The correct option is C 2
Given that the sum of coefficients
2n=64⇒n=6
The term with greatest binomial coefficient is T4,
Given condition is
T4−T3=(n−1)=6−1=5⇒ 6C3(35x/4)3(3−x/4)3− 6C2(35x/4)4(3−x/4)2=5⇒20(33x)−15(39x/2)=5⇒33x[4−3(33x/2)]=1
Assuming 33x/2=t
⇒t2[4−3t]=1⇒3t3−4t2+1=0
By observation, t=1 is a root of the equation,
⇒(t−1)(3t2−t−1)=0⇒t=1,3t2−t−1=0
Now, checking the discriminant, we get
D=1+12=13>0
Now,
t=1,1±√136
As 33x/2>0, ∀x∈R
So,
t=1,1+√136
Hence, there are 2 possible values of x.