Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2024 · Mathematics · Basics of Mathematics

For x ≥ y > 1 , let _x((x)/(y))+ _y((y)/(x))=k , then the value of...

For x≥y>1x \geq y > 1, let log⁡x(xy)+log⁡y(yx)=k\log _x\left(\frac{x}{y}\right)+\log _y\left(\frac{y}{x}\right)=k, then the value of kk can never be equal to:

Options

  1. A-1
  2. B−12-\frac{1}{2}
  3. C0
  4. D1

Correct answer

D. 1

Step-by-step solution

log⁡x(xy)+log⁡y(yx)=k\log _x\left(\frac{x}{y}\right)+\log _y\left(\frac{y}{x}\right)=k⇒log⁡xx−log⁡xy+log⁡yy−log⁡yx=k⇒1−log⁡xy+1−1log⁡xy=k Let, log⁡xy=t∴2−t−1t=k⇒2t−t2−1=kt⇒t2+(k−2)t+1=0 For  solution, D=b2−4ac≥0(k−2)2−4≥0(k−2)2≥0k−2≥2k−2≤0k≥4k≤0∴k≥4 or k≤0\begin{aligned} & \Rightarrow \quad \log _x x-\log _x y+\log _y y-\log _y x=k \\ & \Rightarrow \quad 1-\log _x y+1-\frac{1}{\log _x y}=k \\ & \text{ Let, } \quad \log _x y=t \\ & \therefore \quad 2-t-\frac{1}{t}=k \\ & \Rightarrow \quad 2 t-t^2-1=k t \\ & \Rightarrow \quad t^2+(k-2) t+1=0 \\ & \text{\text{ For } solution, } \quad \mathrm{D}=b^2-4 a c \geq 0 \\ & (k-2)^2-4 \geq 0 \quad(k-2)^2 \geq 0 \\ & k-2 \geq 2 \quad k-2 \leq 0 \\ & k \geq 4 \quad k \leq 0 \\ & \therefore k \geq 4 \text{ or } k \leq 0 \end{aligned}$Hence,Hence,k \neq 1$.
Practice Basics of Mathematics on Quantrex Academy →

More from Basics of Mathematics

All Basics of Mathematics questions Full Basics of Mathematics list Basics of Mathematics JEE Maths PYQsJEE Mathematics PYQs All NDA PYQs