If x-1-a-a2- and y-1-b-b2- then the value of 1-ab-a2b2- is0 a 1- 0 b 1

X=1+a+a^2+....∞= 11 a ⇒ x ax=1 ⇒ x 1=ax ⇒ a=x 1x Similarly, b=y 1y Now, 1+ab++a^2b^2+....∞ = nbsp;11 ab =11 (x 1x)×(y 1y) =xyxy (x 1)(y 1) =xyx+y 1 nbs ...

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Byju's Answer

Standard XIII

Mathematics

Geometric Progression

If x=1+a+a2+....

Question

If $x = 1 + a + a^{2} + . . . . \infty$ and $y = 1 + b + b^{2} + . . . . \infty,$ then the value of $1 + a b + + a^{2} b^{2} + . . . . \infty$ is
$(0 < a < 1, 0 < b < 1)$

\frac{x y}{1 - x - y}

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\frac{x y}{x + y + 1}

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\frac{x y}{2 x y + x + y - 1}

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\frac{x y}{x + y - 1}

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Solution

The correct option is D $\frac{x y}{x + y - 1}$
$x = 1 + a + a^{2} + . . . . \infty = \frac{1}{1 - a} \Rightarrow x - a x = 1 \Rightarrow x - 1 = a x \Rightarrow a = \frac{x - 1}{x}$
Similarly, $b = \frac{y - 1}{y}$

Now, $1 + a b + + a^{2} b^{2} + . . . . \infty$
$= \frac{1}{1 - a b} = \frac{1}{1 - (\frac{x - 1}{x}) \times (\frac{y - 1}{y})} = \frac{x y}{x y - (x - 1) (y - 1)} = \frac{x y}{x + y - 1}$

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g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 (x + 1), & - \infty < x \leq - 1 \sqrt{1 - x^{2}}, & - 1 < x < 1 ∣ ∣ ∣ ∣ | x | - 1 ∣ ∣ - 1 ∣ ∣, & 1 \leq x < \infty \end{matrix} .

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x^{2} / a, 0 \leq x < 1 a, 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{2}}, \sqrt{2} \leq x < \infty \end{matrix}

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Geometric Progression

Standard XIII Mathematics

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If x=1+a+a2+....∞ and y=1+b+b2+....∞, then the value of 1+ab++a2b2+....∞ is (0<a<1, 0<b<1)

The correct option is D xyx+y−1 x=1+a+a2+....∞=11−a⇒x−ax=1⇒x−1=ax⇒a=x−1x Similarly, b=y−1y Now, 1+ab++a2b2+....∞ =11−ab=11−(x−1x)×(y−1y)=xyxy−(x−1)(y−1)=xyx+y−1

If $x = 1 + a + a^{2} + . . . . \infty$ and $y = 1 + b + b^{2} + . . . . \infty,$ then the value of $1 + a b + + a^{2} b^{2} + . . . . \infty$ is
$(0 < a < 1, 0 < b < 1)$