JEE Mathematics 2023 Question with Solution

Given: $a,b,c,d$ are positive real numbers and $a+b+c+d=11$.

Find: The value of $\beta$ if the maximum value of $a^5b^3c^2d$ is $3750\beta$.

Use the A.M. (\geq) G.M. inequality on the $11$ terms

$$\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{2},\frac{c}{2},d$$

Then,

$$\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11} \geq \sqrt[11]{\left(\frac{a}{5}\right)^5\left(\frac{b}{3}\right)^3\left(\frac{c}{2}\right)^2 d}$$

Since

$$\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d=a+b+c+d=11$$

we get

$$\frac{11}{11} \geq \sqrt[11]{\frac{a^5b^3c^2d}{5^5\cdot 3^3\cdot 2^2}}$$

So,

$$a^5b^3c^2d \leq 5^5\cdot 3^3\cdot 2^2 = 337500$$

Now compare with

$$3750\beta = 337500$$

Hence,

$$\beta = \frac{337500}{3750} = 90$$

Therefore, the correct value is $90$. The solution concludes the numerical value $\beta=90$, which corresponds to option C. The page label saying option B is inconsistent with the working and the listed options.

Given: The exponents in $a^5b^3c^2d$ are $5,3,2,1$ and their sum is $11$.

Find: The maximum product under the constraint $a+b+c+d=11$.

The exponents suggest splitting the variables into repeated equal parts:

• $a$ into $5$ equal parts of size $\frac{a}{5}$
• $b$ into $3$ equal parts of size $\frac{b}{3}$
• $c$ into $2$ equal parts of size $\frac{c}{2}$
• $d$ into $1$ part

This creates exactly $11$ terms whose sum is $11$. By A.M. (\geq) G.M., the product is maximum when all these $11$ terms are equal. That yields the upper bound

$$\left(\frac{a}{5}\right)^5\left(\frac{b}{3}\right)^3\left(\frac{c}{2}\right)^2 d \leq 1$$

Multiplying by $5^5\cdot 3^3\cdot 2^2$ gives

$$a^5b^3c^2d \leq 5^5\cdot 3^3\cdot 2^2 = 337500$$

Then

$$337500 = 3750\beta$$

so

$$\beta = 90$$

Thus the correct option from the listed choices is C.