The solution of this division 1-12-14-1-13-15- is equal to

The correct option is C e^2+1e^2 1 The required expression #160; #160; =(1+12!+14!—)(1+13!+15!—) =e+e^ 12e e^ 12 .... [e=1+11!+12!+.... ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

The solution ...

Question

The solution of this division $(1 + \frac{1}{2!} + \frac{1}{4!} + \dots .) \div (1 + \frac{1}{3!} + \frac{1}{5!} + \dots . .)$ is equal to

\frac{e - 1}{e + 1}

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\frac{e + 1}{e - 1}

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\frac{e^{2} - 1}{e^{2} + 1}

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\frac{e^{2} + 1}{e^{2} - 1}

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Solution

The correct option is C $\frac{e^{2} + 1}{e^{2} - 1}$
The required expression $= \frac{(1 + \frac{1}{2!} + \frac{1}{4!} - - -)}{(1 + \frac{1}{3!} + \frac{1}{5!} - - -)}$
$= \frac{\frac{e + e^{- 1}}{2}}{\frac{e - e^{- 1}}{2}}$ .... $[e = 1 + \frac{1}{1!} + \frac{1}{2!} + . . . .]$ and $[e^{- 1} = 1 - \frac{1}{1!} + \frac{1}{2!} - . . . .]$
$= \frac{e^{2} + 1}{e^{2} - 1}$

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The solution of this division (1+12!+14!+….)÷(1+13!+15!+…..) is equal to

The correct option is C e2+1e2−1The required expression =(1+12!+14!−−−)(1+13!+15!−−−)=e+e−12e−e−12 .... [e=1+11!+12!+....] and [e−1=1−11!+12!−....]=e2+1e2−1

The solution of this division $(1 + \frac{1}{2!} + \frac{1}{4!} + \dots .) \div (1 + \frac{1}{3!} + \frac{1}{5!} + \dots . .)$ is equal to