Let f(x) = x^2 - x + k - 2,k in R. Then the c | vedclass

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

Question

(D) The function $y = |f(|x|)|$ is non-derivable at $5$ points if the quadratic equation $x^2 - x + k - 2 = 0$ has two distinct positive roots.For the

Vedclass · Accepted Answer

$(2, 9/4)$

Vedclass · Answer

(D) The function $y = |f(|x|)|$ is non-derivable at $5$ points if the quadratic equation $x^2 - x + k - 2 = 0$ has two distinct positive roots.For the roots to be real and distinct,the discriminant $D > 0$:$D = (-1)^2 - 4(1)(k - 2) > 0$$1 - 4k + 8 > 0$$9 - 4k > 0 \Rightarrow k For the roots to be positive,the product of roots $\alpha\beta > 0$ and the sum of roots $\alpha + \beta > 0$:Product $\alpha\beta = k - 2 > 0 \Rightarrow k > 2$.Sum $\alpha + \beta = 1 > 0$ (which is always true).Combining these conditions,we get $2 Thus,the set of values is $(2, 9/4)$.

Let $f(x) = x^2 - x + k - 2$,$k \in R$. Then the complete set of values of $k$ for which $y = |f(|x|)|$ is non-derivable at $5$ distinct points is:

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

If $2 + i$ is a root of the equation ${x^3} - 5{x^2} + 9x - 5 = 0$,then the other roots are

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+4x+4=0$
$II.$ $y^{2}-8y+16=0$

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is

If the equation $\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} = 3x^3$ has $k$ real roots,then $k$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module