Miscellaneous Question 28
- $a>b>0$
For all the integral values of $a$ and $b$
$x=\frac{a^{3}+b^{3}}{\left(a^{2}-b^{2}\right){(a-b)^{2}+a b }}$
Quantity I : $x$
Quantity II : 1
(1) I $<$ II
(2) I $\leq$ II
(3) I $>$ II
(4) $I=I$
(5) No relation
Show Answer
Correct Answer: 28. (1)
Solution: 28. (1) Given:
$x=\frac{a^{3}+b^{3}}{\left(a^{2}-b^{2}\right)\left\{(a-b)^{2}+a b\right\}}$
$\Rightarrow x=\frac{(a+b)\left(a^{2}-a b+b^{2}\right)}{(a+b)(a-b)\left\{\left(a^{2}-2 a b+b^{2}\right)+a b\right\}}$
$\Rightarrow x=\frac{a^{2}-a b+b^{2}}{(a-b)\left(a^{2}-a b+b^{2}\right)}$
$\Rightarrow x=\frac{1}{a-b}$
Since, both ’ $a$ ’ and ’ $b$ ’ are positive integers and also $a>b$, $\Rightarrow a-b>1$
$\Rightarrow \frac{1}{a-b}<1$
$\therefore x<1$
BETA
