The curve y=ax^2+ bx passes through the point (1 ,2 ) and lies above the X-axis for 0 x 8. If the area enclosed by this curve, the X-aixs and the line x=6 is 108 square units, then 2b-a=

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Correct answer: B. (b) Since, Given curve y=ax^2+ bx passes through point (1 ,2 ). Hence, 2=a (1 )^2+b (1 ) ⇒a+b=2....(i) Given, area under curve = _0^6 0.05em(ax^2+ bx) dx=108 ⇒( ax^33+ bx^22)_0^6=108 ⇒4a+b=6 … . . ( ii ) By solving eqs. (i) and (ii), we get a= 43 and b= 23 ∴2b-a=2 23- 43=0

$The curve y = a x^{2} + bx passes through the point (1, 2) and lies above the X -axis for 0 \leq x \leq 8 . If the area enclosed by this curve, the X -aixs and the line x = 6 is 108 square units, then 2 b - a =$

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