Question

Assertion :If the series represented by function $f\left(x\right)=x^2+x^4+x^6+x^8+...$ converges, then function $g\left(x\right)=\left[x\right]$ (where [.] denotes the greatest integer function) is continuous at one fixed point of $f\left(x\right)$. Reason: $f\left(x\right)=x\Rightarrow x^2+x-1=0$ which gives two fixed points.

• A. If both assertion and reason are correct and reason is the correct explanation of the assertion
• B. If both assertion and reason are correct but reason is not correct explanation of the assertion
• C. If assertion is correct, but reason is incorrect
• D. If assertion is incorrect, but reason is correct

Answer 1

The correct option is C If assertion is correct, but reason is incorrect

Given infinite series is $f\left(x\right)=x^2+x^4+x^6+x^8+...=\frac{x^2}{1-x^2}$

$(\because$ series converges $\Rightarrow \left|x^2\right|<1\Rightarrow -1<x<1)$

Now at fixed points $f\left(x\right)=x$

$\Rightarrow \frac{x^2}{1-x^2}=x$$\Rightarrow x^2=x-x^3$$\Rightarrow x^3+x^2-x=0$$\Rightarrow x(x^2+x-1)=0$

$\Rightarrow x=0$ or $x^2+x-1=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$

$\Rightarrow x=\frac{\sqrt{5}-1}{2}(\because \frac{-1-\sqrt{5}}{2}<-1)$

$\therefore$ there are two fixed points $x=\frac{\sqrt{5}-1}{2}\notin Z$ and $x=0.$

$\therefore g\left(x\right)=\left[x\right]$ is continuous only at one fixed point $\frac{\sqrt{5}-1}{2}$ and discontinuous at $x=0$

Therefore Assertion and Reason are correct but the Reason is not the correct explanation for Assertion.