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

(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.

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

(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) 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

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

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

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,

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Class 10 Chapter 2(Poloynomials)

Solution:

Case Based 5:

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