If the 9 text th term in the expansion of ((1/x2)+(x/2) log 2 x)12 is equal to 495 , then sum of all possible values of x is

Question

If the 9 text th term in the expansion of ((1/x2)+(x/2) log 2 x)12 is equal to 495 , then sum of all possible values of x is (A) (9/2) (B) (9/4) (C) 4 (D) (17/4)

Tardigrade Admin · Accepted Answer

T9= 12 C8((1/x2))4((x log 2 x/2))8= 12 C4 (( log 2 x)8/28)=495. Since, 12 C4=495,( log 2 x)8=28 and log 2 x= ± 2 giving x=4, (1/4)

Q. If the $9^{th}$ term in the expansion of $(\frac{1}{x ^{2}} + \frac{x}{2} lo g_{2} x)^{12}$ is equal to 495 , then sum of all possible values of $x$ is

$\frac{9}{2}$

$\frac{9}{4}$

4

$\frac{17}{4}$

$T_{9} =^{12} C_{8} (\frac{1}{x ^{2}})^{4} (\frac{x l o g _{2} x}{2})^{8} =^{12} C_{4} \frac{( l o g _{2} x ) ^{8}}{2 ^{8}} = 495$ .
Since, $^{12} C_{4} = 495, (lo g_{2} x)^{8} = 2^{8}$ and $lo g_{2} x = \pm 2$ giving $x = 4, \frac{1}{4}$

Q. If the 9th term in the expansion of (x21​+2x​log2​x)12 is equal to 495 , then sum of all possible values of x is

29​

49​

4

417​

T9​=12C8​(x21​)4(2xlog2​x​)8=12C4​28(log2​x)8​=495. Since, 12C4​=495,(log2​x)8=28 and log2​x=±2 giving x=4,41​

Q. If the $9^{th}$ term in the expansion of $(\frac{1}{x ^{2}} + \frac{x}{2} lo g_{2} x)^{12}$ is equal to 495 , then sum of all possible values of $x$ is

$\frac{9}{2}$

$\frac{9}{4}$

$\frac{17}{4}$

$T_{9} =^{12} C_{8} (\frac{1}{x ^{2}})^{4} (\frac{x l o g _{2} x}{2})^{8} =^{12} C_{4} \frac{( l o g _{2} x ) ^{8}}{2 ^{8}} = 495$ .
Since, $^{12} C_{4} = 495, (lo g_{2} x)^{8} = 2^{8}$ and $lo g_{2} x = \pm 2$ giving $x = 4, \frac{1}{4}$