(C) Define a new function $g(x) = f(x) - x$ for $x \in [0, 1]$.Since $f : [0, 1] \to [0, 1]$,we have $0 \leq f(x) \leq 1$ for all $x \in [0, 1]$.At $x

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

(C) Define a new function $g(x) = f(x) - x$ for $x \in [0, 1]$.Since $f : [0, 1] \to [0, 1]$,we have $0 \leq f(x) \leq 1$ for all $x \in [0, 1]$.At $x = 0$,$g(0) = f(0) - 0 = f(0) \geq 0$.At $x = 1$,$g(1) = f(1) - 1 \leq 0$ (since $f(1) \leq 1$).If $g(0) = 0$ or $g(1) = 0$,then $x = 0$ or $x = 1$ is a solution to $f(x) = x$.If $g(0) > 0$ and $g(1) Therefore,the equation $f(x) = x$ must have at least one solution in $[0, 1]$.

Class 12 Mathematics Continuity and Differentiation Continuity

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 $[0,1]$
C
must have atleast one solution in $[0,1]$
D
must have atleast two solutions in $[0,1]$

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Continuity and Differentiation

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Continuity

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Easy

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Check the continuity of the function $f$ given by $f(x) = 2x + 3$ at $x = 1$.

Easy

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If $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}, & -2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{cases}$ is continuous on $[-2, 2]$,then $c$ is equal to

MediumAP EAMCET 2014

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If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

DifficultAP EAMCET 2006

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Let $f(x) = \begin{cases} x^p \sin \left( \frac{1}{x} \right) + x|x^3|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then the complete set of values of $p$ for which $f''(x)$ is continuous at $x = 0$ is:

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

Similar Questions

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Check the continuity of the function $f$ given by $f(x) = 2x + 3$ at $x = 1$.

If $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}, & -2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{cases}$ is continuous on $[-2, 2]$,then $c$ is equal to

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Let $f(x) = \begin{cases} x^p \sin \left( \frac{1}{x} \right) + x|x^3|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then the complete set of values of $p$ for which $f''(x)$ is continuous at $x = 0$ is:

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