The number of integers k for which the equati | vedclass

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Question

(b)

We have,

$x^3-27 x+k=0$

$f(x) =x^3-27 x$

$f^{\prime}(x) =3 x^2-27=3\left(x^2-9\right)$

Now, sum of roots of $x^3-27 x+k=0$ is Zero.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b)

We have,

$x^3-27 x+k=0$

$f(x) =x^3-27 x$

$f^{\prime}(x) =3 x^2-27=3\left(x^2-9ight)$

Now, sum of roots of $x^3-27 x+k=0$ is Zero.

$	herefore$ Third root is also integer.

Now, put $x=6 t$

$216 t^3-27(6 t)+k=0$

$54\left(4 t^3-3 tight) =-k$

$54 \cos 3 	heta =-k \quad[\because t=\cos 	heta]$

$\cos 3 	heta =-\frac{k}{54}$

Now, we get integral solution on $	heta=0, \pi$

$	herefore$ T'wo values of $k$ are possible.

The number of integers $k$ for which the equation $x^3-27 x+k=0$ has at least two distinct integer roots is

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