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Vector Algebra

Question
CBSEENMA12036038
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The area (in sq units) of the region described by {x,y): y2 ≤ 2x and y ≥ 4x-1} is

  • 7/32

  • 5/64

  • 15/64

  • 9/32

Solution

D.

9/32

Given region is {x,y): y2 ≤ 2x and y ≥ 4x-1} y2 ≤ 2x  represents a region inside the parabola,
 y2 = 2x   .. (i)
and y ≥ 4x-1 represents a region to the left of the line
 y = 4x-1 .... (ii)
The point of intersection of the curve (i) and (ii) is 
(4x-1)2  = 2x
⇒ 16x2 + 1-8x = 2x
16x2-10x+1 = 0
 x = 1/2, 1/8
therefore, the points where these curves intersects, are 
open parentheses 1 half comma 1 close parentheses space space and space open parentheses 1 over 8 comma negative 1 half close parentheses
Hence, required area,
integral subscript negative 1 divided by 2 end subscript superscript 1 open parentheses fraction numerator straight y plus 1 over denominator 4 end fraction minus straight y squared over 2 close parentheses space dy space equals space 1 fourth open parentheses straight y squared over 2 plus straight y close parentheses subscript negative 1 divided by 2 end subscript superscript 1 space minus space 1 over 6 left parenthesis straight y cubed right parenthesis to the power of 1 subscript negative 1 divided by 2 end subscript equals space 1 fourth open curly brackets open parentheses 1 half plus 1 close parentheses minus open parentheses 1 over 8 minus 1 half close parentheses close curly brackets minus 1 over 6 open curly brackets 1 plus 1 over 8 close curly brackets space equals space 1 fourth space open curly brackets 3 over 2 space plus 3 over 8 close curly brackets minus 1 over 6 open curly brackets 9 over 8 close curly brackets space equals space 1 fourth space straight x space 15 over 8 space minus space 3 over 16 space equals space 9 over 32

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