INTRODUCTION TO MATHEMATICAL METHODS

QUIZ TWO JULY INTAKE 2024

OPEN DISTANCE LEARNING (ODL)

DATE : 18th OCTOBER, 2024

Question 1:

Let f(x)={4−x2,x<−3x+4,−3≤x≤3,x2+8,x>3f(x) = \begin{cases} 4 - x^2, & x < -3 \\ x + 4, & -3 \leq x \leq 3, \\ x^2 + 8, & x > 3 \end{cases} find f(10).

A. −96-96
B. 1414
C. 108108
D. 84728472
E. None of the above

Answer: C. 108108

Explanation:
To find f(10)f(10), evaluate the correct piece of the piecewise function based on the value of x=10x = 10. Since 10>310 > 3, use the third piece:

f(x)=x2+8forx>3f(x) = x^2 + 8 \quad \text{for} \quad x > 3

Substitute x=10x = 10:

f(10)=(10)2+8=100+8=108f(10) = (10)^2 + 8 = 100 + 8 = 108

Thus, the answer is 108108.

Question 2:

Determine whether the function f(x)=x4−xf(x) = x^4 - x is odd or even.

A. It is odd
B. It is even
C. It is neither odd nor even
D. It is both odd and even
E. None of the above

Answer: C. It is neither odd nor even

Explanation:

  • A function is even if f(−x)=f(x)f(-x) = f(x) for all xx.
  • A function is odd if f(−x)=−f(x)f(-x) = -f(x) for all xx.

Compute f(−x)f(-x):

f(−x)=(−x)4−(−x)=x4+xf(-x) = (-x)^4 - (-x) = x^4 + x

  • Compare to f(x)=x4−xf(x) = x^4 - x: f(−x)=x4+xvs.f(x)=x4−x⇒f(−x)≠f(x)f(-x) = x^4 + x \quad \text{vs.} \quad f(x) = x^4 - x \quad \Rightarrow \quad f(-x) \neq f(x) So, it is not even.
  • Compare to −f(x)=−(x4−x)=−x4+x-f(x) = -(x^4 - x) = -x^4 + x: f(−x)=x4+xvs.−f(x)=−x4+x⇒f(−x)≠−f(x)f(-x) = x^4 + x \quad \text{vs.} \quad -f(x) = -x^4 + x \quad \Rightarrow \quad f(-x) \neq -f(x) So, it is not odd.

Thus, f(x)f(x) is neither odd nor even.

Question 3:

Which of the following function is not one to one?

A. f(x)=x3f(x) = x^3
B. f(x)=x−7x+4f(x) = \frac{x-7}{x+4}
C. f(x)=x2−2f(x) = x^2 - 2
D. f(x)=x7−3f(x) = x^7 - 3
E. None of the above

Answer: C. f(x)=x2−2f(x) = x^2 - 2

Explanation:
A function is one-to-one (injective) if different inputs produce different outputs, i.e., if f(a)=f(b)f(a) = f(b) implies a=ba = b.

  • Option A: f(x)=x3f(x) = x^3
    This is strictly increasing (derivative 3x2≥03x^2 \geq 0), so it is one-to-one.
  • Option B: f(x)=x−7x+4f(x) = \frac{x-7}{x+4}
    Assume f(a)=f(b)f(a) = f(b): a−7a+4=b−7b+4  ⟹  (a−7)(b+4)=(b−7)(a+4)\frac{a-7}{a+4} = \frac{b-7}{b+4} \implies (a-7)(b+4) = (b-7)(a+4) Expand: ab+4a−7b−28=ab−7a+4b−28ab + 4a - 7b - 28 = ab - 7a + 4b - 28
    Simplify: 4a−7b=−7a+4b  ⟹  11a=11b  ⟹  a=b4a - 7b = -7a + 4b \implies 11a = 11b \implies a = b.
    Thus, it is one-to-one.
  • Option C: f(x)=x2−2f(x) = x^2 - 2
    This is a parabola opening upward. For example: f(1)=(1)2−2=−1,f(−1)=(−1)2−2=−1f(1) = (1)^2 - 2 = -1, \quad f(-1) = (-1)^2 - 2 = -1 So f(1)=f(−1)f(1) = f(-1) but 1≠−11 \neq -1, so it is not one-to-one.
  • Option D: f(x)=x7−3f(x) = x^7 - 3
    This is strictly increasing (derivative 7x6≥07x^6 \geq 0), so it is one-to-one.

Thus, f(x)=x2−2f(x) = x^2 - 2 is not one-to-one.

Question 4:

What is the solution of ∣x∣<∣2x−3∣|x| < |2x - 3| ?

A. 1<x<31 < x < 3
B. 1≤x≤31 \leq x \leq 3
C. x<1 and x>3x < 1 \, \text{and} \, x > 3
D. x≤1 and x≥3x \leq 1 \, \text{and} \, x \geq 3
E. None of the above

Answer: C. x<1 and x>3x < 1 \, \text{and} \, x > 3

Explanation:
Solve the inequality ∣x∣<∣2x−3∣|x| < |2x - 3| by considering critical points where the expressions inside absolute values are zero: x=0x = 0 and x=32=1.5x = \frac{3}{2} = 1.5. This divides the real line into three intervals: x<0x < 0, 0≤x<1.50 \leq x < 1.5, and x≥1.5x \geq 1.5.

  • Case 1: x<0x < 0
    Here, ∣x∣=−x|x| = -x and ∣2x−3∣=−(2x−3)=−2x+3|2x - 3| = -(2x - 3) = -2x + 3 (since x<0<1.5x < 0 < 1.5).
    Inequality: −x<−2x+3-x < -2x + 3
    Solve: −x+2x<3  ⟹  x<3-x + 2x < 3 \implies x < 3.
    Since x<0x < 0 and x<3x < 3, the solution is x<0x < 0.
  • Case 2: 0≤x<1.50 \leq x < 1.5
    Here, ∣x∣=x|x| = x and ∣2x−3∣=−2x+3|2x - 3| = -2x + 3 (since x<1.5x < 1.5).
    Inequality: x<−2x+3x < -2x + 3
    Solve: x+2x<3  ⟹  3x<3  ⟹  x<1x + 2x < 3 \implies 3x < 3 \implies x < 1.
    Combining with 0≤x<1.50 \leq x < 1.5, the solution is 0≤x<10 \leq x < 1.
  • Case 3: x≥1.5x \geq 1.5
    Here, ∣x∣=x|x| = x and ∣2x−3∣=2x−3|2x - 3| = 2x - 3 (since x≥1.5x \geq 1.5).
    Inequality: x<2x−3x < 2x - 3
    Solve: 0<x−3  ⟹  x>30 < x - 3 \implies x > 3.
    Combining with x≥1.5x \geq 1.5, the solution is x>3x > 3.

Combine all solutions:

  • From Case 1: x<0x < 0
  • From Case 2: 0≤x<10 \leq x < 1 (which is equivalent to x<1x < 1 for x≥0x \geq 0)
  • From Case 3: x>3x > 3
    Thus, the solution is x<1x < 1 or x>3x > 3, written as x<1 and x>3x < 1 \, \text{and} \, x > 3 (meaning the union of the sets).

Verification:

  • At x=0x = 0 (which is <1< 1): ∣0∣=0|0| = 0, ∣0−3∣=3|0 - 3| = 3, 0<30 < 3 → true.
  • At x=2x = 2 (which is not in solution): ∣2∣=2|2| = 2, ∣4−3∣=1|4 - 3| = 1, 2<12 < 1 → false.
  • At x=4x = 4 (which is >3> 3): ∣4∣=4|4| = 4, ∣8−3∣=5|8 - 3| = 5, 4<54 < 5 → true.

Option C matches the solution.

Question 5:

What is the nature of roots for a quadratic equation x2+4=0x^2 + 4 = 0 ?

A. It has two real distinct roots
B. It has two real equal roots
C. It has two complex roots
D. It has no solutions
E. None of the above

Answer: C. It has two complex roots

Explanation:
Solve the equation:

x2+4=0  

⟹  x2=−4  

⟹  x=±−4

=±2ix^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm \sqrt{-4} = \pm 2i

The roots are complex (specifically, imaginary). Option D is incorrect because solutions exist in the complex numbers. Thus, the nature is two complex roots.

Question 6:

Write f(x)=x2−2x+8f(x) = x^2 - 2x + 8 in the completed square form.

A. f(x)=(x−1)2+7f(x) = (x - 1)^2 + 7
B. f(x)=(x−1)2+9f(x) = (x - 1)^2 + 9
C. f(x)=(x+1)2+7f(x) = (x + 1)^2 + 7
D. f(x)=(x+1)2+9f(x) = (x + 1)^2 + 9
E. None of the above

Answer: A. f(x)=(x−1)2+7f(x) = (x - 1)^2 + 7

Explanation:
Complete the square:

f(x)=x2−2x+8f(x) = x^2 - 2x + 8

  • Coefficient of xx is −2-2, so half is −1-1, square is 11.
  • Add and subtract 11: f(x)=(x2−2x+1)+8−1
  • =(x−1)2+7f(x) = (x^2 - 2x + 1) + 8 - 1 = (x - 1)^2 + 7

Thus, the completed square form is (x−1)2+7(x - 1)^2 + 7.

For questions 7 to 9:

Let α\alpha and β\beta be the roots of the quadratic equation x2−7x+8=0x^2 - 7x + 8 = 0.

Question 7:
Calculate the value of αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}.

A. 338\frac{33}{8}
B. 23 and −23\frac{2}{3} \, \text{and} \, -\frac{2}{3}
C. ±78(17)\pm \frac{7}{8} (\sqrt{17})
D. 78 and −78\frac{7}{8} \, \text{and} \, -\frac{7}{8}
E. None of the above

Answer: A. 338\frac{33}{8}

Explanation:
For the quadratic equation x2−7x+8=0x^2 - 7x + 8 = 0:

  • Sum of roots: α+β=7\alpha + \beta = 7
  • Product of roots: αβ=8\alpha \beta = 8

Now, compute:

αβ+βα=α2+β2αβ\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}


First, α2+β2=(α+β)2−2αβ

=(7)2−2(8)

=49−16

=33\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (7)^2 - 2(8) = 49 - 16 = 33.
Then,

α2+β2αβ=338\frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{33}{8}

Thus, the value is 338\frac{33}{8}.

Question 8:

Find the value of α2β+αβ2\alpha^2 \beta + \alpha \beta^2.

A. 56
B. -56
C. 8\sqrt{33}
D. 18
E. None of the above

Answer: A. 56

Explanation:

α2β+αβ2=αβ(α+β)\alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta)

Substitute known values:

αβ=8,α+β=7  

⟹  8×7=56\alpha \beta = 8, \quad \alpha + \beta = 7 \implies 8 \times 7 = 56

Thus, the value is 56.

Question 9:

Write down new equation whose roots are 1α\frac{1}{\alpha} and 1β\frac{1}{\beta}.

A. 8x2−7x+1=08x^2 - 7x + 1 = 0
B. 8x2+7x−1=08x^2 + 7x - 1 = 0
C. x2+78x+18=0x^2 + \frac{7}{8}x + \frac{1}{8} = 0
D. 1x2−17x+18=0\frac{1}{x^2} - \frac{1}{7x} + \frac{1}{8} = 0
E. None of the above

Answer: A. 8x2−7x+1=08x^2 - 7x + 1 = 0

Explanation:
The new roots are 1α\frac{1}{\alpha} and 1β\frac{1}{\beta}. For a quadratic equation with roots rr and ss, it is x2−(r+s)x+rs=0x^2 - (r+s)x + rs = 0.

  • Sum: 1α+1β=α+βαβ=78\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{7}{8}
  • Product: 1α⋅1β=1αβ=18\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{8}

So the equation is:

x2−(78)x+18=0x^2 - \left(\frac{7}{8}\right)x + \frac{1}{8} = 0

Multiply through by 8 to clear denominators:

8x2−7x+1=08x^2 - 7x + 1 = 0

Thus, the equation is 8x2−7x+1=08x^2 - 7x + 1 = 0.

Question 10:

The product of two numbers is 350 and their sum is 57. The two numbers are.

A. 7 and 50
B. 10 and 35
C. 47 and 8
D. 18 and 39
E. None of the above

Answer: A. 7 and 50

Explanation:
Let the numbers be xx and yy. Then:

x+y=57,xy=350x + y = 57, \quad xy = 350

They satisfy the quadratic equation t2−(x+y)t+xy=0t^2 - (x+y)t + xy = 0, so:

t2−57t+350=0t^2 - 57t + 350 = 0

Solve using quadratic formula:

t=57±(−57)2−4(1)(350)2

=57±3249−14002

=57±18492t = \frac{57 \pm \sqrt{(-57)^2 - 4(1)(350)}}{2} = \frac{57 \pm \sqrt{3249 - 1400}}{2} = \frac{57 \pm \sqrt{1849}}{2}

Since 1849=43\sqrt{1849} = 43 (as 432=184943^2 = 1849):

t=57±432t = \frac{57 \pm 43}{2}

So:

t=57+432=1002=50,t=57−432=142=7t = \frac{57 + 43}{2} = \frac{100}{2} = 50, \quad t = \frac{57 - 43}{2} = \frac{14}{2} = 7

Thus, the numbers are 7 and 50.

Question 11:

Which of the following is a sextic polynomial?

A. 3x9−4x7−5x5+3x3+2x3x^9 - 4x^7 - 5x^5 + 3x^3 + 2x
B. 3x2−4x6−5x5+3x3+2x3x^2 - 4x^6 - 5x^5 + 3x^3 + 2x
C. 3x9−4x8−5x7+3x6+2x3x^9 - 4x^8 - 5x^7 + 3x^6 + 2x
D. −5x5+3x3+2x-5x^5 + 3x^3 + 2x
E. None of the above

Answer: B. 3x2−4x6−5x5+3x3+2x3x^2 - 4x^6 - 5x^5 + 3x^3 + 2x

Explanation:
A sextic polynomial has degree 6 (highest exponent of xx is 6).

  • Option A: Highest exponent is 9 → degree 9.
  • Option B: Highest exponent is 6 → degree 6.
  • Option C: Highest exponent is 9 → degree 9.
  • Option D: Highest exponent is 5 → degree 5.
    Thus, only option B is sextic.

Question 12:

What is the remainder when p(x)=3x3+x2+3x+5p(x) = 3x^3 + x^2 + 3x + 5 is divided by x+2x + 2?

A. −2-2
B. 39
C. 2
D. −21-21
E. None of the above

Answer: D. −21-21

Explanation:
Use the Remainder Theorem: when dividing by x−cx - c, the remainder is p(c)p(c). Here, x+2=x−(−2)x + 2 = x - (-2), so c=−2c = -2.
Compute p(−2)p(-2):

p(−2)=3(−2)3+(−2)2+3(−2)+5

=3(−8)+4+(−6)+5

=−24+4−6+5p(-2) = 3(-2)^3 + (-2)^2 + 3(-2) + 5 = 3(-8) + 4 + (-6) + 5 = -24 + 4 - 6 + 5

Calculate step-by-step:

−24+4=−20,−20−6=−26,−26+5=−21-24 + 4 = -20, \quad -20 - 6 = -26, \quad -26 + 5 = -21

Thus, the remainder is −21-21.

Question 13:

Let x+2x + 2 be a factor of f(x)=x3+3x2+3x+kf(x) = x^3 + 3x^2 + 3x + k. What is the value of k?

A. 26
B. 2
C. -2
D. -26
E. None of the above

Answer: B. 2

Explanation:
If x+2x + 2 is a factor, then f(−2)=0f(-2) = 0.
Compute f(−2)f(-2):

f(−2)=(−2)3+3(−2)2+3(−2)+k

=−8+3(4)+(−6)+k

=−8+12−6+kf(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + k = -8 + 3(4) + (-6) + k = -8 + 12 - 6 + k

Simplify:

(−8−6)+12+k

=−14+12+k

=−2+k(-8 - 6) + 12 + k = -14 + 12 + k = -2 + k

Set equal to 0:

−2+k=0  ⟹  k=2-2 + k = 0 \implies k = 2

Thus, k=2k = 2.

Question 14:

According the methods of partial fractions, 11−x(x−3)(x+4)=Ax−3+Bx+4\frac{11 - x}{(x - 3)(x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4}. The value of B is.

A. 1
B. -2
C. 87\frac{8}{7}
D. −157-\frac{15}{7}
E. None of the above

Answer: D. −157-\frac{15}{7}

Explanation:
Set up the equation:

11−x(x−3)(x+4)=Ax−3+Bx+4\frac{11 - x}{(x - 3)(x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4}

Multiply both sides by (x−3)(x+4)(x - 3)(x + 4):

11−x=A(x+4)+B(x−3)11 - x = A(x + 4) + B(x - 3)

Solve for AA and BB by substituting suitable values of xx.

  • Substitute x=3x = 3: 11−3=A(3+4)+B(3−3)  
  • ⟹  8=7A  
  • ⟹  A=8711 - 3 = A(3 + 4) + B(3 - 3) \implies 8 = 7A \implies A = \frac{8}{7}
  • Substitute x=−4x = -4: 11−(−4)=A(−4+4)+B(−4−3)  
  • ⟹  15=B(−7)  
  • ⟹  B=−15711 - (-4) = A(-4 + 4) + B(-4 - 3) \implies 15 = B(-7) \implies B = -\frac{15}{7}

Thus, B=−157B = -\frac{15}{7}.

Question 15:

When expressed into partial fractions xx2−16\frac{x}{x^2 - 16} will be.

A. 12(x+4)−12(x−4)\frac{1}{2(x + 4)} - \frac{1}{2(x - 4)}
B. 12(x+4)+12(x−4)\frac{1}{2(x + 4)} + \frac{1}{2(x - 4)}
C. 1(x+4)−1(x−4)\frac{1}{(x + 4)} - \frac{1}{(x - 4)}
D. 1(x+4)+1(x−4)\frac{1}{(x + 4)} + \frac{1}{(x - 4)}
E. None of the above

Answer: B. 12(x+4)+12(x−4)\frac{1}{2(x + 4)} + \frac{1}{2(x - 4)}

Explanation:
Factor the denominator: x2−16=(x−4)(x+4)x^2 - 16 = (x - 4)(x + 4).
Set up partial fractions:

x(x−4)(x+4)=Ax−4+Bx+4\frac{x}{(x - 4)(x + 4)} = \frac{A}{x - 4} + \frac{B}{x + 4}

Multiply both sides by (x−4)(x+4)(x - 4)(x + 4):

x=A(x+4)+B(x−4)x = A(x + 4) + B(x - 4)

Solve for AA and BB:

  • Substitute x=4x = 4: 4=A(4+4)+B(4−4)  
  • ⟹  4=8A  
  • ⟹  A=48=124 = A(4 + 4) + B(4 - 4) \implies 4 = 8A \implies A = \frac{4}{8} = \frac{1}{2}
  • Substitute x=−4x = -4: −4=A(−4+4)+B(−4−4)  
  • ⟹  −4=B(−8)
  •   
  • ⟹  B=−4−8=12-4 = A(-4 + 4) + B(-4 - 4) \implies -4 = B(-8) \implies B = \frac{-4}{-8} = \frac{1}{2}

So:

xx2−16=12x−4+12x+4

=12(x−4)+12(x+4)\frac{x}{x^2 - 16} = \frac{\frac{1}{2}}{x - 4} + \frac{\frac{1}{2}}{x + 4} = \frac{1}{2(x - 4)} + \frac{1}{2(x + 4)}

Thus, option B is correct.

For questions 16 to 18:

The polynomial p(x)=(x2−4)(x2−9)(x+7)p(x) = (x^2 - 4)(x^2 - 9)(x + 7) is given.

Question 16:
What is the degree of p(x)p(x)?

A. 1
B. 2
C. 4
D. 5
E. None of the above

Answer: D. 5

Explanation:
The degree of a polynomial is the highest power of xx when expanded.

  • x2−4x^2 - 4 has degree 2.
  • x2−9x^2 - 9 has degree 2.
  • x+7x + 7 has degree 1.
    The degree of the product is the sum of the degrees: 2+2+1=52 + 2 + 1 = 5.
    Thus, degree is 5.

Question 17:

If we are to sketch p(x)p(x), what is our y-intercept?

A. -9
B. 252
C. -252
D. -28
E. None of the above

Answer: B. 252

Explanation:
The y-intercept occurs when x=0x = 0.

p(0)=(02−4)(02−9)(0+7)

=(−4)(−9)(7)p(0) = (0^2 - 4)(0^2 - 9)(0 + 7) = (-4)(-9)(7)

Compute:

(−4)×(−9)=36,

36×7=252(-4) \times (-9) = 36, \quad 36 \times 7 = 252

Thus, the y-intercept is 252.

Question 18:

What are the roots of p(x)p(x)?

A. 4, 9 and -7
B. 2, 3 and 7
C. 2, -2, -3, 3 and 7
D. -2, 2, -3, 3 and -7
E. None of the above

Answer: D. -2, 2, -3, 3 and -7

Explanation:
Set each factor to zero:

x2−4=0  

⟹  x2=4  

⟹  x=±2

andx^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2

x2−9=0  

⟹  x2=9  

⟹  x=±3

andx^2 - 9 = 0 \implies x^2 = 9 \implies x = \pm 3

x+7=0  

⟹  x=−7x + 7 = 0 \implies x = -7

Thus, the roots are x=−2,2,−3,3,−7x = -2, 2, -3, 3, -7.

Question 19:

Divide the polynomial p(x)=2x4−3x3+3x2−x+2p(x) = 2x^4 - 3x^3 + 3x^2 - x + 2 by x−2x - 2.

A. q(x)=2x3−7x2+17x−35q(x) = 2x^3 - 7x^2 + 17x - 35, r(x)=72r(x) = 72
B. q(x)=2x3+x2+5x+9q(x) = 2x^3 + x^2 + 5x + 9, r(x)=20r(x) = 20
C. q(x)=x3−2x2−3x+2q(x) = x^3 - 2x^2 - 3x + 2, r(x)=−3r(x) = -3
D. 2x3−52x^3 - 5, r(x)=x2+2r(x) = x^2 + 2
E. None of the above

Answer: B. q(x)=2x3+x2+5x+9q(x) = 2x^3 + x^2 + 5x + 9, r(x)=20r(x) = 20

Explanation:
Use synthetic division with root c=2c = 2 (since divisor is x−2x - 2).
Coefficients of p(x)p(x): 2,−3,3,−1,22, -3, 3, -1, 2.

Synthetic Division:

  • Bring down 2.
  • Multiply by 2: 2×2=42 \times 2 = 4.
  • Add to next coefficient: −3+4=1-3 + 4 = 1.
  • Multiply by 2: 1×2=21 \times 2 = 2.
  • Add to next: 3+2=53 + 2 = 5.
  • Multiply by 2: 5×2=105 \times 2 = 10.
  • Add to next: −1+10=9-1 + 10 = 9.
  • Multiply by 2: 9×2=189 \times 2 = 18.
  • Add to last: 2+18=202 + 18 = 20.

The quotient coefficients are 2,1,5,92, 1, 5, 9 (degree 3), so q(x)=2x3+x2+5x+9q(x) = 2x^3 + x^2 + 5x + 9.
The remainder is 20, so r(x)=20r(x) = 20.
Thus, option B is correct.

Question 20:

Let f(x)=x3+2x2−3x−4f(x) = x^3 + 2x^2 - 3x - 4 and g(x)=5x2−6g(x) = 5x^2 - 6 be two polynomials. Simplify f(x)+g(x)f(x) + g(x).

A. 6x3−4x2−3x−46x^3 - 4x^2 - 3x - 4
B. x2+8x+45x^2 + 8x + 45
C. x3+7x−9x^3 + 7x - 9
D. x3+7x2−3x−10x^3 + 7x^2 - 3x - 10
E. None of the above

Answer: D. x3+7x2−3x−10x^3 + 7x^2 - 3x - 10

Explanation:
Add the polynomials term by term:

f(x)+g(x)=(x3+2x2−3x−4)+(5x2−6)

=x3+(2x2+5x2)+(−3x)+(−4−6)f(x) + g(x) = (x^3 + 2x^2 - 3x - 4) + (5x^2 - 6) = x^3 + (2x^2 + 5x^2) + (-3x) + (-4 - 6)

=x3+7x2−3x−10= x^3 + 7x^2 - 3x - 10

Thus, the sum is x3+7x2−3x−10x^3 + 7x^2 - 3x - 10.

Question 21:

What is the symmetry of the quadratic function f(x)=2x2−12x+3f(x) = 2x^2 - 12x + 3?

A. x=3x = 3
B. x=6x = 6
C. x=−6x = -6
D. x=−3x = -3
E. None of the above

Answer: A. x=3x = 3

Explanation:
The axis of symmetry for a quadratic ax2+bx+cax^2 + bx + c is x=−b2ax = -\frac{b}{2a}.
Here, a=2a = 2, b=−12b = -12:

x=−−122×2=124=3x = -\frac{-12}{2 \times 2} = \frac{12}{4} = 3

Thus, the axis of symmetry is x=3x = 3.

Question 22:

Which of the following gives the turning point of a quadratic function?

A. (−ba,f(−ba))\left( -\frac{b}{a}, f \left( -\frac{b}{a} \right) \right)
B. (−b2a,f(−b2a))\left( -\frac{b}{2a}, f \left( -\frac{b}{2a} \right) \right)
C. (ba,f(ba))\left( \frac{b}{a}, f \left( \frac{b}{a} \right) \right)
D. (b2a,f(b2a))\left( \frac{b}{2a}, f \left( \frac{b}{2a} \right) \right)
E. None of the above

Answer: B. (−b2a,f(−b2a))\left( -\frac{b}{2a}, f \left( -\frac{b}{2a} \right) \right)

Explanation:
The turning point (vertex) of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c is at:

x=−b2a,y=f(−b2a)x = -\frac{b}{2a}, \quad y = f\left(-\frac{b}{2a}\right)

Thus, the coordinates are (−b2a,f(−b2a))\left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right).

Question 23:

Which of the following is true for the function f(x)=−2x2−3x+4f(x) = -2x^2 - 3x + 4?

A. It does not intercept x-axis
B. It has minimum value at turning point
C. It has maximum value at turning point
D. It does not intercept y-axis
E. None of the above

Answer: C. It has maximum value at turning point

Explanation:

  • The leading coefficient is a=−2<0a = -2 < 0, so the parabola opens downwards, meaning it has a maximum at the turning point.
  • Option A: Discriminant b2−4ac=(−3)2−4(−2)(4)=9+32=41>0b^2 - 4ac = (-3)^2 - 4(-2)(4) = 9 + 32 = 41 > 0, so it intercepts the x-axis at two points.
  • Option D: At x=0x = 0, f(0)=4f(0) = 4, so it intercepts the y-axis.
    Thus, only option C is true.

Question 24:

Which of the following are the solutions of x2−12=0x^2 - 12 = 0?

A. x=±12x = \pm 12
B. x=±4x = \pm 4
C. x=±32x = \pm 3\sqrt{2}
D. x=±23x = \pm 2\sqrt{3}
E. None of the above

Answer: D. x=±23x = \pm 2\sqrt{3}

Explanation:
Solve:

x2−12=0  

⟹  x2=12  

⟹  x=±12

=±4×3

=±23x^2 - 12 = 0 \implies x^2 = 12 \implies x = \pm \sqrt{12} = \pm \sqrt{4 \times 3} = \pm 2\sqrt{3}

Thus, the solutions are x=±23x = \pm 2\sqrt{3}.

Question 25:

Which of the following defines a one-to-one function?

A. f(a)=f(b)f(a) = f(b) if a=ba = b
B. f(a)=f(b)f(a) = f(b) for a,b∈f(x)a, b \in f(x)
C. f(x)=f(a)=f(b)f(x) = f(a) = f(b) at any point
D. f(a)=f(b), ∀ a,b∈Rf(a) = f(b), \, \forall \, a, b \in R
E. None of the above

Answer: E. None of the above

Explanation:
A function is one-to-one (injective) if f(a)=f(b)f(a) = f(b) implies a=ba = b.

  • Option A: "f(a)=f(b)f(a) = f(b) if a=ba = b" is true for all functions (by definition), but it does not ensure that different inputs give different outputs.
  • Option B: "f(a)=f(b)f(a) = f(b) for a,b∈f(x)a, b \in f(x)" is unclear; it may imply f(a)=f(b)f(a) = f(b) for all a,ba,b in the range, which is not relevant.
  • Option C: "f(x)=f(a)=f(b)f(x) = f(a) = f(b) at any point" is ambiguous and incorrect.
  • Option D: "f(a)=f(b)f(a) = f(b) for all a,b∈Ra, b \in \mathbb{R}" implies ff is constant, which is not one-to-one.
    None of the options correctly define a one-to-one function. Thus, E is correct.

@Dr. Microbiota

End of Solutions