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(D) a = 2, b = – 6

Explanation:

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.

2 and - 3 are the zeroes of the polynomial p(x) = x² + (a + 1)x + b

i.e. p(2) = 0 and p(- 3) = 0

p(2) = (2)² + (a + 1)(2) + b = 0

= 4 + 2a + 2 + b = 0

= 6 + 2a + b = 0 (1)

P(- 3) = (- 3)² + 9 + (a + 1)(- 3) + b = 0

= 9 - 3a - 3 + b = 0

= 6 - 3a + b = 0 (2)

Equating (1) = (2), as both the equations are equal to zero. Hence both equations are equal to each other.

6 + 2a + b = 6 - 3a + b

= 5a = 0

⇒ a = 0

Putting the value of ‘a’ in (1)

6 + 2(0) + b = 0

⇒ b = - 6

OR

The equation of a quadratic polynomial is given by x² - (sum of the zeroes) x + (product of the zeroes)

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

⇒ sum of the zeroes = - (a + 1)

= 2 - 3 = - a - 1

= - 1 + 1 = - a

= - a = 0

⇒ a = 0

Product of the zeroes = constant term ÷ coefficient of x²

⇒ b = product of the zeroes

= 2(- 3)

= 6

Given: (x + a) is a factor of polynomial 2x² + 2ax + 5x + 10.

So, we have

x + a = 0

or x = –a , will satisfy the given polynomial.

Therefore, we will have

2 (–a)² + 2a(–a) + 5(–a) + 10 = 0

2a² –2a² – 5a + 10 = 0

– 5a = – 10

a = 2

The value of a is 2.

A Quadratic Equation will have equal roots if it satisfies the condition:

b² – 4ac = 0

Given equation is x² + kx + k = 0

a = 1, b = k, x = k

Substituting in the equation we get,

k² – 4 ( 1 ) ( k ) = 0

k² – 4k = 0

k ( k – 4 ) = 0

k = 0 , k = 4

But in the question, it is given that k is greater than 1.

Hence the value of k is 4 if the equation has common roots.

Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

α + β = -b/a and αβ = c/a

Given roots are: x = 2/3 and x = -3

and quadratic equation ax² + 7x + b = 0

Since x = 2/3 and x = -3 are roots of the above quadratic equation

Hence, will satisfy the given equation.

Step 1: At x = 2/3

a(2/3)² + 7(2/3) + b=0

4/9 a + 14/3 + b = 0

4a + 42 + 9b = 0 ………….equation (1)

Step 2: At x = –3

a(-3)² + 7(-3)+b=0

9a – 21 + b = 0 ………….equation (2)

Step 3: Solving equation (1) and equation (2), we get

a = 3, b = –6

Given: α and β are zeroes of f(x) = 6x² + x – 2

To find: (α/ β + β / α)

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x²) = -1/6

α β = Product of zeros = (constant term)/(coefficient of x²) = -1/3

Now,

(α/ β + β / α) = {(α + β)² – 2αβ) / αβ }

= (1/36 + 2/3) / (-1/3)

= -25/12

Given: α and β are the zeros of polynomial 2x² + 7x + 5

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x²) = -7/2

αβ = Product of zeros = (constant term)/(coefficient of x²) = 5/2

α + β + αβ = (α + β) + αβ 

= -7/2 + 5/2 

= -1

From the question, it’s given that: 

α and β are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1 

Sum of the roots = α+β = \(\frac{-b}{a}\)

= – \(\frac{(-7)}{5}\) 

= \(\frac{7}{5}\) 

Product of the roots = αβ 

= \(\frac{c}{a}\) 

= \(\frac{1}{5}\) 

\(\frac{1}{α} +\frac{1}{β}\) 

⇒ \(\frac{(α +β)}{ αβ}\)

⇒ \(\frac{7}{5} \over \frac{1}{5}\)

= 7

Correct option is C) 3

Correct option is (D) 1

If polynomial has only one zero then it cuts X-axis at exactly one point.

Correct option is D) 1

(1 + x) (1 + x) 

= (1 + x)² = 1² + 2 (1) (x) + x² 

[∵(x + y)² = x² + 2xy + y²] 

= 1 + 2x + x²

= (30 + 0.5) (30 – 0.5) 

= 30² – (0.5)²

= 900 – 0.25 

= 899.75

(x – 5) (x – 5) 

= (x – 5)² = x² – 2(x) (5) + 5² 

[ ∵(x – y)² = x² – 2xy + y² ] 

= x² – 10x + 25

(x + 5) (x + 2) 

= x² + (5 + 2)x + 5 x 2 

[ ∵ (x + a) (x + b) = x² + (a + b) x + ab] 

= x² + 7x + 10

(3x + 2) (3x – 2) = (3x)² – (2)² 

[∵ (x + y) (x – y) =x – y ] 

= 9x² – 4

999 x 999 

= 999² 

= (1000 – 1)² 

= 1000² – 2 x (1000) x 1 + 1² 

= 1000000-2000 + 1 

= 998001

If  x + 2 = 0

x = -2

f(x) = 2x⁴ – 6x³ + 2x² – x + 2, [By remainder theorem]

f(x) = 2(-2)⁴ – 6(-2)³ + 2(- 2)² – (- 2) + 2

= 2(16) – 6(- 8) + 2(4) + 2 + 2

= 32 + 48 + 8 + 2 + 2 

= 92

Hence, required remainder = 92.

Given , Polynomial f(x) = 2x⁴ - 2x² - x + 2 

g (x) = x + 2 = 0        x = -2 

Putting value of x in equation ,

f(-2) = 2x4 – 6x3 + 2x2 – x + 2, 

f(-2) = 2(-2)4 – 6(-2)3 + 2(- 2)2 – (- 2) + 2 

= 2(16) – 6(- 8) + 2(4) + 2 + 2 

= 32 + 48 + 8 + 2 + 2  = 92  

Hence, remainder = 92

(102)³ = (100 + 2)³

= (100)³ + (2)³ + 3(100) (2) (100 + 2)

= 1000000 + 8 + 600 (102)

= 1000000 + 8 + 61200

= 1061208

Correct option is A) ax + b

9x² - 12 xy + 4

= (3x)² - 2 x 3x x 2 + (2)²

= (3x - 2)²

= (3x  - 2) (3x  - 2)

Given (a + b + c) = 9 and (ab + bc + ca) = 26 

(a + b + c)² = 9² 

⇒ a² + b² + c² + 2(ab + bc + ca) = 81 

⇒ a² + b² + c² + 2(26) = 81 

⇒ a² + b² + c² + 52 = 81 

⇒ a² + b² + c²  = 81 − 52 = 29

Correct option is (D) 1

\((x – 3)^2\) = 0

\(\Rightarrow\) x - 3 = 0 or x - 3 = 0

\(\Rightarrow\) x = 3 or x = 3

Number of different zeros = 1

Correct option is D) 1

Correct option is (C) (1/2, 0)

At x-axis, functional value (y-value) is zero.

\(\therefore\) 2x - 1 = 0

\(\Rightarrow\) \(x=\frac{1}{2}\)

Hence, polynomial 2x - 1 cut x-axis at \((\frac{1}{2}\,,0).\)

Correct option is C) ( 1/2, 0)

(x – 3) (x – 4)² (x – 5) – 6 

= (x – 3) (x – 5) (x – 4)² – 6 

= (x² – 5x – 3x + 15) (x² – 8x + 16) – 6 

= (x² – 8x + 15) (x² – 8x + 16) – 6 

= (m + 15) (m+ 16) – 6 … [Putting x² – 8x = m]

 = m (m + 16) + 15 (m + 16) – 6 

= m² + 16m + 15m + 240 – 6 

= m² + 31m + 234 

= m² + 18m + 13m + 234

= m(m + 18) + 13(m + 18) 

= (m + 18) (m + 13)

= (x² – 8x + 18) (x² – 8x + 13) … [Replacing m = x² – 8x]

(i) p(x) = x² + x + k 

g(x) = x – 1 k = ? 

If x – 1 = 0, then x = 1 

p(x) = x² + x + k 

p(1) = (1)² + 1 + k 

p( 1) = 1 + 1 + k 

p( 1) = 2 + k 

If g(x) is a factor, then we have r(x) = 0 

∴ p(1) = 0 

2 + k= 0 

∴ k = 0 – 2 

k = -2 

(ii) p(x) = 2x² + kx + √2

g(x) = x – 1 k = ? 

If x – 1 = 0, then x = 1 

p(x) = 2x² + kx + √2

p(1) = 2(1)² + k(1) + √2 

= 2(1) + k(l) + √2

p(1) = 2 + k + √2

If (x – 1) is the factor of p(x), then we have p(1) = 0. 

∴ 2 + k + = 0 k = -2 – √2 = 0

(iii) p(x) = kx² – √2x + 1 

g(x) = x – 1 k = ? 

If x – 1 = 0, then x = 1 

p(x) = kx² – √2x + 1 

p(1) = k(1)²– √2(1) + 1 

= k( 1) – √2 + 1 

p(1) = k - √2 + 1 

If (x – 1) is the factor of p(x) then we have p(1) = 0. 

∴ p(1) = k – √2 + 1 = 0 

∴ k= √2 – 1 

(iv) p(x) = kx² – 3x + k 

g(x) = x – 1 k = ? 

If x – 1 = 0, then x – 1 

p(x) = kx² – 3x + k 

p(1) = k(1)² – 3(1) + k 

= k(1) – 3(1) + k 

= k – 3 + k p(1) 

= 2k – 3 

If (x – 1) is the factor of p(x), then we have p(1) = 0.

p(x) = 2x² – 3x + 1

p(0) = 2(0)² – 3(0) + 1 = 1

p(1) = 2(1)² – 3(1) +1 = 2 – 3 + 1 = 0

p(-1) = 2(-1)² – 3(-1) + 1 = 2 + 3 + 1 = 6

Correct option is (C) a = 2, b = 12

Given that polynomial \((x^4+ax^3-7x^2-8x+b)\) is completely divisible by \((x^2+5x+6).\)

If implies that \((x^2+5x+6)\) is a factor of \((x^4+ax^3-7x^2-8x+b).\)

\(\Rightarrow\) (x + 2) & (x + 3) are factors of \((x^4+ax^3-7x^2-8x+b)\)

\(\Rightarrow\) x = -2 & x = -3 are zeros of polynomial \((x^4+ax^3-7x^2-8x+b)\)

\(\therefore(-2)^4+a(-2)^3-7(-2)^2-8(-2)+b=0\)

and \((-3)^4+a(-3)^3-7(-3)^2-8(-3)+b=0\)

\(\Rightarrow\) 16 - 8a - 28 + 16 + b = 0 and 81 - 27a - 63 + 24 + b = 0

\(\Rightarrow\) 8a - b = 4 and 27a - b = 42

\(\Rightarrow\) (27a - b) - (8a - b) = 42 - 4

\(\Rightarrow\) 19a = 38

\(\Rightarrow\) a = \(\frac{38}{19}\) = 2

\(\therefore\) b = 8a - 4 = 16 - 4 = 12

Correct option is C) a = 2, b = 12

Correct option is (D) a = -1, b = -2

(x - 1) and (x + 1) are factors of \((x^3+2x^2+ax+b).\)

\(\Rightarrow\) x = 1 and x = -1 are zeros of \((x^3+2x^2+ax+b).\)

\(\therefore\) 1+2+a+b = 0 and -1+2-a+b = 0

\(\Rightarrow\) a+b = -3 and a - b = 1

\(\Rightarrow\) (a+b) + (a - b) = -3+1

\(\Rightarrow\) 2a = -2

\(\Rightarrow\) a = \(\frac{-2}2\) = -1

Then b = - 3 - a = -3 - (-1)

= -3+1 = -2

Correct option is D) a = -1, b = -2

Correct option is (A) g(1) = -17, g(2) = 19, g(3) = 759

\(\because g(x)=x^6+3x^4-24x^2+3\)

\(\therefore\) g(1) = 1+3-24+3 = -17,

\(g(2)=2^6+3.2^4-24.2^2+3\)

= 64 + 48 - 96 + 3

= 115 - 96 = 19 and

\(g(3)=3^6+3.3^4-24.3^2+3\)

= 729 + 243 - 216 + 3

= 975 - 216

= 759

Correct option is A) g(1) = -17, g(2) = 19, g(3) = 759