If the function $f(x)=\left\{\begin{array}{cc}\frac{2}{x}\{\sin \left(k_{1}+1\right) x+\sin \left(k_{2}-1\right) x\} & , x 0\end{array}\right.$ is continuous at $\mathrm{x}=0$, then $\mathrm{k}_{1}{ }^{2}+\mathrm{k}_{2}{ }^{2}$ is equal to

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Continuity and Differentiability

If the function

f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x} 4 \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x < 0, x = 0, x > 0

is continuous at

x = 0

, then

k_{1}^{2} + k_{2}^{2}

is equal to

Answer: (D)

Continuity at

x = 0

requires the left-hand and right-hand limits both to equal

f (0) = 4

.

Left-hand limit, using

sin θ \approx θ

for small

θ

x \to 0^{-} lim \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x} = 2 [(k_{1} + 1) + (k_{2} - 1)] = 2 (k_{1} + k_{2})

Setting this equal to

4

k_{1} + k_{2} = 2

.

Right-hand limit, using

ln (1 + u) \approx u

for small

u

x \to 0^{+} lim \frac{2}{x} ln (\frac{2 + k _{1} x}{2 + k _{2} x}) = x \to 0^{+} lim \frac{2}{x} ln (1 + \frac{( k _{1} - k _{2} ) x}{2 + k _{2} x}) = k_{1} - k_{2}

Setting this equal to

4

k_{1} - k_{2} = 4

.

Solving the two equations:

k_{1} = 3

k_{2} = - 1

k_{1}^{2} + k_{2}^{2} = 9 + 1 = 10

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option