Download App

STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
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 $-\frac{1}{432}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let the tangent to the circle $C _{1}: x^{2}+y^{2}=2$ at the point $M (-1,1)$ intersect the circle $C _{2}$ : $( x -3)^{2}+(y-2)^{2}=5$, at two distinct points $A$ and $B$. If the tangents to $C _{2}$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $ANB$ is equal to
→
If position vectors of a point $A$ is  $a + 2b$  and a divides  $ AB $ in the ratio $2:3$, then the position vector of $ B$  is
→
If $AB = A$ and $BA = B$, then
→
The polynomial equation $x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$ has
→
If in an infinite $G.P.$ first term is equal to the twice of the sum of the remaining terms, then its common ratio is
→
The centre of the circle passing through the point $(0, 1)$ and touching the curve $y = {x^2}$ at $(2, 4)$ is
→
Statement $-1:$ The number of common solutions of the trigonometric equations $2\,sin^2\,\theta - cos\,2\theta  = 0$ and $2 \,cos^2\,\theta - 3\,sin\,\theta  = 0$ in the interval $[0, 2\pi ]$ is two.

Statement $-2:$ The number of solutions of the equation, $2\,cos^2\,\theta  - 3\,sin\,\theta  = 0$ in the interval $[0, \pi ]$ is two.

→
The solution of equation $\frac{{p + q - x}}{r} + \frac{{q + r - x}}{p}$+$\frac{{r + p - x}}{q}$+$\frac{{3x}}{{p + q + r}} = 0$ is
→
If $\frac{1^3+2^3+3^3+\ldots \ldots \text {.upto } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots \text { upto } n \text { terms }}=\frac{9}{5}$, then the value of $n$ is
→
If $\alpha $ and $\beta $ are the roots of the equation ${x^2} - 4x + 1 = 0$ the value of ${\alpha ^3} + {\beta ^3}$is
→