Let f(x)=4 x³-11 x²+8 x-5, x {R}. Then f : [30-Jun-2022-Shift-1]

Correct Answer is : is decreasing in ( 12, 43)≤ft(\;1/2, \;4/3)(21,34)

is increasing in ≤ft(\;1/2, \;3/4)

is decreasing in ≤ft(\;1/2, \;4/3)

Correct Answer is : is decreasing in ( 12, 43)≤ft(\;1/2, \;4/3)(21,34)

Test Index

Application of Derivatives Part 2

Section: Mathematics

Question : 48 of 100

Marks: +1, -0

Let

f(x)=4 x^{3}-11 x^{2}+8 x-5, x \in \mathbb{R}

f

Solution: 👈: Video Solution

Given,

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

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

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

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

=2[2 x(3 x-4)-1(3 x-4)]

=2[(3 x-4)(2 x-1)]

f'(x)

is positive before

\;\frac{1}{2}

means slope of

f(x)

is positive before

\;\frac{1}{2}

and

f'(x)

is negative after

\;\frac{1}{2}

means slope of

f(x)

is negative after

\;\frac{1}{2}

\therefore

\;\frac{1}{2}, f(x)

is local maximum

Also,

f'(x)

is negative before

\;\frac{4}{3}

means slope of

f(x)

is negative before

\;\frac{4}{3}

and

f'(x)

is positive after

\;\frac{4}{3}

means slope of

f(x)

is positive after

\;\frac{4}{3}

\therefore

\;\frac{4}{3}, f(x)

is local minimum

From wavy curve method,

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

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

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

\therefore f(x) \;\text{ decreasing in }\; x \in \left(\;\frac{1}{2}, \;\frac{4}{3}\right)

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