GATE Mechanical 2016 Set 2 | GA | Question: 10

Question

Which of the following curves represents the function $ln(|e^{[|\sin(|x|)]}|)$ for $\mid x\mid< 2\pi$? Here, $x$ represents the abscissa and $y$ represents the ordinate.

• A.
• B.
• C.
• D.

Comments

answer is c by transformation of graphs

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Put X= -π/2 then option B & Option D eliminated.

Put X=3π/2 then Option A eliminated.

So, Option  C is answer.

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we know the property ln(e^x)=x

so in the given question the given graph will be same as graph of |sinx|

https://www.mathway.com/popular-problems/Precalculus/454635

option c is graph of |sinx|

Answer 1

$f(x) = \large \ln \left( |e^{\left [ \;\; |{\color{blue}{\sin}} \left ( {\color{red}{|x|}} \right ) | \;\; \right ]}| \right )$

$1. \qquad {\color{red}{\bf |x|}}\rightarrow \;\; f(x)\;\;\; \text{is Even }\rightarrow \quad \text{option b not possible}$

$2. \qquad m = |\;{\color{blue}{\sin}} \left ( {\color{red}{\bf |x|}} \right ) | \;\; \geq \;\; 0 \quad \rightarrow {\color{blue}{e^{\bf m} \;\; \geq \;\; 1}}$

$3.\Rightarrow f(x) = \ln\left ( | \color{blue}{e^{\bf m}}| \right ) = \ln\left ( \color{blue}{e^{\bf m}}\right ) \geq 0 \quad \left \{ \text{from 2} \right\} \;\; \rightarrow \text{option a not possible}$

$4. \text{ and } f(x)_{x=0} = 0 \;\; \rightarrow \text{option d not possible}$

$\Rightarrow \text{answer C}$

Comments

how do i crack these kind of question?????????????????/

suggest plx

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@Debashish

I still don't understand. Please help.

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Don't try to draw curve for this function.just know the nature of curve and then eliminate options.

Nature 1) Here we have mod to x : $|x|$ , which means graph would be identical in both $+ve$ x-axis  and $-ve$ x-axis. in other words mirror image with respect to y-axis. All options are satisfying this property so know some other nature.

Nature 2) at $x=0$ $y$ would become $0$ so in that way eliminate option B and D.

Nature 3) See if y can be negative or not. To get $-ve$$y$ the power of $e$ must be negative.But see power of $e$ is function of mod , and Mod always result in $+ve$ value so this shows $y$ can't be negative so eliminate A.

in that way only C left , hence answer.

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I think we can't eliminate option b based on nature 2)you mentioned..

Because the curve passes through (0,0).

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But option b can be eliminated at 3π/2

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Option b can be eliminated through nature 1 as it is not symmetric to the y-axis. It is symmetric to the origin.

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Also at x = 3π/2, y = 1. So option C) answer, all other eliminated.

Answer 2

Function Y value will be positive only as Mod is given , A B eliminated now sinPi=0 D eliminated so C is answer