MCQ 11 Mark
Statement-1 (A): $a+b+c=6$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}$, then $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=6$
Statement-2 $(R): (a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Statement-2 $(R): (a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
- AStatement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
- ✓Statement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
- CStatement-1 is True, Statement-2 is False.
- DStatement-1 is False, Statement-2 is True.
Answer
View full question & answer→Correct option: B.
Statement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
(b)
Statement-2 is true.
We have,$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2} \Rightarrow 2(a b+b c+c a)=3 a b c \Rightarrow a b+b c+c a=\frac{3}{2} a b c$
Thus, we have,
$a+b+c=6$ and $a b+b c+c a=\frac{3}{2} a b c$
$\Rightarrow(a+b+c)(a b+b c+c a)=6 \times \frac{3}{2} a b c$
$\Rightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+3 a b c=9 a b c \Rightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)=6 a b c$
$\therefore \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)=\frac{a^2+b^2}{a b}+\frac{a^2+c^2}{c a}+\frac{b^2+c^2}{b c}$
$=\frac{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}{a b c}=\frac{6 a b c}{a b c}=6$
So, statement-1 is true. But, statement-2 is not a correct explanation for statement-1
Hence, option (b) is correct
Statement-2 is true.
We have,$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2} \Rightarrow 2(a b+b c+c a)=3 a b c \Rightarrow a b+b c+c a=\frac{3}{2} a b c$
Thus, we have,
$a+b+c=6$ and $a b+b c+c a=\frac{3}{2} a b c$
$\Rightarrow(a+b+c)(a b+b c+c a)=6 \times \frac{3}{2} a b c$
$\Rightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+3 a b c=9 a b c \Rightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)=6 a b c$
$\therefore \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)=\frac{a^2+b^2}{a b}+\frac{a^2+c^2}{c a}+\frac{b^2+c^2}{b c}$
$=\frac{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}{a b c}=\frac{6 a b c}{a b c}=6$
So, statement-1 is true. But, statement-2 is not a correct explanation for statement-1
Hence, option (b) is correct