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

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-11x+24=0$
$II.$ $2y^{2}-9y+9=0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $17 x^{2} + 48 x = 9$
$II.$ $13 y^{2} = 32 y - 21$

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$,then the equation $(x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6) = 0$ has :-

If the difference between the corresponding roots of $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is the same and $a \neq b$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-16x+63=0$
$II.$ $y^{2}-2y-35=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module