Let f x - x 4-8 x 3- ax 2- bx -16 - If all the roots of f x -0 lies between -0-20 -2- then which of the following is-are correct-A- a - b 2 is a constant and is equal to 64 -B- For particular values of a and b- 2 is one of the root- C- f3-4D- Roots are always in A-P-

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Higher Order Equations

Let f x = x 4...

Question

Let $f (x) = x^{4} - 8 x^{3} + a x^{2} - b x + 16$ . If all the roots of $f (x) = 0$ lies between $[0, 20 \sqrt{2}]$ , then which of the following is/are correct ?

(a - b)^{2}

is a constant and is equal to

64

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For particular values of

a

and

b

2

is one of the root.

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f (3) = 4

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Roots are always in A.P.

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Solution

The correct options are
A $(a - b)^{2}$ is a constant and is equal to $64$ .
B For particular values of $a$ and $b$ , $2$ is one of the root.
D Roots are always in A.P.
$x^{4} - 8 x^{3} + a x^{2} - b x + 16 = 0$
$x_{1} + x_{2} + x_{3} + x_{4} = 8 . . . (1)$
$\sum x_{i} x_{j} = a . . . (2)$
$\sum x_{i} x_{j} x_{k} = b . . . (3)$
$x_{1} . x_{2} . x_{3} . x_{4} = 16 . . . (4)$

$A.M. of x_{1}, x_{2}, x_{3}, x_{4} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}$
$= 2$
$G.M. of x_{1}, x_{2}, x_{3}, x_{4} = \sqrt[4]{x_{1} . x_{2} . x_{3} . x_{4}}$
$= \sqrt[4]{16}$
$= 2$

$∵ A.M. of x_{1}, x_{2}, x_{3}, x_{4} = G.M. of x_{1}, x_{2}, x_{3}, x_{4}$
Hence, the roots are equal.
$\Rightarrow x_{1} = x_{2} = x_{3} = x_{4} = 2$
From eqn(2), $a = 24$
From eqn(3), $b = 32$

$(a - b)^{2} = (24 - 32)^{2} = 64$

$f (x) = x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 16$
$= (x - 2)^{4}$
$f (3) = 1$

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Let f(x)=x4−8x3+ax2−bx+16. If all the roots of f(x)=0 lies between [0,20√2], then which of the following is/are correct ?

Let $f (x) = x^{4} - 8 x^{3} + a x^{2} - b x + 16$ . If all the roots of $f (x) = 0$ lies between $[0, 20 \sqrt{2}]$ , then which of the following is/are correct ?