1 votes 1 votes In this won’t A and C both be correct answers as A can be obtained by considering only central part of C? Also, I tried this by brute force by putting some values; is there any better method for solving these? Written Exam iisc cds mtech written-test + – o asked Apr 29, 2022 edited May 4, 2022 by o o 427 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply ankitgupta.1729 commented May 3, 2022 reply Follow Share yes, Considering it as a rough graph (since scale is not given), both A and C are correct but C gives more information than A. Given function is an odd function since $f(-x) = -f(x)$ So, graph should be symmetric about origin, so you can eliminate options (B) and (D). One more way tp eliminate option (D) as: Since in (D) $f(x)$ is tending to zero when $x \rightarrow -\infty$ but $\lim_{x \rightarrow - \infty} x^2 sin(2x)$ Does not exist because sin oscillates between +1 and -1. There is no simple trick to plot the graph of product of two functions as you can do for $f(x) \pm a$ or $f(x \pm a)$ etc. But by observing something you can try to plot it like finding whether the function goes when x tends to +infinity or -infinity, finding maxima, minima and here sin oscillates between values -1 and +1, so you can say $f(x)$ oscillates between $\pm x^2$ values. You can plot graph of sin(2x) easily and then observe for what values of x, it is going from -1 to +1 and for what values of x , it goes from +1 to -1 and then you can say, for what values of $x$, $x^2 sin(2x)$ goes from -1 to +1 and +1 to -1. 1 votes 1 votes o commented May 4, 2022 reply Follow Share These are great insights to tackle such problems. Thank you so much 1 votes 1 votes Please log in or register to add a comment.