Q. $If\ x = \frac{\begin{bmatrix} 729 + 6(2)(243) + 15(4)(81) \\ + 20(8)(27) + 15(16)(9) \\ + 6(32)3 + 64 \end{bmatrix}}{1 + 4(4) + 6(16) + 4(64) + 256},\ then\ \sqrt{x} - \frac{1}{\sqrt{x}}\ isequal\ to$

A) 0.2
B) 4.8
C) 1.02
D) 5.2
Explanation

(b)
We have,
$x = \frac{\begin{bmatrix}
729 + 6(2)(243) + 15(4)(81) \\
+ 20(8)(27) + 15(16)(9) + 6(32)(3) + 64
\end{bmatrix}}{1 + 4(4)6(16) + 4(64) + 256}$
$= \frac{\begin{bmatrix}
\ ^{6}C_{0}(3)^{6} + \ ^{6}C_{1}3^{5}2 + \ ^{6}C_{2}3^{4}2^{2} \\
+ \ ^{6}C_{3}3^{3}\ 2^{3} + \ ^{6}C_{4}3^{2}2^{4} + \ ^{6}C_{5}{3\ 2}^{5} + \ ^{6}C_{6}2^{6}
\end{bmatrix}}{\ ^{4}C_{0} + \ ^{4}C_{1}4 + \ ^{4}C_{2}4^{2} + \ ^{4}C_{3}4^{3} + \ ^{4}C_{4}4^{4}}$
$\Longrightarrow x = \frac{(3 + 2)^{6}}{(1 + 4)^{4}} = \frac{5^{6}}{5^{4}}$
$\Longrightarrow x = 5^{2}$
$\therefore\ \sqrt{x} - \frac{1}{\sqrt{x}} = 5 - \frac{1}{5} = 4.8$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.