GATE Electrical Engineering (EE) 2020 Solved Paper

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Question : 52 of 65

Marks: +1, -0

For real numbers

x

and

y

, with

y=3 × 2+3 x+1

, the maximum and minimum value of

x ∈

[-2,0]

are respectively,

$7$ and $1 / 4$
$7$ and $1$
$-2$ and $-1 / 2$
$1$ and $1 / 4$

Solution:

y=3 x^{2}+3 x+1

\;{d y}/{d x}=6 x+3=0

x=\;{-1}/{2}

\;{d^{2} y}/{d x^{2}}=6(+v e)

so it is minimum
Minimum value of

y

at

x=-1 / 2

y=3[\;{-1}/{3}]^{2}+3[\;{-1}/{2}]+1

=\;{3}/{4}-\;{3}/{2}+1=\;{3-6+4}/{4}=\;{1}/{4}

y_{\min }=\;{1}/{4}

for maximum value, calculate

y

at boundary

y|_{x=0}=3(0)^{2}+3(0)+1=1

y|_{x=-2}=3[-2]^{2}+3[-2]+1

=12-6+1

y_{\max }=7

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