Question

Assertion (A) : If $a, b, c$ are distinct positive real numbers such that $a^{2} + b^{2} + c^{2} = 1$, then$\Sigma ab =2$
Reason (R) : $A.M. \ge G.M$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true

Answer 1

Answer: (D)

$A.M. \ge G.M$ as $a, b, c$ are distinct real numbers
$\therefore \frac{a^{2} +b^{2}}{2} >\sqrt{a^{2} b^{2}} i e a^{2} +b^{2} >2ab ...\left(i\right)$
Similarly $b^{2} +c^{2} >2bc .....\left(ii\right)$
$\& c^{2} + a^{2} >2ac ....\left(iii\right)$
Now by adding (i), (ii) (iii), we get
$2\left(a^{2} +b^{2} +c^{2}\right) >2 \left(ab +bc +ca\right)$
$\Rightarrow ab +bc +ca < 1$ (using $a^{2} +b^{2} +c^{2} = 1$)
$\therefore $ Assertion (A) is false & Reason (R) is true.