The graph of the function f x -x- 1 8 sin 2- x - 0- x- 1 is shown below- Define f 1 x -f x - f n-1 x -f f n x - for n- 1-Which of the following statements are true-I- There are infinitely many x- 0-1 - for which lim n- f n x -0-II- There are infinitely many x- 0-1 - for which lim n- f n x - 1 2 -III- There are infinitely many x- 0-1 - for which lim n- f n x -1-IV- There are infinitely many x- 0-1 - for which lim n- f n x does not exist-

The correct option is B IV only f(x)=x+1/8sin (2nπ) lim_n→∞f_n(x)=∞ Therefore there are infinitely many solutions and #160; f_n(x) #160 ...

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Byju's Answer

Standard XII

Mathematics

Local Maxima

The graph of ...

Question

The graph of the function $f (x) = x + \frac{1}{8} sin (2 π x), 0 \leq x \leq 1$ is shown below. Define $f_{1} (x) = f (x), f_{n + 1} (x) = f (f_{n} (x)), f o r n \geq 1.$
Which of the following statements are true?
I. There are infinitely many $x \in [0, 1]$ for which $lim n \to \infty f_{n} (x) = 0.$
II. There are infinitely many $x \in [0, 1]$ for which $lim n \to \infty f_{n} (x) = \frac{1}{2} .$
III. There are infinitely many $x \in [0, 1]$ for which $lim n \to \infty f_{n} (x) = 1.$
IV. There are infinitely many $x \in [0, 1]$ for which $lim n \to \infty f_{n} (x)$ does not exist.

I

and

I I I

only

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I V

only

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I, I I, I I I

only

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I, I I, I I I

and

I V

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Solution

The correct option is B $I V$ only
$f (x) = x + \frac{1}{8} sin (2 n π)$

${lim}_{n \to \infty} f_{n} (x) = \infty$

Therefore there are infinitely many solutions and $f_{n} (x)$ does not exist.

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The correct option is B IV onlyf(x)=x+18sin(2nπ)limn→∞fn(x)=∞Therefore there are infinitely many solutions and fn(x) does not exist.

The correct option is B $I V$ only
$f (x) = x + \frac{1}{8} sin (2 n π)$

${lim}_{n \to \infty} f_{n} (x) = \infty$

Therefore there are infinitely many solutions and $f_{n} (x)$ does not exist.