Fx- 1- -cos x- x in 0 - x -2

The correct option is C Is continuous in [ 0,π/2 ] Since, f(x)=1 in 0 lt; x lt; π/2 ( as [ cos x] = 0 ) ∴ f(x) is continuous in #16 ...

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Test for Monotonicity in an Interval

fx= 1+ [cos ...

Question

$f (x) = 1 + [c o s x] x$ in $0 < x \leq \frac{π}{2}$

Has a minimum value 0

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Has a maximum value 2

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Is continuous in

[0, \frac{π}{2}]

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Is not differentiable at

x = \frac{π}{2}

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 2 sin x, & - π \leq x \leq \frac{- π}{2} a cos (x) + b, & \frac{- π}{2} < x < 0 - 2 + cos x, & 0 \leq x \leq \frac{π}{2} c sin x, & \frac{π}{2} < x \leq π \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

[- π, π],

f (x)

f (x) = \frac{1 - cos (x - π)}{(π - x)^{2}}, x \neq π

x = 0

f (π)

- \infty < x < \infty

f (x) = 1

- \infty < x < 0

f (x) = 1 + sin x

0 \leq x \leq π / 2

f (x) = 2 + {(x - π / 2)}^{2}

π / 2 \leq x < \infty .

The maximum value of $s i n (x + \frac{π}{6}) + c o s (x + \frac{π}{6})$ in the

interval $(0, \frac{π}{2})$ is attained at

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