Let f : R → R be a function defined by f ( x )= log _{√{m}}{√{2}(sin x-cos x)+m-2} , for some

English

Gujarati

Hindi

1.Relation and Function

hard

Let $f : R \rightarrow R$ be a function defined by $f ( x )=$ $\log _{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is $............$

$5$

$3$

$2$

$4$

(JEE MAIN-2023)

Solution

Since,

$-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2}$

$\therefore-2 \leq \sqrt{2}(\sin x-\cos x) \leq 2$

$\quad \text { Assume } \sqrt{2}(\sin x-\cos x)=k)$

$-2 \leq k \leq 2 \quad \ldots( i )$

$f(x)=\log _{\sqrt{m}}( k + m -2)$

$\text { Given, }$

$0 \leq f( x ) \leq 2$

$0 \leq \log _{\sqrt{ m }}( k + m -2) \leq 2$

$\leq k + m -2 \leq m$

$- m +3 \leq k \leq 2 \ldots \text { (ii) }$

From eq. $(i)$ and $(ii)$, we get $- m +3=-2$

$\Rightarrow m=5$

Standard 12

Mathematics

Range of the function

$f(x) = \sqrt {\left| {{{\sin }^{ – 1}}\left| {\sin x} \right|} \right| – {{\cos }^{ – 1}}\left| {\cos x} \right|} $ is

normal

Show that the Modulus Function $f : R \rightarrow R$ given by $(x)=|x|$, is neither one – one nor onto, where $|x|$ is $x$, if $x$ is positive or $0$ and $| X |$ is $- x$, if $x$ is negative.

medium

The domain of the function $f(x) =\frac{{\,\cot^{-1} \,x}}{{\sqrt {{x^2}\,\, – \,\,\left[ {{x^2}} \right]} }}$ , where $[x]$ denotes the greatest integer not greater than $x$, is :

normal

If the domain of the function $f(\mathrm{x})=\frac{\cos ^{-1} \sqrt{x^{2}-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}$ is the interval $(\alpha, \beta]$, then $\alpha+\beta$ is equal to:

hard

(JEE MAIN-2021)

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
{3 + x;\,\,\,\,\,x \geqslant 0} \\
{2 – 3x;\,\,\,\,\,x < 0}
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to –

normal

Let $f : R \rightarrow R$ be a function defined by $f ( x )=$ $\log _{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is $............$

Solution

Similar Questions

Range of the function $f(x) = \sqrt {\left| {{{\sin }^{ – 1}}\left| {\sin x} \right|} \right| – {{\cos }^{ – 1}}\left| {\cos x} \right|} $ is

Show that the Modulus Function $f : R \rightarrow R$ given by $(x)=|x|$, is neither one – one nor onto, where $|x|$ is $x$, if $x$ is positive or $0$ and $| X |$ is $- x$, if $x$ is negative.

The domain of the function $f(x) =\frac{{\,\cot^{-1} \,x}}{{\sqrt {{x^2}\,\, – \,\,\left[ {{x^2}} \right]} }}$ , where $[x]$ denotes the greatest integer not greater than $x$, is :

If the domain of the function $f(\mathrm{x})=\frac{\cos ^{-1} \sqrt{x^{2}-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}$ is the interval $(\alpha, \beta]$, then $\alpha+\beta$ is equal to:

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}} {3 + x;\,\,\,\,\,x \geqslant 0} \\ {2 – 3x;\,\,\,\,\,x < 0} \end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to –

Start a Free Trial Now

Range of the function

$f(x) = \sqrt {\left| {{{\sin }^{ – 1}}\left| {\sin x} \right|} \right| – {{\cos }^{ – 1}}\left| {\cos x} \right|} $ is

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
{3 + x;\,\,\,\,\,x \geqslant 0} \\
{2 – 3x;\,\,\,\,\,x < 0}
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to –