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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 71 of 160

Marks: +1, -0

The absolute minimum value of

x^{4}-x^{2}-2x+5

is ______

5
3
7
Does not exist

Solution:

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

(say)

Then,

f'(x)=4x^{3}-2x-2

Equate

f'(x)=0 \Rightarrow 4x^{3}-2x-2=0

\Rightarrow 2x^{3}-x-1=0

\Rightarrow 2x(x^{2}-1)+(x-1)=0

\Rightarrow 2x(x-1)(x+1)+(x-1)=0

\Rightarrow (x-1)[2x(x+1)+1]=0

\Rightarrow (x-1)(2x^{2}+2x+1)=0

\Rightarrow x-1=0

and

2x^{2}+2x+1=0

\Rightarrow x=1

and

x=\frac{-2 \pm \sqrt{4-8}}{4}

(Imaginary)

\Rightarrow x=1

and

x=\frac{-1 \pm 1i}{2}

f''(x)=12x^{2}-2

\frac{f''(x)}{x=1}=10>0

\Rightarrow f(x)

is minimum at

x=1

Now,

f(1)=(1)^{4}-(1)^{2}-2+5=3

∴ Absolute minimum value is 3

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