If the sum of the zeros of the polynomial f(x) = 2x3 − 3kx2 + 4x − 5 is 6,

Solution:

Let $\alpha, \beta$ be the zeros of the polynomial $f(x)=2 x^{3}-3 k x^{2}+4 x-5$ and we are given that

$\alpha+\beta+\gamma=6$

Then,

$\alpha+\beta+\gamma=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=-\frac{(-3 k)}{2}=\frac{3 k}{2}$

It is given that

$\alpha+\beta+\gamma=6$

Substituting $\alpha+\beta+\gamma=\frac{3 k}{2}$, we get

$\frac{+3 k}{2}=6$

$+3 k=6 \times 2$

$+3 k=12$

$k=\frac{12}{+3}$

$k=+4$

The value of $k$ is 4 .

Hence, the correct alternative is $(b)$