F-x- - 1 - - cos-x -x- in-0 lt- x -2 Has a minimum value 0Has a maximum value 2Is continuos in - 0- -2 -Is not differentiable at x - -2

The correct option is C Is continuos in [ 0, π/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 ...

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Standard XII

Mathematics

Test for Monotonicity in an Interval

f(x) = 1 + [...

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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Solution

The correct option is C
Is continuos in $[0, \frac{π}{2}]$

Since f (x) = 1 in 0 < x < $\frac{π}{2}$ (as [cos x]=0)
f (x) is continuos in $[0, \frac{π}{2}]$
Hence (c) is the correct answer.

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Similar questions

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

Q. If the function

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}

is continuous in the interval

[0, π]

, then the values of (a,b) are

Q. Examine the continuity and differentiability in

- \infty < x < \infty

of the following function :

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) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \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} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

is continuous in

[- π, π],

then number of points at which

f (x)

is not differentiable is/are

Q. Let

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

, then which of the following is true ?

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f(x)=1+[cos x]x,in 0<x⩽π2

The correct option is C Is continuos in [0,π2] Since f (x) = 1 in 0 < x < π2 (as [cos x]=0) f (x) is continuos in [0,π2] Hence (c) is the correct answer.

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

The correct option is C
Is continuos in $[0, \frac{π}{2}]$

Since f (x) = 1 in 0 < x < $\frac{π}{2}$ (as [cos x]=0)
f (x) is continuos in $[0, \frac{π}{2}]$
Hence (c) is the correct answer.