Polynomials - Test Papers
CBSE Test Paper 01
CH-2 Polynomials
- If p(x) = x + 3, then p(x) + p(-x) is equal to
- 2x
- 3
- 0
- 6
- If x+y+z = 0, then
- 3xyz
- xyz
- 2xyz
- 0
- The degree of constant function is
- 0
- 3
- 1
- 2
- (x + 1) is a factor of the polynomial
- The coefficient of ‘x’ in the expansion of is
- 1
- 27
- 9
- 18
- Fill in the blanks: A polynomial containing one non-zero term is called a ________.
- Fill in the blanks: The highest power of the variable in a polynomial is called the ________ of the polynomial.
- Write (-3x + y + z)2 in the expanded form:
- Evaluate the following by using identities: 0.54 0.54 - 0.46 0.46
- Evaluate: 185 185 - 15 15
- Evaluate 105 108 without multiplying directly.
- Simplify (x + y)3 - (x - y)3 - 6y(x + y) (x - y).
- Factorize : (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3
- If both x - 2 and are factors of px2 + 5x + r, show that p = r.
- If the polynomials az3 + 4z2 + 3z - 4 and z3 - 4z + a leave the same remainder when divided by z - 3, find the value of a.
CBSE Test Paper 01
CH-2 Polynomials
Solution
- (d) 6
Explanation:
p(x) = x + 3
And p(-x) = -x + 3
Then, p(x) + p(-x)
= x + 3 - x + 3
= 6 - (a) 3xyz
Explanation: => => => If x+y+z = 0, then 3xyz - (a) 0
Explanation: The degree of any constant term 5 (say)
We can write 5 as 5 x 1 = 5xo [Since ao= 1]
Therefore the degree of any constant term is 0 - (b)
Explanation: = = - (b) 27
Explanation: = = = Therefore, the coefficient of , in the expansion of is 27. - monomial
- degree
- We have,
(-3x + y + z)2
Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= (-3x)2 + y2 + z2 + 2 (-3x) y + 2 y z + 2 z (-3x)
= 9x2 + y2 + z2 - 6xy + 2yz - 6zx - We have,
0.54 0.54 - 0.46 0.46
= (0.54)2 - (0.46)2 = (0.54 + 0.46)(0.54 - 0.46) = 1 0.08 = 0.08 - 185 185 - 15 15
[Using ]
200 170 = 34000 -
Using identity
We get, 105 108 = 1002 + (5 + 8)100 + 5 8 - (x + y)3 - (x - y)3 - 6y(x + y)(x - y)
= (x + y)3 - (x - y)3 - 3.2y(x + y)(x - y)
= (x + y)3 - (x - y)3 - 3 (x + y - x + y)(x + y)(x - y)
= [x + y - x + y)]3 [ a3 - b3 - 3ab(a - b) = (a - b)3]
=(2y)3 = 8y3 - Let x = a2 - b2, y = b2 - c2 and z = c2 - a2. Then,
x + y + z = a2 - b2 + b2 - c2 + c2 - a2 = 0
x3 + y3 + z3 = 3xyz
(a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
(a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)
(a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(b + c)(c + a)(a - b)(b - c)(c - a) - Let f(x) = px2 + 5x + r be the given polynomial. Since x - 2 and are factors of f(x).
f(2) = 0 and = 0
p 22 + 5 2 + r = 0 and p()2 + 5 + r = 0
4p + 10 + r = 0 and + r = 0
4p + r = -10 and
4p + r = -10 and p + 4r + 10 = 0
4p + r = -10 and p + 4r = -10
4p + r = p + 4r [RHS of the two equations are equal]
3p = 3r p = r - Let p(z) = az3 + 4z2 + 3z - 4
And q(z) = z3 - 4z + a
As these two polynomials leave the same remainder, when divided by z - 3, then p(3) = q(3).
p(3) = a(3)3 + 4(3)2 + 3(3) - 4
= 27a + 36 + 9 - 4
Or p(3) = 27a + 41
And q(3) = (3)3 - 4(3) + a
= 27 - 12 + a = 15 + a
Now, p(3) = q(3)
27a + 41 = 15 + a
26a = -26; a = -1
Hence, the required value of a = -1.