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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Let $f:(0,1) \rightarrow R$ be the function defined as $f(x)=[4 x]\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right)$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then which of the following statements is(are) true?

($A$) The function $f$ is discontinuous exactly at one point in $(0,1)$

($B$) There is exactly one point in $(0,1)$ at which the function $f$ is continuous but $NOT$ differentiable

($C$) The function $\mathrm{f}$ is $NOT$ differentiable at more than three points in $(0,1)$

($D$) The minimum value of the function $f$ is $-\frac{1}{512}$

  • A
    $B,C$
  • ✓
    $A,B$
  • C
    $B,A$
  • D
    $B,C,D$
✓

Answer

Correct option: B.
$A,B$
b
$f(x)=\left\{\begin{array}{cc}0 & ; \quad 0 < x < \frac{1}{4} \\ \left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) & ; \quad \frac{1}{4} \leq x < \frac{1}{2} \\ 2\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) & ; \quad \frac{1}{2} \leq x < \frac{3}{4} \\ 3\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) & ; \quad \frac{3}{4} \leq x < 1\end{array}\right.$

$\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=\frac{3}{4}$ only

$f^{\prime}(x)=\left\{\begin{array}{cc}0 & ; 0 < x < \frac{1}{4} \\ 2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4} < x < \frac{1}{2} \\ 4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{2} < x < \frac{3}{4} \\ 6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{3}{4} < x < 1\end{array}\right.$

$\mathrm{f}(\mathrm{x})$ is non-differentuable at $\mathrm{x}=\frac{1}{2}$ and $\frac{3}{4}$ minimum values of $\mathrm{f}(\mathrm{x})$ occur at $\mathrm{x}=\frac{5}{12}$ whose value is $-\frac{1}{432}$

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