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In the interval $[0, \pi]$ the equation $x=\cos x$ has

1. No solution

2. Exactly one solution

3. Exactly two solutions

4. An infinite number of solutions

in Calculus
edited | 1.1k views

Looking at the diagram it is clear that at a single point, $x$ and $\cos x$ intersect.

Therefore Answer is $B$.

PS: Even for any interval, we will have only one point of intersection for $x = \cos x.$ You can see that in the plot, $f(x)$ is slanting more vertically, and hence it will not meet the $\cos x$ graph even when $x$ becomes negative.

by Boss (41.1k points)
edited by
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wowwwwwwwwwwwwwww... awesome soln
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waah akash sir
+2
Even in case of interval -pi to pi , we will have only one solution

The graph of y=cos x will intersect the graph of y=x only once.
0

in -pi to pi ...how many solns??

+1
ans is B.

if you consider $x=0$ then $\cos x=1$

now if $x=\frac{\pi}{4} = 0.785$ then $\cos x=0.7071$

for some $x$ value $x=\cos x$

after this x is increasing and cosx is decreasing. so we can say exactly one solution.

EDIT-

It is very easy to show that the equation $x=\cos x$ has a unique solution. For example take $f(x) = x - \cos x$ and notice that $f'(x) = 1+\sin x \ge 0$ (equality holding in isolated points) so $f(x)$ is strictly increasing and hence the equation can have at most one solution.

At $x=0$, $f(x)$ is $\lt 0$ and at $x=\frac{\pi}{2}$, $f(x)$ is $\gt 0$, and function is continious (difference of two continuous functions is continuous). Therefore there is solution in $x \in \left [ 0,\frac{\pi}{2} \right ]$, hence there is a solution in $[0. \pi]$
by Loyal (8.1k points)
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