The graph of $y = x + cos x$ is

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None of these

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Solution

The correct option is B
Given : $y = x + cos x, y (0) = 1$
$\Rightarrow y^{'} = 1 - sin x \geq 0$
So, $y$ is always increasing
Now, $y^{''} = - cos x$
$\Rightarrow y^{''} (0) = 0, x = (2 n + 1) \frac{π}{2}, n \in Z$ (points of inflection)
For, $y^{''} > 0, cos x < 0$
$\Rightarrow (4 n + 1) \frac{π}{2} < x < (4 n + 3) \frac{π}{2}$ (concave up)
and for $y^{''} < 0, cos x > 0$
$\Rightarrow (4 n - 1) \frac{π}{2} < x < (4 n + 1) \frac{π}{2}$ (concave down)
Also, $y (- \frac{3 π}{2}) = - \frac{3 π}{2}, y (- \frac{π}{2}) = - \frac{π}{2},$
$y (\frac{π}{2}) = \frac{π}{2}, y (\frac{3 π}{2}) = \frac{3 π}{2}$
Now, plotting the graph :

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