Question

Let $\rm{P(x)}$ be a nonzero polynomial such that $\rm (x-1)P(x+1)=(x+2)P(x)$ for every real x, and $\rm{(P(2) {)}^{2} =P(3). }$ Then $\rm P( \frac{7}{2} ) = \frac{m}{n} ,$ where m and n are relatively prime positive integers. Find m+n.​

Answer 1

P(x) is non –zero polynomial and P(1+x)=P(1−x) for all x

Differentiate w.r.t x

P(1+x)=−P(1−x)

Put x=0,P(1)=−P(1)

⇒P(1)+P(1)=0

⇒2P′(1)=0

⇒P(1)=0

andP(1)=0⇒P(x) touches the x− axis at x=1

⇒P(x)=(x−1)²Q(x)

⇒m=2 such that (x−1)m divides P(x) for all such P(x)

Answer 2

$\large\underline{\sf{Solution-}}$

Given that, P(x) be a non zero polynomial, such that

$\rm \: (x-1)P(x+1)=(x+2)P(x) - - - (i)$

On substituting x = 1, we get

$\rm \: (1-1)P(1+1)=(1+2)P(1)$

$\rm \: 3P(1) = 0$

$\red{\rm\implies \:P(1) = 0 \: }$

$\red{\rm\implies \:x - 1 \: is \: a \: factor \: of \: P(x) \: }$

On substituting x = 0, we get

$\rm \: (0-1)P(0+1)=(0+2)P(0)$

$\rm \: - P(1) = 2P(0)$

$\rm \: 0 = 2P(0)$

$\red{\rm\implies \:P(0) = 0 \: }$

$\red{\rm\implies \:x \: is \: a \: factor \: of \: P(x) \: }$

On substituting x = - 1, we get

$\rm \: ( - 1-1)P( - 1+1)=( - 1+2)P( - 1)$

$\rm \: P( - 1) = - 2P(0)$

$\red{\rm\implies \:P( - 1) = 0 \: }$

$\red{\rm\implies \:x + 1 \: is \: a \: factor \: of \: P(x) \: }$

On substituting x = 2, we get

$\green{\rm \: (2-1)P(2+1)=(2+2)P(2)}$

$\green{\rm \: P(3)=4P(2)}$

$\green{\rm \: {[P(2)]}^{2} =4P(2)}$

$\green{\rm \: {[P(2)]}^{2} - 4P(2) = 0}$

$\green{\rm \: P(2)[P(2) - 4] = 0}$

$\green{\rm \: P(2) = 0 \: \: or \: \: \: P(2) - 4 = 0}$

$\green{\rm \: P(2) = 0 \: \{rejected \} \: \: or \: \: \: P(2) = 4}$

Now, from above 3 conditions, we concluded that

$\rm \: P(x) = (x - 1)x(x + 1)f(x) - - - (1)$

Replace x by x + 1, we get

$\rm \: P(x + 1) = x(x + 1)(x + 2)f(x + 1)$

On multiply by x - 1, we get

$\rm \: (x - 1)P(x + 1) = x(x + 1)(x + 2)(x - 1)f(x + 1)$

From given equation (i),

$\rm \: (x + 2)P(x) = x(x + 1)(x + 2)(x - 1)f(x + 1)$

$\rm \: P(x) = x(x + 1)(x - 1)f(x + 1)$

From equation (1), we have

$\rm \: x(x + 1)(x - 1)f(x) = x(x + 1)(x - 1)f(x + 1)$

$\rm\implies \:f(x) = f(x + 1)$

$\rm\implies \:f(x) \: is \: constant$

Let f(x) = c, where c is constant.

So, equation (1) can be rewritten as

$\rm \: P(x) = (x - 1)x(x + 1)c - - - (2)$

On substituting x = 3, we get

$\rm \: P(3) = (3 - 1)3(3 + 1)c$

$\rm \: 16 = 24c$

$\rm\implies \:c = \dfrac{2}{3}$

So, equation (2) can be rewritten as

$\rm \: P(x) = \frac{2}{3} (x - 1)x(x + 1) - - - (3)$

Now,

$\rm \: P\bigg(\dfrac{7}{2} \bigg) = \frac{2}{3} \bigg(\dfrac{7}{2} - 1 \bigg)\bigg(\dfrac{7}{2} \bigg)\bigg(\dfrac{7}{2} + 1 \bigg)$

$\rm \: P\bigg(\dfrac{7}{2} \bigg) = \frac{2}{3} \bigg(\dfrac{5}{2} \bigg)\bigg(\dfrac{7}{2} \bigg)\bigg(\dfrac{9}{2} \bigg)$

$\rm \: P\bigg(\dfrac{7}{2} \bigg) = \frac{105}{4}$

$\rm\implies \:m \: = \: 105 \: \: and \: \: n \: = \: 4$

$\red{\rm\implies \boxed{ \rm{ \:\:m + n \: = \: 105 + 4 = 109} \: \: }}$