Square root of frac{(0.75)^3}{1-(0.75)}+left[0.75+(0.75)^2+1right] is

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Basic of Logarithms

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

A

$1$

B

$2$

C

$3$

D

$4$

(KVPY-2012)

Solution

(b)

Let $x=0.75$

According to the question, $\frac{x^3}{1-x}+\left(x+x^2+1\right)$

$=\frac{x^3+(1-x)\left(1+x+x^2\right)}{1-x}$

$=\frac{x^3+1-x^3}{1-x}=\frac{1}{1-x}$

Now, put the value of $x$

$\frac{1}{1-0.75} =\frac{1}{0.25}$

$=\frac{100}{25}=4$

So, square root of the equation $=\sqrt{4}=2$

Standard 11

Mathematics

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