Let f ( x )= x 2+ x 4+ x 6+ x 8+ ldots ldots . . . ∞ for all real x such that the sum converges. Number of real x for which the equation f ( x )- x =0 holds, is

Question

Let f ( x )= x 2+ x 4+ x 6+ x 8+ ldots ldots . . . ∞ for all real x such that the sum converges. Number of real x for which the equation f ( x )- x =0 holds, is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

f(x)=( x 2/1- x 2) when | x |<1 ∴ ( x 2/1- x 2)= x ⇒ x =0 or x =1- x 2 x2+x-1=0 ⇒ x=(-1+√5/2), (-1-√5/2) is to be rejected

Q. Let $f (x) = x^{2} + x^{4} + x^{6} + x^{8} + \dots\dots ...\infty$ for all real $x$ such that the sum converges. Number of real $x$ for which the equation $f (x) - x = 0$ holds, is

0

1

2

3

$f (x) = \frac{x ^{2}}{1 - x ^{2}}$ when $∣ x ∣ < 1$
$∴ \frac{x ^{2}}{1 - x ^{2}} = x \Rightarrow x = 0$
or $x = 1 - x^{2}$
$x^{2} + x - 1 = 0 \Rightarrow x = \frac{- 1 + 5}{2}, \frac{- 1 - 5}{2}$ is to be rejected

Q. Let f(x)=x2+x4+x6+x8+……...∞ for all real x such that the sum converges. Number of real x for which the equation f(x)−x=0 holds, is

0

1

2

3

f(x)=1−x2x2​ when ∣x∣<1 ∴1−x2x2​=x⇒x=0 or x=1−x2 x2+x−1=0⇒x=2−1+5​​,2−1−5​​ is to be rejected

Q. Let $f (x) = x^{2} + x^{4} + x^{6} + x^{8} + \dots\dots ...\infty$ for all real $x$ such that the sum converges. Number of real $x$ for which the equation $f (x) - x = 0$ holds, is

$f (x) = \frac{x ^{2}}{1 - x ^{2}}$ when $∣ x ∣ < 1$
$∴ \frac{x ^{2}}{1 - x ^{2}} = x \Rightarrow x = 0$
or $x = 1 - x^{2}$
$x^{2} + x - 1 = 0 \Rightarrow x = \frac{- 1 + 5}{2}, \frac{- 1 - 5}{2}$ is to be rejected