Q.

Let P(x) be a polynomial of least degree whose graph has three points of inflextion (–1, –1), (1,1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissas at an angle of 60∘

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a

1

b

2−1

c

22−1

d

none of these

answer is D.

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Detailed Solution

we have P′′(x)=a(x+1)x(x−1)=ax3−x⇒P′(x)=a14x4−12x2+b⇒P(x)=a120x5−16x3+bx+cAs−1=a−120+16−b+c1=a120−16+b+c⇒c=0and 1=−760a+bAlso 3=b⇒a=60(3−1)/7we have ∫01 P(x)dx=−130a+12b=27(1−3)+32
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Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
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