Verified Solution • 2025–26 Session moderate

QJEEM25Ma6cd0: JEE Main 2025 Mathematics - Class 12 JEEM25Ma6cd0 Solution: For some let Then is equal to :

For some a,b,a, b, let f(x)=∣a+sin⁡xxamp;1amp; baamp;1+sin⁡xxamp; baamp;1amp; b+sin⁡xx∣,x≠0,lim⁡x→0f(x)=λ+μa+νb.f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}\right|, x \neq 0, \lim \limits_{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}. Then (λ+μ+v)2(\lambda+\mu+v)^2 is equal to :

Multiple Choice Options

  • A 25
  • B 16
  • C 9
  • D 36

Step-by-Step Verified Solution

amp;lim⁡x→0f(x)=∣a+1amp;1amp;baamp;1+1amp;baamp;1amp;b+1∣amp;=(a+1)(2(b+1)−b)+1(ab−a(b+1))+baamp;=(a+1)(b+2)−a+abamp;=b+a+2=λ+μa+vbamp;λ=2,μ=1,v=1⇒(λ+μ+v)2=16\begin{aligned}& \lim _{x \rightarrow 0} f(x)=\left|\begin{array}{ccc}a+1 & 1 & b \\a & 1+1 & b \\a & 1 & b+1\end{array}\right| \\& =(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a \\& =(a+1)(b+2)-a+a b \\& =b+a+2=\lambda+\mu a+v b \\& \lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)^2=16\end{aligned}

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