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26 votes
26 votes

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

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3 Answers

Best answer
40 votes
40 votes

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. 

edited by
14 votes
14 votes
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 $[0. \pi]$
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0 votes
0 votes
Given function $f(x)= cosx – x$ is continous in the given interval $\left [ 0, \pi \right ]$

$f(0) = 1$

$f(\pi) = -(1+\pi)$

According to Mean value theorem, since $f(\pi) = -(1+\pi) \leq f(x)=0 \leq f(x)=1$

atleast one root exists in the given interval $\left [ 0, \pi \right ]$

$f'(x) = -sinx-1$ is negative in interval $\left [ 0, \pi \right ]$

Slope of $f(x)$ is negative in interval $\left [ 0, \pi \right ]$

$f(x)$ is monotonically decreasing function in interval $\left [ 0, \pi \right ]$

Hence only one root exists for  $0= cosx – x$ in interval $\left [ 0, \pi \right ]$

Option $B$ is correct
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