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QJEEAdv23Ma079a: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma079a Solution: Let be the function defined as , where denotes the greatest…

Let f:(0,1)→Rf:(0,1) \rightarrow \mathbb{R} be the function defined as f(x)=[4x](x−14)2(x−12)f(x)=[4 x]\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) , where [x][x] denotes the greatest integer less than or equal to xx . Then which of the following statements is(are) true?

Multiple Choice Options

  • A The function ff is discontinuous exactly at one point in (0,1)(0,1)
  • B There is exactly one point in (0,1)(0,1) at which the function ff is continuous but NOT differentiable
  • C The function ff is NOT differentiable at more than three points in (0,1)(0,1)
  • D The minimum value of the function ff is −1512-\frac{1}{512}

Step-by-Step Verified Solution

f(x)={0amp;;0lt;xlt;14(x−14)2(x−12);amp;14≤xlt;122(x−14)2(x−12);amp;12≤xlt;343(x−14)2(x−12);amp;34≤xlt;1f(x)=\left\{\begin{array}{cc}0 & ; 0 < x <\frac{1}{4} \\\\\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{4} \leq x<\frac{1}{2} \\\\2\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{2} \leq x<\frac{3}{4} \\\\3\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{3}{4} \leq x<1\end{array}\right.

f(x) is discontinuous at x=34 only f(x) \text { is discontinuous at } x=\frac{3}{4} \text { only }

f′(x)={0amp;;0lt;xlt;142(x−14)(x−12)+(x−14)2amp;;14lt;xlt;124(x−14)(x−12)+2(x−14)2amp;;12lt;xlt;346(x−14)(x−12)+3(x−14)2amp;;34lt;xlt;1f^{\prime}(x)=\left\{\begin{array}{cc}0 & ; 0 < x <\frac{1}{4} \\\\2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4}< x <\frac{1}{2} \\\\4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \frac{1}{2} < x <\frac{3}{4} \\\\6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \frac{3}{4} < x < 1\end{array}\right.

f(x)f(x) is non-differentiable at x=12x=\frac{1}{2} and 34\frac{3}{4}

Minimum values of f(x)f(x) occur at x=512x=\frac{5}{12} whose value is −1432-\frac{1}{432}

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