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This is one of our many quizzes on Exponential functions. Click on this to access more such competitive quizzes on Differential Calculus.

Exponential functions Quiz 2

mathematics

 The asymptotes of the exponential function $f\left( x \right) = 5{e^{x + 2}}$ is

 Check for the conditions of the asymptotes for the given function.

This function does not have any vertical asymptote. $ \\ $ For horizontal asymptote, $ \\ y = \mathop {\lim }\limits_{x \to  \pm \infty } 5{e^{x + 2}} = 0 \\ $ This is the x-axis.

 The solution for ${2^{2{x^2}}} - {2^{{x^2} + x + 7}} + {2^{2x + 12}} = 0$ is

 Convert the given equation in the form of${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$.

We can convert this in the form of ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$ $ \\  {2^{2{x^2}}} - {2^{{x^2} + x + 7}} + {2^{2x + 12}} = 0 \\   \Rightarrow {\left( {{2^{{x^2}}}} \right)^2} - {2.2^{{x^2}}}{.2^{\left( {x + 6} \right)}} + {\left( {{2^{\left( {x + 6} \right)}}} \right)^2} = 0 \\   \Rightarrow {\left( {{2^{{x^2}}} - {2^{\left( {x + 6} \right)}}} \right)^2} = 0 \\   \Rightarrow {2^{{x^2}}} = {2^{x + 6}} \\   \Rightarrow {x^2} = x + 6 \\   \Rightarrow {x^2} - x - 6 = 0 \\   \Rightarrow {x^2} - 3x + 2x - 6 = 0 \\   \Rightarrow x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0 \\   \Rightarrow \left( {x + 2} \right)\left( {x - 3} \right) = 0 \\  $  Thus, $x =  - 2\;\;or\;\;3$

 If money is invested at a rate of $7\% $ compounded annually, how long will it take to double the money? $\text{(Given }{\log _2}1.07 = 0.0976)$

 Use the formula for the amount in compound interest given as $A = P{\left( {1 + r} \right)^t}$.

The amount in compound interest is given as, $ \\ A = P{\left( {1 + r} \right)^t} \\   \Rightarrow 2P = P{\left( {1 + 0.07} \right)^t} \\   \Rightarrow 2 = {1.07^t} \\   \Rightarrow \log 2 = t\log 1.07 \\   \Rightarrow t = \dfrac{{\log 2}}{{\log 1.07}} = \dfrac{1}{{{{\log }_2}1.07}} = \dfrac{1}{{0.0976}} \approx 10.25\;years \\  $ 

 Which of the following is correct regarding ${e^x}$ and ${x^e}$?

 Assume a function $f\left( x \right) = {e^x} - {x^e}$ and use a derivative test to find the minimum value of the function.

Let $f\left( x \right) = {e^x} - {x^e}$ $ \\ $ Then $f'\left( x \right) = {e^x} - e{x^{e - 1}}$ $ \\ $  And $f''\left( x \right) = {e^x} - e\left( {e - 1} \right){x^{e - 2}}$ $ \\ $ At $f'\left( x \right) = 0$ at $x = e$. $ \\ $ At $x = e$, $f''\left( e \right) = {e^e} - e\left( {e - 1} \right){e^{e - 2}} = {e^{e - 1}} > 0$ $ \\ $ Thus, $f\left( x \right)$ is minimum at $x = e$. And $f\left( e \right) = 0$. $ \\ $ Thus, $ \\ f\left( x \right) \geqslant 0 \\   \Rightarrow {e^x} - {x^e} \geqslant 0 \\   \Rightarrow {e^x} \geqslant {x^e} \\  $ 

 The total number of cases due to a virus in a city is given as $f\left( t \right) = 20{\left( {\dfrac{5}{4}} \right)^{\dfrac{{t - 1}}{5}}}$ where $t = 1$ corresponds to the day when the first case was detected in the city. The death rate due to the virus in the city is $2.5\% $. The herd immunity is attained when $70\% $ of the population has been infected. If the population of the city is $5\;mn$, calculate the number of days after which herd immunity will be attained. Also calculate the number of deaths these days. $($Given $\log 2 = 0.3010$, $\log 5 = 0.6989$, $\log 7 = 0.8450$ $)$

 Form the equation for total cases required to reach herd immunity and solve the equation.

$70\%$ of population $= 0.7 \times 5 = 3.5\;mn \\ 20{\left( {\dfrac{5}{4}} \right)^{\dfrac{{t - 1}}{5}}} = 3.5 \times {10^6} \\   \Rightarrow {\left( {\dfrac{5}{4}} \right)^{\dfrac{{t - 1}}{5}}} = \dfrac{{3.5 \times {{10}^6}}}{{20}} = \dfrac{7}{4} \times {10^5} \\   \Rightarrow \dfrac{{t - 1}}{5}\log \dfrac{5}{4} = \log \left( {\dfrac{7}{4} \times {{10}^5}} \right) = \log {10^5} + \log 7 - \log 4 \\   \Rightarrow \dfrac{{t - 1}}{5} = \dfrac{{5 + 0.8450 - 0.6020}}{{0.6989 - 0.6020}} = \dfrac{{5.243}}{{0.0969}} \approx 54 \\   \Rightarrow t = \left( {54 \times 5} \right) + 1 = 271 \\  $ Total deaths at $2.5\% $ rate, $ = 0.025 \times 3.5 \times {10^6} = 8.75 \times {10^4}$