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If a , b , c are real numbers such that a ≥ b , c > 0, then
- a)ac < bc
- b)ac > bc
- c)ac ≥ bc
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If a , b , c are real numbers such that a≥b,c>0, thena)ac<bcb...
We know that $a$, $b$, and $c$ are real numbers such that $a< />
We can use the fact that $a< />
Now we can rewrite $ac+bc$ as $c(a+b)=c(a+\frac{a+c}{2})=\frac{3}{2}ac+\frac{1}{2}c^2$. Since $a+c=4$, we can substitute $c=4-a$ to get $ac+bc=\frac{3}{2}a(4-a)+\frac{1}{2}(4-a)^2=\frac{1}{2}(a-2)^2+\frac{7}{2}$.
The minimum value of $(a-2)^2$ is 0, which occurs when $a=2$. Therefore, the minimum value of $ac+bc$ is $\frac{7}{2}$, which occurs when $a=2$, $b=3$, and $c=4$.
We can use the fact that $a< />
Now we can rewrite $ac+bc$ as $c(a+b)=c(a+\frac{a+c}{2})=\frac{3}{2}ac+\frac{1}{2}c^2$. Since $a+c=4$, we can substitute $c=4-a$ to get $ac+bc=\frac{3}{2}a(4-a)+\frac{1}{2}(4-a)^2=\frac{1}{2}(a-2)^2+\frac{7}{2}$.
The minimum value of $(a-2)^2$ is 0, which occurs when $a=2$. Therefore, the minimum value of $ac+bc$ is $\frac{7}{2}$, which occurs when $a=2$, $b=3$, and $c=4$.
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If a , b , c are real numbers such that a≥b,c>0, thena)ac<bcb...
Ac≥bc because in both of them 'c' is common and as a≥b thus ac≥bc.
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