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Math Olympiad Test: Polynomials - 4 - Class 10 MCQ


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10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Polynomials - 4

Math Olympiad Test: Polynomials - 4 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Polynomials - 4 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Polynomials - 4 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Polynomials - 4 below.
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Math Olympiad Test: Polynomials - 4 - Question 1

A polynomial of the form ax5 + bx3 + cx2 + dx + e has atmost _______ zeroes.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 1

Since, degree of given polynomial is 5,
so ax5 + bx3 + cx2 + dx + e has atmost 5 zeroes.

Math Olympiad Test: Polynomials - 4 - Question 2

If one zero of the polynomial f (x) = (k2 + 4) x2 + 13x + 4k is reciprocal of the other, then k is equal to _______.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 2

f(x) = (k2 + 4)x2 + 13x + 4k
Now, let a and b be the roots, then according to the question, α = 1/β ⇒ αβ = 1
Now, we know that αβ = 4k/(k2 + 4)

Math Olympiad Test: Polynomials - 4 - Question 3

For x2 + 2x + 5 to be a factor of x4 + αx2 + β, the values of α and β should respectively be _______.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 3

For x2 + 2x + 5 to be a factor of x4 + αx2 + β, remainder should be zero.

Now, remainder should be equal to zero.
∴ α – 6 = 0 and β – 25 = 0
⇒ α = 6 and β = 25

Math Olympiad Test: Polynomials - 4 - Question 4

If α and β are the roots of the equation 2x2 – 7x + 8 = 0, then the equation whose roots are(3α – 4β) and (3β – 4α) is _______.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 4

2x2 – 7x + 8 = 0
α, β are roots of above equation (ax2 + bx + c = 0)
∴ α + β = −b/a​ = 7/2 -(1)
αβ = 8/2​(c/a​) = 4

Roots of other eqn ⇒ (3α − 4β), (3β − 4α)

⇒ sum of roots = 3α − 4β + 3β − 4α
= −α − β = −(α + β)
=(−7/2​) (from (1))
⇒ product of roots ⇒ (3α − 4β)(3β − 4α)
= 9αβ − 12α2 − 12β2 + 16αβ
= 25αβ − 12(α2 + β2)
= 25αβ − 12((α + β)2 − 2αβ)
= 25αβ − 12(α + β)2 + 21αβ
= 49αβ − 12(α + β)2
⇒ 49(4) − 12(7​/2)2
⇒ 49(4) − 12 × 49/4 ​= 49(4 − 3)
⇒ (49)
⇒ quadratic poly ⇒ x2 - (sum of roots)x +(product of roots)
⇒ x2 − (2 − 7​)x + 49
⇒ 2x2 + 7x + 98

Math Olympiad Test: Polynomials - 4 - Question 5

Area of a triangular field is (x4 – 6x3 – 26x2 + 138x – 35) m2 and base of the triangular field is (x2 – 4x + 1) m. Find the height of the triangular field.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 5

Base of the triangular field = (x2 – 4x + 1) m
Area of the triangular field = (1/2) x Base x Height
Now, x4 – 6x3 – 26x2 + 138x – 35 = 1/2 x (x2 − 4x + 1) × Height
⇒ Height = 

∴ Height = 2(x2 – 2x – 35) m

Math Olympiad Test: Polynomials - 4 - Question 6

If α, β be two zeroes of the quadratic polynomial ax2 + bx – c = 0, then find the value of 

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 6

Since, a and b are the zeroes of quadratic equation ax2 + bx – c = 0

Now, 

Math Olympiad Test: Polynomials - 4 - Question 7

Raghav had ₹(6x3 + 2x2 + 3x) and he bought (4x2 + 3) shirts. The price of each shirt is ₹(x + 5). How much money is left with Raghav?

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 7

Total amount Raghav had = ₹(6x3 + 2x2 + 3x)
Cost of one shirt = ₹(x + 5)
Number of shirts he bought = 4x2 + 3
∴ Amount spent by him = ₹(x + 5)(4x2 + 3)
= ₹(4x3 + 20x2 + 3x + 15)
Hence, money left with Raghav
= ₹(6x3 + 2x2 + 3x – 4x3 – 20x2 – 3x – 15)
= ₹(2x3 – 18x2 – 15)

Math Olympiad Test: Polynomials - 4 - Question 8

A rectangular garden of length (2x3 + 5x2 – 7) m has the perimeter (4x3 – 2x2 + 4) m. Find the breadth of the garden.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 8

Length of the garden = (2x3 + 5x2 – 7) m
Perimeter of the garden = 2 × (length + breadth)
∴ 4x3 - 2x2 + 4 = 2(2x3 + 5x2 – 7 + breadth)
⇒ 2x3 – x2 + 2 = (2x3 + 5x2 – 7) + breadth
So, breadth of the rectangle
= 2x3 – x2 + 2 – 2x3 – 5x2 + 7 = (–6x2 + 9) m

Math Olympiad Test: Polynomials - 4 - Question 9

Length and breadth of a rectangular park are (3x2 + 2x) m and (2x3 – 3) m respectively. Find the area of the park, when x = 3.

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 9

Length of rectangular park = (3x2 + 2x) m
Breadth of rectangular park = (2x3 – 3) m
Area of park = length × breadth
= (3x2 + 2x) (2x3 – 3) = (6x5 + 4x4 – 9x2 – 6x) m
For x = 3, 6x5 + 4x4 – 9x2 – 6x
= 6 × 243 + 4 × 81 – 9 × 9 – 6 × 3 = 1683
Hence, area of park = 1683 m2

Math Olympiad Test: Polynomials - 4 - Question 10

Two different container contains (2x3 + 2x2 + 3x + 3) L and (4x3 – 2x2 + 6x – 3) L water. What is biggest measure that can measure both quantities exactly?

Detailed Solution for Math Olympiad Test: Polynomials - 4 - Question 10

Capacity of both the containers is (2x3 + 2x2 + 3x + 3) L and (4x3 – 2x2 + 6x – 3) L
i.e., (2x2 + 3)(x + 1) L and (2x2 + 3)(2x – 1) L
Required measure is the H.C.F. of capacity of both the containers i.e., (2x2 + 3) L

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