MCQ
If $\text{a},\text{b},\text{c}\in+\text{R}$ such that $\lambda\text{ abc}$ is the minimum value of a(b2 + c2) + b(c2 + a2) + c(a2 + b2), then $\lambda=$
- A1
- B3
- C4
- DNone of the above.
Solution:
We know that $ \text{A}.\text{M}.\geq\text{G}.\text{M}.$
Therefore, $ \frac{{\text{b}^2+\text{c}^2}}{2}\geq\sqrt{\text{b}^2\text{c}^2}$
$\Rightarrow\text{b}^{2}+\text{c}^{2}\geq2\text{bc}$
Multiplying a on both sides doesn’t change the inequality.
Since, given that a is positive.
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(1)$
Similarly, $\text{b}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(2)$
and $\text{c}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(3)$
adding (1), (2) and (3) we get
$\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{c}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq2\text{abc}+2\text{abc}+2\text{abc}$
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{b}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq6\text{abc}$
Therefore $\lambda$ is 6
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$P$ (computer turns out to be defective given that it is produced in plant $T_1$ )
$=10 P\left(\right.$ computer turns out to be defective given that it is produced in plant $\left.T_2\right)$,
where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is