0 votes 0 votes The graph of a cubic polynomial $f(x)$ is shown below. If $k$ is a constant such that $f(x)=k$ has three real solutions, which of the following could be a possible value of $k$? $3$ $0$ $-7$ $-3$ Quantitative Aptitude isi2017-mma general-aptitude quantitative-aptitude + – go_editor asked Sep 15, 2018 edited Nov 11, 2019 by go_editor go_editor 699 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes at y=-3, f(x) cuts the line3 times. So D is the answer. chiku_cr7 answered Sep 18, 2020 chiku_cr7 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes As we clearly see that f(x) is -3 for 3 values of x i.e. k=(-3) answer is D. jjayantamahata answered Mar 29, 2018 jjayantamahata comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes At which point the curve f(x) cuts the X axis will be a solution. Here it is clearly seen the curve X axis at some point (unknown Let x1) with Y value as 0 Hence we can say f(x1) = 0 So 0 is definitely a possible value for k. Ans: B indranil21 answered Sep 17, 2020 indranil21 comment Share Follow See all 0 reply Please log in or register to add a comment.