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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

[x]

denotes the integral part of

x

, then domain of the function

f(x)=\dfrac{\sqrt{3-x}}{(x-1)(x-2)(x-3)}+\sin^{-1}\!\left[\dfrac{3x-2}{2}\right]

is:

[0,2)

[0,2)-\{1\}

correct

(-\infty,3)-\{1,2\}

DNone of these

Solution

Condition (i):

\sin^{-1}

requires

\left[\dfrac{3x-2}{2}\right]\in[-1,0,1]

-1\le\dfrac{3x-2}{2}<2\;\Rightarrow\;-2\le 3x-2<4\;\Rightarrow\;0\le 3x<6\;\Rightarrow\;0\le x<2\quad\ldots(1)

Condition (ii):

\sqrt{3-x}

requires

3-x\ge 0\;\Rightarrow\;x\le 3

and

x\neq 1,2,3\quad\ldots(2)

Step 2: From (1) and (2):

0\le x<2

and

x\neq 1

. Domain

=[0,1)\cup(1,2)=[0,2)-\{1\}

Answer: (2)

[0,2)-\{1\}

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