## Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers

Extra Questions for Class 10 Maths Chapter 2 Polynomials. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

### Polynomials Class 10 Extra Questions Very Short Answer Type

Question 1.

The graph of a quadratic polynomial x^{2} – 3x – 4 is a parabola. Determine the opening of parabola.

Answer:

∵ In x^{2} – 3x – 4, the Coefficient of x^{2} is 1 and 1 > 0.

∴ The parabola opens upwards.

Question 2.

If p(x) = x^{2} + 5x + 2, then find p(3) + p(2) + p(0).

Answer:

p(3) = 3^{2} + 5(3)+ 2 = 26

p(2) = 2^{2} + 5(2) + 2 = 16

p(0) = 0^{2} + 5(0) + 2 = 2

⇒ p(3) + p(2) + p(0) = 26 + 16 + 2 = 44

Question 3.

The graph of y = p(x) is shown in the figure below. How many zeroes does p(x) have?

Answer:

Since, the curve (graph) of p(x) is intersecting the x-axis at three points, y = p(x) has 3 zeroes.

Question 4.

The coefficient of x and the constant term in a linear polynomial are 5 and – 3 respectively, find its zero.

Answer:

∵ The zero of a linear polynomial = – \(\frac{\text { Constant term }}{\text { Coefficient of } x}\)

∴ The zero of the given linear polynomial

= – \(\frac{(-3)}{5}=\frac{3}{5}\)

Question 5.

What is the value of p(x) = x^{2} – 3x – 4 at x = -1 ?

Answer:

We have: p(x) = x^{2} – 3x – 4

∴ P(- 1) = (- 1)^{2} – {3(- 1)} – 4 = 1 + 3 – 4

= 0

Question 6.

If the polynomial p(x) is divisible by (x – 4) and 2 is a zero of p(x), then write the corresponding polynomial.

Answer:

Here, p(x) is divisible by (x – 4) and also 2 is a zero of p(x), therefore p(x) is divisible by (x – 4) and (x – 2)

Thus, the required polynomial p(x) = (x – 4) (x – 2) = x^{2} – 6x + 8.

Question 7.

If (α – β), α, (α + β) are zeroes of the polynomial p(x) = 2x^{3} – 16x^{2} + 15x – 2, then find the value of α

Answer:

Sum of zeroes = – \(\frac{\text { Coeff. of } x^2}{\text { Coeff. of } x^3}\)

⇒ (α – β), α, (α + β) = \(\frac{-(-16)}{2}\)

⇒ 3α = 8

⇒ α = \(\frac{8}{3}\)

Question 8.

What is the zero of 2x + 3?

Answer:

∵ The zero of a linear polynomial = – \(\frac{\text { Constant term }}{\text { Coefficient of } x}\)

∴ The zero of 2x + 3 = – \(\frac{3}{2}\)

Question 9.

Find the value of p for which the poloynomial x3 + 4×2 – px + 8 is exactly divisible by (x – 2).

Here p(x) = x^{3} + 4x^{2} – px + 8

∵ (x – 2) divides p(x), exactly

⇒ P(2) = 0

⇒ (2)^{3} + 4 (2)^{2} – p(2) + 8 = 0

2p = 32 ⇒ p = 16

Question 10.

If α and β are zereos of the polynomial 2x^{2} – 5x + 7, then find the value of α^{-1} + β^{-1}.

Answer:

Here p(x) = 2x^{2} – 5x + 7

α, β are zeroes of p(x)

Question 11.

If p and q are the roots of ax^{2} – bx + c = 0, a ≠ 0, then find the value of p + q.

Answer:

Here, p and q are the roots of ax^{2} – bx + c = 0.

Sum of roots = \(\frac{-b}{a}\)

∴ p + q = \(\frac{-b}{a}\)

Question 12.

If – 1 is a zero of quadratic polynomial, p(x) = kx^{2} – 5x – 4, then find the value of k.

Answer:

Here p(x) = kx^{2} – 5x – 4

Since – 1 is a zero of p(x)

∴ P(- 1) = o

⇒ k (-1)^{2} – 5(- 1) – 4 = 0

⇒ k + 5 – 4 = 0

⇒ k = – 1

### Polynomials Class 10 Extra Questions Short Answer Type-1

Question 1.

Find the quadratic polynomial whose sum of zeroes is 8 and their product is 12. Hence find zeroes of polynomial.

Answer:

Let, α, β be zereos of polynomial.

Now, here α + β = 8, αβ = 12.

Required polynomial

p(x) = k{x^{2} – (α + β) x + α β),

k is a constant.

⇒ p(x) = k{x^{2} – 8x + 12}

In particular, taking k = 1

Reqd. polynomial = x^{2} – 8x + 12

Now, p(x) = x^{2} – 8x + 12

= x^{2} – 6x – 2x + 12

= x(x – 6) – 2(x – 6)

= (x – 6) (x – 2)

∴ p(x) = o

⇒ x = 6, 2

Thus zeroes of polynomial are 6 and 2.

Question 2.

Check whether x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

Answer:

Let p(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2,

g(x) = x^{2} + 3x + 1

Next we divide p(x) by g(x).

Using division algorithm

3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 = (x^{2} + 3x + 1) (3x^{2} – 4x + 2) + 0

= (x^{2} + 3x + 1) (3x^{2} – 4x + 2)

Clearly, as remainder is 0, so the divisor x^{2} + 3x + 1 appear on R.H.S as factor of 3x^{4} + 5x^{3} 7x^{2} + 2x + 2.

Question 3.

If one zero of the polynomial (a^{2} + 9)x^{2} + 13x + 6a is reciprocal of the other, find the value of a.

Answer:

Let α, \(\frac{1}{\alpha}\) be the zeroes of (a^{2} + 9)x^{2} + 13x + 6a

Question 4.

If α and β are zeroes of x^{2} + 7x + 12, then find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) – 2αβ

Answer:

Here α + β = – 7, αβ = 12

Question 5.

Find α^{-1} + β^{-1}, if α and β are zereos of the polynomial 9x^{2} – 3x – 2.

Answer:

Since α and β are zereos of p(x) = 9x^{2} – 3x – 2

Question 6.

Find whether 2x^{3} – 1 is a factor of 2x^{5} + 10x^{4} + 2x^{2} + 5x + 1 or not.

Answer:

Since r(x) ≠ 0

∴ 2x^{3} – 1 is not a factor of given polynomial.

Question 7.

If α, β, γ are zeroes of the polynomial f(x) = x^{3} – 3x^{2} + 7x – 12, then find the value of (αβ)^{-1} + (βγ)^{-1} + (γα)^{-1}.

Answer:

Question 8.

For what value of k is the polynomial x^{3} + kx^{2} + 3x – 18 is exactly divisible by (x – 3).

Answer:

If p(x) = x^{3} + kx^{2} + 3x – 18 is exactly divisible by (x – 3)

⇒ p(3) = 0 ⇒ (3)^{3} + k(3)^{2} + 3(3) – 18 = 0

⇒ 9k = – 18 ⇒ k = – 2

### Polynomials Class 10 Extra Questions Short Answer Type-2

Question 1.

Find all the zeroes of 2x^{4} – 13x^{3} + 19x^{2} + 7x – 3, if you know that two of its zeroes are 2 + √3 and 2 – √3. [C.B.S.E. 2019]

Answer:

Given, x = (2 + √3) and x = (2 – √3) are zeroes of p(x) = 2x^{4} – 13x^{3} + 19x^{2} + 7x – 3

∴ (x – (2 + √3)) (x – (2 – √3)) is factor of p(x).

⇒ (x – 2 – √3) (x – 2 + √3) is factor of p(x).

= (x – 2)^{2} – (√5)^{2}

= x^{2} – 4x + 4 – 3

= x^{2} – 4x + 1

Now, we divide p(x) by x^{2} – 4x + 1

Now p(x) = (x^{2} – 4x + 1) (2x^{2} – 5x – 3)

∴ Other zeroes are given by

2x^{2} – 5x – 3 = 0

⇒ 2x^{2} – 6x + x – 3 = 0

⇒ 2x (x – 3) + 1 (x – 3) = 0

⇒ (2x + 1) (x – 3) = 0

⇒ 2x + 1 = 0 or x – 3 = 0

x = – \(\frac{1}{2}\), 3

∴ Zeroes of given polynomial are

– \(\frac{1}{2}\), 3, (2 + √3), (2 – √3)

Question 2.

Find all the zeroes of 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2, if it is given that two of its zeroes are 1 and \(\frac{1}{2}\). [CBSE 2019 (C) Set-B]

Answer:

Given: x = 1, x = \(\frac{1}{2}\) are zeroes of p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2

∴ (x – 1) and (x – \(\frac{1}{2}\)) or (2x – 1) are factor of p(x).

⇒ (x – 1) (2x – 1) = 2x^{2} – 3x + 1 is a factor of p(x).

Next, we divide p(x) by 2x^{2} – 3x + 1.

∴ p(x) = (2x^{2} – 3x + 1) (x^{2} – 2)

∴ Other zeroes are given by (x^{2} – 2) = 0

⇒ x = ±√2

∴ Zeroes of p(x) are – √2, \(\frac{1}{2}\), 1, √2.

Question 3.

Find all zeros of the polynomial 3x^{3} + 10x^{2} – 9x – 4, if one of its zero is 1. [CBSE 2019]

Answer:

Let p(x) = 3x^{3} + 10x^{2} – 9x – 4

Since, 1 is a zero of p(x)

Therefore, (x – 1) is a factor of p(x)

Dividing p(x) by (x – 1), we have:

∴ By Division Algorithm,

p(x) = (3x^{2} + 13x + 4)(x – 1)

Zeroes of p(x) are given by p(x) = 0

⇒ (3x^{2} + 13x + 4)(x – 1) = 0

⇒ (3x^{2} + 12x + x + 4)(x – 1) = 0

⇒ {3x(x + 4) + 1 (x + 4)}(x – 1) = 0

⇒ (x + 4 )(3x + 1) (x – 1) = 0

⇒ x + 4 = 0

or 3x + 1 = 0 or x – 1 = 0

⇒ x = – 4, – \(\frac{1}{3}\), 1

∴ Zeroes of p(x) are – 4, – \(\frac{1}{3}\), 1.

Question 4.

Divide the polynomial 3x^{3} – 6x^{2} – 20x + 14 by the polynomial x^{2} – 5x + 6 and verify the division algorithm. [CBSE Delhi 2016]

Answer:

By division algorithm,

3x^{3} – 6x^{2} – 20x + 14 = (x^{2} – 5x + 6)(3x + 9) + (7x – 40)

or, p(x) = q(x) g(x) + r(x)

Question 5.

If α and β are the zeroes of a quadratic polynomial x^{2} – x – 2 then find the value of \(\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\).

Answer:

Comparing x^{2} – x – 2 with ax^{2} + bx + c, we have a = 1, b = – 1, c = -2

[∵ (α – β)^{2} = (α + β)^{2} – 4αβ

= (1)^{2} – 4 (- 2) = 1 + 8 = 9

∴ α – β = √9

⇒ (α – β) = ± 3]

Question 6.

On dividing p(x) by a polynomial x – 1 – x^{2}, the quotient and remainder were (x – 2) and 3 respectively. Find p(x).

Answer:

Here,

dividend = p(x)

Divisor, g(x) = (x – 1 – x^{2})

Quotient, q(x) = (x – 2)

Remainder, r(x) = 3

∵ Dividend = [Divisor × Quotient] + Remainder

∴ P(x) = [g(x) × q(x)] + r(x)

= [(x – 1 – x^{2}) (x – 2)] + 3

= [x^{2} – x – x^{3} – 2x + 2 + 2x^{2}] + 3

= 3x^{2} – 3x – x^{3} + 2 + 3

= – x^{3} + 3x^{2} – 3x + 5

Question 7.

Find the zeroes of the quadratic polynomial 5x^{2} – 4 – 8x and verify the relationship between the zeroes and the coefficients of the polynomial.

Answer:

p(x) = 5x^{2} – 4 – 8x = 5x^{2} – 8x – 4

= 5x^{2} – 10x + 2x – 4

= 5x(x – 2) + 2(x – 2)

= (x – 2) (5x + 2)

= 5(x – 2) (x + \(\frac{2}{5}\))

∴ Zeroes of p(x) are 2 and – \(\frac{2}{5}\)

Relationship Verification:

### Polynomials Class 10 Extra Questions Long Answer Type 1

Question 1.

If the polynomial x^{4} + 2x^{3} + 8x^{2} + 12x + 18 is divided by another polynomial x^{2} + 5, the remainder comes out to be px + q. Find the values of p and q.

Answer:

∴ Remainder = 2x + 3

Comparing 2x + 3 with px + q, we have

p = 2 and q = 3

Question 2.

If α, β, γ are the zeroes of p(x) = x^{3} – 7x^{2} + 11x – 7 find the value of

(i) α^{2} + β^{2} + γ^{2}

(ii) α^{3} + β^{3} + γ^{3}

Answer:

Comparing p(x) = x^{3} – 7x^{2} + 11x – 7 with standard cubic polynomial ax^{3} + bx^{2} + cx + d, we have

a = 1, b = – 7, c = 11, d = – 7

(i) Now, since (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2(αβ + βγ + γα)

⇒ α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} – 2(αβ + βγ + γα)

= (7)^{2} – 2(11) = 49 – 22 = 27

(ii) Since

α^{3} + β^{3} + γ^{3} – 3αβγ = (α^{2} + β^{2} + γ^{2} – αβ – βγ – γα)

(α + β + γ)

= {(α + β + γ)^{2} – 2(αβ + βγ + γα) – (αβ + βγ + γα)}(α + β + γ)

⇒ α^{3} + β^{3} + γ^{3} = {(α + β + γ)^{2} – 3(αβ + βγ + γα)} (α + β + γ) + 3αβγ

= {7^{2} – 3(11)} 7 + 3 × 7

= (49 – 33) 7 + 21

= 16 × 7 + 21 = 133.

Question 3.

Find the quadratic polynomial whose zeroes are 1 and – 3. Verify the relation between the coefficients and the zeroes of the polynomial.

Answer:

∵ The given zeroes are 1 and – 3.

∴ Sum of the zeroes = 1 + (- 3) = – 2

Product of the zeroes = 1 × (- 3) = – 3

A quadratic polynomial p(x) is given by

x^{2} – (sum of the zeroes) x + (product of the zeroes)

∴ The required polynomial is

x^{2} – (- 2)x + (-3)

⇒ x^{2} + 2x – 3

Verification of relationship

∵ Sum of the zeroes = – \(\frac{\text { [coeff. of } x]}{\text { Coeff. of } x^2}\)

∴ 1 + (- 3) = \(\frac{-[2]}{1}\)

⇒ – 2 = – 2

i.e., LHS = RHS

⇒ The sum of zeroes is verified.

∵ Product of the zeroes = \(\frac{\text { [Constant term] }}{\text { Coefficient of } x^2}\)

∴ 1 × (- 3) = \(\frac{[-3]}{1}\)

⇒ – 3 = – 3

i.e., L.H.S. = R.H.S.

⇒ The product of zeroes is verified.

Question 4.

What must be added to 6x^{5} + 5x^{4} + 11x^{3} – 3x^{2} + x + 1, so that the polynomial so obtained is exactly divisible by 3x^{2} – 2x + 4?

Answer:

Therefore, we must add -(- 17x + 13),

i.e., 17x – 13.

Question 5.

Find the value of b for which the polynomial 2x^{3} + 9x^{2} – x – b is divisible by 2x + 3.

Answer:

If the polynomial 2x^{3} + 9x^{2} – x – b is divisible by 2x + 3, then the remainder must be zero.

So, 15 – b = 0

⇒ b = 15

Question 6.

Find the zeroes of a cubic polynomial p(x) = 3x^{3} – 5x^{2} – 11x – 3 when it is given that product of two of its zeroes is -1.

Answer:

Here, p(x) = 3x^{3} – 5x^{2} – 11x – 3

On comparing p(x) with ax^{3} + bx^{2} + cx + d, we have;

a = 3, b = -5, c = -11, d = -3

Let α β γ be the zeroes of the given polynomial.

⇒ 3α^{2} – 3 = 8α

⇒ 3α^{2} – 8α – 3 = 0

⇒ 3α^{2} – 9α + α – 3 = 0

⇒ 3α(α – 3) + 1 (α – 3) = 0

⇒ (α – 3) (3α + 1) = 0

### Polynomials Class 10 Extra Questions HOTS

Question 1.

If p and q are the zeroes of x^{2} + px + q, then find the values of p and q.

Answer:

p, q are the zeroes of x^{2} + px + q.

⇒ p + q = – p …………… (i)

and pq = q ……… (ii)

⇒ pq – q = 0

⇒ (p – 1)q = 0

⇒ either p = 1 or q = 0

When p = 1

⇒ 1 + q = – 1

⇒ q = – 2

When 9 = 0

⇒ p + 0 = – p

⇒ 2p = 0

⇒ p = 0

∴ p = 1, q = – 2 or p = q = 0.

Question 2.

If the zeroes of p(x) = ax^{3} + 3bx^{2} + 3cx + d are in A.P., prove that 2b^{3} – 3abc + a^{2}d = 0

Answer:

Let α β γ be zeroes of p(x) = ax^{3} + 3bx^{2} + 3cx + d

But α, β, γ are in A.P.

So, take α = m – n, β = m, γ = m + n.

Multiplying throughout by a^{3}, we get – b^{3} + 3b (b^{2} – ac) = – a^{2}d

⇒ – b^{3} + 3b^{3} – 3abc + a^{2}d = 0

or 2b^{3} – 3abc + a^{2}d = 0

Note: Three numbers are said to be in Arithmetic Progression (A.P.). If each number is obtained by adding a constant to the previous number e.g. 3, 7, 11 are in A.P. Generally we take three numbers in A.P. as a – d, a, a + d.

It would be discussed in detail in Second term.

Question 3.

Find a relation between p and q, if one zero of x^{2} + px + q is 37 times the other.

Answer:

Let α, β be the zeroes of the polynomial

x^{2} + px + q

then α + β = – p, αβ = q

α = 37β (given)

∴ α + β = – p

⇒ 38β = – p

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.

The graph of p(x) is shown alongside. The number of zeroes of p(x) are:

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(a) 1

Question 2.

If one zero of polynomial (k^{2} + 16) x^{2} + 13x + 8k is reciprocal of the other then k is equal to

(a) – 4

(b) +4

(c) 8

(d) 2

Answer:

(b) +4

Question 3.

If α and β are zeroes of the polynomial p(x) = x^{2} + mx + n, then a polynomial whose zeroes are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is given by

(a) nx^{2} + mx + 1

(b) mx^{2} + x + n

(c) x^{2} + nx + m

(d) x^{2} – mx + n

Answer:

(a) nx^{2} + mx + 1

Question 4.

If the graph of y = p(x) does not cut the x-axis at any point, then polynomial has

(a) one zero

(b) two zeroes

(c) no zeroes

(d) infinite no. of zeroes

Answer:

(c) no zeroes

Question 5.

Sum of the zeroes of the polynomial p(x) = – 3x^{2} + k is

(a) \(\frac{k}{3}\)

(b) \(\frac{-k}{3}\)

(c) 0

(d) k

Answer:

(c) 0

Question 6.

If x – 1 is a factor of p(x) = kx^{2} – √2x + 1, then the value of k is

(a) √2 – 1

(b) – √2 + 1

(c) – 1 – √2

(d) 1 + √2

Answer:

(a) √2 – 1

Question 7.

Number of zeroes of a polynomial of degree n is

(a) equal to n

(b) less than n

(c) greater than n

(d) less than or equal to n

Answer:

(d) less than or equal to n

Question 8.

Zeroes of the polynomial p(x) = 2x^{2} – 9 – 3x are

(a) 3, \(\frac{3}{2}\)

(b) –\(\frac{3}{2}\) , 3

(c) 2, 3

(d) 9, \(\frac{3}{2}\)

Answer:

(b) –\(\frac{3}{2}\) , 3

Question 9.

If (α – β), α, (α + β) are zeroes of the polynomial p(x) = 2x^{3} – 16x^{2} + 15x – 2, value of α is

(a) 8

(b) 0

(c) \(\frac{3}{8}\)

(d) \(\frac{8}{3}\)

Answer:

(d) \(\frac{8}{3}\)

Question 10.

If n represents number of real zeroes for the polynomial ax3 + bx2 + cx + d then which of the following inequality is valid

(a) 0 < n < 3

(b) 0 < n < 3

(c) 0 < n < 3

(d) 0 < n < 3

Answer:

(c) 0 < n < 3

Question 11.

Number of quadratic polynomials having – 2 and – 5 as their two zeroes is :

(a) one

(b) two

(c) three

(d) infinite

Answer:

(d) infinite

Question 12.

If α, β, γ are zeroes of the polynomial p (x) such that α + β + γ = 2, αβ + βγ + γα = 5, αβγ = – 7, then p(x) is:

(a) x^{3} – 2x^{2} + 5x – 7

(b) x^{3} + 2x^{2} – 5x + 7

(c) x^{3} – 2x^{2} – 5x – 7

(d) x^{3} – 2x^{2} + 5x + 7

Answer:

(d) x^{3} – 2x^{2} + 5x + 7

Question 13.

If the sum of products of zeroes taken two at a time of polynomial p(x) = x3 – 5×2 + cx + 8 is 2 then the value of c is

(a) 2

(b) – 2

(c) 8

(d) – 5

Answer:

(a) 2

Question 14.

The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x) there are polynomial q(x) and r(x) such that p(x) = g(x) q{x) + r(x), where r(x) is

(a) either = 0 or deg. r(x) < deg. g(x) (b) either = 0 or deg. r(x) > deg. g(x)

(c) a linear polynomial or deg. r(x) = deg. g(x)

(d) either = 0 or deg. r(x) < deg. g(x)

Answer:

(d) either = 0 or deg. r(x) < deg. g(x)

Question 15.

If divisor, quotient and remainder are x +

(a) 3x^{2} + x + 1

(b) 3x^{2} – x – 1

(c) 3x^{2} + x – 1

(d) 3x^{2} – x + 1

Answer:

(c) 3x^{2} + x – 1

Fill in the blanks

Question 1.

The zeroes of the polynomial x^{2} – 49 are ______________ .

Answer:

± 7

Question 2.

The quadratic polynomial, whose sum and product of zeroes are 4 and – 5 respectively is _____________ .

Answer:

k(x^{2} – 4x – 5)

Question 3.

The value of the polynomial p(x) = 4x^{2} – 7 at x = – 2 is _______________ .

Answer:

9

Question 4.

Product of zeroes of a polynomial p(x) = 6x^{2} – 7x – 3 is _______________ .

Answer:

– \(\frac{1}{2}\)

Question 5.

If one zero of 3x^{2} – 8x +2k + 1 is seven times the other, then k is _________________.

Answer:

\(\frac{2}{3}\)

Question 6.

The degree of the constant polynomial is _______________ .

Answer:

zero

Question 7.

A real number k is a zero of the polynomial p(x) if and only if _______________.

Answer:

p(k) = 0

Question 8.

The shape of the graph of a cubic polynomial is _______________ .

Answer:

not fixed

Question 9.

If α, β and γ are the zeroes of the cubic polynomial px^{3} + qx^{2} + rx + s; a ≠ 0, then α + β + γ = ____________ , αβ + βγ + γα = _____________ and αβγ = ___________.

Answer:

\(\frac{-q}{p}, \frac{r}{p}, \frac{-s}{p}\)

Question 10.

The standard form of the polynomial x^{3} – x^{6} + x^{5} + 2x^{2} – x^{4} – 5 is ______________ .

Answer:

x^{6} + x^{5} – x^{4} + x^{3} + 2x^{2} – 5 or – 5 + 2x^{2} + x^{3} – x^{4} + x^{5} – x^{6}

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