The values of a for which 2x^2 - 2(2a + 1)x + | vedclass

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

Question

(D) Let $f(x) = 2x^2 - 2(2a + 1)x + a(a + 1)$.For one root to be less than $a$ and the other root to be greater than $a$,the value of the quadratic fu

Vedclass · Accepted Answer

$a > 0 	ext{ or } a < -1$

Vedclass · Answer

(D) Let $f(x) = 2x^2 - 2(2a + 1)x + a(a + 1)$.For one root to be less than $a$ and the other root to be greater than $a$,the value of the quadratic function at $x = a$ must be negative,i.e.,$f(a) Substituting $x = a$ into the function:$f(a) = 2(a)^2 - 2(2a + 1)(a) + a(a + 1)$$f(a) = 2a^2 - 4a^2 - 2a + a^2 + a$$f(a) = -a^2 - a$Setting $f(a) $-a^2 - a $a^2 + a > 0$$a(a + 1) > 0$This inequality holds when $a > 0$ or $a Since the discriminant $D = 4(2a + 1)^2 - 8a(a + 1) = 8(a^2 + a + 1/2) = 8((a + 1/2)^2 + 1/4) > 0$ for all real $a$,the roots are always real. Thus,the condition $f(a) < 0$ is sufficient.

The values of $a$ for which $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ may have one root less than $a$ and other root greater than $a$ are given by

Similar Questions

$\frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}=?$

If the ratio of the roots of the equation $ax^2 + bx + c = 0$ is $p:q$,then:

If the roots of $x^2 - 7x + 6 = 0$ are $\alpha$ and $\beta$,then $\frac{1}{\alpha} + \frac{1}{\beta} = $

If the roots of the equation $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$ are equal but opposite in sign,then the value of $m$ is:

The number of solutions of the equation $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module