Chapter 2 Exponents of Real Numbers R.D. Sharma Solutions for Class 9th Math MCQ Exercise
MCQ
1. If α, β are the zeros of the polynomial f(x) = x2 + x + 1 , then 1/α + 1/ β =
(a) 1
(b) -1
(c) 0
(d) None of these
Solution
Since α and β are the zeros of the quadratic polynomial f(x) = x2 + x + 1
Solution
Since α and β are the zeros of the quadratic polynomial f(x) = x2 + x + 1
α + β = - coefficient of x/coefficient of x2
= -1/1 = 1
α × β = constant term / coefficient of x2
= 1/1 = 1
We have
1/α + 1/β =
= β+α/ αβ
= -1/1
= -1
The value of 1/α + 1/β is -1
Hence , the correct choice is b .
2. If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7, then 1/α + 1/β is equal to
2. If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7, then 1/α + 1/β is equal to
(a) 7/3
(b) – 7/3
(c) 3/7
(d) -3/7
Solution
Since α and β are the zeros of the quadratic polynomial p(x) = 4x2 + 3x + 7
α + β = -coefficient of x / coefficient of x2
= -3/4
= αβ = Constant term / Coefficient of x2
= 7/4
We have
![]()
Solution
We are given f(x) = (k2 + 4)x2 13x + 4k then
α + β = -coefficient of x / coefficient of x2
= -3/4
= αβ = Constant term / Coefficient of x2
= 7/4
We have
3. If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x + 4k is reciprocal of the other, then k =
(a) 2
(b) -2
(c) 1
(d) -1
Solution
α + β = -Coefficient of x/ Coefficient of x2
= -13/k2 + 4
α × β = Constant term / Coefficient of x2
= 4k/k2 + 4
One root of the polynomial is reciprocal of the other. Then , we have
α × β = 1
= 4k/k2+4
= 1
= k2 – 4k + 4 = 0
= (k-2)2 = 0
= k = 2
Hence the correct choice is a .
4. If the sum of the zeros of the polynomial f(x) = 2x3 − 3kx2 + 4x − 5 is 6, then the value of k is
4. If the sum of the zeros of the polynomial f(x) = 2x3 − 3kx2 + 4x − 5 is 6, then the value of k is
(a) 2
(b) 4
(c) −2
(d) −4
(b) 4
(c) −2
(d) −4
Solution
Let α, β be the zeros of the polynomial f(x) = 2x3 – 3kx2 + 4x -5 and we are that α + β + γ = 6 .
Then,
α + β + γ = -Coefficient of x/Coefficient of x2
= - (-3k)/2 = 3k/2
It is given that
α + β + γ = 6
Substituting α + β + γ = 3k/2 , we get
+3k/2 = 6
+ 3k = 6 × 2
+3k = 12
k = 12/+3
k = +4
The value of k is 4 .
Hence, the correct alternative is b .
5. If α and β are the zeros of the polynomial f(x) = x2 + px + q, then a polynomial having 1/α and 1/ β is its zero is
(a) x2 + qx + p
(b) x2 – px + q
(c) qx2 + px + 1
(d) px2 + qx + 1
Solution
Let α and β are the zeros of the polynomial f(x) = x2 + px + q . Then ,
Let α and β are the zeros of the polynomial f(x) = x2 + px + q . Then ,
α + β = -Coefficient of x/ Coefficient of x2
= -p/1
= -p
And
αβ = Constant term/Coefficient of x2
= q/1
= q
Let S and R denote respectively the sum and the product of the zeros of a polynomial
Whose zeros are 1/α and 1/β . then
S = 1/α + 1/β
= α+ β/ αβ
= -p/q
R = 1/α × 1/β
= 1/ αβ
= 1/q
Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by
x2 - Sx + R = 0
x2 + P/q x + 1/q = 0
qx2 + Px + 1/q = 0
qx2 + Px + 1
So g(x) = qx2 + Px + 1
Hence , the correct choice is c .
6. If α,β are the zeros of polynomial f(x) = x2 – p(x+1) – c , then (α + 1) (β + 1) =
(a) c – 1
(b) 1 – c
(c) c
(d) 1 + c
Solution
Since, α and β are the zeros of quadratic polynomial
Since, α and β are the zeros of quadratic polynomial
f(x) = x2 – p(x+1)-c
= x2
α + β = -Coefficient of x/Coefficient of x2
= -(-p/1)
= p
α×β = Constant term/Coefficient of x2
= -p-c/1
= -p-c
We have
(α +1)( β+1)
= αβ + β + α + 1
= αβ + (α+β) + 1
= - p - c + (p) + 1
= - p - c + (p) + 1
= -c + 1
= 1 –c
The value of (α +1)( β+1) is 1 – c .
Hence, the correct choice is b .
7. If α, β are the zeros of the polynomial f(x) = x2 − p(x + 1) − c such that (α +1) (β + 1) = 0, then c =
7. If α, β are the zeros of the polynomial f(x) = x2 − p(x + 1) − c such that (α +1) (β + 1) = 0, then c =
(a) 1
(b) 0
(c) −1
(d) 2
(b) 0
(c) −1
(d) 2
Solution
Since, α and β are the zeros of quadratic polynomial
f(x) = x2 – p(x+1) – c
f(x) = x2 – px – p – c
α+β = -Coefficient of x/Coefficient of x2
= -(-p/1)
= p
αβ = Constant term/Coefficient of x2
= -p - c/1
= -p-c
We have
0 = (α+1)(β+1)
0 = αβ + (α+β) + 1
0 = -p-c+p+1
0 = -c + 1
c = 1
The value of c is 1 .
Hence, the correct alternative is a.
8. If f(x) = ax2 + bx + c has no real zeros and a + b + c < 0, then
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these
Solution
If f(x) = ax2 + bx + c has no real zeros and a + b + c < 0 then c < 0
If f(x) = ax2 + bx + c has no real zeros and a + b + c < 0 then c < 0
Hence, the correct choice is c .
9. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax2 + bx + c, then
![]()
(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0
Solution
Solution
9. If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax2 + bx + c, then
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0
Solution
Clearly, f(x) = ax2 + bx + c represent a parabola opening upwar .
![]()
Therefore, f(x) = ax2 + bx + c cuts Y axis at P which lies on OY. Putting x = 0 in y = ax2 + bx + c, we get y = c. So the coordinates of P is (0,c). Clearly , P lies on OY. Therefore c > 0
Hence , the correct choice is a .
Therefore, f(x) = ax2 + bx + c cuts Y axis at P which lies on OY. Putting x = 0 in y = ax2 + bx + c, we get y = c. So the coordinates of P is (0,c). Clearly , P lies on OY. Therefore c > 0
Hence , the correct choice is a .
10. Figure 2.23 show the graph of the polynomial f(x) = ax2 + bx + c for which
![]()
(a) a < 0, b > 0 and c > 0
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0
Solution
Clearly, f(x) = ax2 + bx + c represent a parabola opening downwards. Therefore, a<0
y = ax2 + bx + c cuts y - axis at P which lies on OY. Putting x = 0 in y = ax2 + bx + c, we get y = c.So the coordinates P are (0,c). Clearly , P lies on OY. Therefore c > 0
The vertex (-b/2a, -D/4a) of the parabola is in the second quadrant. Therefore -b/2a<0, b<0
![]()
Therefore a<0,b<0, and c>0
Hence , the correct choice is b
11. If the product of zeros of the polynomial f(x) ax3 − 6x2 + 11x − 6 is 4, then a =
(a) 3/2
(b) -3/2
(c) 2/3
(d) -2/3
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0
Solution
Clearly, f(x) = ax2 + bx + c represent a parabola opening downwards. Therefore, a<0
y = ax2 + bx + c cuts y - axis at P which lies on OY. Putting x = 0 in y = ax2 + bx + c, we get y = c.So the coordinates P are (0,c). Clearly , P lies on OY. Therefore c > 0
The vertex (-b/2a, -D/4a) of the parabola is in the second quadrant. Therefore -b/2a<0, b<0
Therefore a<0,b<0, and c>0
Hence , the correct choice is b
11. If the product of zeros of the polynomial f(x) ax3 − 6x2 + 11x − 6 is 4, then a =
(a) 3/2
(b) -3/2
(c) 2/3
(d) -2/3
Solution
Since α and β are the zeros of quadratic polynomial f(x) = ax3 - 6x2 + 11x - 6
αβ = -Constant term/Coefficient of x2
So we have
4 = -(-6/a)
4 = 6/a
4a = 6
a = 6/4
a = 3×2/2×2
a = 3/2
The value of a is 3/2
Hence, the correct alternative is a .
αβ = -Constant term/Coefficient of x2
So we have
4 = -(-6/a)
4 = 6/a
4a = 6
a = 6/4
a = 3×2/2×2
a = 3/2
The value of a is 3/2
Hence, the correct alternative is a .
12. If zeros of the polynomial f(x) = x3 − 3px2 + qx − r are in A.P., then
(a) 2p3 = pq − r
(b) 2p3 = pq + r
(c) p3 = pq − r
(d) None of these
Solution
Let a-d,a,a+d be the zeros of the polynomial f(x) = x3 - 3px2 + qx - r then
Sum of zeros = -Coefficient ofx2/Coefficient of x3
(a-d)+a+(a+d) = -(-3p)/1
a-d + a + a + d = 3p
3a = 3p
a = 3/3p
a = p
Since a is a zero of the polynomial f(x)
Therefore,
f(a) = 0
a3 - 3pa3 + qa - r = 0
Substituting a = p . we get
p3 - 3p(p)2 + q×p-r=0
p3 - 3p3 + qp - r = 0
-2p3 + qp - r = 0qp - r = 2p3Hence, the correct choice is a
13. If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 − 4x + 9 is 3, then its third zero is
(a) 3/2
(b) -3/2
(c) 9/2
(d) -9/2
Solution
Let α,β,γ be the zeros of polynomial f(x) = 2x3 + 6x2 - 4x + 9 such that αβ = 3
We have,
αβγ = Constant term/Coefficient of x3
= -9/2
Putting αβ = 3 in αβγ = -9/2, we get
αβγ = -9/2
3γ = -9/2
γ = -9/2 × 1/3
γ = -3/2
Therefore, the value of third zero is -3/2
Hence, the correct alternative is b.
14. If the polynomial f(x) = ax3 + bx − c is divisible by the polynomial g(x) = x2 + bx + c, then ab =
(a) 1
(b) 1/c
(c) -1
(d) -1/c
Solution
We have to find the value of ab
Given f(x) = ax3 + bx - c is divisible by the polynomial g(x) = x2 + bx + c
We must have
bx - acx + ab2x + abc - c = 0, for all x
So put x = 0 in this equation
x(b-ac+ab2)+c(ab-1) = 0
c(ab-1) = 0
Since c≠0 , so
ab - 1 = 0
⇒ ab = 1
Hence, the correct alternative is a.
15. If the polynomial f(x) = ax3 + bx − c is divisible by the polynomial g(x) = x2 + bx + c, then c =
(a) b
(b) 2b
(c) 2b2
(d) −2b
Solution
We have to find the value of c
Given f(x) = ax3 + bx -c is divisible by the polynomial g(x) = x2 + bx + c
We must have,
bx - acx + ab2x + abc - c = 0 for all x
x(b-ac+b2) + c(ab - 1) = 0 ...(1)
c(ab-1) = 0
Since c≠0, so
ab - 1 = 0
ab = 1
Now in the equation (1) the condition is true for all x. So put x =1 and also we have ab = 1
Therefore we have
b-ac+ab2=0
b+ab2 -ac=0
b(1+ab)-ac=0
Substituting a = 1/b and ab = 1 we get,
b(1+1) - 1/b×c=0
2b - 1/b×c = 0
-1/b×c = -2b
c=2b×b/1
c=2b2
Hence, the correct alternative is c .
16. If one root of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other, then the value of k is
(a) 0
(b) 5
(c) 1/6
(d) 6
Solution
If one zero of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other . So β = 1/α ⇒ αβ = 1
Now we have
α×β = Constant term/Coefficient of x2
= k/5
Since αβ = 1
Therefore we have
αβ = k/5
1= k/5
⇒ k = 5
Hence the correct choice is b .
17. If α,β,γ are the zeros of the polynomial f(x) ax3 + bx2 + cx + d, then 1/α + 1/β + 1/γ =
(a) -b/d
(b) c/d
(c) -c/d
(d) -c/a
Solution
We have to find the value of 1/α + 1/β + 1/γ =
Given α,β,γ be the zeros of the polynomial f(x) = ax3 + bx2 + cx + d
we know that
αβ +βγ + γα = coefficient of x/coefficient of x3
= c/a
αβγ = -Constant term/Coefficient of x3
= -d/a
So
1/α + 1/β + 1/γ = βγ + αγ + αβ/αβγ
1/α + 1/β + 1/γ = c/a/-d/a
1/α + 1/β + 1/γ = c/a × (-a/d)
1/α + 1/β + 1/γ = -c/d
Hence, the correct choice is c.
18. If α, β, γ are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d, then α2 + β2 + γ2 =
(a) b2-ac/a2
(b) b2-2ac/a
(c) b2 + 2ac/b2
(d) b2 - 2ac/a2
Solution
We have to find the value of α2 + β2 + γ2
Given α,β,γ be the zeros of the polynomial f(x) - ax3 + bx2 + cx + d
α+β+γ = -Coefficient of x2/Coefficient of x3
= -b/a
αβ + βγ + γα = Coefficient of x/Coefficient of x3
= c/a
Now
α2 + β2 + γ2 = (α+β+γ)-2(αβ + βγ + γα)
α2 + β2 + γ2 = (-b/a)2 -2(c/a)
α2 + β2 + γ2 = b2/α2 - 2c/a
α2 + β2 + γ2 = b2/α2 - 2ca/α2
α2 + β2
+ γ2 = b2-2ac/a2
The value of α2 + β2 + γ2 = b2-2ac/a2
Hence, the correct choice is d.
19. If α, β, γ are the zeros of the polynomial f(x) = x3 - px2 + qx - r , then 1/αβ + 1/βγ + 1/γα =
(a) r/p
(b) p/r
(c) -p/r
(d) -r/p
Solution
If α, β, γ are the zeros of the polynomial f(x) = x3 - px2 + qx - r , then 1/αβ + 1/βγ + 1/γα =
We have to find the value of 1/αβ + 1/βγ + 1/γα
Given α,β,γ be the zeros of the polynomial f(x) = x3 - px2 + qx - r
α+β+γ = -Coefficient of x2/Coefficient of x3
= -(p)/1
= p
αβγ = -Constant term/Coefficient of x3
= -(r)/1
= r
Now we calculate the expression
1/αβ + 1/βγ + 1/γα = γ/αβγ + α/αβγ + β/αβγ
1/αβ + 1/βγ + 1/γα = α+γ+β/αβγ
1/αβ + 1/βγ + 1/γα = p/r
Hence, the correct choice is b.
20. If α,β are the zeros of the polynomial f(x) = ax2 + bx + c, then 1/α + 1/β =
(a) b2-2ac/a2
(b) b2-2ac/c2
(c)b2+2ac/a2
(d) b2+2ac/c2
Solution
We have to find the value of 1/α2+ 1/β2
Given α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c
α+β = -Coefficient of x/Coefficient of x3
= -b/a
αβ = Constant term/Coefficient of x3
= c/a
We have,
21. If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are each equal to zero, then the third zero is
(a) -d/a
(b) c/a
(c) -b/a
(d) b/a
Solution
Let α=0, β=0 and γ be the zeros of the polynomial
f(x) = ax3 + bx2 + cx + d
Therefore
α+β+γ = -Coefficient of x2/Coefficient of x3
= -(b/a)
α+β+γ = -b/a
0+0+γ = -b/a
γ = - b/a
The value of γ = -b/a
Hence, the correct choice is c.
22. If two zeros x3 + x2 - 5x - 5 are √5 and -√5, then its third zero is
(a) 1
(b) -1
(c) 2
(d) -2
Solution
Let α = √5 and β = -√5 be the given zeros and γ be the third zero of x3 + x2 - 5x - 5 = 0 then
By using α+β+γ = -Coefficient of x2/Coefficient of x3
α+β+γ = +(+1)/1
α+β+γ = -1
By substituting α = √5 and β = -√5 in α+β+γ = -1
√5 - √5 + γ = -1
γ = -1
Hence, the correct choice is b .
23. The product of the zeros of x3 + 4x2 + x − 6 is
(a) −4
(b) 4
(c) 6
(d) −6
Solution
Given α,β,γ be the zeros of the polynomial f(x) = x3 + 4x2 + x -6
Product of the zeros = Constant term/Coefficient of x3 = -(-6)/1 =6
The value of product of the zeros is 6.
24. What should be added to the polynomial x2 − 5x + 4, so that 3 is the zero of the resulting polynomial?
(a) 1
(b) 2
(c) 4
(d) 5
Solution
If x = a, is a zero of a polynomial then x - a is a factor of f(x)
Since 3 is the zero of the polynomial f(x) = x2 - 5x + 4,
Therefore x -3 is a factor of f(x)
Now we divide f(x) = x2 - 5x + 4 by (x-3) we get
Therefore we should add 2 to the given polynomial
Hence, the correct is b.
25. What should be subtracted to the polynomial x2 − 16x + 30, so that 15 is the zero of the resulting polynomial ?
(a) 30
(b) 14
(c) 15
(d) 16
Solution
We know that, if x = a, is zero of a polynomial then x -a is a factor of f(x)
Since 15 is zero of the polynomial f(x) = x2 - 16x + 30, therefore (x-15) is a factor of f(x)
Now, we divide f(x) = x2 - 16x + 30 by (x-15) we get
Thus we should subtract the remainder 15 from x2 - 16x + 30 .
Hence, the correct choice is c.
26. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is
(a) x2 − 9
(b) x2 + 9
(c) x2 + 3
(d) x2 − 3
Solution
Since α and β are the zeros of the quadratic polynomials such that
0 = α+β
If one of zero is 3 then
α+β = 0
3+β = 0
β = 0-3
β = -3
Substituting β = -3 in α + β = 0 we get
α - 3 = 0
α = 3
Let S and P denote the sum and the product of the zeros of the polynomial respectively then
S = α+β
S = 0
P = αβ
P = 3×-3
P = -9
Hence, the required polynomial is
= (x2 -Sx + P)
= (x2 - 0x - 9)
= x2 - 9
Hence, the correct choice is a .
27. If two zeroes of the polynomial x3 + x2 − 9x − 9 are 3 and −3, then its third zero is
(a) −1
(b) 1
(c) −9
(d) 9
Solution
Let α = 3 and β = -3 be the given zeros and γ be the third zero of the polynomial x3 + x2 - 9x -9 then,
By using α+β+γ = -Coefficient of x2/Coefficient of x3
α+β+γ = -1/1
α+β+γ = -1
Substituting α = 3 and β = -3 in α+β+γ = -1, we get
3-3+γ = -1
γ = -1
Hence the correct choice is a .
28. If √5 and -√5 are two zeroes of the polynomial x3 + 3x2 - 5x - 15, then its third zero is
(a) 3
(b) -3
(c) 5
(d) -5
Solution
Let α = √5 and β = -√5 be the given zeros and γ be the third zero of the polynomial x3 + 3x2 - 5x - 15. Then,
By using α+β+γ = -Coefficient of x2/Coefficient of x3
α+β+γ = -3/1
α+β+γ = -3
Substituting α = √5 and β = -√5 in α+β+γ = -3
We get
√5 - √5 + γ = -3
√5 - √5 + γ = -3
γ = -3
Hence, the correct choice is b.
29. If x + 2 is a factor of x2 + ax + 2b and a + b = 4, then
(a) a = 1, b = 3
(b) a = 3, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1
Solution
Given that x+2 is a factor of x2 + ax + 2b and a + b = 4
f(x) = x2 + ax + 2b
f(-2) = (-2)2 + a(-2) + 2b
0 = 4 - 2a + 2b
-4 = -2a + 2b
By solving -4 and -2a+2b and a+b = 4 by elimination method we get
Multiply a+b = 4 by 2 we get,
2a + 2b = 8. So
-4 = -2a + 2b
+8 = +2a + 2b
4 = 4b
4/4 = b
1 = b
By substituting b =1 in a+b = 4 we get
a+1 = 4
a = 4-1
a = 3
Then a = 3, b = 1
Hence, the correct choice is b.
30. The polynomial which when divided by - x2 + x - 1 gives a quotient x -2 and remainder 3, is
(a) x3 - 3x2 + 3x - 5
(b) -x3 - 3x2 -3x - 5
(c) -x3 + 3x2 -3x + 5
(d) x3 -3x2 -3x + 5
Solution
We know that,
f(x) = g(x)q(x)+r(x)
= (-x2+x-1)(x-2)+3
= -x3 + x2 - x + 2x2 - 2x + 2 + 3
= -x3 + x2 + 2x2 - x - 2x + 2 + 3
= -x3 + 3x2 - 3x + 5
Therefore,
The polynomial which when divide by -x2 + x - 1 gives a quotient x-2 and remainder 3, is x3+3x2-3x+5
Hence, the correct choice is c .
31. The number of polynomials having zeros -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) More than 3.
Solution
Polynomials having zeros -2 and 5 will be of the form
p(x) = a(x+2)n(x-5)m
Here, n and m can take any value from 1,2,3,...
Thus, the number of polynomials will be more than 3.
Hence, the correct answer is option D.
32. If one of the zeroes of the quadratic polynomial (k-1)x2 + kx + 1 is -3, then the value of k is
(a) 4/3
(b) -4/3
(c) 2/3
(d) -2/3
Solution
The given polynomial is f(x) = (k-1)x2 + kx + 1.
Since -3 is one of the zeros of the given polynomial, so f(-3) = 0
(k-1)(-3)2 + k(-3) + 1 = 0
⇒ 9(k-1)-3k + 1 = 0
⇒ 9k - 9 - 3k + 1 = 0
⇒ 6k - 8 = 0
⇒ k = 4/3
Hence, the correct answer is option A.
33. The zeroes of the quadratic polynomial x2 + 99x + 127 are
(a) Both positive
(b) Both negative
(c) Both equal
(d) One positive and one negative
Solution
Let f(x) = x2 + 99x + 127.
Product of the zeroes of f(x) = 127 × 1 = 127 [Product of zeroes = c/a when f(x) = ax2 + bx + c]
Since the product of zeroes is positive , we can say that it is only possible when both zeroes are positive or both zeroes are negative.
Also, sum of the zeroes = -99 [sum of zeroes = b/a when f(x) = ax2 + bx + c]
The sum being negative implies that both zeroes are positive is not correct.
So, we conclude that both zeroses are negative.
Hence, the correct answer is option B.
(a) a = -7,b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6
Solution
The given quadratic equation is x2 + (a+1)x+b = 0.
Since the zeroes of the given equation are 2 and -3.
So,
α = 2 and β = -3
Now,
Sum of zeroes = -Coefficient of x/Coefficient of x2
⇒ 2+(-3)= -(a+1)/1
⇒ -1 = -a-1
⇒ a = 0
Product of zeroes = Constant term/Coefficient of x2
⇒ 2 × (-3) = b/1
⇒ b = -6
So, a = 0 and b =-6
Hence, the correct answer is option D.
35. Given that one of the zeroes of the cubic polynomials ax3 + bx2 + cx + d is zero, the product of the other two zeroes is
(a) -c/a
(b) c/a
(c) 0
(d) -b/a
Solution
Let p(x) = ax3 + bx2 + cx + d.
Now 0 is the zero of the polynomial .
So, p(0) = 0
⇒ d = 0
So,
p(x) = ax3 + bx2 + cx = x(ax2 + bx + c)
Putting p(x) = 0, we get
x = 0 or ax2 + bx + c = 0 ...(1)
Let α,β be the zeroes of ax3 + bx + c = 0
So,αβ = c/a
Hence, the correct answer is option B.
36. The zeroes of the quadratic polynomial x2 + ax + a,a 0,
(a) Cannot both be positive
(b) Cannot both be negative
(c) Are always unequal
(d) Are always equal
Solution
Let f(x) = x2 + ax + a.
Product of the zeroes of f(x) = a [Product of zeroes = c/a when f(x) = ax2 + bx + c]Since the product of zeroes is positive, so the zeroes must be either both positive or both negative.
Also, sum of the zeroes = -a [sum of zeroes =b/a when f(x) = ax2 + bx + c]
So, the sum of the zeroes is negative, so the zeroes cannot be both positive.
Hence, the correct answert is option B.
37. If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is -1, then the product of other two zeroes is
(a) b-a+1
(b) b-a-1
(c) a-b+1
(d) a-b-1
Solution
Let p(x) = x3 + ax2 + bx + c
Now, -1 is a zero of the polynomial .
So,p(-1) = 0.
⇒ (-1)3 + a(-1)2 + b(-1) + c = 0
⇒ -1 + a - b + c = 0
⇒ a - b + c = 1
⇒ c = 1 - a + b
Now, if α,β,γ are the zeroes of the cubic polynomial ax3 + bx2 + cx + d, then product of zeroes is given by
αβγ = -d/a
So, for the given polynomial, p(x) = x3 + ax2 + bx + c
αβ(-1) = -c/1 = -1(-a+b)/1
⇒ αβ = 1 - a + b
Hence, the correct answer is option A.
38. Given that two of zeros of the cubic polynomial ax3 + bx2 + cx + d are 0, the third zero is
(a) -(b/a)
(b) b/a
(c) c/a
(d) -(d/a)
Solution
Let the polynomial be f(x) = ax3 + bx2 + cx + d.
Suppose the two zeroes of f(x) are α = 0 and β = 0.
We know that,
Sum of the zeros,
α+β+γ = -b/a
⇒ 0+0+γ = -b/a
⇒ γ = -b/a
Hence, the correct answer is option A.
39. If one zero of the quadratic polynomial x2 + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5
Solution
Let the given polynomial be f(x) =x2 + 3x + k.
Since 2 is on e of the zero of the given polynomial, so(x-2) will be a factor of the given polynomial .
Now, f(2) = 0
⇒ 22 + 3 × 2 + k = 0
⇒ 4 + 6 + k = 0
⇒ k = -10
Hence, the correct answer is option B.
40. If the zeros of the quadratic polynomial of the form ax2 + bx + c, c ≠ 0 are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign
Solution
Let the given quadratic polynomial be f(x) = ax2 + bx + c.
Suppose α and β be the zeros of the given polynomial .
Since α and β are equal so they will have the same sign i.e. either both are positive or both are negative .
So, αβ > 0
But, αβ = c/a
∴ c/a > 0, which is possible only when both have same sign
Hence, the orrect answer is option c.
41. If one of the zeros of a quadratic polynomial of the form x2 + ax + b is the negative of the other, then it
(a) Has no linear term and constant term is negative
(b) Has no linear term and the constant term is positive .
(c) Can have a linear term but the constant term is negative .
(d) Can have a linear term but the constant term is positive .
Solution
It is given that, for quadratic polynomial
Solution
It is given that, for quadratic polynomial
x2 + ax + b . One zero is negative of other
Let one zero be α, then another.
which is negative of first is - α
We know
Sum of zeros = (-Cofficient of x)/(Cofficient of x2)
But, here sum of zeros
= α - α = 0
So,
for this case
0 = -a/1 (a is cofficient of x)
a = 0
a is cofficient of x, or linear term
which is zero .
So, linear term is zero
Now, if constant term is positive
Then, the expression, will be
x2 + bx + c = x2 + c
In order to find zero .
x2 + c = 0
⇒ x2 = -c
⇒ x2 = (same positive number)
⇒ x2 is negative which is not possible for real number x.
So, C cannot be positive
C is negative
C is constant term here
So constant term is negative and linear term is zero so, option (a) is correct .
42. Which of the following is not the graph of a quadratic polynomial ?
Solution
For a quadratic polynomial, ax2+bx+c, the zeros are precisely the x-coordinates of the points where the graph representing y = ax2+bx+c intersects the x-axis.
The graph has one of the two shapes either open upwards like ∪ (parabolic shape) or open downwards like ∩ (parabolic shape) depending on whether a > 0 or a < 0.
Three cases are thus possible:
a) graph cuts x-axis at two distinct points (two zeroes)
b) graph cuts the x-axis at exactly one point (one zero)
c) the graph is either completely above the x-axis or completely below the x-axis (no zeroes)
In option (d), the graph is cutting the x-axis at three distinct points and it is not a parabola opening either upwards or downwards.
So, option (d) does not represent the graph of a quadratic polynomial.
Hence, the correct answer is option D.