The number of solutions of the equation sin e x -5 x -5- x isA- 0B- 2C- 1D- infinitely many

The correct option is A 0sin(e^x)= 5x+5^ x, let 5^x = t, then sin(e^x)= t + 1/t = ( √(t) 1/√(t) ) + 2 [∵ 5^x gt; 0] ⇒ sin(e^x) ≥ 2, which is not p ...

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Standard XII

Mathematics

Particular Solution of a Differential Equation

The number of...

Question

The number of solutions of the equation $s i n (e^{x}) = 5^{x} + 5^{- x}$ is

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infinitely many

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Solution

$s i n (e^{x}) = 5 x + 5^{- x}, l e t 5^{x} = t, t h e n s i n (e^{x}) = t + \frac{1}{t} = (\sqrt{t} - \frac{1}{\sqrt{t}}) + 2 [∵ 5^{x} > 0] \Rightarrow s i n (e^{x}) \geq 2$ , which is not possible for any real x.

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The number of solutions of the equation sin (ex)=5x+5−x is

sin(ex)=5x+5−x, let 5x=t, thensin(ex)=t+1t=(√t−1√t)+2 [∵5x>0]⇒sin(ex)≥2, which is not possible for any real x.

The number of solutions of the equation $s i n (e^{x}) = 5^{x} + 5^{- x}$ is

$s i n (e^{x}) = 5 x + 5^{- x}, l e t 5^{x} = t, t h e n s i n (e^{x}) = t + \frac{1}{t} = (\sqrt{t} - \frac{1}{\sqrt{t}}) + 2 [∵ 5^{x} > 0] \Rightarrow s i n (e^{x}) \geq 2$ , which is not possible for any real x.