MCQ
The set of all real values of $\lambda$ for which the quadratic equations, $\left(\lambda^{2}+1\right) x ^{2}-4 \lambda x +2=0$ always have exactly one root in the interval $(0,1)$ is
- A$(-3,-1)$
- ✓$(1,3]$
- C$(0,2)$
- D$(2,4]$
✓
Answer
Correct option: B.
$(1,3]$
b
If exactly one root in (0,1) then
If exactly one root in (0,1) then
$\Rightarrow \quad f (0) \cdot f (1)<0$
$\Rightarrow \quad 2\left(\lambda^{2}-4 \lambda+3\right)<0$
$\Rightarrow \quad 1<\lambda<3$
Now for $\lambda=1,2 x ^{2}-4 x +2=0$
$(x-1)^{2}=0, x=1,1$
So both roots doesn't lie between (0,1)
$\therefore \lambda \neq 1$
Again for $\lambda=3$
$10 x^{2}-12 x+2=0$
$\Rightarrow \quad x=1, \frac{1}{5}$
so if one root is 1 then second root lie between (0,1)
so $\lambda=3$ is correct.
$\therefore \quad \lambda \in(1,3]$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $3{p^2} = 5p + 2$ and $3{q^2} = 5q + 2$, where $p \ne q$, then $pq$ is equal to
→If $\sin A,\sin B,\cos A$ are in $G.P.$, then roots of ${x^2} + 2x\cot B + 1 = 0$ are always
→$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin (x + a) + \sin (a - x) - 2\sin a}}{{x\sin x}}} \right] = $
→A dice is rolled three times, find the probability of getting a larger number than the previous number each time ?
The sum of first two terms of a $G.P.$ is $1$ and every term of this series is twice of its previous term, then the first term will be
→The solution set of $x-y \geq 0$ is shown graphically by
→The value of $\cos A - \sin A$ when $A = \frac{{5\pi }}{4},$ is
→What is number of outcomes when tossing two coins together?
${ }^x C_7-{ }^x C_5=0$, then $x=$ :
→Let $0 < \theta < \frac{\pi }{2}$. If the eccentricity of the hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\,\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\,\theta }} = 1$ is greater than $2$, then the length of its latus rectum lies in the interval
→