Question:
If $\alpha,\beta$ $(\alpha<\beta)$ are the roots of
\[
2x^2-x-6=0
\]
and
\[
\alpha x^2+kx-\beta\leq0 \quad \forall x\in\mathbb{R},
\]
then the number of integral values $k$ takes is
If $\alpha,\beta$ $(\alpha<\beta)$ are the roots of
\[
2x^2-x-6=0
\]
and
\[
\alpha x^2+kx-\beta\leq0 \quad \forall x\in\mathbb{R},
\]
then the number of integral values $k$ takes is
Show Hint
For quadratic inequalities valid for all real numbers, always check:
\[
a<0,\quad D\leq0
\]
for expressions of the form
\[
ax^2+bx+c\leq0.
\]
Updated On: Jun 17, 2026
- $24$
- $15$
- $6$
- $7$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
For a quadratic expression:
\[
ax^2+bx+c\leq0
\quad \forall x\in\mathbb{R},
\]
the necessary conditions are:
\[
a<0
\]
and
\[
D\leq0
\]
where $D$ is the discriminant.
Step 1: Find the roots $\alpha$ and $\beta$.
Given: \[ 2x^2-x-6=0 \] Factorizing: \[ 2x^2-4x+3x-6=0 \] \[ 2x(x-2)+3(x-2)=0 \] \[ (x-2)(2x+3)=0 \] Hence: \[ \alpha=-\frac{3}{2}, \qquad \beta=2 \]
Step 2: Apply the condition for negativity.
The quadratic is: \[ \alpha x^2+kx-\beta \] Substituting values: \[ -\frac{3}{2}x^2+kx-2\leq0 \] Since: \[ -\frac{3}{2}<0, \] the leading coefficient condition is satisfied. Now discriminant: \[ D=k^2-4\left(-\frac{3}{2}\right)(-2) \] \[ =k^2-12 \] For the inequality to hold for all real $x$: \[ D\leq0 \] Hence: \[ k^2\leq12 \] \[ -2\sqrt{3}\leq k\leq2\sqrt{3} \] Integral values of $k$ are: \[ -3,-2,-1,0,1,2,3 \] Total values: \[ 7 \] Hence, \[ \boxed{7} \]
Step 1: Find the roots $\alpha$ and $\beta$.
Given: \[ 2x^2-x-6=0 \] Factorizing: \[ 2x^2-4x+3x-6=0 \] \[ 2x(x-2)+3(x-2)=0 \] \[ (x-2)(2x+3)=0 \] Hence: \[ \alpha=-\frac{3}{2}, \qquad \beta=2 \]
Step 2: Apply the condition for negativity.
The quadratic is: \[ \alpha x^2+kx-\beta \] Substituting values: \[ -\frac{3}{2}x^2+kx-2\leq0 \] Since: \[ -\frac{3}{2}<0, \] the leading coefficient condition is satisfied. Now discriminant: \[ D=k^2-4\left(-\frac{3}{2}\right)(-2) \] \[ =k^2-12 \] For the inequality to hold for all real $x$: \[ D\leq0 \] Hence: \[ k^2\leq12 \] \[ -2\sqrt{3}\leq k\leq2\sqrt{3} \] Integral values of $k$ are: \[ -3,-2,-1,0,1,2,3 \] Total values: \[ 7 \] Hence, \[ \boxed{7} \]
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