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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Advanced · Mathematics · Basics of Mathematics

Let y_i(i=1 to n) are the solution satisfying the equation y^ _3( 3...

Let yi(i=1y_i(i=1 to n)n) are the solution satisfying the equation ylog⁡3(3y)=ylog⁡3(3y)−6y^{\log _3(\sqrt{3 y})}=y^{\log _3(3 y)}-6 where y>0y>0. Then:

Options

  1. A∑i=1nyi=289\sum_{i=1}^n y_i=\frac{28}{9}
  2. B∏i=1nyi=13\prod_{i=1}^n y_i=\frac{1}{3}
  3. CValue of nn is equal to 3.
  4. D∑i=1n1yi=9\sum_{i=1}^n \frac{1}{y_i}=9

Correct answer

B. ∏i=1nyi=13\prod_{i=1}^n y_i=\frac{1}{3}

Step-by-step solution

Let ylog⁡3(3y)=t\quad y^{\log _3(\sqrt{3 y})}=tt=t2−6⇒t2−t−6=0⇒(t−3)(t+2)=0t=3\begin{aligned} & t=t^2-6 \quad \Rightarrow \quad t^2-t-6=0 \quad \Rightarrow \quad(t-3)(t+2)=0 \\ & t=3\end{aligned}$\begin{aligned} \therefore \quad & y^{\log _3(\sqrt{3 y})}=3 \\ & \frac{1}{2} \log _3 3 y \cdot \log _e y=1\end{aligned}12(1+a)a=1⇒a2+a−2=0⇒(a+2)(a−1)=0a=1,−2y=3,1/9y1=1/9 and y2=3\begin{aligned} & \frac{1}{2}(1+a) a=1 \quad \Rightarrow \quad a^2+a-2=0 \quad \Rightarrow \quad(a+2)(a-1)=0 \\ & \begin{array}{l}a=1,-2 \\ y=3,1 / 9 \\ y_1=1 / 9 \quad \text { and } y_2=3\end{array}\end{aligned}
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