Fx-1-cos x- x- in 0-x -2A- Is continuos in -0- -2-B- Has a minimum value 0C- Has a maximum value 2D- Is not differentiable at x-2

The correct option is A Is continuos inSince f (x) = 1 in 0 lt; x lt; π/2 nbsp;(as [cos x]=0) f (x) is continuos in [ 0, π/2 ] nbsp; nbsp; Hence (c ...

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Standard XII

Mathematics

Test for Monotonicity in an Interval

fx=1+[cos x] ...

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

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Is not differentiable at x =

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Solution

The correct option is C
Is continuos in

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

0 < x \leq \frac{π}{2}

Q. The function

f (x) = \frac{1 + sin x - cos x}{1 - sin x - cos x}

is not defined at

x = 0

. The value of

f (0)

so that

f (x)

is continuous at

x = 0

, is

f (x) = cos x

and

g (x) = {\begin{matrix} m i n {f (t) : 0 \leq t \leq x}, & 0 \leq x \leq π sin x - 1, & x > π \end{matrix}

g (x)

continuous in

(0, \infty)

f (x) = 1 + [cos x]

, in

0 < x \leq \frac{π}{2}

(where

[.]

denotes greatest integer function)

Q. At

x = 0

f (x) = cos x - 1 + \frac{x^{2}}{2} - \frac{x^{3}}{6}

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f(x)=1+[cos x]x,in 0<x⩽π2

The correct option is C Is continuos inSince 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

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.