Least integer in the range of $f(x)$=$\sqrt {(x + 4)(1 - x)} - {\log _2}x$ is
$-2$
$-1$
$0$
$1$
$f(x)=\sqrt{(x+4)(1-x)}+\log _{1 / 2} x$
$\because$ $'f'$ is decreasing
minimum value is $f(1)=0.$
If $f(x) = \cos (\log x)$, then $f({x^2})f({y^2}) – \frac{1}{2}\left[ {f\,\left( {\frac{{{x^2}}}{2}} \right) + f\left( {\frac{{{x^2}}}{{{y^2}}}} \right)} \right]$ has the value
For $x\,\, \in \,R\,,x\, \ne \,0,$ let ${f_0}(x) = \frac{1}{{1 – x}}$ and ${f_{n + 1}}(x) = {f_0}({f_n}(x)),$ $n\, = 0,1,2,….$ Then the value of ${f_{100}}(3) + {f_1}\left( {\frac{2}{3}} \right) + {f_2}\left( {\frac{3}{2}} \right)$ is equal to
Let $f(x) = {\cos ^{ – 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) + {\sin ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right)$ then the value of $f(1) + f(2)$, is –
Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3 .$ If $\sum \limits_{i=1}^{n} f(i)=363,$ then $n$ is equal to
Statement $-1$ : The equation $x\, log\, x = 2 – x$ is satisfied by at least one value of $x$ lying between $1$ and $2$
Statement $-2$ : The function $f(x) = x\, log\, x$ is an increasing function in $[1, 2]$ and $g (x) = 2 -x$ is a decreasing function in $[ 1 , 2]$ and the graphs represented by these functions intersect at a point in $[ 1 , 2]$
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