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Question

Let f(x) is a cubic polynomial such that f(1)=1,f(2)=2,f(3)=3 and f(4)=16, then f(0) is equal to

A
6
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B
6
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C
12
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D
12
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Solution

The correct option is D 12
Let us assume:

f(x)=ax3+bx2+cx+d

f(1)=a+b+c+d=1 ………….(1)

f(2)=8a+4b+2c+d=2 ………..(2)

f(3)=27a+9b+3c+d=3 …………(3)

f(4)=64a+16b+4c+d=16 ……….(4)

Solving equation (1) & (2), we get

7a+3b+c=1 ………..(5)

Solving equation (1) & (3), we get

26a+8b+2c=2 …….(6)

Solving equation (1) & (4), we get

63a+15b+3c=15 ……..(7)

Solving (5) and (6),

6a+b=0

Solving (5) and (7)

7a+b=2

This gives a=2,b=12,c=23,d=12

f(x)=2x312x2+23x12

f(0)=2(0)312(0)3+23(0)12

f(0)=12

The option [D] is right.

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