AP EAPCET 2025 - 23rd May Evening Shift | Application of Derivatives Question 16 | Mathematics

If $m$ and $M$ are the absolute minimum and absolute maximum values of the function $f(x)=2 \sqrt{2} \sin x-\tan x$ in the interval $[0, \pi / 3]$, then $m+M=$

-1

AP EAPCET 2025 - 23rd May Morning Shift

MCQ (Single Correct Answer)

-0

If $\frac{1}{2} \leq \frac{x^2+x+a}{x^2-x+a} \leq 2 \forall x \in R$, then $a=$

$\frac{3}{4}$

$\frac{-3}{4}$

$\frac{9}{4}$

$\frac{-9}{4}$

AP EAPCET 2025 - 23rd May Morning Shift

MCQ (Single Correct Answer)

-0

$P$ and $Q$ are the ends of a diameter of the circle $x^2+y^2=a^2\left(a>\frac{1}{\sqrt{2}}\right) . s$ and $t$ are the lengths of the perpendiculars drawn from $P$ and $Q$ onto the line $x+y=1$ respectively. When the product st is maximum, the greater value among $s, t$ is

$a+\sqrt{2}$

$a+\frac{1}{\sqrt{2}}$

$a-\frac{1}{\sqrt{2}}$

$a-\sqrt{2}$

AP EAPCET 2025 - 23rd May Morning Shift

MCQ (Single Correct Answer)

-0

Let $P(x)=x^4+a x^3+b x^2+c x+d$ be such that $x=0$ is the only real root of $P^1(x)=0$. If $P(-1)

$P(-1)$ is not minimum of $P(x)$, but $P(1)$ is the maximum of $P(x)$

$P(-1)$ is minimum of $P(x)$, but $P(1)$ is not the maximum of $P(x)$

Neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of $P(x)$

$P(-1)$ is the minimum and $P(1)$ is the maximum of $P(x)$

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