Question

The necessary and sufficient condition for the equation \[\left( 1-{{a}^{2}} \right){{x}^{2}}+2ax-1=0\] to have roots lying in the interval (0, 1) is:
(a) a > 0
(b) a < 0
(c) a > 2
(d) None of these

Answer 1

Hint: First find the roots of the equation in terms of a. Now, apply the condition for the roots in the given range. First, take the maximum value and then take the minimum value. Find the condition on a in both the cases. Try to get a generalized condition satisfying both the cases which is our required result.

Complete step-by-step answer:
Given the quadratic equation in the question is written as:
\[\left( 1-{{a}^{2}} \right){{x}^{2}}+2ax-1=0\]
A quadratic equation can be written in terms of p, q, r is:
\[p{{x}^{2}}+qx+r=0\]
The roots of the above quadratic equation can be defined as:
\[x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}\]
By comparing our equation to this equation, we get,
\[p=\left( 1-{{a}^{2}} \right),q=2a,r=-1\]
By substituting these values in our roots, we get,
\[x=\dfrac{-2a\pm \sqrt{{{\left( 2a \right)}^{2}}+4\left( 1-{{a}^{2}} \right)}}{2\left( 1-{{a}^{2}} \right)}\]
By simplifying, we can write the roots in terms of a as:
\[x=\dfrac{-2a\pm \sqrt{4{{a}^{2}}+4-4{{a}^{2}}}}{2-2{{a}^{2}}}\]
By further simplifying, we can write as,
\[x=\dfrac{-2a\pm 2}{2-2{{a}^{2}}}\]
By taking 2 common and canceling, we can write the equation as
\[x=\dfrac{-a\pm 1}{1-{{a}^{2}}}\]
The given interval of x can be written as (0, 1). So, the roots must satisfy the inequality given by:
\[0<\dfrac{-a\pm 1}{1-{{a}^{2}}}<1\]
By taking the maximum value, we can write it as
\[\dfrac{-a\pm 1}{1-{{a}^{2}}}<1\]
By cross multiplying, we get the inequality as,
\[-a\pm 1<1-{{a}^{2}}\]
By adding \[{{a}^{2}}\] and subtracting 1 on both the sides, we get,
\[{{a}^{2}}-a\pm 1-1<0\]
By simplifying, we can write it as 2 inequalities
\[a\left( a-1 \right)<0;{{a}^{2}}-a-2<0\]
By further simplifying, we can write it as,
\[a\left( a-1 \right)<0;\left( a-2 \right)\left( a+1 \right)<0\]
\[0By taking the minimum value, we get,
\[\dfrac{-a\pm 1}{1-{{a}^{2}}}>0\]
By cross multiplying, we get it as,
\[-a\pm 1>0\]
By simplifying we can get it as,
a < 1, a < – 1
Therefore, the generalized condition from 4 of them will be a > 0 (0 < a < 1).
Hence, option (a) is the right answer.

Note: Be careful while getting the roots in terms of a. Just substitute any value of a ( > 0) just for verification. You will see that you will get the roots in the interval. So, use the interval condition carefully because the whole answer depends on that condition.