If the area (in sq. units) of the region {(x,y):y2≤4x,x+y≤1,x≥0,y≥0} is a√2+b, then a−b is equal to :
A
83
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B
6
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C
103
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D
−23
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Solution
The correct option is B6 C1:y2≤4x C2:x+y≤1 x,y≥0
P is the point of intersection of these two curve y2=4x,x+y=1 y2−4(1−y)=0⇒y2+4y−4=0 y=−4±√16−4(−4)2=−2±2√2 P:(3−2√2,−2+2√2),O:(0,0)R:(1,0)
Hence, the required area=3−2√2∫02√xdx+1∫3−2√2(1−x)dx =43∣∣∣x32∣∣∣3−2√20+∣∣∣x−x22∣∣∣13−2√2 =8√23−103=a√2+b a=83,b=−103 ∴a−b=8+103=6