Description

This mock test of Test: Polynomials - 4 for UPSC helps you for every UPSC entrance exam.
This contains 25 Multiple Choice Questions for UPSC Test: Polynomials - 4 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Polynomials - 4 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Polynomials - 4 exercise for a better result in the exam. You can find other Test: Polynomials - 4 extra questions,
long questions & short questions for UPSC on EduRev as well by searching above.

QUESTION: 1

Form a quadratic polynomial, sum of whose zeroes is 3 and product of whose zeroes is 2 solution

Solution:

Sum of zeros = 3/1

-b/a = 3/1

Product of zeros = -2/1

c/a = -2/1

This gives

a = 1

b = -3

c = -2,

The required quadratic equation is

ax^{2}+bx+c

So, x^{2}-3x-2

QUESTION: 2

If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is

Solution:

Given ,g(x)=x+2=0

x=-2

P(x)=x^{3}-2ax^{2}+16

p(-2)=(-2^{3})-2×a×(-2^{2})+16

= -8 - 8a +16

= 8 - 8a

⇒ -8a = -8

⇒ a=1

Hence, value of a=1

So option B is correct answer.

QUESTION: 3

If 3 + 5 – 8 = 0, then the value of (3)^{3} + (5)^{3} – (8)^{3} is

Solution:

Given :

3 + 5 + (-8) = 0.

Then,

3³ + 5³ + (-8)³ = ?

Using, identity...

If a+b+c=0, then a³ + b ³ + c³ = 3abc

Let a = 3, b = 5 and c = -8.

Here, a + b + c = 3 + 5 + (-8) = 0.

Then, a³ + b³ + c³ = 3³ + 5³ + (-8)³

= 3× 3 × 5 × -8

= 9 × -40

= -360

QUESTION: 4

If one of the factors of x^{2} + x – 20 is (x + 5), then other factor is

Solution:

Using mid-term splitting,

x^{2}+x-20=x^{2}+5x-4x-20=x(x+5)-4(x+5)

Taking common x+5

(x+5)(x-4) , so the other factor is x-4

QUESTION: 5

If α,β be the zeros of the quadratic polynomial 2x^{2} + 5x + 1, then value of α + β + αβ =

Solution:

P(x) = 2x² + 5x + 1

Sum of roots = -5/2

Product of roots = 1/2

Therefore substituting these values,

α + β +αβ

=(α + β) + αβ

= -5/2 + 1/2

= -4/2

= -2

QUESTION: 6

If α,β be the zeros of the quadratic polynomial 2 – 3x – x^{2}, then α + β =

Solution:

We have Sum of zeros = -b/a

+=-b/a=-(-3)/(-1)=-3

Which is none of the above.

QUESTION: 7

Quadratic polynomial having sum of it's zeros 5 and product of it's zeros – 14 is –

Solution:

The quadratic equation is of the form x^{2 }- (sum of zeros) x + (product of zeros)

=x^{2 }- 5x - 14

QUESTION: 8

If x = 2 and x = 3 are zeros of the quadratic polynomial x^{2} + ax + b, the values of a and b respectively are :

Solution:

Zeros of the polynomials are the values which gives zero when their value is substituted in the polynomial

When x=2,

x^{2}+ax+b =(2)^{2}+a*2+b=0

4+2a+b=0

b=-4-2a ….1

When x=3,

(3)^{2}+ 3a + b=0

9 + 3a + b=0

Substituting

9 + 3a - 4 - 2a =0

5 + a =0

a = -5

b = 6

QUESTION: 9

If 3 is a zero of the polynomial f(x) = x^{4} – x^{3} – 8x^{2} + kx + 12, then the value of k is –

Solution:

QUESTION: 10

The sum and product of zeros of the quadratic polynomial are – 5 and 3 respectively the quadratic polynomial is equal to –

Solution:

QUESTION: 11

On dividing x^{3} – 3x^{2} + x + 2 by polynomial g(x), the quotient and remainder were x – 2 and 4 – 2x respectively then g(x) :

Solution:

We have remainder theorem as

x^{3}-3x^{2}+x+2=q(x)*g(x)+r(x)

x^{3}-3x^{2}+x+2=(x-2)g(x)+(4-2x)

g(x)=

So by division method,

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

QUESTION: 12

If the polynomial 3x^{2} – x^{3} – 3x + 5 is divided by another polynomial x – 1 – x^{2}, the remainder comes out to be 3, then quotient polynomial is –

Solution:

QUESTION: 13

If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is

Solution:

QUESTION: 14

If is the zero of the cubic polynomial f(x) = 3x^{3} – 5x^{2} – 11x – 3 the other zeros are :

Solution:

QUESTION: 15

If p and q are the zeroes of the polynomial x^{2}- 5x - k. Such that p - q = 1, find the value of K

Solution:

QUESTION: 16

Let p(x) = ax^{2} + bx + c be a quadratic polynomial. It can have at most –

Solution:

QUESTION: 17

The graph of the quadratic polynomial ax^{2} + bx + c, a ≡ 0 is always –

Solution:

We have a=0 So,

ax^{2 }+ bx + c = 0*x^{2 }+ bx + c = bx + c

Which is a linear equation in one variable

Linear equation as the name suggests is an equation which when plotted gives a straight line.

QUESTION: 18

If 2 and as the sum and product of its zeros respectively then the quadratic polynomial f(x) is –

Solution:

QUESTION: 19

If α and β are the zeros of the polynomial f(x) = 16x^{2} + 4x – 5 then is equal to –

Solution:

QUESTION: 20

If α and β are the zeros of the polynomial f(x) = 15x^{2} – 5x + 6 then is equal to –

Solution:

**The correct answer is a.**

**(1+1/α) (1+1/β) **

**= (α+1/α) (β+1/β)**

**=αβ + α + β + 1 / αβ**

**=( 2/5 + 1/3+ 1) / 2/5**

**=26/15 X 5/2 = 13/3**

QUESTION: 21

If the sum of the two zeros of x^{3} + px^{2} +qx + r is zero, then pq =

Solution:

Sum of two solutions is zero ,so both are inverse of each other. So we have the solutions as a,-a,b.

(x-a)(x+a)(x+b)=(x^{2}-a^{2})(x+b)=x^{3}+bx^{2}-a2x-a2b

pq=-a^{2}b=r

QUESTION: 22

Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when divided respectively by x + a and x – a, the remainder when p(x) is divided by x^{2} – a^{2 }is

Solution:

Let P(x)=(x−a)(x+a)Q(x)+rx+s

When we divide by x+a, we get

P(−a)=(−a−a)(−a+a)Q(−a)+r(−a)+s=a

⇒s−ra=a ...(1)

And when we divide by x−a, we get

P(a)=(a−a)(a+a)Q(a)+ra+s=−a

⇒s+ra=−a ...(2)

Solving (1) and (2), we get

s=0,r=−1

Hence, P(x)=(x−a)(x+a)Q(x)−x

And when we divide by x2-a2, it leaves a remainder −x.

QUESTION: 23

If one root of the polynomial x^{2} + px + q is square of the other root, then –

Solution:

Let one root be a, other root is a^{2}.

(x-a)(x-a^{2})=x^{2}+(-a^{2}-a)x+a^{3}

p^{3}=(-a^{2}-a)^{3}=-a^{6}-3a^{5}-3a^{4}-a^{3}

-q(3p-1)= -a^{3}(3(-a^{2}-a)-1)=3a^{5}+3a^{4}+a^{3}

q^{2}=a^{6}

Substituting the values

-a^{6}-3a^{5}-3a^{4}-a^{3}+3a^{5}+3a^{4}+a^{3}+a^{6}=0

QUESTION: 24

If α,β are the zeros of x2 + px + 1 and γ,δ be those of x2 + qx + 1, then the value of (α–γ) (β–γ) (α+δ) (β+δ) =

Solution:

Alpha(a) and beta(b) are roots of x^2 + px + 1

This implies that sum of roots= a+b = -p/1=-p

And the product of roots = ab = 1/1=1

Similarly ,

Gamma(c) and delta(d) are roots of x^2 + qx + 1

So c+d=-q and cd =1.

The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (x-p)(x-q)=0 where a and b are the roots .

So x^2 -(p+q)x + pq =0

Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get

Sum of the roots =p+q= -b/a

And product of the roots = pq = c/a}

RHS=(a-c)(b-c)(a+d)(b+d)

=(c^2-ac-bc +ab)(d^2 +bd +ad + ab)

We know ab=1

So RHS= (c^2-ac-bc +1)(d^2 +bd +ad + 1)

= (c^2)(d^2) +(a+b)c^2(d) + c^2 -(d^2)c(a+b) -(a+b)^2(cd) -(a+b)c + d^2 + bd + ad + 1

= 1 + ac+bc + c^2 - da-db - (a^2 + b^2 + 2(1)) -ac -bc + bd + ad +1

Cancelling off all the common terms,

We get c^2 +d^2-a^2-b^2

= c^2+d^2-a^2-b^2

=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}

=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab

LHS=(c+d)^2-(a+b)^2

Therefore,

But we know -p= a+b and -q = c+d

LHS= q^2-p^2

QUESTION: 25

The quadratic polynomial whose zeros are twice the zeros of 2x^{2} – 5x + 2 = 0 is –

Solution:

Let α and β be the roots of the given equation.

Then, α + β = 5/2 and αβ = 2/2 = 1

∴ 2α + 2β

∴ 2(α + β)

∴ 5,(2α)(2β) = 4

So, the requiared equation is :

x^{2}−5x+4=0

### Practice Test on Polynomials, Class 10

Doc | 2 Pages

### Evolution Test-4

Doc | 1 Page

### Solution- Matrices Test- 4

Doc | 5 Pages

### Test Paper 4 Solution

Doc | 8 Pages

- Test: Polynomials - 4
Test | 25 questions | 25 min

- Test: Division Algorithm For Polynomials
Test | 20 questions | 20 min

- Polynomials - MCQ Test - 1
Test | 25 questions | 25 min

- Test: Polynomials - 3
Test | 25 questions | 25 min

- Polynomials - Practice Test
Test | 10 questions | 20 min