Test 02 || Class 10th Maths (2) || Ch. 02 Polynomials

Test 02 || Class 10th Maths (2) || Ch. 02 Polynomials

Q. 1. If one zero of a quadratic polynomial P(x) = kx² + 3x + k is 2, then what is the value of k.

Sol. We have (Px) = kx² + 3x + k

Since, 2 is a zero of the quadratic polynomial

(Px) = kx² + 3x + k

p(2) = 0

k(2)²+ 3 (2) + k = 0

4k+6+ k = 0

5k =− 6 

k = 6/–5

Q.2. Verify whether 2, 3 and ½ are the zeroes of the polynomial p(x)= 2x³–11x² +17x– 6.

Sol.

If 2, 3 and ½ are the zeroes of the polynomial p(x), then these must satisfy p(x) = 0

p(x)= 2x³–11x² +17x– 6.

(1) If x =2

p(2)= 2(2)³–11(2)² +17(2)– 6.

        = 2(8)–11(4) + 34– 6.

        = 16 –44 + 28.

        = –28 + 28.

        = 0

(2) If x =3

p(2)= 2(3)³–11(3)² +17(3)– 6.

        = 2(27)–11(9) + 51– 6.

        = 54 –99 + 45.

        = –45 + 45.

        = 0

(3) If x =½

p(2)= 2(½)³–11(½)² +17(½)– 6.

        = 2(⅛)–11(¼) + 51– 6.

        = 1/4 –11/4 + 17/2–24/4.

        =(1–11+34–24) /4

        = (35–35)/4

        = 0/4

        = 0

Hence, 2, 3, and ½ are the zeroes of p(x).

Q.3. Find the value for k for which x⁴ + 10x³ + 25x² +15x + k is exactly divisible by x + 7.

Sol. 

We have 

f(x) = x⁴ + 10x³ + 25x² +15x + k

If x + 7 is a factor then -7 is a zero of f(x) and x =− 7 satisfy f(x) = 0.

Thus substituting x =− 7 in f(x) and equating to zero then we have,

f(x) = 0

x⁴ + 10x³ + 25x² +15x + k =0

(-7)⁴+10(-7)³+25(-7)²+15(-7)+ k= 0 

2401+10(–7)³+25(–7)²+105 + k= 0 

2401–10(343) + 25(49) –105 + k = 0

2401–3430+1225–105 + k = 0

3626–3535  + k = 0

91 + k = 0

k =− 91

Q.4. The graph of a polynomial is shown in Figure, then the number of its zeroes is

(a) 3 (b) 1 (c) 2 (d) 4

Ans : [Board 2020 Delhi Basic]

Since, the graph cuts the x -axis at 3 points, the number of zeroes of polynomial p(x) is 3.

Q.5. Which of the following is not the graph of a quadratic polynomial?

Sol. 

As the graph of option (d) cuts x -axis at three points.

So, it does not represent the graph of quadratic polynomial.

Thus (d) is correct option.

Q.6. If α and β are the roots of ax²+ bx +c = 0 (a ≠ 0), then calculate α+β and α.β

Sol. 

We know that

Sum of the roots = – (coefficient of x)/(coefficient of x²)

Thus 

α + β =–b/a 

Product of the roots = Constent c)/(coefficient of x²)

Thus 

α × β =c/a 

Q.7. If one zero of the polynomial 3x²+8x+k is the reciprocal of the other, then value of k is

(a) 3 (b) -3 (c) 1/3 (d) –1/3 

Ans : [Board 2020 OD Basic]

Let the zeroes be 

α = α  

β = 1/ α . 

Product of zeroes, 

α×β = α ×1/ α 

c/a =1

k/3=1

k = 3

Thus (a) is correct option.

Q.8. If one root of the equation (k–1)x²+8x+k=0 is the reciprocal of the other then the value of k is ......... 

Ans : 

We have (k–1)x²+8x+k=0

Let one root be α , then another root will be 1/α 

Let the zeroes be 

α = α  

β = 1/ α . 

Product of zeroes, 

α×β = α ×1/ α 

c/a =1

3/(k–1)=1/1

k – 1 = 3 

k = 3 +1

k = 4 

Q.9. If zeroes of the polynomial x² + 4x + 2a are a and, 2/a then find the value of a .

Sol. 

Let one root be α , then another root will be 1/α 

Let the zeroes be 

α = a 

β = 2/a

Product of zeroes, 

α×β = c/a

a ×2/a = 2a/1 

2a=2

a=1

Thus a = 1

Q.10. Write the quadratic polynomial, the sum of whose zeroes is -5 and their product is 6.

Ans : [Board 2020 Delhi Standard]

Let α and β be the zeroes of the quadratic polynomial, 

then we have

α + β + =− 5

and α.β = 6

Now  

p (x) = x²–(α+β)x + (α.β)

p (x) = x²− (–5)x + (6)

p(x) = x² + 5x + 6

Q.11. Find a quadratic polynomial, the sum and product of whose zeroes are 6 and 9 respectively. Hence find the zeroes.

Sol.

Sum of zeroes, 

α +β  = 6

Product of zeroes 

α.β = 9

Now  

p (x) = x²–(α+β)x + (α.β)

p (x) = x²− (6)x + (9)

p(x) = x² – 6x + 9

Q.12. Find the quadratic polynomial whose sum and product of the zeroes are 21/8 and 5/16 respectively.

Sol.

Sum of zeroes, 

α + β =21/ 8

Product of zeroes 

α×β =5/16

Now 

p(x) = x²–(α+β)x + (α.β)

p(x) = x²− (21/8)x + (5/16)

p(x) = x²− 21x/8 + 5/16

p(x) =16/16 [x²− 21x/8 + 5/16]

p(x) =1/16 [16x²−16×21x/8 +16×5/16]

p(x) =1/16 [16x²−43x +5]

Q.13. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x)= ax² + bx + c, (a≠0), (c≠0).

Sol.

Let α and β be zeros of the given polynomial ax² + bx + c

α + β =–b/a and 

α×β = c/a

Let 1/α and 1/β be the zeros of new polynomial then we have

Sum of zeros, 

S = 1/α + 1/β = (β + α)/αβ 

= (–b/a)/(c/a) =–b/c

Product of zeros, 

P =1α×1/β = 1/αβ =1/(c/a)=a/c

Required polynomial,

g(x) = x² + Sx + P 

g(x) = x² + (–b/c)x + a/c 

g(x) = x² –bx/c + a/c 

g(x) = c/c[x² –bx/c + a/c] 

g(x) = 1/c[cx² –c.bx/c + c.a/c] 

g(x) = 1/c[cx² –bx + a] 

Q.14. The maximum number of zeroes a cubic polynomial can have, is

(a) 1 (b) 4 (c) 2 (d) 3

Sol.

A cubic polynomial has maximum 3 zeroes because its degree is 3.

Thus (d) is correct option.

Q.15. If one zero of the quadratic polynomial x²+ 3x+ k is 2, then the value of k is

(a) 10    (b) -10    (c) -7    (d) -2

Ans : [Board 2020 Delhi Standard]

We have p(x)=x²+ 3x+ k 

If 2 is a zero of p(x), then we have

p(2)=0

x²+ 3x+ k=0

(2)²+ 3(2)+ k=0

4+6+k=0

10 + k = 0 

k =− 10

Thus (b) is correct option

Q.17. The zeroes of the polynomial x²–3x –m (m+3) are

(a) m, m + 3 

(b) −m, m+3

(c) m,  m −(m + 3) 

(d) − m, –(m + 3)

Sol. 

We have p(x) = x²–3x –m (m+3)

Substituting x =− m in p (x) 

we have

p(–m) = (–m)²–3(–m) –m (m+3)

            = m²+3m–m²–3m

            = 0

Thus x =− m is a zero of given polynomial.

Now substituting x = (m + 3) in given polynomial we have

p(x) = (m + 3)²–3(m + 3) –m (m+3)

        = (m + 3)[(m+3)–3 –m]

        = (m + 3)[m + 3–3 –m]

        = (m + 3)[0]

        = 0

Thus x = m + 3 is also a zero of given polynomial.

Hence, -m and m + 3 are the zeroes of given polynomial.

Thus (b) is correct option.

Q.18. If the square of difference of the zeroes of the quadratic polynomial x² + px + 45 is equal to 144, then the value of p is

(a) ±9 (b) ±12 (c) ±15 (d) +18

Sol.

We have f(x) = x² + px + 45 

Then, α + β = –p/1 

           α + β = –p

and α×β = 45/1

        α×β = 45

According to given condition, we have

(α – β)² = 144

(α + β)² –4αβ = 144

(–p)² –4(45) = 144

(–p)² –180 = 144

p² = 144 +180 

p² = 324

p = √324 

p = ± 18

Q.19. If the zeroes of the quadratic polynomial x²+( a +1) x + b are 2 and –3, then

(a) a =− 7  b= –1 

(b) a = 5    b= –1 

(c) a = 2    b= –6 

(d) a = 0   b= –6 

Sol.

If a is zero of the polynomial, then f(a) = 0.

Here, 2 and –3 are zeroes of the polynomial 

x² + ( a +1) x + b 

So, f(2)= 2² + (a+1)2 + b = 0

                4 + 2a + 2 + b = 0

                         2a + b   =− 6       ...(1)

x² + ( a +1) x + b 

So, f(–3)= (–3)² + (a+1)–3 + b = 0

         f(–3)= (–3)² –3(a+1) + b = 0

                9 – 3a – 3 + b = 0

                         –3a + b  =− 6       

                            3a – b  = 6       ...(2)

Adding equations (1) and (2), we get

                         2a + b   =− 6 

                         3a – b   = 6  

                          5a = 0 

                            a = 0

Substituting value of a in equation (1), we get

2a + b   =− 6 

2(0) + b   =− 6 

Hence, a = 0 and b =− 6.

Thus (d) is correct option.

Q.20. The zeroes of the quadratic polynomial x²+99x +127 are

(a) both positive

(b) both negative

(c) one positive and one negative

(d) both equal

Sol.

Let f(x) = x²+99x +127

Comparing the given polynomial with ax²+ bx+ c, we get a = 1, b = 99 and c = 127.

Sum of zeroes 

α+ β =–b/a 

=− 99

Product of zeroes 

α×β =c/a

c = 127

Now, product is positive and the sum is negative, so both of the numbers must be negative.

Alternative Method :

Let f(x) = x²+99x +127

Comparing the given polynomial with ax²+ bx+ c, we get a = 1, b = 99 and c = 127.

Now by discriminant rule,

So, the zeroes of given polynomial,

Now,as 99 > 96.4

So, both zeroes are negative.

Thus (b) is correct option.

Q.21. If α and β are the zeroes of a quadratic polynomial such that α β + = 24 and α β − = 8. Find the quadratic polynomial having α and β as its zeroes.

Sol.

We have 

α + β = 24 ...(1)

α – β  = 8 ...(2)

Adding equations (1) and (2) we have

α + β = 24 

α – β  = 8 

2α = 32 

α = 16

Substute α = 16 in equation (1) 

α + β = 24 

16 + β = 24 

β = 24–16  

β = 8

α = 16 

Hence, the quadratic polynomial

p(x) = x²–(α + β)x + αβ

p(x) = x²–(16 + 8)x + 16×8

p(x) = x²–(24)x + 128

p(x) = x²–24x + 128

Q.22. Polynomial x⁴ + 7x³ + 7x² +px + q is exactly divisible by x² +7x + 12, then find the value of p and q .

Sol.

We have 

f(x) = x⁴ + 7x³ + 7x² +px + q 

Now 

x² +7x + 12 = 0

x² +3x + 4x + 12 = 0

x(x +3) + 4(x+3) = 0

(x +3) (x+4) = 0

(x +3)= 0 and (x+4) = 0

x = –4 and x = –3

x =−4,–3

Since f(x) = x⁴ + 7x³ + 7x² + px + q is exactly divisible by x² +7x + 12, then x =− 4 and x =−3 must be its 

zeroes and these must satisfy 

f(x) = 0

So putting x =− 4 and x =− 3 in f(x) and equating to zero we get

f(x) = 0

x⁴ + 7x³ + 7x² + px + q = 0

x =− 4

(–4)⁴ + 7(–4)³ + 7(–4)² + p(–4)+q = 0

(256) + 7(–64) + 7(16) – 4p + q = 0

256 – 448 – 112 – 4p + q = 0

− 4p + q − 80 = 0

4p – q =− 80     ...(1)

x =− 3

(–3)⁴ + 7(–3)³ + 7(–3)² + p(–3)+q = 0

(81) + 7(–27) + 7(9) – 3p + q = 0

81 – 189 + 63 – 3p + q = 0

− 3p + q − 45 = 0

3p – q =− 45      ...(2)

Subtracting equation (2) from (1) we have

[4p – q =− 80] – [3p – q =− 45]

[4p – q =− 80] – 3p + q =+ 45]

p =− 35

Substituting the value of p in equation (1) we have

4p – q =− 80   

4(–35) – q =− 80   

–140 – q =− 80

–q = – 80 +140

or – q = 60

q =− 60

Hence, p =− 35 and q =− 60.

Q.23. If α and β are the zeroes the polynomial 2x²–4x + 5, find the values of

Sol.

We have p(x) = 2x²–4x + 5

If α and β are then zeroes of p(x) = 2x²–4x + 5

then

α+β =–(–4)/2

α+β = 4/2

α+β = 2

and

α.β = 5/2

(i) α²+ β² =  (α+β)² – 2.αβ

                 =  (2)² – 2.(5/2)

                 =  4 – 5

                 =  – 1