(B) Let $x = 0.75$.The given expression is $\frac{x^3}{1-x} + (x + x^2 + 1)$.We know that $(1-x)(1+x+x^2) = 1-x^3$.So,the expression becomes $\frac{x^

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(B) Let $x = 0.75$.The given expression is $\frac{x^3}{1-x} + (x + x^2 + 1)$.We know that $(1-x)(1+x+x^2) = 1-x^3$.So,the expression becomes $\frac{x^3 + (1-x)(1+x+x^2)}{1-x} = \frac{x^3 + 1 - x^3}{1-x} = \frac{1}{1-x}$.Substituting $x = 0.75$:$\frac{1}{1-0.75} = \frac{1}{0.25} = 4$.The square root of the result is $\sqrt{4} = 2$.

Class 11 Mathematics Basic of Logarithms Mix Examples-Logarithms, Indices and Surds, Partial Fractions

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The square root of $\frac{(0.75)^3}{1-0.75}+[0.75+(0.75)^2+1]$ is

KVPY 2012 All KVPY

A
$1$
B
$2$
C
$3$
D
$4$

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The square root of $\frac{(0.75)^3}{1-0.75}+[0.75+(0.75)^2+1]$ is

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