GATE CSE 1995 | Question: 1.21

Question

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

• A. No solution

• B. Exactly one solution

• C. Exactly two solutions

• D. An infinite number of solutions

Comments

Can anyone take a look at my intuition

The function $x - \cos(x)$ is a continuous function, the value of this function at $x = 0$ is -1 and at $x = \pi$ is $\pi + 1$, as the sign changes, we can confirm a root. As the function is strictly increasing it can intersect the x axis once, so we can say there is only a single root.

Answer 1

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. 

Comments

@Akash Kanase  Sir how to approach this question ? 

$Q:$ How many distinct values of x satisfy the equation $sinx=\dfrac{x}{3}$ where x is in radians.

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As Akash Sir has solved it, plot the graph for y = sinx and y = x/3 and look for the points of intersection. Since, sinx lies between -1 and 1, it should be fairly easy.

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Alternative approach:

At 1st let's divide the range [0,π] into 2 ranges : [0,π/2] and [π/2,π]

Now let G(x) = x - cosx, so

G(0) = 0 - cos0 = -1  and.   G(π/2) = π/2 - cosπ/2 = π/2

Now as G(0) and G(π/2) are of opposite sign and also G(x) is continuous in [0,π/2] and differentiable in (0,π/2) so we can say there must be atleast one x in (0,π/2) such that G(x) = 0. (Intermediate value theorem)

Now G'(x) = 1 + sinx and 1+sinx > 0 for x in [0,π/2]. So G(x) is a positive function in the range [0,π/2] and hence we can conclude that there exist exactly one x in [0,π/2] such that G(x) = 0 (or x-cosx = 0).

Now let's see the range [π/2,π]. In this range G(x) can never be 0 as x is positive and cosx is negative in this range and hence x = cosx is not possible in this range.

So we can conclude that x = cosx has exactly one solution in range [0,π/2] and exactly 0 solution in range [π/2,π] and therefore option B is the answer.

Please correct me if I am wrong.

Answer 2

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 $[0. \pi]$

Comments

How does it guarantee that it does not have another solution at $\left [ \frac{\pi}{2},\pi \right ]$?

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Because cos is negative in that range

Answer 3

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