1 + ax)/x → a.

   lim(x → 0⁺) 2/x[ln(1+(k₁ x)/2) - ln(1+(k₂ x)/2)] = 2(k₁/2 - k₂/2) = k₁ - k₂

4. Solve. Setting the right limit equal to 4 gives k₁ - k₂ = 4. Together with k₁ + k₂ = 2, this gives k₁ = 3 and k₂ = -1.
5. So. k₁² + k₂² = 9 + 1 = 10, which is option (b).

The idea to keep. For continuity puzzles, reduce each piece to (sin ax)/x → a or ln(1+ax)/x → a. Each side then gives you one linear equation in the unknowns.

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