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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2023 · Mathematics · Quadratic Equation

Consider the equation (1-x)^4+(5-x)^4=82 What is the number of real...

Consider the equation (1−x)4+(5−x)4=82(1-x)^4+(5-x)^4=82 What is the number of real roots of the equation?

Options

  1. A0
  2. B2
  3. C4
  4. D8

Correct answer

B. 2

Step-by-step solution

 Given, (1−x)4+(5−x)4=82Let, 3−x=y⇒((y−2)2)2+((y+2)2)2=82⇒(y2−4y+4)2+(y2+4y+4)2=82⇒2[(y2+4)2+(4y)2]=82⇒y4+8y2+16+16y2=41⇒y4+24y2−25=0⇒(y2+25)(y2−1)=0⇒y=±1,±5i\begin{aligned} & \text{ Given, }(1-\mathrm{x})^4+(5-\mathrm{x})^4=82 \\ & \text{Let, } \quad 3-\mathrm{x}=\mathrm{y} \\ \Rightarrow & \left((\mathrm{y}-2)^2\right)^2+\left((\mathrm{y}+2)^2\right)^2=82 \\ \Rightarrow & \left(\mathrm{y}^2-4 \mathrm{y}+4\right)^2+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)^2=82 \\ \Rightarrow & 2\left[\left(\mathrm{y}^2+4\right)^2+(4 \mathrm{y})^2\right]=82 \\ \Rightarrow & \mathrm{y}^4+8 \mathrm{y}^2+16+16 \mathrm{y}^2=41 \\ \Rightarrow & \mathrm{y}^4+24 \mathrm{y}^2-25=0 \\ \Rightarrow & \left(\mathrm{y}^2+25\right)\left(\mathrm{y}^2-1\right)=0 \\ \Rightarrow & \mathrm{y}= \pm 1, \pm 5 \mathrm{i}\end{aligned} If y=1⇒x=2y=1 \Rightarrow x=2 If y=−1⇒x=4y=-1 \Rightarrow x=4 If y=5i⇒x=3−5iy=5 i \Rightarrow x=3-5 i If y=−5i⇒x=3+5iy=-5 i \Rightarrow x=3+5 i So, number of real roots =2=2
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