Polynomials - Test Papers

 CBSE Test Paper 01

CH-2 Polynomials

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1. If p(x) = x + 3, then p(x) + p(-x) is equal to
   1. 2x
   2. 3
   3. 0
   4. 6
2. If x+y+z = 0, thenx3+y3+z3 is $x^3+y^3+z^3\text{ is }$
   1. 3xyz
   2. xyz
   3. 2xyz
   4. 0
3. The degree of constant function is
   1. 0
   2. 3
   3. 1
   4. 2
4. (x + 1) is a factor of the polynomial
   1. x3+x2−x+1$x^3+x^2-x+1$
   2. x3+x2+x+1$x^3+x^2+x+1$
   3. x4+3x3+3x2+x+1$x^4+3x^3+3x^2+x+1$
   4. x4+x3+x2+1$x^4+x^3+x^2+1$
5. The coefficient of ‘x’ in the expansion of (x+3)3$(x+3{)}^3$ is
   1. 1
   2. 27
   3. 9
   4. 18
6. Fill in the blanks: A polynomial containing one non-zero term is called a ________.
7. Fill in the blanks: The highest power of the variable in a polynomial is called the ________ of the polynomial.
8. Write (-3x + y + z)2 in the expanded form:
9. Evaluate the following by using identities: 0.54 ×$\times$ 0.54 - 0.46 ×$\times$ 0.46
10. Evaluate: 185 ×$\times$ 185 - 15 ×$\times$ 15
11. Evaluate 105 ×$\times$ 108 without multiplying directly.
12. Simplify (x + y)3 - (x - y)3 - 6y(x + y) (x - y).
13. Factorize : (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3
14. If both x - 2 and x−12$x-\frac{1}{2}$ are factors of px2 + 5x + r, show that p = r.
15. 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

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Solution

1. (d) 6
   Explanation:
   p(x) = x + 3
   And p(-x) = -x + 3
   Then, p(x) + p(-x)
   = x + 3 - x + 3
   = 6
2. (a) 3xyz
   Explanation: x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)$x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$ =>x3+y3+z3−3xyz=(0)(x2+y2+z2−xy−yz−zx)$x^3+y^3+z^3-3xyz=\left(0\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$ => x3+y3+z3−3xyz=0$x^3+y^3+z^3-3xyz=0$ => x3+y3+z3=3xyz$x^3+y^3+z^3=3xyz$ If x+y+z = 0, thenx3+y3+z3 is $x^3+y^3+z^3\text{ is }$ 3xyz
3. (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
4. (b) x3+x2+x+1$x^3+x^2+x+1$
   Explanation: x3+x2+x+1$x^3+x^2+x+1$ = x3(x+1)+1(x+1)$x^3\left(x+1\right)+1\left(x+1\right)$ = (x3+1)(x+1)$\left(x^3+1\right)\left(x+1\right)$
5. (b) 27
   Explanation:(x+3)3$(x+3{)}^3$   = x3+(3)3+3×x×3(x+3)$x^3+{\left(3\right)}^3+3\times x\times 3\left(x+3\right)$ = x3+27+9x2+27x$x^3+27+9x^2+27x$ = x3+9x2+27x+27$x^3+9x^2+27x+27$ Therefore, the coefficient of x$x$, in the expansion of (x+3)3$(x+3{)}^3$ is 27.
6. monomial
7. degree
8. We have,
   (-3x + y + z)2
   Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
   = (-3x)2 + y2 + z2 + 2 ×$\times$ (-3x) ×$\times$ y + 2 ×$\times$ y ×$\times$ z + 2 ×$\times$ z ×$\times$ (-3x)
   = 9x2 + y2 + z2 - 6xy + 2yz - 6zx
9. We have,
   0.54 ×$\times$ 0.54 - 0.46 ×$\times$ 0.46
   = (0.54)2 - (0.46)2 = (0.54 + 0.46)(0.54 - 0.46) = 1 ×$\times$ 0.08 = 0.08
10. 185 ×$\times$185 - 15 ×$\times$ 15
    (185)2−(15)2$(185{)}^2-(15{)}^2$ [Using a2−b2=(a−b)(a+b)$a^2-b^2=(a-b)(a+b)$]
    (185+15)(185−15)$(185+15)(185-15)$
    200 ×$\times$ 170 = 34000
11. 105$105$×$\times$ 108=(100+5)(100+8)$108=(100+5)(100+8)$
    Using identity (x+a)(x+b)=x2+(a+b)x+ab$(x+a)(x+b)=x^2+(a+b)x+ab$
    We get, 105 ×$\times$ 108 = 1002 + (5 + 8)100 + 5 ×$\times$ 8
    =10000+1300+40=11340$=10000+1300+40=11340$
12. (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 [∴$\therefore$ a3 - b3 - 3ab(a - b) = (a - b)3]
    =(2y)3 = 8y3
13. Let x = a2 - b2, y = b2 - c2 and z = c2 - a2. Then,
    x + y + z = a2 - b2 + b2 - c2 + c2 - a2 = 0
    ∴$\therefore$ x3 + y3 + z3 = 3xyz
    ⇒$\Rightarrow$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
    ⇒$\Rightarrow$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)
    ⇒$\Rightarrow$ (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(b + c)(c + a)(a - b)(b - c)(c - a)
14. Let f(x) = px2 + 5x + r be the given polynomial. Since x - 2 and x−12$x-\frac{1}{2}$are factors of f(x).
    ∴$\therefore$ f(2) = 0 and f(12)$f\left(\frac{1}{2}\right)$ = 0
    ⇒$\Rightarrow$ p ×$\times$ 22 + 5 ×$\times$ 2 + r = 0 and p(12$\frac{1}{2}$)2 + 5 ×12$\times \frac{1}{2}$ + r = 0
    ⇒$\Rightarrow$ 4p + 10 + r = 0 and p4+52$\frac{p}{4}+\frac{5}{2}$ + r = 0
    ⇒$\Rightarrow$ 4p + r = -10 and p+4r+104=0$\frac{p+4r+10}{4}=0$
    ⇒$\Rightarrow$ 4p + r = -10 and p + 4r + 10 = 0
    ⇒$\Rightarrow$ 4p + r = -10 and p + 4r = -10
    ⇒$\Rightarrow$ 4p + r = p + 4r [RHS of the two equations are equal]
    ⇒$\Rightarrow$ 3p = 3r ⇒$\Rightarrow$ p = r
15. 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).
    ∴$\therefore$ 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)
    ⇒$\Rightarrow$ 27a + 41 = 15 + a
    ⇒$\Rightarrow$ 26a = -26; a = -1
    Hence, the required value of a = -1.