Question

Let P, Q, R, S and T are five sets about the quadratic equation
(a – 5)$x^2$ – 2ax + (a – 4) = 0, a $\ne$ 5 such that
P : All values of ‘a’ for which the product of roots of given quadratic equation is positive.
Q : All values of ‘a’ for which the product of roots of given quadratic equation is negative.
R : All values of ‘a’ for which the product of real roots of given quadratic equation is positive.
S : All values of ‘a’ for which the roots of given quadratic are real.
T : All values of ‘a’ for which the given quadratic equation has complex roots.

• A. least positive integer for set R is 2
• B. least positive integer for set R is 3
• C. greatest positive integer for set T is 3
• D. none of the above.

Answer 1

The correct option is B

least positive integer for set R is 3

$\because a-5\ne 0$

$\therefore x^2-\left(\frac{2a}{a-5}\right)x+\left(\frac{a-4}{a-5}\right)=0$

If roots are $\alpha$ and $\beta$, then

For $P:\alpha \beta =\left(\frac{a-4}{a-5}\right)>0$

or $a\in (-\infty ,4)\cup (5,\infty )$

For $Q=\{a:a\in (4,5\left)\right\}$

For R : D $\ge$ 0 and $\alpha \beta >0$

$\Rightarrow \frac{4a^2}{(a-5{)}^2}-\frac{4(a-4)}{(a-5)}\le 0$ and $\left\{\frac{a-4}{a-5}\right\}>0$

$\frac{9a-20}{(a-5{)}^2}\le 0$ and $\left(\frac{a-4}{a-5}\right)>0$

$\therefore a\in [\frac{20}{9},\infty )$ and $a\in (-\infty ,4)\cup (5,\infty )$

$\therefore R=\{a:a\in [\frac{20}{9},4)\cup (5,\infty \left)\right\}$

For S : D $\ge$ 0

$\therefore S=\{a:[\frac{20}{9},\infty )\}$ an

For T :

D < 0

$a<\frac{20}{9}$

$\therefore T=\{a:(-\infty ,\frac{20}{9})\}$

$R=\{a:aϵ[\frac{20}{9},4)\cup (5,\infty \left)\right\}$

$\therefore$ Least positive integer of R is 3.