Class 10 Case based problem of Chapter 2 Polynomials 3

Class 10 Chapter 2(Poloynomials)

Case based 4: The pictures are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures,their curve represents an efficient method of carrying load and so can be found in bridges and in architecture in a variety of forms.(Class 10 Case based problem of Chapter 2 Polynomials 3)

The pictures are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures,their curve represents an efficient method of carrying load and so can be found in bridges and in architecture in a variety of forms.(Class 10 Case based problem of Chapter 2 Polynomials 3)
The pictures are few natural examples of parabolic shape which is represented by a quadratic

The pictures are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures,their curve represents an efficient method of carrying load and so can be found in bridges and in architecture in a variety of forms.(Class 10 Case based problem of Chapter 2 Polynomials 3)

(A) In the standard form of quadratic polynomial, ax² + bx + c, a, b and c are :

(a) All are real numbers.

(b) All are rational numbers.

(c) ‘a’ is a non-zero real number and, b and c are any real numbers.

(d) All are integers.

(B) If the root of quadratic polynomialare equal, where the discrimnant D = b² – 4ac, then:

(a) D > 0                (b) D < 0

(c) D ≥ 0                 (d) D = 0

(C) If α and 1/α are the zeroes of the quadratic polynomial 2x² – x + 8k, then k is:

(a) 4                          (b) 1/4

(c) -1/4                      (d) 2

(D) The graph of x² + 1 = 0:

(a) Intersects x-axis at two distinct points

(b) Touches x – axis at a point.

(c) Neither touches nor intersects x – axis

(d) Either touches or intersects x- axis.

(E) If the sum of the roots is -p and product of the roots is -1/p, then then the quadratic polynomial is

(a)  k(-px² + x/p + 1)

(b)  k(px² – x/p – 1)

(c)  k(x² + px – 1/p)

(d)  k(x² – px + 1/p)

Solution:

(A) Answer  (c) ‘a’ is non-zero real number and b and c are any real numbers.

Explatnation: The standard form of the quadratic polynomial is ax² + bx + c, where a, b, c are real numbers and a ≠ 0

(B) Answer (d) D = 0

(C) Answer(b) k = 1/4

Explanation: 2x² – x + 8k

Roots of polynomial = α, 1/α

Product of the roots = Constant term/ Coefficient of x²

⇒ α× 1/α = 8k/2

⇒ k = 1/4

(D) Answer (c) Neither touches nor intersects x – axis,

Explanation: x² + 1 = 0

⇒ x² = -1

It has no real roots

Hence the graph of equation never touches and intersect the x- axis

(E) Answer (c) k(x² + px – 1/p)

Explanation: sum of the roots = -p

product of the roots = -1/p

Quadratic equation

= k[x² – (Sum of the roots) x + (Productof the roots)]

= k[x² – (-p)x +(-1/p) ]

= k[x² + px -1/p]

Case Based 5:

Government of India allotted some funds to the National Disaster Relief Fund to help the families of flood affected village. The fund alloted is represented by 3x^4 -14x^3+12x^2 +6x+5. The fund is equally divided between each of the families of that village. Each family received an amount of x^2-4x-1. After distribution, the amount left was 32x + 12. Which was used to construct a primary health centre in each village.

The pictures are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures,their curve represents an efficient method of carrying load and so can be found in bridges and in architecture in a variety of forms.(Class 10 Case based problem of Chapter 2 Polynomials 3)

(A) If the product of the zeros of the polynomial ax^2-6x-6 is 4, then then the value of a is:

(a) -\frac{3}{2}          (b) -\frac{2}{3}

(c) -1                                  (d) 0

(B) If α and β are zeroes of the quadratic polynomial f(x) = 2x^2+4x+k such that α² + β² + αβ = 21/4, The value of k is

(a) -2                             (b) 0

(c) -\frac{5}{2}      (d) \frac{5}{2}

(C) If the sum of squares of zeroes of the polynomial x^2-8x+k is 40, the value of k is

(a) 2                             (b) 4

(c) 8                              (d) 12

(D) The zeroes of the quadratic polynomial a(x^2+1)-x(a^2+1) are:

(a) -a, -1/a                   (b) a, 1/a

(c) -1, a                         (d) 1, -a

(E) If one of the zeroes of the quadratic polynomial x^2+3x+a is 2, then the value of a is:

(a) 10                            (b) -10

(c) -7                              (d) -2

Solution:

(A) Answer (a) -\frac{3}{2}

Explanation: Given ax^2-6x-6

Product of the zeroes  = \frac{c}{a}

\Rightarrow 4 = -\frac{6}{a}

\Rightarrow a = -\frac{6}{4}

\Rightarrow a = -\frac{3}{2}

(B) Answer (c) -5/2

Explanation:Given f(x) = 2x^2+4x+k

Since, \alpha +\beta = -\frac{b}{a}= -\frac{4}{2}

=  -2

\alpha.\beta = \frac{c}{a}= \frac{k}{2}

Given that, α² + β² + αβ = 21/4

add an substract αβ

α² + β² + αβ + αβ- αβ = 21/4

⇒α² + β² + 2αβ – αβ = 21/4

⇒(α + β)²-αβ = 21/4

\Rightarrow (-2)^2-\frac{k}{2}=\frac{21}{4}

\Rightarrow 4 -\frac{k}{2} = \frac{21}{4}

\Rightarrow 4-\frac{21}{4} = \frac{k}{2}

\Rightarrow \frac{16-21}{4}=\frac{k}{2}

\Rightarrow -\frac{5}{4} = \frac{k}{2}

\Rightarrow k = -\frac{5}{2}

(C) Answer (d) 12

Explanation: Let zeroes of the equation x^2-8x+k are α and β

Then, α + β = -b/a = -(8)/1 =8

α.β = c/a = k/1 = k

Given that \alpha^2 + \beta^2 =40

\alpha^2+\beta^2 +2\alpha.\beta - 2\alpha.\beta =40

\Rightarrow (\alpha +\beta)^2 -2\alpha.\beta = 40

\Rightarrow (8)^2 -2k =40

\Rightarrow  64- 40 = 2k

\Rightarrow 2k = 24

\Rightarrow k = 12

(D) Answer (b) a, 1/a

Explanation: given that a(x^2+1)-x(a^2+1)

\Rightarrow ax^2+a-a^2x-x

\Rightarrow ax^2-a^2x+a-x

= ax(x-a) - 1(x-a)

= (x-a)(ax-1)

Hence, zeroes  are = a, 1/a

(E)Answer (b)  -10

Let zeroes of the equation x^2+3x+a are α and β

Given \alpha = 2

\alpha +\beta =-\frac{b}{a}= -\frac{3}{1}

\Rightarrow 2 +\beta = -3

\Rightarrow \beta = -5

and \alpha.\beta =\frac{c}{a}

\Rightarrow 2\times -5 =\frac{a}{1}

\Rightarrow -10 =a

\Rightarrow a = -10

Class 10 Case based problem of Chapter 2 Polynomials 1
SLV-3 was successfully launched on july 18, 1980 from Shriharikota Range (SHAR), when Rohini satellite,

Class 10 Case based problem of Chapter 2 Polynomials 2

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