# The number of real solutions of the equation $\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}}$

$

a.0 \\

b.1 \\

c.2 \\

d.{\text{None of these}}{\text{.}} \\

$

Answer

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Hint- sine function is given in the problem and we know sine always lie between $\left[ { - 1,1} \right]$i.e.$ - 1 \leqslant \sin \theta \leqslant 1$

Given equation

$\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}}$

Let ${y_1} = \sin \left( {{e^x}} \right),{\text{ & }}{y_2} = {5^x} + {5^{ - x}}$

Now as we know that $ - 1 \leqslant \sin \theta \leqslant 1$

Now as we know for real solution $0 < {e^x} < \infty $

Therefore $ - 1 \leqslant \sin \left( {{e^x}} \right) \leqslant 1$

Therefore the maximum value of ${y_1}$is 1 and the minimum value of ${y_1}$is -1.

$

\therefore {\left( {{y_1}} \right)_{\min }} = - 1 \\

{\left( {{y_1}} \right)_{\max }} = 1..................\left( 1 \right) \\

$

Now let ${y_2} = {5^x} + {5^{ - x}} = {5^x} + \dfrac{1}{{{5^x}}}$

Let ${5^x} = t$

$\therefore {y_2} = t + \dfrac{1}{t}$

Now we have to find out the maximum and minimum value of above function

So, differentiate above equation w.r.t. $t$ And equate to zero.

$

\Rightarrow \dfrac{d}{{dt}}{y_2} = 1 - \dfrac{1}{{{t^2}}} = 0 \\

\Rightarrow {t^2} = 1 \\

\Rightarrow t = \pm 1 \\

$

Now double differentiate the above equation

$ \Rightarrow \dfrac{{{d^2}}}{{d{t^2}}}{y_2} = \dfrac{d}{{dt}}\left( {1 - \dfrac{1}{{{t^2}}}} \right) = 0 + \dfrac{2}{{{t^3}}}$

Now for $t = 1$

The value of above equation is positive so the function is minimum at $t = 1$

Now for real solution t should be greater than zero$\left( {t > 0} \right)$

Because ${5^x} = t$ for t less than zero (t<0), ${5^x}$ become imaginary hence there is no real solution for t less than zero.

So, at $t = 1,{\text{ }}{y_2} = 1 + 1 = 2$

So, ${\left( {{y_2}} \right)_{\min }} = 2................\left( 2 \right)$

$\therefore {y_2} \geqslant 2$

Now from equation (1) and (2)

${\left( {{y_1}} \right)_{\max }} = 1,{\text{ }}{\left( {{y_2}} \right)_{\min }} = 2$…………….. (3)

But from the given equation

$

\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}} \\

{y_1} = {y_2} \\

$

So, from equation (3) the above condition never holds for real solutions.

So, the number of real solutions of the given equation is zero.

Hence option (a) is correct.

Note- In such types of questions always remember the range of sine function so, in above problem solve L.H.S and R.H.S separately and find out the range of these functions for real solution, then compare their ranges and also check the equality, we will get the required answer.

Given equation

$\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}}$

Let ${y_1} = \sin \left( {{e^x}} \right),{\text{ & }}{y_2} = {5^x} + {5^{ - x}}$

Now as we know that $ - 1 \leqslant \sin \theta \leqslant 1$

Now as we know for real solution $0 < {e^x} < \infty $

Therefore $ - 1 \leqslant \sin \left( {{e^x}} \right) \leqslant 1$

Therefore the maximum value of ${y_1}$is 1 and the minimum value of ${y_1}$is -1.

$

\therefore {\left( {{y_1}} \right)_{\min }} = - 1 \\

{\left( {{y_1}} \right)_{\max }} = 1..................\left( 1 \right) \\

$

Now let ${y_2} = {5^x} + {5^{ - x}} = {5^x} + \dfrac{1}{{{5^x}}}$

Let ${5^x} = t$

$\therefore {y_2} = t + \dfrac{1}{t}$

Now we have to find out the maximum and minimum value of above function

So, differentiate above equation w.r.t. $t$ And equate to zero.

$

\Rightarrow \dfrac{d}{{dt}}{y_2} = 1 - \dfrac{1}{{{t^2}}} = 0 \\

\Rightarrow {t^2} = 1 \\

\Rightarrow t = \pm 1 \\

$

Now double differentiate the above equation

$ \Rightarrow \dfrac{{{d^2}}}{{d{t^2}}}{y_2} = \dfrac{d}{{dt}}\left( {1 - \dfrac{1}{{{t^2}}}} \right) = 0 + \dfrac{2}{{{t^3}}}$

Now for $t = 1$

The value of above equation is positive so the function is minimum at $t = 1$

Now for real solution t should be greater than zero$\left( {t > 0} \right)$

Because ${5^x} = t$ for t less than zero (t<0), ${5^x}$ become imaginary hence there is no real solution for t less than zero.

So, at $t = 1,{\text{ }}{y_2} = 1 + 1 = 2$

So, ${\left( {{y_2}} \right)_{\min }} = 2................\left( 2 \right)$

$\therefore {y_2} \geqslant 2$

Now from equation (1) and (2)

${\left( {{y_1}} \right)_{\max }} = 1,{\text{ }}{\left( {{y_2}} \right)_{\min }} = 2$…………….. (3)

But from the given equation

$

\sin \left( {{e^x}} \right) = {5^x} + {5^{ - x}} \\

{y_1} = {y_2} \\

$

So, from equation (3) the above condition never holds for real solutions.

So, the number of real solutions of the given equation is zero.

Hence option (a) is correct.

Note- In such types of questions always remember the range of sine function so, in above problem solve L.H.S and R.H.S separately and find out the range of these functions for real solution, then compare their ranges and also check the equality, we will get the required answer.

Last updated date: 06th Jun 2023

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Total views: 330.3k

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