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OTI365 - Olympiad -IX Standard
Olympiad Class IX Mathematics - Polynomials Sample Practice Questions with Answer

Olympiad Mathematics - Polynomials

Exam Duration: 45 Mins Total Questions : 30

1 )

Identify one of the factors of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2-2x-\frac { 2 }{ x } \) from the following

(a)
\(x-\frac { 1 }{ x } \)
(b)
\(x+\frac { 1 }{ x } -1\)
(c)
\(x+\frac { 1 }{ x } \)
(d)
\({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \)

Solution :

\({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2-2x-\frac { 2 }{ x } \)
\(=\left( x+\frac { 1 }{ x } \right) \left( x+\frac { 1 }{ x } -2 \right) \)

2 )

If (x²+ 3x + 5) (x²- 3x + 5) = m² - n²,find m.

(a)
x²-3x
(b)
3x+5
(c)
x²+5
(d)
x²-5

Solution :

(x² + 3x + 5) (x² - 3x + 5)
= (x² + 5)² - (3x)²
\(\therefore\) m = x² + 5

3 )

Factorise: a⁴ + 4

(a)
(a²+ 2a - 2)(a² + 2a + 2)
(b)
(a²- 2a - 2)(a²+ 2a - 2)
(c)
(a²+ 2a - 2)(a²- 2a + 2)
(d)
(a²+ 2a + 2)(a²- 2a + 2)

Solution :

a⁴ + 4
= a⁴ + 4a² - 4a² + 4
= (a² + 2a + 2)(a² - 2a + 2)

4 )

If x + Y + z = 0, what is the value of x³+y³+z^3?

(a)
xyz
(b)
2xyz
(c)
3xyz
(d)
0

5 )

Find the value of \({ 2 }^{ \frac { 1 }{ 4 } }.{ 4 }^{ \frac { 1 }{ 8 } }.{ 16 }^{ \frac { 1 }{ 16 } }.{ 256 }^{ \frac { 1 }{ 32 } }\)

(a)
1
(b)
2
(c)
4
(d)
8

Solution :

\({ 2 }^{ \frac { 1 }{ 4 } },{ 4 }^{ \frac { 1 }{ 8 } }.{ 16 }^{ \frac { 1 }{ 16 } }.{ 256 }^{ \frac { 1 }{ 32 } }\)
\(={ 2 }^{ \frac { 1 }{ 4 } }.{ 2 }^{ \frac { 1 }{ 4 } }.{ 2 }^{ \frac { 1 }{ 4 } }.{ 2 }^{ \frac { 1 }{ 4 } }={ 2 }^{ \frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } +\frac { 1 }{ 4 } }={ 2 }^{ 1 }=2\)

6 )

If P = (2 - a), what is a³+ 6ap + p³ - 8?

(a)
1
(b)
0
(c)
2
(d)
3

7 )

Find the value of 'a' if the polynomials 2x³ + ax² + 3x - 5 and x³ + x²- 4x - a leave the same remainder when divided by x -1.

(a)
a = 1
(b)
a = -1
(c)
a = 2
(d)
a = -2

Solution :

2(1)³ + a(1)² + 3(1) - 5 = (1)³ + (1)² - 4(1) - a
2 + a + 3 - 5 = 1 + 1 - 4 - a
\(\Rightarrow\) 2a = -2 \(\Rightarrow\) a = -1

8 )

If y - 2 and y - \(\frac { 1 }{ 2 } \) are the factors of py² + 5y + r, which of the following holds good?

(a)
p > r
(b)
p = r
(c)
p < r
(d)
Both (A) and (C)

Solution :

If y - 2 is a factor of py² + 5y + r,
\(\Rightarrow\) 4p + r + 10 = 0 .........(1)
If \(y-\frac { 1 }{ 2 } \) is a factor of py² + 5y + r,
\(\Rightarrow \quad \frac { p }{ 4 } +\frac { 5 }{ 2 } +r=0\) .......(2)
By solving (1) and (2), we get
p = r = -2

9 )

If x²+ x +1 is a factor of the polynomial 3x³+ 8x² + 8x + 3 + 5k, what is the value of k?

(a)
0
(b)
2/5
(c)
5/2
(d)
-1

Solution :

let p(x) = 3x³ + 8x² + 8x + 3 + 5k
\(\Rightarrow\) -2 + 5k = 0 (Since, it is given that x² + x+ 1 is a factor of 3x³ + 8x² + 8x+3 + 5k, the remainder is equal to 0)
\(\Rightarrow k=\frac { 2 }{ 5 } \)

10 )

If x + \(\frac { 1 }{ x } \) = 3, find the value of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \)?

(a)
9
(b)
11
(c)
7
(d)
8

Solution :

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ 3 }^{ 2 }\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =7\)

11 )

If (x + y)³- (x - y)³- 6y (x²- y²) = ky³,find k.

(a)
1
(b)
2
(c)
4
(d)
8

12 )

Factorise: \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } +2+\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)

(a)
\({ \left( \frac { a }{ b } +\frac { b }{ a } \right) }^{ 2 }\)
(b)
\({ \left( 1+\frac { a }{ b } \right) }^{ 2 }\)
(c)
\({ \left( 1-\frac { b }{ a } \right) }^{ 2 }\)
(d)
\({ \left( \frac { b }{ a } -\frac { a }{ b } \right) }^{ 2 }\)

13 )

Let R₁ and R₂ be the remainders when the polynomials f(x) = 4x³ + 3x² - 12ax -5 and g(x) = 2x³ + ax² - 6x + 2 are divided by (x - 1) and (x - 2) respectively. If 3R₁ + R₂+ 28 = 0, find the value of 'a'.

(a)
0
(b)
-1
(c)
1
(d)
32

14 )

If (x -1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + ....+a₁ x + a₀, what is the value of a₇ + a₆ +a₅ + .....+ a₁ + a₀?

(a)
0
(b)
1
(c)
128
(d)
64

Solution :

Given (x -1)⁷ = a₇x⁷ + a₆x⁶ +............+ a₁x + a₀ Since, the given polynomial terms are in expression, form Pascal's triangle we get the coefficients of 1st four terms symmetrical to those of the last four terms.
\(\Rightarrow\) (x-1)⁷ = x⁷ - 7x⁶ + 31x⁵ - 35x⁴ + 35x³ - 21x² + 7x - 1 = 0
\(\Rightarrow\) a₇ + a₆ + a₅ ..... + a₁ + a₀
= 1 - 7 + 21 - 35 + 35 - 21 + 7 - 1 = 0

15 )

Which of the following are the factors of 4x²- y² + 2x - 2y - 3xy?

(a)
(x + y) and (4x + y - 2)
(b)
(x - y) and (4x - y + 2)
(c)
(x + y) and (4x - y - 2)
(d)
(x - y) and (4x + y + 2)

Solution :

4x² - y² + 2x - 2y - 3xy
= (x - y) (4x + y + 2)

16 )

If p(x) = 4x³ - 3x² + 2x + 1, q(x) = x³- x²+ x + 1 and r(x) = x²- 2x + 1 find the value of 3p(x) + 7q(x) + r(x).

(a)
19x³- 15x² + 11 x + 11
(b)
- 19x³ - 15x² + 11 x - 11
(c)
19x³ - 15x² - 11x + 11
(d)
19x³ - 15x² - 11 x - 11

17 )

If x³ + y³ + z³- 3xyz = k (x + Y + z) {(x - y)² + (y - z)² + (z - x)²} , find k.

(a)
1
(b)
\(\frac { 1 }{ 4 } \)
(c)
\(\frac { 1 }{ 2 } \)
(d)
\(\frac { 1 }{ 3 } \)

18 )

The value of k for which (x + 2) is a factor of (x + 1)⁷ + (3x + k)³ is ______

(a)
-7
(b)
7
(c)
-1
(d)
-6-3^(7/3)

Solution :

Let f(x) = (x + 1)⁷ + (3x + k)³
Since, (x + 2) is a factor of fix). Therefore, by factor theorem, f(-2) = 0
⇒ (-2+1)⁷+(3x(-2)+k)³=0
⇒ (-1)⁷ + (- 6 + k)³ = 0 ⇒ (- 6 + k)³ = 1
⇒ -6+k=1 ⇒ k=7

19 )

The remainder when x⁴-y⁴ is divided by x-y is ______

(a)
0
(b)
x+y
(c)
x²-y²
(d)
2y⁴

Solution :

Let f(x) = x⁴ - y⁴ and g(x) = x _ Y
Now, g(x) = 0 ⇒ x - Y = 0 ⇒ x = y
∴. By remainder theorem, we know that when f(x) is divided by g(x), then the remainder is f(y).
Now, f(y) = y⁴ - y⁴ = 0

20 )

When p(x) = x³ + ax² + 2x + a is divided by (x + a), the remainder is _______

(a)
0
(b)
a
(c)
-a
(d)
2a

Solution :

We have, p(x) = x³ + ax² + 2x + a
By remainder theorem, we know that when p(x) is divided by (x + a), then the remainder is p(-a).
Now, p(- a) = (- a)³ + a(- a)² + 2(- a) + a
= - a³ + a³ - 2a + a = - a

21 )

x¹² - y¹² =_____

(a)
(x - y)(x² + xy + y²)(x + y)(x² - xy + y²) (x² + y²)(x⁴ - x²y² + y⁴)
(b)
(x+y)(x²-xy+y²)(x+y)(x²-xy+y²)(x²+y²)(x⁴-x²y²+y⁴)
(c)
(x+y)(x²+xy-y²)(x+y)(x²-xy+y²)(x²+y²)(x⁴-x²y²+y⁴)
(d)
(x-y)(x²-xy+y²)(x+y)(x²-xy+y²)(x²+y²)(x⁴-x²y²+y⁴)

Solution :

We have, x¹²-y¹²
=(x⁶)²-(y⁶)²=(x⁶-y⁶)(x⁶+y⁶)=[(x³)²-(y³)²][(x²)³+(y²)³]
=(x³+y³)(x³-y³)[(x²+y²)(x⁴+y⁴-x²y²)]
=(x+y)(x²+y²-xy)(x-y)(x²+y²+xy)(x²+y²)(x⁴+y⁴-x²y²)

22 )

If \(x=\frac { a-b }{ a+b } ,y=\frac { b-c }{ b+c } ,z=\frac { c-a }{ c+a } \), then the value of \(\frac { (1+x)(1+y)(1+z) }{ (1-x)(1-y)(1-z) } \) is ________

(a)
abc
(b)
a²b²c²
(c)
1
(d)
-1

Solution :

We have,
\(x=\frac { a-b }{ a+b } ,y=\frac { b-c }{ b+c } ,z=\frac { c-a }{ c+a } \)
Now, 1+x=\(1+\frac { a-b }{ a+b } =\frac { a+b+a-b }{ a+b } =\frac { 2a }{ a+b } \)
1-x=1-\(\frac { (a-b) }{ a+b } =\frac { a+b-a+b }{ a+b } =\frac { 2b }{ a+b } \)
Similarly, \(1+y=\frac { 2b }{ b+c } ,1+z=\frac { 2c }{ a+c } ,1-y=\frac { 2c }{ b+c } ,1-z=\frac { 2a }{ a+c } \)
Now, \(\frac { (1+x)(1+y)(1+z) }{ (1-x)(1+y)(1-z) } =\frac { \left( \frac { 2a }{ a+b } \right) \left( \frac { 2b }{ b+c } \right) \left( \frac { 2c }{ a+c } \right) }{ \left( \frac { 2b }{ a+b } \right) \left( \frac { 2c }{ b+c } \right) \left( \frac { 2a }{ a+c } \right) } \)=1

23 )

Given that x = 2 is a solution of x³-7x+6=0 The other solutions are ________

(a)
-1,3
(b)
1,-3
(c)
1,-2
(d)
-1,-2

Solution :

Let, f(x) = x³ - 7x + 6 and g(x) = x - 2
Now, by long division method, we have

∴ (x - 2) (x² + 2x - 3) = 0
For other solutions, x²+ 2x - 3 = 0
⇒ x² + 3x - x - 3 = 0 ⇒ (x + 3) (x - 1) = 0
⇒ x = -3 or 1

24 )

The product (a + b) (a - b) (a² - ab + b²) (a² + ab + b²) is equal to ______

(a)
a⁶+b⁶
(b)
a⁶-b⁶
(c)
a³-b³
(d)
a³+b³

Solution :

We have, (a + b) (a - b)(a² - ab + b²)(a² + ab + b²)
= (a + b) (a² - ab + b²) (a - b) (a² + ab + b²)
= (a³ + b³)(a³ - b³) = (a³)² - (b³)² = a⁶ - b⁶

25 )

When (x³ - 2x² + px - q) is divided by (x² - 2x - 3), the remainder is (x - 6). The values of p and q respectively are _____

(a)
-2,-6
(b)
2,-6
(c)
-2,6
(d)
2,6

Solution :

Let f(x) = x³ - 2x² + px- q
and g(x) = x²- 2x - 3 = (x - 3) (x + 1)
Now, when g(x) divides f(x), leaves a remainder (x - 6).
Then, we have, f(3) = x - 6 and f(-1) = x - 6
⇒ 27 - 18 + 3p - q = 3 - 6 and - 1 - 2 - P - q = - 1 - 6
⇒ 3p - q = - 12 ....... (i)
and p + q = 4 ........(ii)
Adding (i) and (ii), we get
4p =-8 ⇒ p=-2 and q=3(-2)+12=-6+12=6

26 )

Find the remainder when the expression 3x³ + 8x² - 6x + 1 is divided by x + 3.

(a)
1
(b)
10
(c)
6
(d)
0

Solution :

Let f(x) = 3x³ + 8x² - 6x + 1
We know that when ((x) is divided by x + 3, the remainder is f(-3).
Now, f(-3) = 3(-3)³ + 8(-3)² - 6(-3) + 1
= -81 + 72 + 18 + 1 = 10.

27 )

If x²- 1 is a factor of ax⁴+ bx³ + cx² + dx + e, then

(a)
a + b + e = c + d
(b)
a + b + c = d + e
(c)
b + c + d = a + e
(d)
None of these

Solution :

Let, f(x) = ax⁴ + bx³ + cx² + dx + e
Since, (x² - 1) i.e., (x - 1)(x + 1) is a factor of f(x).
Therefore, by factor theorem,
f(1)=0 and f(-1)=0
⇒ a+b+c+d+e=0 and a-b+c-d+e=0
⇒a+b+c=-(d+e) and a+c+e=b+d

28 )

Area of a rectangular field is (2x³ - 11x² - 4x + 5) sq. units and side of a square field is (2x² + 4) units. Find the difference between their areas (in sq. units).

(a)
4x² - 2x³ - 27x² - 4x + 11
(b)
4x⁴ - 2x³ + 27x² + 4x + 11
(c)
4x⁴ + 27x² + 4x-11
(d)
4x⁴ + 2x³ + 27x² + 4x + 11

Solution :

Area of rectangular field
= (2x³ - 11x² - 4x + 5) sq. units
Side of square field = (2x² + 4) units
∴ Area of square field = (2x² + 4)²
= (4x⁴ + 16 + 16x²) sq. units
∴ Required difference
= 4x⁴ + 16 + 16x² - 2x³ + 11x²+ 4x - 5
= (4x⁴ - 2x³ + 27x² + 4x + 11) sq. units.

29 )

Vikas has RS(x³ + 2ax + b), with this money he can buy exactly (x - 1) jeans or (x + 1) shirts with no money left. How much money Vikas has, if x = 4?

(a)
Rs 80
(b)
Rs 120
(c)
Rs 30
(d)
Rs 60

Solution :

Amount of money Vikas has
=Rs (x³+2ax+b)
Now, he can buy exactly (x - 1) jeans or (x + 1) shirts.
∴ (x - 1) and (x + 1) are factors of x³ + 2ax + b.
∴ (1)³+2a(1)+b=0 ⇒2a+b=-1 ........(i)
and (- 1)³ - 2a + b = 0 ⇒ 2a - b = - 1 ........(ii)
Adding (i) and (ii), we get
4a=-2 ⇒ a=\(\frac { -1 }{ 2 } \)
∴ -1+b=-1 ⇒ b=0
So, amount of money he has = Rs (x³- x)
= Rs(4³- 4) = Rs(64 - 4) = Rs 60.

30 )

Match the following:

Column-I	Column-II
(P) If f(x) = x³ - 6x² + 11x- 6 then f(-1)=_____	(i) -210
(Q) If f(x)=2x³-13x²+17x+12, then f(-3) = ______	(ii) 1
(R) If x=\(\frac{4}{3}\)is a root of f(x)=6x³-11x²+kx-20, then k= _____	(iii) -24
(S) If x=-1 is aroot of f(x)=x¹⁰⁰+2x⁹⁹+k, then k= _____	(iv) 19

(a)
(P) ⟶ (iii); (Q) ⟶ (iv); (R) ⟶ (i); (S) ⟶ (ii)
(b)
(P) ⟶ (ii); (Q) ⟶ (iv); (R) ⟶ (i); (S) ⟶ (iii)
(c)
(P) ⟶ (iii); (Q) ⟶ (i); (R) ⟶ (iv); (S) ⟶ (ii)
(d)
(P)~ (iii); (Q) ~ (ii); (R) ~ (i); (S) ~ (iv)

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Olympiad Class IX Mathematics - Polynomials Sample Practice Questions with Answer

Olympiad Mathematics - Polynomials

Exam Duration: 45 Mins Total Questions : 30

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