CTET 2 Math and Science 3 Jan 2022 Paper

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Question : 38 of 150

Marks: +1, -0

Let

p=3 x^2-6 x-4

{\q}=3+ {\x}-2 {\x}^2

and

r=-4 x+x^2+9

, then

(p-q)+r

is equal to

$6 x^2+11 x-9$
$3 x^2-2 x+7$
$3 x^2-9 x+8$
$6 x^2-11 x+2$

Solution:

Given:

p=3 x^2-6 x-4

q=3+x-2 x^2

r=-4 x+x^2+9

Calculation:

(p-q)+r

(3 x^2-6 x-4-(3+x-2 x^2)+(-4 x+x^2+9).

3 x^2-6 x-4-3-x+2 x^2-4 x+x^2+9

(3 x^2+2 x^2+x^2)+(-6 x-x-4 x)+(-4-3+9)

6 x^2-11 x+2

(p-q)+r=6 x^2-11 x+2

The correct option is 4 i.e.

6 x^2-11 x+2

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