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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.

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33.
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).

the vertices of the triangle be A(–2, 1), B(0, 4) and C(2, 3).
Equation of line segment AB is
$Equation space of space line space segment space AB space is left parenthesis straight y minus 1 right parenthesis space equals space fraction numerator left parenthesis 4 minus 1 right parenthesis over denominator left parenthesis 0 plus 2 right parenthesis end fraction left parenthesis straight x plus 2 right parenthesis rightwards double arrow space straight y space equals space fraction numerator 3 left parenthesis straight x plus 2 right parenthesis over denominator 2 end fraction space plus 1 rightwards double arrow space straight y space equals fraction numerator space 3 straight x space plus space 6 plus 2 over denominator 2 end fraction rightwards double arrow space straight y space equals space fraction numerator 3 straight x plus 8 over denominator 2 end fraction Equation space of space line space segment space BC space is left parenthesis straight y minus 4 right parenthesis space equals space fraction numerator left parenthesis 3 minus 4 right parenthesis over denominator left parenthesis 2 minus 0 right parenthesis end fraction left parenthesis straight X minus 0 right parenthesis rightwards double arrow space straight Y space equals space minus space 1 half space straight x space plus 4 Equation space of space line space segment space AC space is left parenthesis straight y minus 3 right parenthesis space equals space fraction numerator left parenthesis 1 minus 3 right parenthesis over denominator left parenthesis negative 2 minus 2 right parenthesis end fraction left parenthesis straight X minus 2 right parenthesis rightwards double arrow space straight Y space equals space space straight x over 2 space plus 2$

AL, BO and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ABOL) + Area (BOMCB) – Area (ALMCA)
$Area space left parenthesis ABOL right parenthesis space equals space integral subscript negative 2 end subscript superscript 0 open parentheses fraction numerator 3 straight x space plus 8 over denominator 2 end fraction close parentheses space dx space equals space integral subscript negative 2 end subscript superscript 0 space open parentheses fraction numerator 3 straight x over denominator 2 end fraction plus 4 close parentheses space dx space equals fraction numerator 3 straight x squared over denominator 4 end fraction space plus right enclose 4 straight x end enclose subscript negative 2 end subscript superscript 0 space equals space left parenthesis 0 space plus space 0 right parenthesis minus open parentheses 12 over 4 minus 8 close parentheses space equals space 5 space sq space units Area space left parenthesis BOMCB right parenthesis space equals space integral subscript 0 superscript 2 open parentheses 1 half straight x space plus space 4 close parentheses dx space equals space right enclose space open parentheses negative straight x squared over 4 plus 4 straight x squared close parentheses end enclose subscript 0 superscript 2 space equals space left parenthesis negative 1 plus space 8 right parenthesis minus left parenthesis 0 space plus space 0 right parenthesis space equals space 7 space sq space units Area space left parenthesis ALMCA right parenthesis space equals space integral subscript negative 2 end subscript superscript 2 space open parentheses straight x over 2 plus 2 close parentheses dx space equals space space equals space right enclose space open parentheses straight x squared over 4 plus 2 straight x close parentheses end enclose subscript negative 2 end subscript superscript 2 space equals space left parenthesis 1 plus 4 right parenthesis minus left parenthesis 1 minus 4 right parenthesis space equals space 8 space sq space units space Required space area space equals space 4 space units$