Verified Solution • 2025–26 Session moderate

QJEEM24Maa5b1: JEE Main 2024 Mathematics - Class 12 JEEM24Maa5b1 Solution: Let be a non constant twice differentiable function such that . …

Let g:R→Rg: \mathbf{R} \rightarrow \mathbf{R} be a non constant twice differentiable function such that g′(12)=g′(32)\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right) . If a real valued function ff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\frac{1}{2}[g(x)+g(2-x)] , then

Multiple Choice Options

  • A f′′(x)=0f^{\prime \prime}(x)=0 for atleast two xx in (0,2)(0,2)
  • B f′(32)+f′(12)=1f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1
  • C f′′(x)=0f^{\prime \prime}(x)=0 for no xx in (0,1)(0,1)
  • D f′′(x)=0f^{\prime \prime}(x)=0 for exactly one xx in (0,1)(0,1)

Step-by-Step Verified Solution

f′(x)=g′(x)−g′(2−x)2,f′(32)=g′(32)−g′(12)2=0f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0

Also f′(12)=g′(12)−g′(32)2=0,f′(12)=0\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0

⇒f′(32)=f′(12)=0\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0

⇒rootsin⁡(12,1)\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right) and (1,32)\left(1, \frac{3}{2}\right)

⇒f′′(x)\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x}) is zero at least twice in (12,32)\left(\frac{1}{2}, \frac{3}{2}\right)

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