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

Since f (x) = 1 in 0 lt; x lt; π/2 (as [cos x]=0) f (x) is continuos in [ 0, π/2 ] Hence (c) is the correct answer. ...

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Test for Monotonicity in an Interval

fx = 1 + [ co...

Question

$f (x) = 1 + [c o s x] x, i n 0 < x ⩽ \frac{π}{2}$

Has a minimum value 0

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Has a maximum value 2

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Is continuos in $[0, \frac{π}{2}]$

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Is not differentiable at x =

\frac{π}{2}

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Similar questions

- \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 .

f (x) = c o s (\frac{π}{x}), x \neq 0

f (x) = \frac{2}{π} x sin x + x^{3}

x \in [\begin{matrix} 0, \frac{π}{2} \end{matrix}]

f (x) = \frac{π}{2}

x \in [\begin{matrix} 0, \frac{π}{2} \end{matrix}]

f (x) \geq 0

x

[\begin{matrix} 0, \frac{π}{2} \end{matrix}]

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

f (x) = \frac{{cos}^{- 1} (1 - {x}^{2}) {sin}^{- 1} (1 - {x})}{{x} - {x}^{3}}, x \neq 0

{x}

x

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