State true or false.
Let\[f(x) = {x^3} + \cos \pi x + 7\], then \[f(x) = 13\] has exactly one solution in the interval [1,2].
A. True
B. False

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Hint: We will solve this in a simpler way. Here the interval is closed that means for f(x) = 13 the value of x can neither be less than 1 nor be greater than 2. And we will check this by putting values of x as 1 and 2 in the given function and also by using the mean value theorem.

Complete step-by-step answer:
Step-1
We have \[f(x) = {x^3} + \cos \pi x + 7\]
In the interval of [1,2],
Step-2
$f(1) = {1^3} + \cos \pi + 7$
$ \Rightarrow f(1) = 1 + 1 + 7$
$ \Rightarrow f(1) = 9$……………(1)
And $f(2) = {2^3} + \cos 2\pi + 7$
Or, $f(2) = 8 + 1 + 7$
Or, $f(2) = 16$………..(2)
Step-3
From above two value we get to know that $f(1) < f(x) = 13 < f(2)$
So, from this we get to know that \[f(x) = 13\] has a solution in the interval of (1,2).
Step-4
For\[f(x) = {x^3} + \cos \pi x + 7\] to be continuous f’(x) must be greater than 0.
For that,
 $f(c) = \dfrac{{f(2) - f(1)}}{{2 - 1}}$
$f(c) = \dfrac{{16 - 9}}{{2 - 1}}$
$f(c) = 7$
Step-5
Again differentiating f(x) we get,
$f'(x) = 3{x^2} - \pi \sin \pi x$
Step-6
We know that,
$f'(x) = 3{x^2} - \pi \sin \pi x = f(c)$
Or, $f'(x) = 3{x^2} - \pi \sin \pi x = 7$
Or, $f'(x) = 3{x^2} - \pi \sin \pi x > 0$, for all x belongs to closed [1,2]
Step-7
Hence, f(x) is a continuous function at [1,2].
F(x) is a strictly increasing function.
Therefore for \[f(x) = 13\], it has exactly one solution in the closed interval of [1,2].

So, the correct answer is “Option A”.

Note: Mean value theorem- The mean value theorem states that if a function f is continuous on the closed interval of [a, b ] and differentiable on the open interval ( a, b ), then there exist a point C in the interval (a, b ) such that f’(c) is equal to the functions average rate of change over closed interval [a, b ].
Difference between open interval and closed interval is, an open interval does not include its limit points while closed interval does.