CTET 2 Math and Science 3 Jan 2022 Paper

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Question : 37 of 150

Marks: +1, -0

If

{\a}+ {\b}=3

and

{\a}^2+ {\b}^2=89

, then What is the value of

\;{1}/{a}+\;{1}/{b}

?

$\;{3}/{20}$
$-\;{3}/{40}$
$-\;{3}/{20}$
$\;{1}/{8}$

Solution:

(a+b)^2=a^2+2 a b+b^2

Calculation:

a+b=3

Squaring both the side

(a+b)^2=(3)^2

⇒ a^2+2 a b+b^2=9

⇒ a^2+b^2+2 a b=9

⇒ 89+2 a b=9

⇒ 80=-2 a b

⇒ a b=-40

Now,

\;{1}/{a}+\;{1}/{b}

⇒ \;{a+b}/{a b}

⇒-\;{3}/{40}

∴ \;{1}/{a}+\;{1}/{b}=-\;{3}/{40}

The correct option is 3 i.e.

-3 \/ 40

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