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

JEE Main · Mathematics · Basics of Mathematics

If 12^ 4+2 x^2 =(24 3 )^ 3 x^2-2 , then x is equal to

If 124+2x2=(243)3x2−212^{4+2 x^2}=(24 \sqrt{3})^{3 x^2-2}, then xx is equal to

Options

  1. A±1312\pm \sqrt{\frac{13}{12}}
  2. B±145\pm \sqrt{\frac{14}{5}}
  3. C±1213\pm \sqrt{\frac{12}{13}}
  4. D±514\pm \sqrt{\frac{5}{14}}

Correct answer

B. ±145\pm \sqrt{\frac{14}{5}}

Step-by-step solution

 We have, (12)4−2x2=(243)3x2−2(12)4+2x2=(12)32(3x2−2)4+2x2=32(3x2−2)8+4x2=9x2−69x2−4x2=8+6⇒5x2=14⇒x2=145∴x=±145\begin{aligned} & \text { We have, }(12)^{4-2 x^2}=(24 \sqrt{3})^{3 x^2-2} \\ & \begin{array}{l}(12)^{4+2 x^2}=(12)^{\frac{3}{2}}\left(3 x^2-2\right)\end{array} \\ & 4+2 x^2=\frac{3}{2}\left(3 x^2-2\right) \\ & 8+4 x^2=9 x^2-6 \\ & 9 x^2-4 x^2=8+6 \\ & \Rightarrow 5 x^2=14 \quad \Rightarrow x^2=\frac{14}{5} \quad \therefore \quad x= \pm \sqrt{\frac{14}{5}}\end{aligned}
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 99 Percentile Qs Bank for JEE Main PYQs