JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $a, b, c$ be three distinct positive real numbers such that $(2 a)^{l o g_{e} a} = (b c)^{l o g_{e} b}$ and $b^{l o g_{e} 2} = a^{l o g_{e} c}$ Then $6a + 5bc$ is equal to _____.

Choose the correct answer:

Explanation

Given that $(2a)^{\log_e a} = (bc)^{\log_e b}$

\implies \log_e a (\log_e 2 + \log_e a) = \log_e b (\log_e b + \log_e c) \dots (i)

And $b^{\log_e 2} = a^{\log_e c}$

\implies \log_e 2 \cdot \log_e b = \log_e c \cdot \log_e a \implies \log_e 2 = \frac{\log_e c \cdot \log_e a}{\log_e b}

Putting in equation $(i)$ , we get:

\log_e a \left( \frac{\log_e c \cdot \log_e a}{\log_e b} + \log_e a \right) = \log_e b (\log_e b + \log_e c)

(\log_e a)^2 (\log_e b + \log_e c) - (\log_e b)^2 (\log_e b + \log_e c) = 0

\implies \log_e bc \{ (\log_e a)^2 - (\log_e b)^2 \} = 0

\implies \log_e bc = 0 \text{ or } \log_e a = \log_e b

\implies bc = 1 \text{ or } a = b

If $bc = 1$ then $(2a)^{\log_e a} = (1)^{\log_e b} = 1$

\implies (2a)^{\log_e a} = 1 \implies a = 1 \text{ or } a = \frac{1}{2}

Now,

If $a = 1$ and $bc = 1$ then $6a + 5bc = 11$

And if $a = \frac{1}{2}$ and $bc = 1$ then $6a + 5bc = 8$