Animated Solution for Mathematics - Quadratic Equations: The set of all real values of $\lambda$ for which the quadratic equations, $({\lambda }^2+1)x^2-4\lambda x+2=0$ always have exactly one root in the interval (0,1) is :

A Journey into Quadratic Behavior

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving an equation; we are exploring the geometric soul of a quadratic function.

When you see a problem asking for the set of all real values of $\lambda$ for which the quadratic equations have exactly one root in the interval $(0,1)$, do not rush to the discriminant. Instead, pause, visualize, and imagine you are standing on the coordinate plane, watching a parabola dance.

The Visual Foundation

We define our function as $f(x)=({\lambda }^2+1)x^2-4\lambda x+2$.

The first thing to notice is the coefficient of $x^2$, which is ${\lambda }^2+1$. Since ${\lambda }^2$ is always non-negative for any real $\lambda$, ${\lambda }^2+1$ is strictly positive. This is our anchor: the parabola always opens upwards.

Now, look at $f(0)$. When we plug in $x=0$, the $x$ terms vanish, leaving us with $f(0)=2$.

Our parabola starts at the point $(0,2)$, firmly above the x-axis. This is a crucial piece of information. If the curve starts above the x-axis and must cross it exactly once in the interval $(0,1)$, it must dive downwards and cross the axis before it reaches $x=1$.

The Logic of Crossing

This brings us to the Intermediate Value Theorem. If a continuous function starts positive at $x=0$ and must cross the x-axis at some point $x_0$ where $0<x_0<1$, it must necessarily be negative at $x=1$.

Mathematically, this is expressed as $f(0)\cdot f(1)<0$. Since we know $f(0)=2$, which is positive, the condition simplifies beautifully to $f(1)<0$.

Let us calculate $f(1)$ by substituting $x=1$ into our function:

Simplifying this, we get $f(1)={\lambda }^2-4\lambda +3$. We need this to be less than zero:

Factorizing this quadratic, we find $(\lambda -1)(\lambda -3)<0$. The solution to this inequality is $\lambda \in (1,3)$.

The JEE Trap - Boundary Analysis

Many students stop here, but this is where the JEE Advanced examiners test your depth. Is the inequality $f(1)<0$ sufficient? What if the root lands exactly on the boundary?

We must manually check the endpoints $\lambda =1$ and $\lambda =3$.

First, let us test $\lambda =3$. Substituting this into our original equation, we get:

This simplifies to $2(5x-1)(x-1)=0$. The roots are $x=1/5$ and $x=1$.

Since $1/5$ lies perfectly within our interval $(0,1)$, $\lambda =3$ is a valid solution.

Next, let us test $\lambda =1$. The equation becomes:

This simplifies to $2(x-1{)}^2=0$, yielding the root $x=1$. Does $x=1$ lie in the open interval $(0,1)$? No, it does not. Therefore, $\lambda =1$ must be rejected.

The Final Set

By combining our initial interval $(1,3)$ with the valid boundary point $\lambda =3$, we arrive at our final answer:

This journey shows that math is not just about formulas; it is about checking the boundaries, respecting the definitions, and visualizing the behavior of the function. Keep this mindset, and you will conquer any problem the exam throws at you.