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QUESTION: 1

If α,β and γ are the zeros of the polynomial 2x^{3} – 6x^{2} – 4x + 30, then the value of (αβ + βγ + γα) is

Solution:

We have a 3 degree polynomial here of form

ax^{3} + bx^{2} + cx +d = 0

Polynomial p(x) = 2x^{3} – 6x^{2} – 4x + 30

The roots of this polynomial are α, β, γ

We have formula for products of zeroes taking 2 at a time = αβ + βγ + αγ = c/a

αβ + βγ + αγ = -4/2 = -2

QUESTION: 2

The quadratic polynomial whose sum of zeroes is 3 and product of zeroes is –2 is :

Solution:

Sum of zeroes = 3

Product of zeroes = -2

Let the quadratic polynomial be ax^{2} + bx + c

Sum of zeroes = 3

-b/a = 3/1

So a = 1, b = -3

Product of zeroes = -2

c/a = -2/1

So a = 1, c = -2

Hence the quadratic polynomial will be x^{2} - 3x - 2

QUESTION: 3

If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is

Solution:

Given ,g(x)=x+2=0

x=-2

P(x)=x^{3}-2ax^{2}+16

p(-2)=(-2^{3}) -2 × a × (-2^{2})+16

=-8 - 8a +16

=8 - 8a

=>-8a = -8

=>a=1

QUESTION: 4

If α,β and γ are the zeros of the polynomial f(x) = x^{3} + px^{2} – pqrx + r, then =

Solution:

QUESTION: 5

If the parabola f(x) = ax^{2} + bx + c passes through the points (–1, 12), (0, 5) and (2, –3), the value of a + b + c is –

Solution:

f(x)=ax^{2}+bx+c

It is given that f(0)=5, f(−1)=12 and f(2)=−3

f(0)=c=5 ... (i)

f(−1)=a−b+c=12

∴a−b=12−c

∵c=5

∴a−b=12−5 =7 ... (ii)

f(2)=4a+2b+c=−3

⇒4a+2b=−3−c

as c=5

⇒4a+2b=−8

⇒2a+b=−4 ... (iii)

Solving (ii) and (iii), we get

a=1

b=−6

Hence, a=1,b=−6,c=5

Thus, a+b+c =1−6+5 =0.

QUESTION: 6

If a, b are the zeros of f(x) = x^{2} + px + 1 and c, d are the zeros of f(x) = x^{2} + qx + 1 the value of E = (a – c) (b – c) (a + d) (b + d) is –

Solution:

The given polynomials are

f(x) = x² + px + 1

g(x) = x² + qx + 1

Since a, b are the zeroes of f(x),

a + b = - p ..... (1)

ab = 1 ..... (2)

Since c, d are the zeroes of g(x),

c + d = - q ..... (3)

cd = 1 ..... (4)

Now, E = (a - c) (b - c) (a + d) (b + d)

= {ab - (a + b)c + c²} {ab + (a + b)d + d²}

= (1 + pc + c²) (1 - pd + d²) [by (1) & (2)]

= 1 - pd + d² + pc - p²cd + pcd² + c² - pc²d + c²d²

= 1 - pd + d² + pc - p²(cd) + p(cd)d + c² - pc(cd) + (cd)²

= 1 - pd + d² + pc - p² + pd + c² - pc + 1 [by (4)]

= 2 + c² + d² - p²

= 2 + (c + d)² - 2(cd) - p²

= 2 + (- q)² - 2 - p² [by (3) & (4)]

= q² - p²

QUESTION: 7

If α,β are zeros of ax^{2 }+ bx + c then zeros of a3x^{2} + abcx + c^{3} are –

Solution:

QUESTION: 8

Let α,β be the zeros of the polynomial x^{2} – px + r and be the zeros of x^{2} – qx + r. Then the value

of r is –

Solution:

Since a and b are roots of x^{2} – px + r = 0

We can say a+b=p....say equation 1 and ab=r

Also a/2 and 2b are roots of x^{2} – qx + r

So, a/2 +2b=q ..... say equation 2 and a/2*2b=ab=r

2(equation1)-equation 2 = 2a -a/2 +2b-2b = 2p-q

so 3a/2=2p-q

a=2(2p-q)/3

Substituting in equation 1

We get b=p-4p/3+2q/3=(2q-p)/3

So we know ab=r=2(2p-q)(2q-p)/9

QUESTION: 9

When x^{200} + 1 is divided by x^{2} + 1, the remainder is equal to –

Solution:

When x^{200 }+ 1 is divided by x^{2} + 1, the remainder is equal to

x²⁰⁰ + 1 = x²⁰⁰ + (x¹⁹⁸ - x¹⁹⁸ - x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴ - x¹⁹⁴ - x¹⁹² + x¹⁹² +x¹⁹⁰ - x¹⁹⁰ +..................-x⁴ + x⁴ + x² - x² - 1 + 1 )+ 1

=> x²⁰⁰ + 1 = x²⁰⁰ + x¹⁹⁸ - x¹⁹⁸ - x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴ - x¹⁹⁴ - x¹⁹² + x¹⁹² +x¹⁹⁰ - x¹⁹⁰ +..................-x⁴ + x⁴ + x² - x² - 1 + 1 + 1

=> x²⁰⁰ + 1 = x¹⁹⁸(x² + 1) - x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1) - x¹⁹²(x² +1) + x¹⁹⁰(x² +1) - x¹⁸⁸(x²+1) +..................+ x²(x² + 1) - (x² + 1) + 1 + 1

=> x²⁰⁰ + 1 = x¹⁹⁸(x² + 1) - x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1) - x¹⁹²(x² +1) + x¹⁹⁰(x² +1) - x¹⁸⁸(x²+1) +..................+ x²(x² + 1) - 1.(x² + 1) + 2

=> x²⁰⁰ + 1 = (x² + 1)(x¹⁹⁸- x¹⁹⁶ + x¹⁹⁴ - x¹⁹² + x¹⁹⁰ - x¹⁸⁸+..................+ x² - 1) + 2

=> when x²⁰⁰ + 1 is divided by x² + 1 the reminder = 2

QUESTION: 10

If a (p + q)^{2} + 2bpq + c = 0 and also a(q + r)^{2} + 2bqr + c = 0 then pr is equal to –

Solution:

QUESTION: 11

If a,b and c are not all equal and α and β be the zeros of the polynomial ax^{2} + bx + c, then value of (1 + α + α^{2}) (1 + β + β^{2}) is :

Solution:

QUESTION: 12

Two complex numbers α and β are such that α + β = 2 and α^{4} + β^{4} = 272, then the polynomial whose zeros

are α and β is –

Solution:

QUESTION: 13

If 2 and 3 are the zeros of f(x) = 2x^{3} + mx^{2} – 13x + n, then the values of m and n are respectively –

Solution:

QUESTION: 14

If α,β are the zeros of the polynomial 6x^{2} + 6px + p^{2}, then the polynomial whose zeros are (α + β)^{2} and (α - β)^{2} is

Solution:

QUESTION: 15

If c, d are zeros of x^{2} – 10ax – 11b and a, b are zeros of x^{2} – 10cx – 11d, then value of a + b + c + d is –

Solution:

QUESTION: 16

If the ratio of the roots of polynomial x^{2} + bx + c is the same as that of the ratio of the roots of x^{2} + qx + r, then –

Solution:

QUESTION: 17

If the roots of the polynomial ax^{2} + bx + c are of the form and then the value of (a + b + c)^{2} is–

Solution:

QUESTION: 18

If α, β and γ are the zeros of the polynomial x^{3} + a_{0}x^{2} + a_{1}x + a_{2}, then (1 – α^{2}) (1 – β^{2}) (1 – γ^{2}) is

Solution:

QUESTION: 19

If α,β,γ are the zeros of the polynomial x^{3} – 3x + 11, then the polynomial whose zeros are (α + β), (β + γ) and

(γ + α) is –

Solution:

QUESTION: 20

If α,β,γ are such that α + β + γ = 2, α^{2} + β^{2} + γ^{2} = 6, α^{3} + β^{3} + γ^{3} = 8, then α^{4} + β^{4} + γ^{4} is equal to–

Solution:

(α+β+γ)^{2}=α^{2}+β^{2}+γ^{2}+2(αβ+βγ+αγ)

⇒4=6+2(αβ+βγ+αγ)

⇒αβ+βγ+αγ=−1

(α^{3}+β^{3}+γ^{3}−3αβγ)=(α+β+γ)(α^{2}+β^{2}+γ^{2}−αβ−βγ−αγ)

⇒8−3αβγ=2(6+1)

⇒αβγ=−2

(α^{2}+β^{2}+γ^{2})^{2}=α^{4}+β^{4}+γ^{4}+2(α^{2}β^{2}+β^{2}γ^{2}+α^{2}γ^{2})

⇒α^{4}+β^{3}+γ^{4}+2[(αβ+βγ+αγ)^{2}−2αβγ(α+β+γ)]

⇒36=α^{4}+β^{4}+γ^{4}+2[(−1)^{2}−2(−2(2)]

⇒α^{4}+β^{4}+γ^{4}=36−18

⇒18

Hence (C) is the correct answer

QUESTION: 21

If α,β are the roots of ax^{2} + bx + c and α + k, β + k are the roots of px^{2} + qx + r, then k =

Solution:

QUESTION: 22

If α,β are the roots of the polynomial x^{2} – px + q, then the quadratic polynomial, the roots of which are (α^{2}– β^{2}) (α^{3} – β^{3}) and (α^{3}β^{2} + α^{2}β^{3}) :

Solution:

If α,β are the roots of the polynomial x^{2} - px + q.

α + β = p and αβ = q

if the roots are (α^{2} -β^{2})(α^{3} - β^{3}) and (α^{3}β^{2} + α^{2}β^{3})

= (α^{2} -β^{2})(α^{3} - β^{3}) + (α^{3}β^{2} + α^{2}β^{3})

= (α +β)(α -β)2(α^{2} + αβ+ β^{2}) + α^{2}β^{2}(α + β).

= (α +β)[ (α +β)^{2} - 4αβ ] [ (α +β)^{2} - αβ ] + α^{2}β^{2}(α + β).

= p( p^{2} - 4q)( p^{2} - q) + q^{2}p

= p( p4 - p^{2}q -4p^{2}q + 4q^{2}) + pq^{2}

= p^{5} - 5p^{3}q + 5pq^{2} .

⇒ (α^{2} -β^{2})(α^{3} - β^{3})(α^{3}β^{2} + α^{2}β^{3}) = (p^{6}q^{2}-5p^{4}q^{3}+4p^{2}q^{4})

∴ x^{2}- (p^{5} - 5p^{3}q + 5pq^{2} ) x + ( p^{6}q^{2} - 5p^{4}q^{3} + 4p^{2}q^{4} ) = 0

QUESTION: 23

The condition that x^{3} – ax^{2} + bx – c = 0 may have two of the roots equal to each other but of opposite signs is:

Solution:

QUESTION: 24

If the roots of polynomial x^{2} + bx + ac are α,β and roots of the polynomial x^{2} + ax + bc are α,γ then the values of α,β,γ respectively are –

Solution:

α+β=−b

αβ=ac

Similarly,

α+γ=−a

αγ=bc

Now α is the common root.

Therefore, from the equations,

αβ=ac and αγ=bc

We get the common root α=c

Therefore

β=a and γ=b

Since α+β=−b

Hence

c+a=−b

a+b+c=0

QUESTION: 25

If one zero of the polynomial ax^{2} + bx + c is positive and the other negative then (a,b,c εR, a = 0)

Solution:

if α and β are two roots of equation ax^{2}+bx+c=0

product of roots α*β = c/a

now if one root is negative and other positive, then product of the roots should be negative.

i.e c/a is negative which means both a and c are of opposite signs.

QUESTION: 26

If α,β are the zeros of the polynomial x^{2} – px + q, then is equal to –

Solution:

QUESTION: 27

If α,β are the zeros of the polynomial x^{2} – px + 36 and α^{2} + β^{2} = 9, then p =

Solution:

sum of roots α+β= -b/a = p

product of roots αβ = c/a= 36

α^{2}+β^{2}= 9

α^{2}+β^{2 }+2αβ - 2αβ = 9

(α+β)^{2 }- 2αβ = 9

p^{2}= 9 + 2*36

p^{2 }= 81

p = 9

QUESTION: 28

If α,β are zeros of ax^{2} + bx + c, ac = 0, then zeros of cx^{2} + bx + a are –

Solution:

α+β = -b/a (eq1) αβ = c/a (eq 2)

let the second equation cx^{2} + bx + a has roots x and y

x+y = -b/c (eq 3) xy = a/c (eq 4)

divide eq 1 by eq 3

α+β / x+y = (-b/a) / (-b/c)

= c/a

= αβ

α+β / αβ = x+y

x+y = 1/α + 1/β

QUESTION: 29

A real number is said to be algebraic if it satisfies a polynomial equation with integral coefficients. Which of the following numbers is not algebraic :

Solution:

a **transcendental number** is a complex number that is not algebraic—that is, not a root (i.e., solution) of a nonzero polynomial equation with integer or equivalently rational coefficients. The most popular transcendental numbers are π and *e*

QUESTION: 30

The bi-quadraic polynomial whose zeros are 1, 2, 4/3, - 1 is :

Solution:

As the roots are 1,2,4/3,-1

Therefore the equation will be

(x-1)(x-2)(x-4/3)(x+1)

If we will solve this thing then we will get

3x^{4} – 10x^{3} + 5x^{2} + 10x – 8

QUESTION: 31

The cubic polynomials whose zeros are 4,3/2 and – 2 is :

Solution:

roots are 4,3/2 and – 2 so we can form equation as

(x-4)(x-3/2)(x-2)=0

which will give 2x^{3} – 7x^{2} – 10x + 24

QUESTION: 32

If the sum of zeros of the polynomial p(x) = kx^{3} – 5x^{2} – 11x – 3 is 2, then k is equal to :

Solution:

Let α,β,γ be the zeroes of given polynomial p(x)......

then,

α+β+γ= -b/a = 2 (given)

=> 2 = -(-5)/k

=> k =5/2

QUESTION: 33

If f(x) = 4x^{3} – 6x^{2} + 5x – 1 and α, β and γ are its zeros, then αβγ =

Solution:

for a cubic equation product of roots is α,β,γ = -d/a

= -(-1)/4 = 1/4

QUESTION: 34

Consider f(x) = 8x^{4} – 2x^{2} + 6x – 5 and α,β,γ,δ are it's zeros then α + β + γ + δ =

Solution:

for a equation of 4 roots ax^{4}+bx^{3}+cx^{2}+dx+e=0

sum of roots is -b/a

in our equation b = 0 (since there is no term of x^{3})

so sum of roots is -b/a=0

QUESTION: 35

If x^{2} – ax + b = 0 and x^{2} – px + q = 0 have a root in common and the second equation has equal roots, then –

Solution:

x^{2} – ax + b = 0 and x^{2} – px + q = 0 are the equations

second equation has two equal roots let the roots be k

2k = p => k = p/2 and k^{2}=q

now both equations has one common root k

for first equation k+ l = a => l = a- p/2 and kl = b

put k and l

p/2 (a-p/2) = b

ap/2 - p^{2}/4 = b

ap/2 = k^{2}+b

ap/2 = q+b

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