Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by
$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest
integer $\leq \mathrm{x} .$ Then the range of $f$ is
Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by
$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest
integer $\leq \mathrm{x} .$ Then the range of $f$ is
- [JEE MAIN 2020]
- A
$\left(\frac{3}{5}, \frac{4}{5}\right)$
- B
$\left(\frac{2}{5}, \frac{3}{5}\right] \cup\left(\frac{3}{4}, \frac{4}{5}\right)$
- C
$\left(\frac{2}{5}, \frac{4}{5}\right]$
- D
$\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$
Similar Questions
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Let $f(x) = \frac{{x\,\, - \,\,1}}{{2\,{x^2}\,\, - \,\,7x\,\, + \,\,5}}$ . Then :
Consider a function $f:\left[0, \frac{\pi}{2}\right]$ $ \rightarrow$ $R$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] $ $\rightarrow$ $R$ given by $g(x)=\cos x .$ Show that $f$ and $g$ are one-one, but $f\,+\,g$ is not one-one.
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Let $R$ be the set of all real numbers and $f(x)=\sin ^{10} x\left(\cos ^8 x+\cos ^4 x+\cos ^2 x+1\right)$ $x \in R$. Let $S=\{\lambda \in R$ there exists a point $c \in(0,2 \pi)$ with $\left.f^{\prime}(c)=\lambda f(c)\right\}$ Then,
Let $R$ be the set of all real numbers and $f(x)=\sin ^{10} x\left(\cos ^8 x+\cos ^4 x+\cos ^2 x+1\right)$ $x \in R$. Let $S=\{\lambda \in R$ there exists a point $c \in(0,2 \pi)$ with $\left.f^{\prime}(c)=\lambda f(c)\right\}$ Then,
- [KVPY 2020]
The domain of $f(x) = [\sin x] \cos \left( {\frac{\pi }{{[x - 1]}}} \right)$ is (where $[.]$ denotes $G.I.F.$)
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If $f(x)$ and $g(x)$ are functions satisfying $f(g(x))$ = $x^3 + 3x^2 + 3x + 4$ $f(x)$ = $log^3x + 3$, then slope of the tangent to the curve $y = g(x)$ at $x = \ -1$ is
If $f(x)$ and $g(x)$ are functions satisfying $f(g(x))$ = $x^3 + 3x^2 + 3x + 4$ $f(x)$ = $log^3x + 3$, then slope of the tangent to the curve $y = g(x)$ at $x = \ -1$ is