Question

Let polynomial $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$ have integral coefficient (where $a > 0$ ). If there exist four distinct integers $α_{1}, α_{2}, α_{3}, α_{4} (α_{1} < α_{2} < α_{3} < α_{4})$ such that $f (α_{1}) = f (α_{2}) = f (α_{3}) = f (α_{4}) = 5$ and equation $f (x) = 9$ has integral roots, then find

$(i)$ $f (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4})$

$(ii)$ $f^{'} (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4})$

$(iii)$ Range of $f (x)$ in $[α_{2}, α_{3}]$

$(iv)$ Difference of largest and smallest root of equation $f (x) = 9$

Accepted Answer

$f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$ .

Since, $f (α_{1}) = f (α_{2}) = f (α_{3}) = f (α_{4}) = 5 \Rightarrow α_{1}, α_{2}, α_{3}, α_{4}$ are the roots of equation $f (x) = 5$ .

So it can be written as $f (x) - 5 = a (x - α_{1}) (x - α_{2}) (x - α_{3}) (x - α_{4}) . . . . . . . . . (1)$

Let for $x = β, f (β) = 9 (β \in I)$

$\Rightarrow f (β) - 5 = a (β - α_{1}) (β - α_{2}) (β - α_{3}) (β - α_{4})$

$f (α_{1}) = f (α_{2}) = f (α_{3}) = f (α_{4}) = 5$

Since, $α_{1}, α_{2}, α_{3}, α_{4}, β$ all are integer $\Rightarrow β - α_{1}, β - α_{2}, β - α_{3}, β - α_{4}$ are also integers.

Also, $α_{1} < α_{2} < α_{3} < α_{4} \Rightarrow β - α_{1} > β - α_{2} > β - α_{3} > β - α_{4}$ .

$\Rightarrow β - α_{1} = 2, β - α_{2} = 1, β - α_{3} = - 1, β - α_{4} = - 2, a = 1$ (By checking the factors)

$\Rightarrow α_{1} = β - 2, α_{2} = β - 1, α_{3} = β + 1, α_{4} = β + 2$

$\Rightarrow α_{1} + α_{2} + α_{3} + α_{4} = 4 β . . . . . . . . . (2)$

Replacing values of $α_{1}, α_{2}, α_{3}, α_{4}$ in equation $(1)$ , we get

$f (x) - 5 = \{x - (β - 2)\} \{x - (β - 1)\} \{x - (β + 1)\} \{x - (β + 2)\}$

$\Rightarrow f (x) = (x - β + 2) (x - β + 1) (x - β - 1) (x - β - 2) + 5$

$\Rightarrow f (x) = \{{(x - β)}^{2} - 4\} \{{(x - β)}^{2} - 1\} + 5$

$\Rightarrow f (x) = {(x - β)}^{4} - 5 {(x - β)}^{2} + 9$

Replacing $x$ with $x + β$ , we get,

$f (x + β) = x^{4} - 5 x^{2} + 9$

$(i)$ $f (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4})$

From equation $(2)$ $\Rightarrow f (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4}) = f (β)$

Since, $f (x + β) = x^{4} - 5 x^{2} + 9$ .

$\Rightarrow$ For $x = 0, f (β) = 9$

$(ii)$ $f^{'} (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4})$

From equation $(2)$ $\Rightarrow f^{'} (\frac{α_{1} + α_{2} + α_{3} + α_{4}}{4}) = f^{'} (β)$

Since, $f (x + β) = x^{4} - 5 x^{2} + 9 \Rightarrow f^{'} (x + β) = 4 x^{3} - 10 x$ .

$\Rightarrow$ For $x = 0, f^{'} (β) = 0$

$(iii)$ Range of $f (x)$ in $[α_{2}, α_{3}]$

Replacing values of $α_{2} = β - 1, α_{3} = β + 1$

$\Rightarrow$ Range of $f (x)$ in $[β - 1, β + 1]$ .

Range of $f (x)$ in $[β - 1, β + 1]$ is equivalent to range of $f (x + β)$ in $[- 1, 1]$ .

Range of $x^{4} - 5 x^{2} + 9$ in $[- 1, 1]$

$\Rightarrow f^{'} (x + β) = 4 x^{3} - 10 x = 0$

$\Rightarrow x = 0, \pm \sqrt{\frac{5}{2}}$

$0$ is the only value which lie in $[- 1, 1]$ .

Value at $x = 0$ is $9$ .

Value at $x = 1$ is $5$ .

Value at $x = - 1$ is $5$ .

Range is $[5, 9]$ .

$(iv)$ Difference of largest and smallest root of equation $f (x) = 9$ .

$\Rightarrow f (x) = {(x - β)}^{4} - 5 {(x - β)}^{2} + 9 = 9$

$\Rightarrow {(x - β)}^{2} ({(x - β)}^{2} - 5) = 0$

$\Rightarrow {(x - β)}^{2} = 0, {(x - β)}^{2} - 5 = 0$

$\Rightarrow x = β, β + \sqrt{5}, β - \sqrt{5}$

Difference of largest and smallest root of equation $= (β + \sqrt{5}) - (β - \sqrt{5}) = 2 \sqrt{5}$ .

Important Questions on Quadratic Equations

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