(B) Given that $p(x) = kx^2 - \sqrt{2}x + 1$ and $(x - 1)$ is a factor of $p(x)$.According to the Factor Theorem, if $(x - 1)$ is a factor of $p(x)$,

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Given that $p(x) = kx^2 - \sqrt{2}x + 1$ and $(x - 1)$ is a factor of $p(x)$.According to the Factor Theorem, if $(x - 1)$ is a factor of $p(x)$, then $p(1) = 0$.Substituting $x = 1$ in the polynomial:$p(1) = k(1)^2 - \sqrt{2}(1) + 1 = 0$$k - \sqrt{2} + 1 = 0$Solving for $k$:$k = \sqrt{2} - 1$.