Verified Solution • 2025–26 Session moderate

QJEEM25Ma6a61: JEE Main 2025 Mathematics - Class 12 JEEM25Ma6a61 Solution: If the function is continuous at , then is equal to :

If the function

f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},xlt;04,x=02xlog⁡e(2+k1x2+k2x),xgt;0f(x)=\left\{\begin{array}{l}\frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x<0 \\4, \quad x=0 \\\frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x>0\end{array}\right.

is continuous at x=0x=0 , then k12+k22k_1^2+k_2^2 is equal to :

Multiple Choice Options

  • A 5
  • B 10
  • C 20
  • D 8

Step-by-Step Verified Solution

amp;lim⁡x→0−2x{sin⁡(k1+1)x+sin⁡(k2−1)x}=4amp;⇒2(k1+1)+2(k2−1)=4amp;⇒k1+k2=2amp;⇒lim⁡x→0+2xln⁡(2+k1x2+k2x)=4amp;⇒lim⁡x→0+1xln⁡(1+(k1−k2)x2+k2x)=2amp;⇒k1−k22=2amp;⇒k1−k2=4amp;∴k1=3,k2=−1amp;k12+k22=9+1=10\begin{aligned}& \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\& \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\& \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\& \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\& \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\& \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\& \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\& \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\& \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10\end{aligned}

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