QJEEM22Ma7dac: JEE Main 2022 Mathematics - Class 12 JEEM22Ma7dac Solution: Let . Then f :

Given,

$f(x) = 4{x^3} - 11{x^2} + 8x - 5$

$\therefore$ $f'(x) = 12{x^2} - 22x + 8$

$= 2(6{x^2} - 11x + 4)$

$= 2(6{x^2} - 8x - 3x + 4)$

$= 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$

$= 2\left[ {(3x - 4)(2x - 1)} \right]$

$f'(x)$ is positive before ${1 \over 2}$ means slope of f(x) is positive before ${1 \over 2}$ and f'(x) is negative after ${1 \over 2}$ means slope of f(x) is negative after ${1 \over 2}$ .

$\therefore$ At ${1 \over 2}$ , f(x) is local maximum

Also, $f'(x)$ is negative before ${4 \over 3}$ means slope of f(x) is negative before ${4 \over 3}$ and f'(x) is positive after ${4 \over 3}$ means slope of f(x) is positive after ${4 \over 3}$ .

$\therefore$ At ${4 \over 3},\,f(x)$ is local minimum

From wavy curve method,

f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)

$\therefore$ f(x) increasing in $x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$

f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)

$\therefore$ f(x) decreasing in $x \in \left( {{1 \over 2},{4 \over 3}} \right)$