Question

Q.No: 1. If the zeroes of the quadratic polynomial x² + (a+1)x + b are 2 , -3,then the values of a & b are -------
1. -7 , -1
2. 5 , -1
3. 2 , -6
4. 0 , -6

Q.No: 2. Zeroes of a polynomial p(x) = x² - 2x + 1 are ------------
1. 1,2
2. -1,-2
3. 1 ,1
4. -1,-1

Q.No: 3. If α and β are the zeroes of the polynomial p(x) = 2x² - 7x + 3, then α² + β² = -----------
1. 4/37
2. 37/4
3. -37/4
4. -4/37

Q.No: 4. If α and β are the zeroes of the polynomial p(x) = kx² + 3x +2 , then 1/α + 1/β = ---------------
1. 3/2
2. -3/2
3. 2/3
4. -2/3

Q.No: 5. Which of the following is a quadratic polynomial whose zeroes are reciprocal of zeroes of the polynomial p(x) = x² - 4x -5 ?
1. x² - 4x/5 - 1/5
2. x² + 4x/5 + 1/5
3. x² + 4x/5 - 1/5
4. x² - 4x/5 + 1/5

Q.No: 6. Which of the following is a quadratic polynomial whose zeroes are 5 + √2 and 5 - √2 ?
1. x² - 10x - 23
2. x² + 10x +23
3. x² - 10x + 23
4. x² + 10x - 23

Q.No: 7. If α , β are the zeroes of the polynomial ax² - 5x+ c such that α + β = α β = 10, then the values of a and c are ------------
1. 2 , 5
2. 1/2 , 5
3. 1/2 , -5
4. -2 , 5

Q.No: 8. If zeroes of the polynomial p(x) = 2 x² - x + k are reciprocal of each other, then the value of k is ---------
1. -1
2. 0
3. 2
4. None of these

Q.No: 9. ANSWER FROM IMAGE

1. ANSWER 1
2. ANSWER 2
3. ANSWER 3
4. ANSWER 4

Q.No: 10. If zeroes of the polynomial x² + 4x + 2a are α and 2/α , then the value of a is ---------
1. 0
2. 2
3. -1
4. 1

Q.No: 11. Which of the following algebraic expression is not a polynomial?
1. x² + √3 x + 7
2. 7
3. x - 3 + x²/4
4. 3 + 1/x

Q.No: 12. The graph of a polynomial p(x) does not intersects the x axis but intersects y axis at one point, then the number of zeroes of p(x) is -----------
1. 0
2. 1
3. 2
4. None of these

Q.No: 13. Which of the following are the zeroes of the polynomial p(x) = √3x² - 8x + 4√3
1. 6/√3 , 2/√3
2. -2/√3 , -6/√3
3. -2√3 , -6√3
4. 2√3 , 6√3

Q.No: 14. If the product of zeroes of x² - 3kx + 2 k² - 1 is 7, then the values of k are ----------
1. ±1
2. ±2
3. ±4
4. ±3

Q.No: 15. If zeroes of p(x) = ax² + bx + c are negative reciprocal of each other, then their product is ----------
1. 1
2. 0
3. -1
4. 1/2

Q.No: 16. Polynomial (x - 1) (x + 2) (x + 3 ) has
1. 3 real zeroes
2. no real zeroes
3. 2 real zeroes
4. 4 real zeroes

Q.No: 17. Zeroes of a polynomial p(x) = 6x² - 24 is
1. √2 , -√2
2. 2 , -2
3. 4 , -4
4. None of these

Q.No: 18. If sum of zeroes of the polynomial p(x) = x² + kx + 1 is zero, then k= -------------
1. 0
2. 1
3. -1
4. ±1

Q.No: 19. If one zero of the polynomial x² - 4x + 1 is 2 + √3, then the other zero is -----------
1. √3 - 2
2. √3 + 2
3. 2 - √3
4. √2 - 3

Q.No: 20. Which of the following is a quadratic polynomial whose zeroes are -2 and 5?
1. x² - 3x - 10
2. 3x² - 9x - 30
3. (x²/2) - (3x/2) - 5
4. All the above

Q.No: 21. The zeroes of the quadratic polynomial x² + 99x + 127 are
1. both positive
2. both negative
3. one positive & one negative
4. None of these

Q.No: 22. For what value of p, -4 is a zero of the polynomial x² - 2x - ( 7p + 3 ) ?
1. 3
2. -3
3. 7
4. -7

Q.No: 23. The graph of a cubic polynomial intersects x axis at ------------ points.
1. exactly 3
2. atleast 3
3. atmost 3
4. more than 3

Q.No: 24. For polynomial ( k² - 25 ) x² + 3x + 8 to be quadratic, which of the following is true?
1. k ≠ 5
2. k ≠ -5
3. k ≠ ± 5
4. None of these

Q.No: 25. A polynomial of degree zero is known as------
1. Linear polynomial
2. zero polynomial
3. constant polynomial
4. none of these

Answer 1

Step-by-step explanation:

Solution1:

Put x=2

$4 + 2(a + 1) + b = 0 \\ \\ 2a + b = - 6$

Put x=-3

$9 - 3(a + 1) + b = 0 \\ \\ - 3a + b = - 6$

Subtract both equations,

$5a = 0 \\ a = 0 \\ \\ 2a + b = - 6 \\ 2(0) + b = - 6 \\ \\ b = - 6$

Thus,Option 4 is correct ;(0,-6)

Solution2:

${x}^{2} - 2x + 1 = 0 \\ \\ {x}^{2} - x - x + 1 = 0 \\ \\ x(x - 1) - 1(x - 1) = 0 \\ \\ (x - 1)(x - 1) = 0 \\ x = - 1$

Both the zeros are -1.

Option 4 is correct.

Solution3:

$\alpha + \beta = \frac{7}{2} ...eq1 \\ \\ \alpha \beta = \frac{3}{2} \\ \\ squaring \: eq1 \: both \: sides \\ \\ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{49}{4} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{49}{4} - 2 \times \frac{3}{2} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{49}{4} - 3 = \frac{49 - 12}{4} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{37}{4}$

Option 2 is correct.

Solution4:

$\alpha + \beta = \frac{ - 3}{k} \\ \\ \alpha \beta = \frac{2}{k} \\ \\ divide \: eq2 \: by \: eq1 \\ \\ \frac{ \alpha + \beta }{ \alpha \beta } = \frac{ - 3}{2} \\ \\ or \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ - 3}{2}$

Option 2 is correct.

Solution 5:

${x}^{2} - 4x - 5 \\ \\ {x}^{2} - 5x + x - 5 = 0 \\ \\ x(x - 5) + 1(x - 5) = 0 \\ \\ (x - 5)(x + 1) = 0 \\ \\ zeroes \: are \\ x = 5 \\ x = - 1$

Reciprocal of zeros

$\frac{1}{5} \: - 1$

Equation is

$(x - \frac{1}{5} )(x + 1) = 0 \\ \\ {x}^{2} + \frac{4}{5} x - \frac{1}{5}$

Option 3 is correct.

Solution 6:

$x = 5 + \sqrt{2} \\ \\ factor \: (x - 5 - \sqrt{2} ) \\ \\ x = 5 - \sqrt{2} \\ \\ factor \: (x - 5 + \sqrt{2} ) \\ \\ equation \: is \\ (x - 5 - \sqrt{2} )(x - 5 + \sqrt{2} ) \\ \\ ( {x - 5)}^{2} - ( { \sqrt{2} )}^{2} \\ \\ {x}^{2} - 10x + 25 - 2 \\ \\ {x}^{2} - 10x + 23$

Option 3 is correct.

Solution 7:

$\alpha + \beta = \frac{5}{a} = 10 \\ \\ a = \frac{5}{10} \\ \\ a = \frac{1}{2} \\ \\ \alpha \beta = \frac{c}{a} = 10 \\ \\ \frac{c}{ \frac{1}{2} } = 10 \\ \\ c = \frac{10}{2} \\ \\ c = 5$

Option 2 is correct.

Solution 8:

Let one zero is alpha then other is 1/alpha

$\alpha \times \frac{1}{ \alpha } = \frac{k}{2} \\ \\ \frac{k}{2} = 1 \\ \\ k = 2$

Option 3 is correct.

Solution9: No image is present

Solution 10:

Product of zeros

$\alpha\times\frac{2}{\alpha}=2a\\\\2a=2\\\\a=1$

Option 4 is correct.

Solution 11: Option 4 is correct.

3+1/x; is not a polynomial

Solution 12:0

Option 1 is correct.

Solution 13:

p(x) = √3x² - 8x + 4√3

$x_{1,2}=\frac{8±\sqrt{(64-48)}}{2\sqrt{3}}\\\\x_{1,2}=\frac{8±4}{2\sqrt{3}}\\\\x_1=\frac{6}{\sqrt{3}}\\\\x_2=\frac{2}{\sqrt{3}}$

Option 1 is correct.

Solution 14:x² - 3kx + 2 k² - 1

let zeros are alpha and beta

$\alpha\times\beta=2k^{2}-1=7\\\\2k^{2}=8\\\\k^{2}=4\\\\k=±2$

Option 2 is correct.

Solution 15:-1,Option 3 is correct

Solution 16: Zeroes are 1,-2 and -3

three real zeroes

Option 1 is correct.

Solution 17:p(x) = 6x² - 24

6(x² -4)=0

x² =4

x=±2

Option 2 is correct

Solution 18:

$\alpha + \beta =-k=0(given)\\\\k=0$

Option 1 is correct

Solution 19:Option 3; (2-√3) is correct

these irrational zeros are occurs in pairs.

Solution20: All have these zeros,so option 4(All the above),is correct

Solution 21:Find D

$D=b^2-4ac\\\\=9801-508>0$

Real and distinct zeros,but for finding sign,one have to find the zeroes

Now find the zeros

$x_{1,2}=\frac{-99±\sqrt{(9293)}}{2}\\\\x_{1,2}=\frac{-99±96.40}{2}\\\\x_1=-1.29\\\\x_2=-97.70$

Both zeros are negative

Option 2 is correct.

Q.No: 22. For what value of p, -4 is a zero of the polynomial x² - 2x - ( 7p + 3 ) ?

1. 3

2. -3

3. 7

4. -7

Solution22: Put x= -4

16+8-7p-3=0

-7p=-21

p=3

Option 1 is correct.

Q.No: 23. The graph of a cubic polynomial intersects x axis at ------------ points.

1. exactly 3

2. at least 3

3. at most 3

4. more than 3

Solution : Exactly 3 points

Q.No: 24. For polynomial ( k² - 25 ) x² + 3x + 8 to be quadratic, which of the following is true?

1. k ≠ 5

2. k ≠ -5

3. k ≠ ± 5

4. None of these

Solution:k ≠ ± 5

Option 3 is correct

Q.No: 25. A polynomial of degree zero is known as------

1. Linear polynomial

2. zero polynomial

3. constant polynomial

4. none of these

Solution: Constant polynomial

Option 3 is correct

Hope it helps you.