Question

If the quadratic expression
$a{x}^2+(a-2+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}})x+(5^{{\log }_{5}3}-3^{{\log }_{3}5})$ is negative for exactly two integral values of $x$, then the possible value(s) of $a$ is/are

• A. $-\frac{2}{3}$
• B. $1$
• C. $2$
• D. $-\frac{1}{2}$

Answer 1

The correct option is B $1$

Given : $a{x}^2+(a-2+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}})x+(5^{{\log }_{5}3}-3^{{\log }_{3}5})$
$=a{x}^2+(a-2+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}})x+(3-5)$

Let $y=a^{\sqrt{{\log }_{a}b}};$ where $a,b$ are in the domain of $\log$ function.
$\Rightarrow {\log }_{a}y=\sqrt{{\log }_{a}b}\Rightarrow \frac{\ln y}{\ln a}=\sqrt{\frac{\ln b}{\ln a}}\Rightarrow \ln y=\sqrt{\ln a\times \ln b}\Rightarrow y=e^{\sqrt{\ln a\times \ln b}}$

$3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}}=e^{\sqrt{\ln 3\times \ln 5}}-e^{\sqrt{\ln 5\times \ln 3}}=0$

So,
$a{x}^2+(a-2+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}})x+(3-5)=a{x}^2+(a-2)x-2=a{x}^2+ax-2x-2=a(x+1)(x-\frac{2}{a})$
Now, $a<0$ not possible because it is a downward parabola which can be negative for infinite integral values of $x.$

So, $a>0$
As one root is $-1$ and the given expression is negative for exactly two integers, so


$1<\frac{2}{a}\le 2\Rightarrow \frac{1}{2}\le \frac{a}{2}<1\Rightarrow a\in [1,2)$

Hence, $a\in [1,2)$