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Question

Answers

$

A.{\text{ }}x \geqslant 1 \\

B.{\text{ }}x \geqslant 2 \\

C.{\text{ }}x < 2 \\

D.{\text{ }}x < 1 \\

$

Answer
Verified

Hint: In this question L.H.S of the given equation is a square of real numbers, so the square of real number is always positive or non-negative so, R.H.S part of the given equation should be positive, so use this concept to get the solution of the question.

Given equation is

${y^2} = \left( {x - 1} \right){\left( {x - 2} \right)^2}$

We have to prove for what value of x the above function is not defined.

As we know that the square of any real number is non-negative.

So, above curve is not defined if R.H.S is negative

So, R.H.S of above curve is

$\left( {x - 1} \right){\left( {x - 2} \right)^2}$

$ \Rightarrow \left( {x - 1} \right){\left( {x - 2} \right)^2} < 0$â€¦â€¦â€¦â€¦â€¦â€¦. (1), (Condition of not defined)

As we see from above equation

${\left( {x - 2} \right)^2} \geqslant 0$ for any value of x.

i.e. ${\left( {x - 2} \right)^2} \geqslant 0{\text{ }}\forall \left( {x \in R} \right)$ , where R is a real number.

Therefore from equation (1)

$\left( {x - 1} \right) < 0$

$\therefore x < 1$

Therefore for x is less than 1 curve is not defined.

Say $x = 0.9$,

Therefore given equation converts

$

{y^2} = \left( {0.9 - 1} \right){\left( {0.9 - 2} \right)^2} = \left( { - 0.1} \right){\left( { - 1.1} \right)^2} = - 0.1\left( {1.21} \right) = - 0.121 \\

\Rightarrow y = \sqrt { - 0.121} \\

$

So, as we see, the square root of a negative number is not defined.

Hence option (D) is correct.

Note: Whenever we face such types of questions always remember the condition of square of any real number which is square of any real number is non negative so, find out for what values of x R.H.S part of given equation is negative as above which is the required answer for which the given function is not defined.

Given equation is

${y^2} = \left( {x - 1} \right){\left( {x - 2} \right)^2}$

We have to prove for what value of x the above function is not defined.

As we know that the square of any real number is non-negative.

So, above curve is not defined if R.H.S is negative

So, R.H.S of above curve is

$\left( {x - 1} \right){\left( {x - 2} \right)^2}$

$ \Rightarrow \left( {x - 1} \right){\left( {x - 2} \right)^2} < 0$â€¦â€¦â€¦â€¦â€¦â€¦. (1), (Condition of not defined)

As we see from above equation

${\left( {x - 2} \right)^2} \geqslant 0$ for any value of x.

i.e. ${\left( {x - 2} \right)^2} \geqslant 0{\text{ }}\forall \left( {x \in R} \right)$ , where R is a real number.

Therefore from equation (1)

$\left( {x - 1} \right) < 0$

$\therefore x < 1$

Therefore for x is less than 1 curve is not defined.

Say $x = 0.9$,

Therefore given equation converts

$

{y^2} = \left( {0.9 - 1} \right){\left( {0.9 - 2} \right)^2} = \left( { - 0.1} \right){\left( { - 1.1} \right)^2} = - 0.1\left( {1.21} \right) = - 0.121 \\

\Rightarrow y = \sqrt { - 0.121} \\

$

So, as we see, the square root of a negative number is not defined.

Hence option (D) is correct.

Note: Whenever we face such types of questions always remember the condition of square of any real number which is square of any real number is non negative so, find out for what values of x R.H.S part of given equation is negative as above which is the required answer for which the given function is not defined.

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