Question

In the expansion of ${(3^{5x/4}+3^{-x/4})}^n$ the sum of binomial coefficient is $64$. If the term with greatest binomial coefficient exceeds the third term by $(n-1),$ then the number of value(s) of $x$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given that the sum of coefficients
$2^n=64\Rightarrow n=6$
The term with greatest binomial coefficient is $T_4$,
Given condition is
$T_4-T_3=(n-1)=6-1=5\Rightarrow {}^6{C}_3{\left(3^{5x/4}\right)}^3{\left(3^{-x/4}\right)}^3-{}^6{C}_2{\left(3^{5x/4}\right)}^4{\left(3^{-x/4}\right)}^2=5\Rightarrow 20\left(3^{3x}\right)-15\left(3^{9x/2}\right)=5\Rightarrow 3^{3x}[4-3(3^{3x/2}\left)\right]=1$
Assuming $3^{3x/2}=t$
$\Rightarrow t^2[4-3t]=1\Rightarrow 3t^3-4t^2+1=0$
By observation, $t=1$ is a root of the equation,
$\Rightarrow (t-1)(3t^2-t-1)=0\Rightarrow t=1,3t^2-t-1=0$
Now, checking the discriminant, we get
$D=1+12=13>0$
Now,
$t=1,\frac{1\pm \sqrt{13}}{6}$
As $3^{3x/2}>0,\forall x\in R$
So,
$t=1,\frac{1+\sqrt{13}}{6}$

Hence, there are $2$ possible values of $x$.