Let f : [0,1] [0,1] be a continuous function, then the equation f… - Vidyadip

MCQ

Let $f : [0,1] \to [0,1]$ be a continuous function, then the equation $f(x) = x$

A
may not have any solution in $[0,1]$
B
must have exactly one solution in $[0,1]$
✓
must have atleast one solution in $[0,1]$
D
must have atleast two solutions in $[0,1]$

✓

Answer

Correct option: C.

must have atleast one solution in $[0,1]$

c
Define new function $\mathrm{g}(\mathrm{x})=f(\mathrm{x})-\mathrm{x}$ in $[0,1]$

then $g(1)=f(1)-1 \leq 0 $ and $ g(0)=f(0) \geq 0$

$\therefore $ equation $\mathrm{g}(\mathrm{x})=0$ has atleast one root in $[0,1]$

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

More "M.C.Q (1 Marks)" from 5. continuity and differentiation→5. continuity and differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{\,0}^{\,\pi } {\sqrt {\frac{{1 + \cos 2x}}{2}} \,dx} $ is equal to

If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$, then adj $A$ is equal to

If for the matrix $A=\left[\begin{array}{cc}\tan x & 1 \\ -1 & \tan x\end{array}\right], A+A^{\prime}=2 \sqrt{3}$, , then the value of $x \in\left[0, \frac{\pi}{2}\right]$ is :

The probability of a man hitting a target is $\frac{1}{10}$. The least number of shots required, so that the probability of his hitting the target at least once is greater than $\frac{1}{4},$ is

Let $f : R \rightarrow R$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ for all $x, y \in R$ Then, the value of $\log _c(f(4))$ is. . . . . .

Let $y=y(x)$ be the solution of the differential equation $\sec \mathrm{x} d \mathrm{y}+\{2(1-\mathrm{x}) \tan \mathrm{x}+\mathrm{x}(2-\mathrm{x})\}$ $\mathrm{dx}=0$ such that $\mathrm{y}(0)=2$. Then $\mathrm{y}(2)$ is equal to :

Suppose $X =\left[x_{i j}\right]$ a matrix, where
$
X=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]
$
Then matrix $Y =\left[m_{i j}\right]$, where $m_{i j}=$ minor of $x_{i j}$ :

Let $ABC$ be a triangle such that $\overrightarrow{ BC }=\overrightarrow{ a }, \overrightarrow{ CA }=\overrightarrow{ b }$, $\overrightarrow{ AB }=\overrightarrow{ c },|\overrightarrow{ a }|=6 \sqrt{2}, \quad|\overrightarrow{ b }|=2 \sqrt{3}$ and $\overrightarrow{ b } \cdot \overrightarrow{ c }=12$ Consider the statements.

$( S 1):|(\overrightarrow{ a } \times \overrightarrow{ b })+(\overrightarrow{ c } \times \overrightarrow{ b })|-|\overrightarrow{ c }|=6(2 \sqrt{2}-1)$

$( S 2): \angle ABC =\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then

The value of the integral $\int_0^\pi(1-|\sin 8 x|) d x$ is

If $f(x) = {x^n},$ then the value of $f(1) - \frac{{f'(1)}}{{1!}} + \frac{{f''(1)}}{{2!}} - \frac{{f'''(1)}}{{3\,!}} + ...... + \frac{{{{( - 1)}^n}{f^n}(1)}}{{n!}}$ is