C-11-Q096mediumsingle_mcq

By componendo-dividendo on $\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}$ , the next simplified form is—

a $\dfrac{2}{2bx} = \dfrac{2(1 + a^2x^2)}{4ax}$
b $\dfrac{1}{bx} = ax$
c $\dfrac{1+bx}{1+ax} = 1$
d $bx = ax$

ব্যাখ্যা

Applying componendo–dividendo to each side, the left becomes $\frac{2}{2bx}$ , while the right's numerator $(1+ax)^2+(1-ax)^2=2(1+a^2x^2)$ and denominator $(1+ax)^2-(1-ax)^2=4ax$ . This gives $\frac{2}{2bx}=\frac{2(1+a^2x^2)}{4ax}$ .

এই অধ্যায়ের আরও প্রশ্ন

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Algebraic Ratio and Proportion — সব প্রশ্ন দেখুন (180টি) →

By componendo-dividendo on $\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}$ , the next simplified form is—

\dfrac{2}{2bx} = \dfrac{2(1 + a^2x^2)}{4ax}

\dfrac{1}{bx} = ax

\dfrac{1+bx}{1+ax} = 1

bx = ax

By componendo-dividendo on 1+bx1−bx=(1+ax)2(1−ax)2\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}1−bx1+bx​=(1−ax)2(1+ax)2​, the next simplified form is—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

By componendo-dividendo on 1+bx1−bx=(1+ax)2(1−ax)2\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}1−bx1+bx​=(1−ax)2(1+ax)2​, the next simplified form is—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

By componendo-dividendo on $\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}$ , the next simplified form is—

By componendo-dividendo on $\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}$ , the next simplified form is—