RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomials |

E×ercise |
Miscellaneous Exercise |

Number of Questions Solved |
23 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Miscellaneous Exercise

**Multiple Choice Questions**

Question 1.

If one zero of polynomial f(x) = 5x^{2} + 13x + k is inverse of (RBSESolutions.com) other, then k will be :

(A) 0

(B) \(\frac { 1 }{ 5 }\)

(C) 5

(D) 6

Solution

Question 2.

Zeros of (RBSESolutions.com) polynomial x^{2} – x – 6 are :

(A) 1, 6

(B) 2, -3

(C) 3, -2

(D) 1, -6

Solution

f(x) = x^{2} – x – 6

= x^{2} – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3)(x + 2)

for zeros f(x) = 0

⇒ (x – 3)(x + 2) = 0

⇒ x = 3 or x = -2

Hence, option (C) is correct.

Question 3.

If 3 is a zero of polynomial 2x^{2} + x + k, then (RBSESolutions.com) value of k will be

(A) 12

(B) 21

(C) 24

(D) -21

Solution

Let f(x) = 2x^{2} + x + k

One zero is 3

f(3) = 0

⇒ 2(3)^{2} + 3 + k = 0

⇒ 2 × 9 + 3 + k = 0

⇒ 18 + 3 + k = 0

⇒ k = -21

Hence, option (D) is correct.

Question 4.

If α, β are zeros of (RBSESolutions.com) polynomial x^{2} – p(x + 1) – c such that (α + 1)(β + 1) = 0, then c will

(A) 0

(B) -1

(C) 1

(D) 2

Solution

Question 5.

If roots of quadratic (RBSESolutions.com) equation x^{2} – kx + 4 = 0 then k will be

(A) 2

(B) 1

(C) 4

(D) 3

Solution

Given equation x^{2} – kx + 4 = 0

Comparing it with ax^{2} + bx + c = 0

We get a = 1, b= -k and c = 4

D = b^{2} – 4 ac = (-k)^{2} – 4 × 1 × 4 = k^{2} – 16

Roots are same

D = 0

⇒ k^{2} – 16 = 0

⇒ k^{2} = 16

⇒ k = ±√16

⇒ k = ± 4

Hence, option (C) is correct.

Question 6.

If x = 1, is common root of (RBSESolutions.com) equation ax^{2} + ax + 3 = 0 and x^{2} + x + b = 0, then ab will be:

(A) 1

(B) 3.5

(C) 6

(D) 3

Solution

Putting x = 1 in equation

ax^{2} + ax + 3 = 0

⇒ a(1)^{2} + a(1) + 3 = 0

⇒ a + a + 3 = 0

⇒ 2a = – 3

⇒ a = \(\frac { -3 }{ 2 }\)

and putting x = 1 in equation

x^{2} + x + b = 0

⇒ (1)^{2} + (1) + b = 0

⇒ 1 + 1 + b = 0

⇒ b = -2

ab = \(\frac { -3 }{ 2 }\) × -2 = 3

Hence, option (D) is correct.

Question 7.

Discriminant of quadratic (RBSESolutions.com) equation 3√3 x^{2} + 10x + √3 = 0

(A) 10

(B) 64

(C) 46

(D) 30

Solution

Comparing 3√3x^{2} + 10x + √3 by ax^{2} + bx + c = 0,

a = √3, b = 10 and c = √3

Discriminant(D) = b^{2} – 4ac

= (10)^{2} – 4 × 3√3 × √3

= 100 – 4 × 3 × 3

= 100 – 36

= 64

Hence, option (B) is correct.

Question 8.

Nature of roots of quadratic (RBSESolutions.com) equation 4x^{2} – 12x – 9 = 0 is :

(A) Real and same

(B) Real and distinct

(C) Imgiary and same

(D) Imaginary and distinct

Solution

Given equations 4x^{2} – 12x – 9 = 0

Where a = 4, b = -12, c = -9

Discriminant (D) = b^{2} – 4ac

= (-12)^{2} – 4 × 4 × (-9)

= 144 + 144

= 288 > 0

Hence, option (B) is correct.

Question 9.

H.C.F. of (RBSESolutions.com) expressions 8a^{2}b^{2}c and 20ab^{3}c^{2} is:

(A) 4ab^{2}c

(B) 4abc

(C) 40a^{2}b^{3}c^{2}

(D) 40abc

Solution

8a^{2}b^{2}c = 2 × 2 × 2 × a^{2} × b^{2} × c = 2^{3} × a^{2} × b^{2 }× c

and 20ab^{3}c^{3} = 2 × 2 × 5 × a × b^{3} × c^{3} = 2^{2} × 5 × a × b^{3} × c^{3}

Product of common least power = 2^{2} × a × b^{2} × c = 4ab^{2}c

Hence, option (A) is correct.

Question 10.

Find L.C. M. of expressions x^{2} – 1 and x^{2} + 2x + 1

(A) x + 1

(B) (x^{2} – 1) (x – 1)

(C) (x – 1) (x + 1)^{2}

(D) (x^{2} – 1) (x + 1)

Solution

x^{2} – 1 = (x – 1) (x + 1)

and x^{2} + 2x + 1 = (x + 1) (x + 1) = (x + 1)^{2}

Product of highest powers = (x – 1)(x + 1)^{2}

Hence, option (C) is correct.

Question 11.

L.C.M. of (RBSESolutions.com) expressions 6x^{2}y^{4} and 10xy^{2}, then 30x^{2}y^{4} H.C.F. will be:

(A) 6x^{2}y^{2}

(B) 2xy^{2}

(C) 10x^{2}y^{2}

(D) 60x^{3}y^{6}

Solution :

Question 12.

To find roots of quadratic (RBSESolutions.com) equation ax^{2} + bx + c = 0 write Shridhar Acharya formula.

Solution

Question 13.

Write nature of roots by general form of discriminant of equation ax^{2} + bx + c = 0.

Solution

Nature of roots of quadratic equation ax^{2} + bx + c = 0, a 0 depends on its discriminant value, D = b^{2} – 4ac

(i) If D = b^{2} – 4ac > 0, then roots will be real and distinct.

If α and β are roots of the equation then

Question 14.

Find zeros of quadratic (RBSESolutions.com) equation 2x^{2} – 8x + 6 and test the relationship between zeros and coefficient.

Solution

Given polynamial

f(x) = 2x^{2} – 8x + 6

= 2x^{2} – 6x – 2x + 6

= 2x(x – 3) – 2(x – 3)

= (x – 3) (2x – 2)

To find zeros of polynomial f(x), f(x) = 0.

⇒ f(x) = 0

⇒ (x – 3)(2x – 2) = 0

⇒ x – 3 = 0 or 2x – 2 = 0

⇒ x – 3 or x = 1

Zeros of polynomial 2x^{2} – 8x + 6 are 1 and 3.

Relation between zeros and coefficient:

Sum of zeros = 1 + 3 = 4

Product of zeros = 1 × 3 = 3

comparing (RBSESolutions.com) polynomial by ax^{2} + bx + c = 0

a = 2, b = -8, c = 6

Thus there is a relationship between zeros and coefficient.

Question 15.

If α and β are zeros of quadratic (RBSESolutions.com) equation f(x) = x^{2} – px + q, then find the values of the following:

(a) α^{2} + β^{2}

(b) \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\)

Solution

Given α and β are zeros quadratic equation f(x) = x^{2} – px + q

Question 16.

If polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 is divided (RBSESolutions.com) by another polynamial x^{2} – 2x + k and remainder obtained x + a, then find k and a.

Solution

Given x^{4} – 6x^{3} + 16x^{2} – 25x + 10 divided by x^{2} – 2x + k obtained remainder is x + a

Question 17.

The area of a rectangular (RBSESolutions.com) plot is 528 m^{2}. Length of a plot (in m) is 1 more than twice of breadth. By forming quadratic equation, find the length and breadth of the plot.

Solution

Let breadth of a plot is x m.

According to question,

Length of plot = (2 × breadth) + l = (2 × x + 1) = (2x + 1) m.

Area of rectangular plot = l × b = (2x + 1) × x = (2x^{2} + x) sq. m.

Given : Area of plot = 528 sq. m.

2x^{2} + x = 528

⇒ 2x^{2} + x – 528 = 0

Required quadratic (RBSESolutions.com) equation is :

2x^{2} + x – 528 = 0

⇒ 2x^{2} + 33x – 32x – 528 = 0

⇒ x(2x + 33) – 16(2x + 33) = 0

⇒ (2x + 33) (x – 16) = 0

⇒ x – 16 = 0 or 2x + 33 = 0

⇒ x = 16 or x = \(\frac { –33 }{ 2 }\) (impossible)

Hence, Length of plot is 2x + 1 = 2 × 16 + 1 = 33 m

and breadth is 16 m.

Question 18.

Solve quadratic (RBSESolutions.com) equation x^{2} + 4x – 5 = 0 by completing the square method.

Solution

Given quadratic equation

x^{2} + 4x – 5 = 0

⇒ x^{2} + 4x = 5

Adding square of half of coefficient of x on both sides

x^{2} + 2 × 2x + (2)^{2} = 5 + (2)^{2}

⇒ (x + 2)^{2} = 5 + 4

⇒ (x + 2)^{2} = 9

⇒ x + 2 = ±√9

⇒ x + 2 = ± 3

Taking + ve sign x = 3 – 2 = 1

Taking – ve sign x = -3 – 2 = – 5

Hence, solution of given quadratic equation are 1 and – 5.

Question 19.

Solve the following (RBSESolutions.com) equations by factorisation method.

Solution

Question 20.

If -5 is a root of quadratic (RBSESolutions.com) equation 2x^{2} + px – 15 = 0 and roots of quadratic equation p(x^{2} + x) + k = 0 are same, then find k.

Solution

Given

One root of quadratic equation 2x^{2} + px – 15 = 0 is -5.

2(-5)^{2} + p(-5) – 15 = 0

⇒ 2 × 25 – 5p – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 5p = 35

⇒ p = 7

Putting this value in (RBSESolutions.com) quadratic equation p(x^{2} + x) + k = 0

7 (x^{2} + x) + k = 0

⇒ 7x^{2} + 7x + k = 0

Comparing by ax^{2} + bx + c = 0

a = 7, b = 7 c = k

root of equation are equal

Discriminant (D) = 0

⇒ b^{2} – 4ac = (7)^{2} – 4 × 7 × k

⇒ 0 = 49 – 28k

⇒ 28k = 49

⇒ k = \(\frac { 49 }{ 28 }\)

Thus, k = \(\frac { 7 }{ 4 }\)

Question 21.

Using, Shridhar Acharya (RBSESolutions.com) formula, solve the following quadratic equations : p^{2}x^{2} + (p^{2} – q^{2}) x – q^{2} = 0

Solution

Given equation p^{2}x^{2} + (p^{2} – q^{2}) x – q^{2} = 0

Comparing by ax^{2} + bx + c = 0, we get

a = p^{2}, b = (p^{2} – q^{2}) and c = -q^{2}

By Shridhar Acharya formula

Question 22.

L.C.M. and H.C.F. of two quadratic (RBSESolutions.com) expression are respectively x^{3} – 7x + 6 and (x – 1). Find the expression.

Solution

Least common multipile (L.C.M.) = x^{3} – 7x + 6

Highest common factor (H.C.F.) = (x – 1)

Factorising expression x^{3} – 7x + 6

putting x = 1 in this expression, we get = (1)^{3} – 7(1) + 6 = 1 – 7 + 6 = 0

At x = 0, expression = 0

(x – 1) is a factor

(x^{3} – 7x + 6)

= (x – 1) (x^{2} + x – 6)

= (x – 1) [x^{2} + 3x – 2x – 6]

= (x – 1)[x(x + 3) – 2(x + 3)]

= (x – 1)(x + 3)(x – 2)

Now L.C.M. = (x – 1) (x – 2) (x + 3)

and H.C.F. = (x – 1)

factror (x – 1) will be (RBSESolutions.com) common in two expressions

first expression = (x – 1) (x – 2) = x^{2} – 3x + 2

and second expression = (x – 1)(x + 3) = x^{2} + 2x – 3

Question 23.

L.C.M. of two polynomials is x^{3} – 6x^{2} + 3x + 10 and H.C.F. is (x + 1) if one polynomial is x^{2} – 4x – 5, then find the other.

Solution

L.C.M. = x^{3} – 6x^{2} + 3x + 10

H.C.F. = (x + 1)

Putting x = 1 in x^{3} – 6x^{2} + 3x + 10

x = 1 putting

= (1)^{3} – 6(1)^{2} + 3(1) + 10

= 1 – 6 + 3 + 10 ≠ 0

putting x = -1

= (-1)^{3} – 6(-1)^{2} + 3(-1) + 10

= -1 – 6 – 3 + 10 = 0

at x = – 1, expression = 0

(x + 1) is a factor of expression

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