A-b-c-d are real number such that a-b - 2-3-b-c - 1-3-c-d - 1-4-e - ac-b2 - c2 -Then value of e is

The correct option is C 1/5 [ e = 2/31/1/3 + 3; #160;= 2/31/1 + 9/3; #160;= 2/3.3/10; e = 1/5 ] ...

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Standard XII

Mathematics

Combination

a,b,c,d are r...

Question

a,b,c,d are real number such that $\frac{a}{b} = \frac{2}{3}, \frac{b}{c} = \frac{1}{3}, \frac{c}{d} = \frac{1}{4}, e = \frac{a c}{b^{2} + c^{2}}$ .Then value of e is

\frac{1}{9}

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\frac{2}{9}

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\frac{1}{5}

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\frac{2}{5}

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Solution

The correct option is C $\frac{1}{5}$
$\begin{matrix} e = \frac{2}{3} \frac{1}{\frac{1}{3} + 3} = \frac{2}{3} \frac{1}{\frac{1 + 9}{3}} = \frac{2}{3} . \frac{3}{10} e = \frac{1}{5} \end{matrix}$

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a,b,c,d are real number such that ab=23,bc=13,cd=14,e=acb2+c2 .Then value of e is

The correct option is C 15e=23113+3=2311+93=23.310e=15

a,b,c,d are real number such that $\frac{a}{b} = \frac{2}{3}, \frac{b}{c} = \frac{1}{3}, \frac{c}{d} = \frac{1}{4}, e = \frac{a c}{b^{2} + c^{2}}$ .Then value of e is

The correct option is C $\frac{1}{5}$
$\begin{matrix} e = \frac{2}{3} \frac{1}{\frac{1}{3} + 3} = \frac{2}{3} \frac{1}{\frac{1 + 9}{3}} = \frac{2}{3} . \frac{3}{10} e = \frac{1}{5} \end{matrix}$