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4-2.Quadratic Equations and Inequations
normal
The number of cubic polynomials $P(x)$ satisfying $P(1)=2, P(2)=4, P(3)=6, P(4)=8$ is
A
$0$
B
$1$
C
more than one but finitely many
D
infinitely many
(KVPY-2019)
Solution
(a)
Let the equation of a cubic polynomial
$P(x)=a x^3+b x^2+c x+d$
Now,
$P(1)=a+b+c+d=2 \ldots (i)$
$P(2)=8 a+4 b+2 c+d=4 \ldots (ii)$
$P(3)=27 a+9 b+3 c+d=6 \ldots (iii)$
$P(4)=64 a+16 b+4 c+d=8 \ldots$ (iv)
From Eqs. $(i)$ and $(ii),$ we get
$7 a+3 b+c=2$
From Eqs.$(ii)$ and $(iii)$, we get
$19 a+5 b+c=2$
From Eqs. $(iii)$ and $(iv)$, we get
$37 a+7 b+c=2$
Now, from Eqs.$(v)$ and $(vi)$, we get
$12 a+2 b=0$
and from Eqs.$(vi)$ and $(vii)$, we get
$18 a+2 b=0$
From Eqs.$(viii)$ and $(ix)$, we get
$a=0 \text { and } b=0 \text {, }$
$c=2 \text { and } d=0 \text {. }$
So, $P(x)=2 x$
$\therefore$ no cubic polynomial is possible.
Standard 11
Mathematics
