Let m be the minimum possible value of \log _3(3^y_1 + 3^y_2 + 3^y_3) , where y_1,y_2,y_3 are real numbers for which y_1 + y_2 + y_3 = 9. Let M be the maximum possible value of (\log _3x_1 + \log _3x_2 + \log _3x_3) , where x_1,x_2,x_3 are positive real numbers for which x_1 + x_2 + x_3 = 9. Then the value of \log _2(m^3) + \log _3(M^2) is ...........

Sequences and Series PYQ — JEE Advanced 2020 · Shift 1 · Q31 with Solution

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Correct answer: 8

We need the minimum value of

\log_3\left(3^{y_1}+3^{y_2}+3^{y_3}\right)

subject to

y_1+y_2+y_3=9.

Since $3^x$ is a convex function, by Jensen's inequality (or AM-GM for exponentials), for fixed sum $y_1+y_2+y_3$ , the quantity

3^{y_1}+3^{y_2}+3^{y_3}

is minimum when

y_1=y_2=y_3=\frac{9}{3}=3.

So,

3^{y_1}+3^{y_2}+3^{y_3}\ge 3\cdot 3^3=81.

Hence,

m=\log_3(81)=4.

We need the maximum value of

\log_3 x_1+\log_3 x_2+\log_3 x_3

subject to

x_1+x_2+x_3=9, \qquad x_1,x_2,x_3>0.

Using log properties,

\log_3 x_1+\log_3 x_2+\log_3 x_3 = \log_3(x_1x_2x_3).

So we need to maximize $x_1x_2x_3$ given $x_1+x_2+x_3=9$ .

By AM-GM,

\frac{x_1+x_2+x_3}{3}\ge \sqrt[3]{x_1x_2x_3}

3\ge \sqrt[3]{x_1x_2x_3}

x_1x_2x_3\le 27,

with equality when

x_1=x_2=x_3=3.

Thus,

M=\log_3(27)=3.

We need

\log_2(m^3)+\log_3(M^2).

Substitute $m=4$ and $M=3$ :

\log_2(4^3)+\log_3(3^2).

Now,

4^3=64 \Rightarrow \log_2 64=6,

and

\log_3 9=2.

Therefore,

\log_2(m^3)+\log_3(M^2)=6+2=8.

Derived answer = $8$ .

Stored correct answer = $8$ .

So the answers agree.