Solveeit Logo
Login

Question

Question: If the 4<sup>th</sup> term in the expansion of \((px + x^{- 1})^{m}\) is 2.5 for all \(x \in R\) the...

If the 4th term in the expansion of (px+x−1)m(px + x^{- 1})^{m} is 2.5 for all x∈Rx \in R then

A

p=5/2,m=3p = 5/2,m = 3

B

p=12,m=6p = \frac{1}{2},m = 6

C

p=−12,m=6p = - \frac{1}{2},m = 6

D

None of these

Answer

p=12,m=6p = \frac{1}{2},m = 6

Explanation

Solution

We have T4=52T_{4} = \frac{5}{2} ⇒ T3+1=52T_{3 + 1} = \frac{5}{2}

⇒ m⥂C3(px)m−3(1x)3=52⇒m⥂C3pm−3xm−6=52m ⥂ C_{3}(px)^{m - 3}\left( \frac{1}{x} \right)^{3} = \frac{5}{2} \Rightarrow^{m} ⥂ C_{3}p^{m - 3}x^{m - 6} = \frac{5}{2} ......(i)

Clearly, R.H.S. of the above equality is independent of x

∴m−6=0\therefore m - 6 = 0, m=6m = 6

Putting m=6m = 6 in (i) we get 6⥂C3p3=52⇒6 ⥂ C_{3}p^{3} = \frac{5}{2} \Rightarrow p=12p = \frac{1}{2}.

Hence p=1/2,m=6p = 1/2,m = 6.