Let z = x+iy be a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation $z {¯ z}^{3} + ¯ z z^{3} = 350$ , is

Q. Consider the polynomial

f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} .

Let

s

be the sum of all distinct real roots of

f (x) = 0

and let

| s | = t .

The real number

s

lies in the interval

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Let f(x)=x12+a11x11+a10x10+⋯+a0 where ai's are integers. The number of polynomials such that all the roots of f(x) are from the set {1,2,3,4} is

The correct option is C 455 f(x)=(x−1)p(x−2)q(x−3)r(x−4)s So p+q+r+s=12 Where p,q,r,s are non-negative integers. Hence the number of polynomial will be = 12+4−1C4−1=455

Let $f (x) = x^{12} + a_{11} x^{11} + a_{10} x^{10} + \dots + a_{0}$ where $a_{i}$ 's are integers. The number of polynomials such that all the roots of $f (x)$ are from the set ${1, 2, 3, 4}$ is

The correct option is C $455$
$f (x) = (x - 1)^{p} (x - 2)^{q} (x - 3)^{r} (x - 4)^{s}$
So
$p + q + r + s = 12$
Where $p, q, r, s$ are non-negative integers.

Hence the number of polynomial will be $=^{12 + 4 - 1} C_{4 - 1} = 455$