Quiz1

Mathematics for Data Science I · Quiz 1 · May 2026

Questions and published explanations below are available without starting a test. Some questions may not have a published solution yet.

Question 2 NAT · 3.0 marks

[[IMAGE:c5a6d7f097a74d94_2_2]]

Published solution

**1. Given:** \(p(x)=(x-2)(3x+1)q(x)\) has zeros \(2,-\tfrac13,4,-2\), each with multiplicity 1. The leading coefficient of \(q(x)\) is 1. **2. Find \(q(x)\):** The factors \((x-2)\) and \((3x+1)\) already give the zeros \(2\) and \(-\tfrac13\). So the remaining zeros \(4\) and \(-2\) must come from \(q(x)\). Each has multiplicity 1, so \(q(x)\) has degree 2. With leading coefficient 1: \[ q(x)=(x-4)(x+2) \] **3. Calculate \(q(1)\):** \[ q(1)=(1-4)(1+2)=(-3)(3)=-9 \] Answer: \(-9\).

Question 3 MCQ · 2.0 marks

Consider the polynomial [[IMAGE:c5a6d7f097a74d94_2_3]] Which of the following best describes the end behavior of [[IMAGE:c5a6d7f097a74d94_3_4]] ?

[[IMAGE:c5a6d7f097a74d94_3_5]] as [[IMAGE:c5a6d7f097a74d94_3_6]] and [[IMAGE:c5a6d7f097a74d94_3_7]] as [[IMAGE:c5a6d7f097a74d94_3_8]]
[[IMAGE:c5a6d7f097a74d94_3_9]] as [[IMAGE:c5a6d7f097a74d94_3_10]] and [[IMAGE:c5a6d7f097a74d94_3_11]] as [[IMAGE:c5a6d7f097a74d94_3_12]]
[[IMAGE:c5a6d7f097a74d94_3_13]] as [[IMAGE:c5a6d7f097a74d94_3_14]] and [[IMAGE:c5a6d7f097a74d94_3_15]] as [[IMAGE:c5a6d7f097a74d94_3_16]]
[[IMAGE:c5a6d7f097a74d94_3_17]] as [[IMAGE:c5a6d7f097a74d94_3_18]] and [[IMAGE:c5a6d7f097a74d94_3_19]] as [[IMAGE:c5a6d7f097a74d94_3_20]]

Published solution

**1. Given:** \(p(x)=-2(x+3)^5(x-1)^4(x+2)^3\). **2. Concept:** For large \(|x|\), a polynomial behaves like its leading term. Its degree and the sign of the leading coefficient decide the end behavior. **3. Find the leading term:** Degree \(=5+4+3=12\) (even). Leading coefficient \(=-2\) (negative). So the leading term is \(-2x^{12}\). **4. Analyze:** Since \(x^{12}\) is positive for both large positive and large negative \(x\), multiplying by \(-2\) makes the value negative on both sides: \[ p(x)\to-\infty \text{ as } x\to\infty,\qquad p(x)\to-\infty \text{ as } x\to-\infty \] Answer: B — \(p(x)\to-\infty\) as \(x\to\infty\) and \(p(x)\to-\infty\) as \(x\to-\infty\). A: **Incorrect:** This needs a positive leading coefficient with even degree, but the leading coefficient is \(-2\). B: **Correct:** Even degree 12 with negative leading coefficient sends \(p(x)\) to \(-\infty\) at both ends. C: **Incorrect:** Opposite ends going to different infinities needs an odd degree, but the degree is 12 (even). D: **Incorrect:** Opposite ends going to different infinities needs an odd degree, but the degree is 12 (even).

Question 4 MSQ · 4.0 marks

Which of the following statements is (are) correct?

[[IMAGE:c5a6d7f097a74d94_3_21]] represents a parabola whose vertex is [[IMAGE:c5a6d7f097a74d94_3_22]]
[[IMAGE:c5a6d7f097a74d94_3_23]] , where a=0 and [[IMAGE:c5a6d7f097a74d94_3_24]] , then [[IMAGE:c5a6d7f097a74d94_3_25]] is a polynomial of degree [[IMAGE:c5a6d7f097a74d94_3_26]]
The lines [[IMAGE:c5a6d7f097a74d94_3_27]] and [[IMAGE:c5a6d7f097a74d94_3_28]] are coincident.
The lines [[IMAGE:c5a6d7f097a74d94_3_29]] and [[IMAGE:c5a6d7f097a74d94_3_30]] are perpendicular to each other.

Published solution

**1. Check the parabola:** The vertex form is \(y-k=a(x-h)^2\) with vertex \((h,k)\). Write \(y+3=5(x-2)^2\) as \(y-(-3)=5(x-2)^2\). So the vertex is \((2,-3)\). **True.** **2. Check the polynomial:** If \(a=0\), then \(p(x)=bx^3+7x-1\). Since \(b\neq0\), the highest power with a nonzero coefficient is \(x^3\). So the degree is 3. **True.** **3. Check the coincident lines:** Divide \(4x-8y+12=0\) by 4: \[ x-2y+3=0 \] This is exactly the second line, so the lines are coincident. **True.** **4. Check perpendicular lines:** \(2x+y-1=0\) has slope \(-2\). \(2x-y+4=0\) has slope \(2\). Lines are perpendicular only if the product of slopes is \(-1\). Here \((-2)(2)=-4\neq-1\). **False.** Answer: A, B, C — The first three statements are correct. A: **Correct:** In vertex form the vertex is \((2,-3)\). B: **Correct:** With \(a=0\), the leading term is \(bx^3\) with \(b\neq0\), so the degree is 3. C: **Correct:** Dividing the first equation by 4 gives the second, so the lines coincide. D: **Incorrect:** The slopes are \(-2\) and \(2\); their product is \(-4\), not \(-1\).

Question 5 MSQ · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_4_31]]

If [[IMAGE:c5a6d7f097a74d94_4_32]] , then both roots of [[IMAGE:c5a6d7f097a74d94_4_33]] are equal.
If [[IMAGE:c5a6d7f097a74d94_4_34]] , then both roots of [[IMAGE:c5a6d7f097a74d94_4_35]] are integers.
There exists a value of [[IMAGE:c5a6d7f097a74d94_4_36]] for which [[IMAGE:c5a6d7f097a74d94_4_37]] is a root of [[IMAGE:c5a6d7f097a74d94_4_38]] .
If [[IMAGE:c5a6d7f097a74d94_4_39]] , then the sum of the roots of [[IMAGE:c5a6d7f097a74d94_4_40]] is [[IMAGE:c5a6d7f097a74d94_4_41]]

Published solution

**1. Factorise:** \(p(x)=x^2-(k+2)x+2k=(x-2)(x-k)\). Check: \((x-2)(x-k)=x^2-(k+2)x+2k\). So the roots are \(2\) and \(k\). **2. Check each statement:** - \(k=1\): roots are \(2\) and \(1\). They are different, so **False**. - \(k=2\): roots are \(2\) and \(2\). Both are integers, so **True**. - \(k=0\): then \(0\) is a root. So such a \(k\) exists, **True**. - \(k=0\): \(p(x)=x^2-2x=x(x-2)\). Roots are \(0\) and \(2\), sum \(=2\). **True.** Answer: B, C, D — The last three statements are correct. A: **Incorrect:** For \(k=1\) the roots are 2 and 1, which are not equal. B: **Correct:** For \(k=2\) both roots equal 2, which is an integer. C: **Correct:** Taking \(k=0\) makes \(0\) a root since the roots are \(2\) and \(k\). D: **Correct:** For \(k=0\) the roots are 0 and 2, and their sum is 2.

Question 6 MSQ · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_4_42]]

[[IMAGE:c5a6d7f097a74d94_4_43]] and [[IMAGE:c5a6d7f097a74d94_4_44]] are perpendicular.
[[IMAGE:c5a6d7f097a74d94_4_45]] and [[IMAGE:c5a6d7f097a74d94_4_46]] are parallel.
[[IMAGE:c5a6d7f097a74d94_4_47]] and [[IMAGE:c5a6d7f097a74d94_4_48]] are parallel.
[[IMAGE:c5a6d7f097a74d94_5_49]] and [[IMAGE:c5a6d7f097a74d94_5_50]] are perpendicular.

Published solution

**1. Concept:** For \(ax+by+c=0\), slope \(=-\tfrac{a}{b}\). Parallel lines have equal slopes. Perpendicular lines have slopes whose product is \(-1\). **2. Find the slopes:** - \(\ell_1: 2x-y+4=0\), slope \(=-\tfrac{2}{-1}=2\) - \(\ell_2: 4x-2y-3=0\), slope \(=-\tfrac{4}{-2}=2\) - \(\ell_3: x+2y-5=0\), slope \(=-\tfrac12\) **3. Check each option:** - \(\ell_1,\ell_2\): slopes \(2\) and \(2\). They are equal, so the lines are parallel, not perpendicular. - \(\ell_2,\ell_3\): slopes \(2\) and \(-\tfrac12\). Not equal, so not parallel. - \(\ell_1,\ell_3\): product \(=2\times(-\tfrac12)=-1\). They are perpendicular. Answer: B, D — \(\ell_1\) and \(\ell_2\) are parallel; \(\ell_1\) and \(\ell_3\) are perpendicular. A: **Incorrect:** Both slopes are 2, so the lines are parallel, not perpendicular. B: **Correct:** Both \(\ell_1\) and \(\ell_2\) have slope 2. C: **Incorrect:** Slopes are 2 and \(-\tfrac12\), which are not equal. D: **Correct:** The product of slopes is \(2\times(-\tfrac12)=-1\).

Question 7 MSQ · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_5_51]]

[[IMAGE:c5a6d7f097a74d94_5_52]] represents an injective function.
[[IMAGE:c5a6d7f097a74d94_5_53]] represents a function.
[[IMAGE:c5a6d7f097a74d94_5_54]] represents a function.
[[IMAGE:c5a6d7f097a74d94_5_55]] represents a relation but not a function.

Published solution

**1. Concept:** A relation is a function if each input \(x\) has exactly one output \(y\). It is injective if different inputs give different outputs. **2. \(R_1\):** \(y=2x+3\). Each \(x\) gives one \(y\), so it is a function. If \(2x_1+3=2x_2+3\), then \(x_1=x_2\). So it is injective. **True.** **3. \(R_2\):** \(x^2+y^2=1\) on integers gives \((0,1),(0,-1),(1,0),(-1,0)\). Input \(0\) has two outputs, \(1\) and \(-1\). So \(R_2\) is a relation but **not** a function. Option 2 is false and option 4 is true. **4. \(R_1\cap R_2\):** We need \(y=2x+3\) and \(x^2+y^2=1\). Substituting: \[ x^2+(2x+3)^2=1\;\Rightarrow\;5x^2+12x+8=0 \] The discriminant is \(144-160=-16<0\), so there is no real (hence no integer) solution. So \(R_1\cap R_2=\varnothing\). The empty relation has no input with two outputs, so it is (vacuously) a function. **True** by this convention. Answer: A, C, D — \(R_1\) is injective, \(R_1\cap R_2\) is a function, and \(R_2\) is a relation but not a function. **Convention note:** Option C is correct only because the empty relation is treated as a (vacuous) function, which agrees with the supplied key. A: **Correct:** \(y=2x+3\) gives one output per input and different inputs give different outputs. B: **Incorrect:** The input 0 is related to both 1 and \(-1\), so \(R_2\) is not a function. C: **Correct:** \(R_1\cap R_2\) is empty, and the empty relation has no input with two outputs, so it is vacuously a function. D: **Correct:** \(R_2\) is a relation, but \(x=0\) has two outputs, so it is not a function.

Question 8 NAT · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_5_56]]

Published solution

**1. Given:** \(x^2-kx+16=0\). One root is four times the other, and both roots are positive. **2. Set up:** Let the roots be \(r\) and \(4r\) with \(r>0\). Product of roots \(=\tfrac{c}{a}=16\): \[ r\cdot4r=16\;\Rightarrow\;4r^2=16\;\Rightarrow\;r^2=4 \] Since \(r>0\), \(r=2\). So the roots are \(2\) and \(8\). **3. Find \(k\):** Sum of roots \(=k\): \[ k=2+8=10 \] Answer: \(10\).

Question 9 NAT · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_6_57]]

Published solution

**1. Concept:** If \(x+a\) is a factor of \(p(x)\), then \(p(-a)=0\) (Factor Theorem). **2. Substitute \(x=-a\):** \[ p(-a)=a^2-(2a-3)a+a^2+3a-4 \] **3. Simplify:** \[ =a^2-2a^2+3a+a^2+3a-4=6a-4 \] **4. Solve:** \(6a-4=0\Rightarrow a=\tfrac23\approx0.667\), which lies in the range 0 to 1. Answer: \(a=\tfrac23\).

Question 10 NAT · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_6_58]]

Published solution

**1. Find \(A\cup B\):** \(\{1,2,3,4,5,6,7,8,10\}\). **2. Find \(A\cup C\):** \(\{1,2,3,4,5,6,7,9\}\). **3. Intersect:** \((A\cup B)\cap(A\cup C)=\{1,2,3,4,5,6,7\}\). (8 and 10 are not in \(A\cup C\), and 9 is not in \(A\cup B\).) **4. Find \(B\cap C\):** \(B\) has only even numbers and \(C\) has only odd numbers, so \(B\cap C=\varnothing\). **5. Subtract:** Removing the empty set changes nothing, so the set is \(\{1,2,3,4,5,6,7\}\), with 7 elements. Answer: \(7\).

Question 11 NAT · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_7_59]]

Published solution

**1. Given:** \(A(0,0)\), \(C(12,6)\), \(D(4,1)\). \(B\) divides \(AC\) internally in the ratio \(k:1\). **2. Section formula:** \[ B=\left(\frac{12k}{k+1},\frac{6k}{k+1}\right)=(12t,6t),\quad t=\frac{k}{k+1} \] **3. Use \(BD=5\):** \[ (12t-4)^2+(6t-1)^2=25 \] \[ 144t^2-96t+16+36t^2-12t+1=25\;\Rightarrow\;180t^2-108t-8=0\;\Rightarrow\;45t^2-27t-2=0 \] **4. Solve for \(t\):** \[ t=\frac{27\pm\sqrt{729+360}}{90}=\frac{27\pm33}{90}\;\Rightarrow\;t=\frac23\text{ or }t=-\frac1{15} \] For internal division with \(k>0\), \(0<t<1\), so \(t=\tfrac23\). **5. Find \(k\):** \(\dfrac{k}{k+1}=\dfrac23\Rightarrow3k=2k+2\Rightarrow k=2\). Check: \(B=(8,4)\), \(BD=\sqrt{16+9}=5\). ✓ Answer: \(2\).

Question 12 NAT · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_7_60]]

Published solution

**1. Concept:** If \(|B|=n\), then \(|B\times B|=n^2\). **2. Find \(|B|\):** \(n^2=16\Rightarrow n=4\). **3. Find the elements:** The pair \((3,5)\) shows \(3,5\in B\). The pair \((7,0)\) shows \(7,0\in B\). That gives 4 distinct elements, so \(B=\{0,3,5,7\}\). **4. Sum:** \(0+3+5+7=15\). Answer: \(15\).

Question 13 MCQ · 4.0 marks

[[IMAGE:c5a6d7f097a74d94_8_61]]

[[IMAGE:c5a6d7f097a74d94_8_62]]
[[IMAGE:c5a6d7f097a74d94_8_63]]
[[IMAGE:c5a6d7f097a74d94_8_64]]
[[IMAGE:c5a6d7f097a74d94_8_65]]

Published solution

**1. Given:** \(\alpha+\beta=7\) and \(\alpha^2+\beta^2=29\). **2. Formula:** \((\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta\). **3. Find the product:** \[ 49=29+2\alpha\beta\;\Rightarrow\;\alpha\beta=10 \] **4. Form the equation:** \(x^2-(\alpha+\beta)x+\alpha\beta=0\) gives \[ x^2-7x+10=0 \] Check: roots are 2 and 5; sum 7 and \(4+25=29\). ✓ Answer: A — \(x^2-7x+10=0\). A: **Correct:** Sum \(=7\) and product \(=10\) match this equation. B: **Incorrect:** Its product of roots is 12, but \(\alpha\beta=10\). C: **Incorrect:** Its product of roots is 14, but \(\alpha\beta=10\). D: **Incorrect:** Its product of roots is 20, but \(\alpha\beta=10\).

Question 14 NAT · 5.0 marks

[[IMAGE:c5a6d7f097a74d94_8_66]]

Published solution

**1. Given:** Parallel lines \(3x+4y+c_1=0\) and \(3x+4y+c_2=0\) are 4 units apart, with \(c_2>c_1>0\). The distance from \((2,3)\) to the first line is 6. **2. Distance between the lines:** Formula \(\dfrac{|c_2-c_1|}{\sqrt{3^2+4^2}}=\dfrac{|c_2-c_1|}{5}\). \[ \frac{c_2-c_1}{5}=4\;\Rightarrow\;c_2-c_1=20 \] **3. Distance from the point to the first line:** \[ \frac{|3(2)+4(3)+c_1|}{5}=6\;\Rightarrow\;|18+c_1|=30 \] So \(c_1=12\) or \(c_1=-48\). Since \(c_1>0\), \(c_1=12\). **4. Find \(c_2\):** \(c_2=12+20=32\). **5. Sum:** \(c_1+c_2=12+32=44\). Answer: \(44\).