10 Question MCQ (including Assertion)

Question 1 of 10

Assertion (A): The rule \[ h(x)=\begin{cases} x^2, & 0\le x\le2,\\ 2x, & 2\le x\le5 \end{cases} \] is not a function because \(x=2\) belongs to both intervals.
Reason (R): At \(x=2\), both formulas give the same value, namely \(4\).

A Both A and R are true, and R is the correct explanation of A. B Both A and R are true, but R is not the correct explanation of A. C A is true, but R is false. D A is false, but R is true.

Select one option. Answers are shown after the test.

Question 2 of 10

The range of \(f:\mathbb R\to\mathbb R\) defined by \(f(x)=x^2+2\) is

A \((-\infty,2]\) B \((2,\infty)\) C \(\mathbb R\) D \([2,\infty)\)

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Question 3 of 10

Assertion (A): A school offers 4 clubs and 3 activity periods. If a student’s choice is recorded as ⟨club, period⟩, then there are 12 possible ordered choices.
Reason (R): Every club can be paired with each of the 3 activity periods.

A Both A and R are true, and R is the correct explanation of A. B Both A and R are true, but R is not the correct explanation of A. C A is true, but R is false. D A is false, but R is true.

Select one option. Answers are shown after the test.

Question 4 of 10

Assertion (A): The domain of the rational function \(f(x) = \frac{1}{x - 2}\) is \(\mathbb{R} - \{2\}\).

Reason (R): A rational function of the form \(\frac{p(x)}{q(x)}\) is defined for all real numbers except those where the denominator \(q(x) = 0\).

A Both A and R are true and R is the correct explanation of A. B Both A and R are true but R is NOT the correct explanation of A. C A is true but R is false. D A is false but R is true.

Select one option. Answers are shown after the test.

Question 5 of 10

Let \(A=\{1,2,3\}\). Which relation from \(A\) to a suitable codomain is a function with domain \(A\)?

A \(\{(1,2),(1,3),(2,4),(3,5)\}\) B \(\{(1,2),(2,2),(3,4)\}\) C \(\{(1,2),(2,3)\}\) D \(\{(1,2),(2,3),(3,4),(3,5)\}\)

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Question 6 of 10

A billing rule is written as \[ f(x)= \begin{cases} x^2,&0\le x\le2,\\ 3x,&2\le x\le5. \end{cases} \] Does this rule define a function on \([0,5]\)?

A No, because \(x=2\) is assigned both \(4\) and \(6\). B Yes, because both formulas are polynomial expressions. C Yes, because the two intervals cover \([0,5]\). D No, because \(f(0)=0\).

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Question 7 of 10

Assertion (A): If a set \(A\) has 3 elements and set \(B\) has 2 elements, the total number of relations from \(A\) to \(B\) is 64.

Reason (R): The total number of relations from a set with \(p\) elements to a set with \(q\) elements is \(2^{pq}\).

A Both A and R are true and R is the correct explanation of A. B Both A and R are true but R is NOT the correct explanation of A. C A is true but R is false. D A is false but R is true.

Select one option. Answers are shown after the test.

Question 8 of 10

Let \(A=\{9,10,11,12,13\}\), and define \(f:A\to\mathbb N\) by \(f(n)=\) the highest prime factor of \(n\). The range of \(f\) is

A \(\{3,5,7,11,13\}\) B \(\{2,3,5,11,13\}\) C \(\{9,10,11,12,13\}\) D \(\{3,5,11,13\}\)

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Question 9 of 10

A sensor relation is represented by the graph below.

Viewed as a relation from \(x\)-values to \(y\)-values, why is it not a function?

A Some \(y\)-values have no preimage. B The graph does not pass through the origin. C At least one \(x\)-value corresponds to two different \(y\)-values. D The graph contains negative \(y\)-values.

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Question 10 of 10

Six customers must each receive exactly one coupon type from \(4\) available types. How many different assignment functions are possible?

A \(6^4\) B \(4^6\) C \(2^{24}\) D \(24\)

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10 Question MCQ (including Assertion)