JEE Advance - Mathematics (2013 - Paper 1 Offline - No. 4)
The number of points in $$\left( { - \infty \,\infty } \right),$$ for which $${x^2} - x\sin x - \cos x = 0,$$ is
6
4
2
0
Explanation
$$\text { We know the graph of } y=x \sin x \text { is }$$
And the graph of $$y=\cos x$$ is
Now add the graph of $$y=x \sin x$$ and $$y=\cos x$$ and also draw the graph of $$y=x^2$$
It is clear that the graph of $$y=x^2$$ and $$y=x \sin x +\cos x$$ intersect at two points
$$\therefore$$ The equation $$x^2=x \sin x+x$$ satisfy for two values of $$x$$.
Hints:
(i) Recall the graph of $$y=x \sin x, y=\cos x$$ and $$y=x^2$$.
(ii) Recall the method of addition of two graphs.
(iii) The number of point of intersection of graphs of $$y=f(x)$$ and $$y=g(x)$$ is equal to the number of solution of $$f(x)=g(x)$$
Comments (0)
