Q.
Let p(x) be a real polynomial of degree 4 having extreme values at x=1 and x=2.If limx→0 p(x)x2=1then p( 4) is equal
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a
0
b
16
c
32
d
64
answer is B.
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Detailed Solution
Let p(x)=ax4+bx3+cx2+dx+e.Then p′(x)=4ax3+3bx2+2cx+d Now, limx→0 p(x)x2=1⇒limx→0 ax2+bx+c+dx+ex2=1⇒c=1,d=e=0∴ p′(x)=4ax3+3bx2+2x It is given that p(x) has extreme values at x =1 and x =2. ∴ p′(1)=0 and p′(2)=0⇒ 4a+3b+2=0 and 32a+12b+4=0 ⇒ a=14and b=−1∴ p(x)=14x4−x3+x2⇒ p(4)=64−64+16=16
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