
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
520.2k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Distinguish between esterification and saponification class 12 chemistry CBSE

Give five points to show the significance of varia class 12 biology CBSE

How was the Civil Disobedience Movement different from class 12 social science CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

The Coordination number of bcc fcc ccp hcp is class 12 chemistry CBSE

Difference between saponification and esterificati class 12 chemistry CBSE
