MCQ
Least integer in the range of $f(x)$=$\sqrt {(x + 4)(1 - x)} - {\log _2}x$ is
- A$-2$
- B$-1$
- ✓$0$
- D$1$
$\because$ $'f'$ is decreasing
minimum value is $f(1)=0.$
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$(1,1)(-1,1),(1,-1),(-1,-1),(-2,1)(2,-1),(-1,2)$ and $(-2,-1)$
($A$) $f$ has a local minimum at $x=2$
($B$) fhas a local maximum at $x=2$
($C$) $f^{\prime \prime}(2)>f(2)$
($D$) $f(x)-f^{\prime \prime}(x)=0$ for at least one $x \in \mathbb{R}$