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CBSE Class 12 Mathematics Solved Question Paper 2017

Long Answer Type

31.

Let space straight a with rightwards arrow on top space equals space straight i with hat on top space plus straight j with hat on top space plus stack straight k comma with hat on top space straight b with rightwards arrow on top space equals space straight i with hat on top space and straight c with rightwards arrow on top space equals space straight c subscript 1 straight i with hat on top space plus straight c subscript 2 straight j with hat on top space plus straight c subscript 3 straight k with hat on top space space then

a) Let c₁ = 1 and c₂ = 2, find c₃ which makes

straight a with rightwards arrow on top space comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top

coplanar.
b) If c₂ = –1 and c₃ = 1, show that no value of c₁ can make

straight a with rightwards arrow on top space comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top

coplanar.

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32.
Find the area bounded by the circle x² + y² = 16 and the line √3y=x in the first quadrant, using integration.

The area of the region bounded by the circle, x²+y²=16, x=√3y, and the x-axis is the area OAB.

Solving x²+y²=16 and x=√3y, we have

(√3y)²+y²=16
⇒3y²+y²=16
⇒4y²=16
⇒y²=4
⇒y=2
(In the first quadrant, y is positive)

When y = 2, x = 2√3

So, the point of intersection of the given line and circle in the first quadrant is (2√3,2).

The graph of the given line and circle is shown below:

Required area = Area of the shaded region = Area OABO = Area OCAO + Area ACB
Area OCAO = 12×2√3×2=2√3 sq units
$Area space ABC space equals space integral subscript 2 square root of 3 end subscript superscript 4 space straight y space dx space equals space integral subscript 2 square root of 3 end subscript superscript 4 square root of 16 minus straight x squared dx end root space equals space open square brackets straight x over 2 square root of 16 minus straight x squared end root space plus 16 over 2 space sin to the power of negative 1 end exponent space straight x over 4 close square brackets subscript 2 square root of 3 end subscript superscript 4 space equals space open square brackets open parentheses 0 plus space 8 space sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses fraction numerator 2 square root of 3 over denominator 2 end fraction space straight x space 2 space plus 8 space straight x space sin to the power of negative 1 end exponent fraction numerator square root of 3 over denominator 2 end fraction close parentheses close square brackets space equals space 8 space straight x space straight pi over 2 minus 2 square root of 3 space minus 8 space straight x space straight pi over 3 space equals open parentheses fraction numerator 4 straight pi over denominator 3 end fraction minus 2 square root of 3 space close parentheses space sq space unit therefore space Required space area space equals space open parentheses fraction numerator 4 straight pi over denominator 3 end fraction minus 2 square root of 3 close parentheses space plus 2 square root of 3 space equals space fraction numerator 4 straight pi over denominator 3 end fraction space sq space units$