JEE MAIN Maxima And Minima Previous Year Questions (PYQs) – Page 1 of 5
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The sum of the absolute maximum and minimum values of the function $f(x)=\lvert x^{2}-5x+6\rvert-3x+2$ in the interval $[-1,3]$ is equal to:
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Solution
Let $a>0$. If the function $f(x)=6x^3-45ax^2+108a^2x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2=54$, then $a+x_1+x_2$ is equal to
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Solution
$f(x)=6x^3-45ax^2+108a^2x+1$
$f'(x)=18x^2-90ax+108a^2$
$=18(x^2-5ax+6a^2)$
So critical points:
$x_1,x_2=2a,3a$
Given $x_1x_2=54$
⇒ $6a^2=54 \Rightarrow a^2=9$
$ \Rightarrow a=3$ (since $a>0$)
$x_1+x_2=5a=15$
$a+x_1+x_2=3+15=18$
$f'(x)=18x^2-90ax+108a^2$
$=18(x^2-5ax+6a^2)$
So critical points:
$x_1,x_2=2a,3a$
Given $x_1x_2=54$
⇒ $6a^2=54 \Rightarrow a^2=9$
$ \Rightarrow a=3$ (since $a>0$)
$x_1+x_2=5a=15$
$a+x_1+x_2=3+15=18$
If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha\log|x|+\beta x^{2}+x$ then
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Solution
Let $f(x) = x^{2025} - x^{2000},; x \in [0,1]$ and maximum value of the function $f(x)$ in the interval $[0,1]$ be $\left(\frac{80}{81}\right)^{n}$, then $n$ is equal to:
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Solution
$f(x) = x^{2025} - x^{2000}$
$f'(x) = 0 \Rightarrow x = \left(\frac{2000}{2025}\right)^{1/25} = \alpha$
$f(0) = 0,; f(1) = 0,; f(\alpha) = \left(\frac{80}{81}\right)^{80} - \left(\frac{80}{81}\right)^{81}$
$= \left(\frac{80}{81}\right)^{80}\left(1 - \frac{80}{81}\right)$
$= \left(\frac{80}{81}\right)^{80}\cdot \frac{1}{81}$
$= \left(\frac{80}{81}\right)^{80} \cdot \left(\frac{80}{81}\right)^{-1}$
$= \left(\frac{80}{81}\right)^{-81}$
$\Rightarrow n = -81$
The set of all real values of $\lambda $ for which thefunction$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$ has exactly one maxima and exactly oneminima, is :
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The maximum value of the function $f(x)=3x^{3}-18x^{2}+27x-40$ on the set $S=\{x\in\mathbb{R}: x^{2}+30\le 11x\}$ is :
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Solution
If a right circular cone, having maximum volume, is inscribed in a sphere of radius $3\ \text{cm}$, then the curved surface area (in $\text{cm}^{2}$) of this cone is :
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Solution
Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^{3}+ax^{2}+b\log_{e}|x|+1,;x\neq0$. Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $\left[-2,-\dfrac{1}{2}\right]$. Then $|M+m|$ is equal to (take $\log_{e}2=0.7$):
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Solution
Let $f:\mathbb{R}\to\mathbb{R}$ be a polynomial of degree four having extreme values at $x=4$ and $x=5$. If $\displaystyle \lim_{x\to 0}\frac{f(x)}{x^{2}}=5$, then $f(2)$ is equal to:
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Solution
The sum of absolute maximum and absolute minimum values of the function $f(x) = |2{x^2} + 3x - 2| + \sin x\cos x$ in the interval [0, 1] is :
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Solution
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