Many functions

Calculus Level 5

$\large f(x) = \begin{cases}{\{ x \} \cdot \sqrt { 4{ x }^{ 2 }-12x+9 } ,} && {1\le\>x\le\>2} \\ {\cos\left( \frac { \pi }{ 2 } (|x|-\{ x\} ) \right) ,} && {\>-1\le\>x<1}\end{cases}$

Consider the piecewise function $f(x)$ above with $\{x\}$ denoting the fractional part of $x$ .

Which of the following is/are true?

$A.$ Range of $f(x)=[0,1]$

$B.$ The number of values of $x$ for which function is continuous but not differentiable is $1$ .

$C.$ $f(x)=1$ has two solutions.

$D.$ Number of values of $x$ for which $f(x)$ is discontinuous is $2$ .

Try my set .

BC ACD AC ABD AB ABCD None

1 solution

Tanishq Varshney
May 30, 2015

@Tanishq Varshney Look's like ABD seems to do well in math as well :P (ABD = Ab De Villiers )

chandrasekhar S - 6 years ago

There are three points of discontinuity, namely at $x = 0$ , 1, and 2, so D is not true.

Jon Haussmann - 6 years ago

well , what i have learnt says that till u dont know the curve for a value a, u cant consider the function discontinuous, I mean to say for $x\leq 2$ we know the curve but not for $x>2$ . Hence we cant say its discontinuous at $2$ same goes for $-1$ . Plz correct me if i am wrong

Tanishq Varshney - 6 years ago

$f(2) = 0$ , but $\lim_{x \to 2^-} f(x) = 1,$ so $f(x)$ is not continuous at $x = 2$ .

Jon Haussmann - 6 years ago

I agree with @Jon Haussmann sir. The function becomes 0 at x=2 whereas its left hand limit is 1. so it is discontinuous at x=2.

Ankush Tiwari - 6 years ago

sir just then one doubt why did u first mark ABD as correct although i provided the none of these option

Tanishq Varshney - 6 years ago

Many functions

Try my set .

1 solution

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