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This is the second of our many quizzes on Graphing trig functions: sin, cos, tan. Click on this to access more such competitive quizzes on Trigonometry.

Graphing trig functions: sin, cos, tan Quiz 2

mathematics

 Check whether the function representing the following graph is periodic or not. If yes, find the period. $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/92884b7b-bbb3-48f9-b4ae-91181944516d1548716249317074884.png"/>

 Check for the repeating portion of the graph. If there is no repeating part then it is not a periodic function.

 The given graph does not have any repeating portion. The domain of $x$ for which the graph of a function repeats is called period. As there is no repeating portion, the given function is not a periodic function.

 Find the domain of the following graph. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/40ad106c-499c-448a-bca9-c076fd882cb61125084355903303642.png"/>

 Check the values in X – axis at which the graph is not defined and then remove those values from real numbers to get the required domain.

 The domain of a function is the values of $x$ at which the function is defined. $ \\ $  In the given graph at every odd multiple of $\dfrac{\pi }{2}$ the function is not defined. $ \\ $ Removing all odd multiples of $\dfrac{\pi }{2}$ from real numbers we get the required domain as $ \\ \Rightarrow D=\mathbb{R}-\left\{ \left( 2n+1 \right)\dfrac{\pi }{2} \right\};n\in \mathbb{Z}$

 Find the number of solutions of $\ln \left( \tan x \right)=x$ in the interval $\left( -\infty  \right.,\left. \pi  \right]$

 Regroup the given equation so that we can draw the graph of function on both sides of the equation. Then check the number of intersections in the given interval and in the domain of the given functions.

 Given, $\ln \left( \tan x \right)=x$ $ \\ \Rightarrow \tan x={{e}^{x}} \\ $ Draw the graph of $y=\tan x\text{ and }y={{e}^{x}} $ $ \\ $ There are infinite number of intersection in $\left( -\infty  \right.,\left. \pi  \right]$ $ \\ $ But for $\ln \left( \tan x \right)$ to be defined $\Rightarrow \tan x > 0\Rightarrow x > 0$ $ \\ $ There is only one intersection for $x > 0$ $ \\ $ So, number of solutions is 1 $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/991419e6-56b8-489c-8fce-0928fa7694b91902357516770944827.png"/>

 If there is a function $f\left( x \right)=\tan x$ made changes in amplitude and vertical shift is applied so that the function converted to $g\left( x \right)$. The graph of $g\left( x \right)$ is shown below then find ${g}\left( \dfrac{\pi }{3} \right) $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3c2e0d73-06e9-464e-8181-7a49ffbecf5d4061294518813446122.png"/>

 Take the modified function in general form as $g\left( x \right)=a\tan x+b$ and find the values of variables from the graph and differentiate it to get the required answer.

 The general form of the function is $g\left( x \right)=a\tan x+b$ $ \\ $ From graph, $g\left( 0 \right)=2$ and from equation $g\left( 0 \right)=b\Rightarrow b=2$ $ \\ $ Substituting $\left( \dfrac{\pi }{6},3 \right)$ in new function then, $ \\ \Rightarrow 3=a\tan \left( \dfrac{\pi }{6} \right)+2 \\ \Rightarrow 1=\dfrac{a}{\sqrt{3}}\Rightarrow a=\sqrt{3} \\ \Rightarrow g\left( x \right)=\sqrt{3}\tan x+2 \\ \Rightarrow {g}\left( x \right)=\sqrt{3}{{\sec }^{2}}x \\ \Rightarrow {g}\left( \dfrac{\pi }{3} \right)=\sqrt{3}{{\left( 2 \right)}^{2}} \\ \Rightarrow {g}\left( \dfrac{\pi }{3} \right)=4\sqrt{3}$

 If there is a function $f\left( x \right)=\sin x$ made changes in amplitude, frequency and vertical shift is applied so that the function converted to $g\left( x \right)$. The graph of $g\left( x \right)$ is shown below then find $\int\limits_{0}^{\dfrac{\pi }{2}}{g\left( x \right).dx}$ $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/32ad437b-7802-47e8-b9e6-eabc592204594723799270022859199.png"/>

 Take the new function as $g\left( x \right)=a\sin \left( bx \right)+c$ and obtain the values of $a,b,c$ from the graph using period and points given. Then apply the integral with limits to get the required answer.

 $\text{The function }g\left( x \right)\text{ will be} \\  \Rightarrow g\left( x \right)=a\sin \left( bx \right)+c \\  \text{The period of }g\left( x \right)\text{ is }\dfrac{2\pi }{3}\text{ but the period of }f\left( x \right)\text{ is }2\pi  \\  \Rightarrow b=3 \\  \Rightarrow f\left( 0 \right)=0\text{ but }g\left( 0 \right)=2 \\  \Rightarrow c=2 \\  \text{Substituting }\left( \dfrac{\pi }{6},4 \right)\text{ in }g\left( x \right) \\  \Rightarrow 4=a\sin \left( 3\times \dfrac{\pi }{6} \right)+2 \\ \Rightarrow 2=a \\  \Rightarrow g\left( x \right)=2\sin \left( 3x \right)+2 \\  \Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{g\left( x \right).dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\sin 3x+2 \right).dx} \\ \Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{g\left( x \right).dx}=2{{\left[ \dfrac{-\cos 3x}{3} \right]}_{0}}^{\dfrac{\pi }{2}}+2{{\left[ 1 \right]}_{0}}^{\dfrac{\pi }{2}} \\ \Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{g\left( x \right).dx}=\dfrac{2}{3}+\pi $