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 The curve y=ax3+bx2+cx+8 touches x-axis at P(2,0) and cuts the y-axis at a point Q where its gradient is 3.    The values of a, b, c are   respectively

a
−1/2,   −3/4,  3
b
3,​​  −1/2,   − 4
c
−1/2,   − 7/4,   2
d
none of these

detailed solution

Correct option is D

dydx=3ax2+2bx+c Since the curve touches x-axis at (−2,0) so dydx(−2,  0)  =   0 ⇒  12a − 4b +c  =  0----iThe curve cut the y-axis at (0, 8)  sodydx(  0,   8)  =   3 ⇒  c  =  3 Also the curve passes through (−2,0) so 0=−8a+4b−2c+8⇒−8a+4b−2=0----ii Solving (i) and  (ii) a=−1/4,b=0

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