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(d) 2, 1

px³ + x² – 2x – q is divisible by (x – 1) and (x + 1) 

⇒ p(1)³ + (1)² – 2(1) – q = 0 ⇒ p – q = 1               ...(i) 

and p(–1)³ + (–1)² – 2(–1) – q = 0 ⇒ p + q = 3        ...(ii) 

Solving (i) and (ii) p = 2, q = 1.

i. 5 + 3x⁴

Here, the highest power of x is 4. 

∴Degree of the polynomial = 4 

ii. 7 = 7x° 

∴ Degree of the polynomial = 0

 iii. ax⁷+ bx⁹ 

Here, the highest power of x is 9.

∴Degree of the polynomial = 9

(d) 6.(2x – 3y)(3y – 4z)(2z – x).

Since (2x – 3y) + (3y – 4z) + (4z – 2x) = 0, therefore, 

(2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³         (Take out 2 common)

= 3.(2x – 3y)(3y – 4z) (4z – 2x) 

= 6.(2x – 3y)(3y – 4z)(2z – x).

(b) – 4.

Let f(x) = 3x³ – kx² + 4x + 16. Then, f(x) will be divisible by \(\big(\)x – \(\frac{k}{2}\)\(\big)\) if f\(\big(\frac{k}{2}\big)\) = 0 

⇒ 3.\(\big(\frac{k}{2}\big)\)³ – k.\(\big(\frac{k}{2}\big)\)² + 4\(\big(\frac{k}{2}\big)\) + 16 = 0

⇒ \(\frac{3k^3}{8}\) - \(\frac{k^3}{4}\) + \(\frac{4k}{2}\) + 16 = 0

⇒ \(\frac{3k^3-2k^3+16k+128}{8}\) = 0

⇒ k³ + 16k + 128 = 0 ⇒ (k + 4) (k² – 4k + 32) = 0 

⇒ k + 4 = 0 ⇒ k = – 4.

Given that,

p() =x² – 4x + 3

According to the question,

When x = 2,

p(x) = p(2)

p() =x² – 4x + 3

Substituting x = 2,

p(2) = (2)² – 4(2) + 3

= 4 – 8 + 3

= – 4 + 3

= – 1

When x = – 1,

p(x) = p(– 1)

p() =x² – 4x + 3

Substituting x = – 1,

p(– 1) = (– 1)² – 4(– 1) + 3

= 1 + 4 + 3

= 8

When x = ½ ,

p(x) = p(½)

p() =x² – 4x + 3

Substituting x = ½,

p(½) = (½)² – 4(½) + 3

= ¼ – 2 + 3

= ¼ + 1

= 5/4

Now,

p(2)− p(−1) + p(½) = – 1 – 8 + (5/4)

= – 9 + (5/4)

= ( – 36 + 5)/4

= – 31/4

(a) p(x) = x³ + 2x² – 3x + 1,

q(x) = x³ – 2x² + 3x + 5

p(x) + q(x)

= x³ + 2x² – 3x + 1 + x³ – 2x² + 3x + 5

= 2x³ + 6

Degree = 3

(b) p(x) – q(x)

= x³ + 2x² – 3x + 1 – (x³ – 2x² + 3x + 5)

= x³ + 2x² – 3x + 1 -x³ + 2x² – 3x – 5

= 4x² – 6x – 4

Degree = 2

(c) (p + q + 1) = 0.

Let the common factor be x – α, then x = α will make the given expressions zero, i.e., 

α² + pα + q = 0

α² + qα + p = 0 

Solving by the rule of cross-multiplication, we have

\(\frac{α^2}{p^2-q^2}\) = \(\frac{α}{q-p}\) = \(\frac{1}{q-p}\)

From, last two relation, α = 1.

⇒ α² = \(\frac{p^2-q^2}{q-p}\) ⇒ 1 = – (p + q) ⇒ p + q + 1 = 0.

Given that,

p(x) = 3x³ – 4x² + 7x – 5

According to the question,

When x = 3,

p(x) = p(3)

p(x) = 3x³ – 4x² + 7x – 5

Substituting x = 3,

p(3)= 3(3)³ – 4(3)² + 7(3) – 5

p(3) = 3(3)³ – 4(3)² + 7(3) – 5

= 3(27) – 4(9) + 21 – 5

= 81 – 36 + 21 – 5

= 102 – 41

= 61

When x = – 3,

p(x) = p(– 3)

p(x) = 3x³ – 4x² + 7x – 5

Substituting x = – 3,

p(– 3)= 3(– 3)³ – 4(– 3)² + 7(– 3) – 5

p(– 3) = 3(–3)³ – 4(–3)² + 7(–3) – 5

= 3(–27) – 4(9) – 21 – 5

= –81 – 36 – 21 – 5

= –143

p(x) = 3x² – ax + 1

p(1) = 3(1)² – (a × 1) + 1 = 3 – a + 1 = 4 – a

Given p(1) = 2

That is, 4 – a = 2 , a = 4 – 2 = 2

Correct option is (B) \(\frac{2(x^2+1)}{x^2-1}\)

Sum of \(\left(\frac{x-1}{x+1}\right)\) and its reciprocal \(=\frac{x-1}{x+1}+\frac{x+1}{x-1}\)

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

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

Correct option is  B) \(\frac{2(x^2+1)}{x^2-1}\)

(a) (x – 1) only

For (x – 1) to be a factor of the given expression, the value of expression at x = 1 is 

(1)²⁹ – (1)²⁴ + (1)¹³ – 1 = 1 – 1 + 1 – 1 = 0 

∴ (x – 1) is a factor of x²⁹ – x²⁴ + x¹³ – 1 

Similarly for (x + 1) to be the factor, the value of expression at x = – 1 is 

(–1)²⁹ – (–1)²⁴ + (–1)¹³ – 1 = – 1 – 1 – 1 – 1 = – 4 ≠ 0 

∴ (x + 1) is not a factor of x²⁹ – x²⁴ + x¹³ – 1. 

Hence, (a) is the correct option.

a. Terms = 7x², -4x², -x, 4

b. Coefficient of x² = -4

c. Constant term = 4

d. Degree of the polynomial = 3

Correct option is (A) 0

\(\because\) (x - 1) & (x - 2) are factors of \((x^3 – 6x^2 + ax + b)\)

\((\because\) (x - 1) (x - 2) \(=x^2-3x+2)\)

\(\therefore\) x = 1 and x = 2 are zeros of \((x^3 – 6x^2 + ax + b)\)

\(\therefore\) 1-6+a+b = 0 and 8-24+2a+b = 0

\(\Rightarrow\) a+b = 5 and 2a+b = 16

\(\Rightarrow\) (2a+b) - (a+b) = 16 - 5

\(\Rightarrow\) a = 11

\(\therefore\) b = 5 - a = 5 - 11 = -6

\(\therefore\) 6a+11b \(=6\times11+11\times-6\)

= 66 - 66 = 0

Correct option is A) 0

Correct option is (C) a = 11, b = -6

\((x^2 – 3x + 2)\) is a factor of \((x^3-6x^2+ax+b)\)

\(\Rightarrow\) (x-1) & (x-2) are factors of \((x^3-6x^2+ax+b)\)

\((\because\) (x-1) (x-2) \(=x^2 – 3x + 2)\)

\(\Rightarrow\) x = 1 & x = 2 are zeros of \((x^3-6x^2+ax+b)\)

\(\therefore\) 1 - 6 + a + b = 0 and 8 - 24 + 2a + b = 0

\(\Rightarrow\) a+b = 5 and 2a+b = 16

\(\Rightarrow\) (2a+b) - (a+b) = 16 - 5

\(\Rightarrow\) a = 11

\(\therefore\) b = 5 - a = 5 - 11 = -6

Correct option is B) a = 11, b = 6

Let f(x) = px³ + 3x² – 3 

g(x) = 2x³ – 5x + p 

When divisible by x – 4, the remainders for the given expressions are f(4) and g(4) respectively. 

f(4) = p(4)³ + 3(4)² – 3 = 64p + 48 – 3 = 64p + 45 

g(4) = 2(4)³ – 5(4) + p = 128 – 20 + p = 108 + p. 

Given, f(4) = g(4) ⇒ 64p + 45 = 108 + p ⇒ 63 p = 63 ⇒ p = 1.

i. (2x + 1) (3x + 4) (4x + 3) (3x + 4)

= (3x + 4) [(2x + 1 + 4x + 3)]

= (3x + 4) [6x + 4]

= (3x x (6x) + (3x) x (4)+ 4x(6x) + 4 x 4

= 18x² + 12x + 24x + 16

= 18x² + 36x + 16

ii. (3x + 4)² – (2x – 1) (3x + 4)

= (3x + 4) [3x + 4 – (2x – 1)]

= (3x + 4)[3x + 4 – 2x + 1] = (3x + 4) (x + 5)

= 3x² + 15x + 4x + 20

= 3x² + 19x + 20

Let f(x) = x⁵ – 9x² + 12x – 14 

f(x) is divisible by (x – 3) so remainder = f(3). 

∴ f(3) = (3)⁵ – 9(3)² + 12(3) – 14 = 243 – 81 + 36 – 14 = 184.

a. p(2) = 2 × 2² – 7 × 2² + k × 2 + 20

= 16 – 28 + 2k + 20 = 8 + 2k

p(3) = 2 × 3³ – 7 × 3² + k × 3 + 20

= 54 – 63 + 3k + 20 = 11 + 3k

P(2) = P(3)

8 + 2k = 11 + 3k

k = -3

b. p(x) = 2x³ – 7x² - 3x + 20

c. p(1) = 2 – 7 – 3 + 20 = 12

Correct option is (A) -101

Let \(p(x)=x^4+15x^3+6x^2+12x+3\)

Remainder when p(x) is divided by x+2 is

\(p(-2)=(-2)^4+15(-2)^3\) \(+6(-2)^2+12(-2)+3\)

= 16 - 120 + 24 - 24 + 3 = -101

Correct option is A) -101

p(x) = 2x³ + 9x² + kx + 3

p(-2) = 2(-2)³ + 9(-2)² + k(-2) + 3

= -16 + 36 – 2k + 3 = 23 – 2k

p(-3) = 2(-3)³ + 9(-3)² + k(-3) + 3

= -54 + 81 – 3k + 3 = 30 – 3k

We have p(-2) = p(-3),

23 -2k = 30 -3k

-2k + 3k = 30 – 23

k = 7

p(x) = 2x³ + ax² – 7x + b

We have p(1) = 3.

p(1) = 2 x 1³ + a x 1² – 7 x 1 + b = 3

2 + a – 7 + b = 3

a + b – 5 = 3

a + b = 8 ………. (1)

We have p(2) = 19.

p(2 ) = 2 x 2³ + a x 2² – 7 x 2 + b = 19

16 + 4a – 14 + b = 19

4a + b + 2 = 19

4a + b = 17…….. (2)

From equation (1), (2)

(2) – (1) 4a + b = 17

\(\frac{a+b=8}{3a=9}\)

a = 9/3 = 3

From equation (1)

a + b = 8

3 + b = 8

b = 5

(i) p(x) = 2x³ + x² – 2x – 1 

g(x) = x + 1 

If x + 1 = 0, then x = -1 

p(x) = 2x³ + x² – 2x – 1 

p(-1)= 2(-1)³ + (-1)² -2(- 1) – 1 

= 2(-1) + (1) + 2 – 1 = -2 + 1 + 2 – 1 

P(-1)= 0 

Here, r(x) = p(a) = 0, 

∴ g(x) is the a factor f(x) 

(ii) p(x) = x³ + 3x² + 3x + 1 

g(x) = x + 2 

If x + 2 = 0, then x = -2 

∴ p(x) = x³ + 3x² + 3x + 1 

p(-2) = (-2)³ + 3(-2)² + 3 (-2) + 1 

= -8 + 3(4) + 3(-2) + 1 

= -8 + 12 – 6 + 1 

= 13 – 24 

p(-2)= -11 

Here we have r(x) = p(a) =-11. 

Value of r(x) is not equal to zero. 

∴ g(x) is not a factor of f(x). 

(iii) p(x) = x³ – 4x² + x + 6 

g(x) = x – 3 

Let, x – 3 = 0, then x = 3 

p(x) = x³ – 4x² + x + 6 

p(3) = (3)² – 4(3)² + 3 + 6 

= 27 – 4(9) + 3 + 6 

= 27 – 36 + 3 + 6 = 36 – 36 

p(3) = 0 

Here, we have r(x) = p(a) = 0 

∴ (x – 3) is the factor of p(x).

(i) x- 1 is a factor of p(x) 

x + 1 = x – a 

a = -1 

For the value of p(a), 

value of r(x) = 0. 

∴ p(x) = x³ + x² + x + 1

p(-1)= (-1)³ + (-1)² + (-1) + 1 

= -1 + 1 -1 + 1 

p(-1)= 0 

Here, p(a) = r(x) = 0 

∴ x + 1 is a factor. 

(ii) If x + 1 = x – a, then a = -1 

p(x) = x⁴ + x³ + x² + x + 1 

p(-1)= (-1)⁴ + (-1)³ +(-1)² + (-1)+ 1 

= 1 – 1 + 1 – 1 + 1 

= 3 – 2 p(-1)= 1 

Here, r(x) = p(a)= 1 

Reminder is not zero. 

∴ x+1 is not a factor. 

(iii) If x + 1 = x – a then a = -1 

p(x) = x⁴ + 3x³ + 3x² + x + 1 

p(-1) = (-1)⁴ + 3(-1 )³ +3(-1 )² + (-1) + 1 

= 1 + 3(-1) + 3(1) + 1(-1) + 1 

= 1- 3 + 3 – 1 + 1 

= 5 – 4 P(-1)= 1 

Here, r(x) = p(a) = 1 

Remainder is not zero 

∴ x+1 is not a factor. 

(iv) If x + 1 = x – a then, 

a = -1 

p(x) = x³ – x² – (2 + √2)x+ p(-1) 

= (-1)³ – (-1)² -(2 + √2)(-1) + √2 

= -1 -(+1) – (2 – √2) + √2

= -1 – 1 + 2 + √2 +√2  

= -2 + 2 + √2 

= + 2√2

p(-1) = 2√2

Here, r(x) = p(a) = 2√2 Value of remainder r(x) is not zero. 

∴ x + 1 is not a factor.

i. p(x) = 3x + 5 

p(0) = 3 × 0 + 5 = 5 

p(1) = 3 × 1 + 5 = 8 

p(-1) = 3 x - 1 + 5 = 2

ii. p(x) = 3x² + 6x + 1 

p(0) = 3 × 0² + 6 × 0 + 1 = 1

p(1) = 3 × 1² + 6 × 1 + 1 = 10 

p(-1) = 3 × (-1)² + 6 × (-1) + 1 = -2

iii. p(x) = 2x² – 3x + 4 

p(0) = 2 × 0² – 3 × 0 + 4 = 4 

p(1) = 2 × 1² – 3 × 1 + 4 = 3 

p(-1) =2 × (-1)² – 3 × (-1) + 4 = 9

iv. p(x) = 4x³ + 2x² + 3x + 7 

p(0) = 4 × 0³ + 2 × 02 + 3 × 0 +7 = 7 

p(1) = 4 × 1³ + 2 × 12 + 3 × 1 + 7 = 16 

p(-1) = 4 × (-1)³ + 2 × (-1)² + 3 × (-1) +7 = 2

v. p(x) = 5x³ – x² + 2x – 3 

p(0) = 5 × 0³ – 0² + 2 × 0 – 3 =-3 

p(1) = 5 × 1³ – 1² + 2 × 1 – 3 = 3 

p(-1) = 5 × (-1)³ – (-1)² + 2 × (-1) – 3 = -11

i. p(x) = 2x + 5 

p(1) = 2 × 1 + 5 = 2 + 5 = 7 

p(10) = 2 × 10 + 5 = 20 + 5 = 25

ii. p(x) = 3x² + 6x + 1 

p(1) = 3 × 12 + 6 x 1 + 1 

= 3 + 6 + 1 = 10 

p(10) = 3 × 10² + 6 × 10 + 1 = 300 + 60 + 1 = 361

iii. p(x) = 4x³ + 2x² + 3x + 7 

p(1) = 4 × 13 + 2 × 12 + 3 × 1 + 7

= 4 + 2 + 3 + 7 = 16 

p(10) = 4 × 10³ + 2 × 10² + 3 × 10 + 7 

= 4000 + 200 + 30 + 7 = 4237

(x² – 6x)² – 8(x² – 6x + 8) – 64 

= m² – 8(m + 8)-64 …[Putting x² – 6x = m] 

= m² – 8m – 64 – 64 

= m² – 8m – 128

= m² – 16m + 8m- 128 

= m(m – 16) + 8(m – 16) 

= (m – 16)(m + 8)

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

= (x² – 8x + 2x – 16) (x² – 4x – 2x + 8) 

= [x(x – 8) + 2(x – 8)] [x(x – 4) – 2(x – 4)] 

= (x – 8) (x + 2) (x – 4) (x – 2)

101 x 99 

= (100 + 1) (100 – 1) 

= 100² – 1² 

= 10000 – 1 

= 9999

(y + 2) (y – 3) (y + 8) (y + 3) + 56 

= (y + 2)(y + 3)(y – 3)(y + 8) + 56 

= (y² + 3y + 2y + 6) (y² + 8y – 3y – 24) + 56 

= (y² + 5y + 6) (y² + 5y – 24) + 56 

= (m + 6) (m – 24) + 56 … [Putting y² + 5y = m]

= m (m – 24) + 6 (m – 24) + 56 

= m² – 24m + 6m – 144 + 56

 = m² – 18m – 88 

= m² – 22m + 4m – 88

 = m(m – 22) + 4(m – 22)

 = (m – 22) (m + 4) 

= (y² + 5y – 22)(y² + 5y + 4) … [Replacing m = y² + 5y] 

= (y² + 5y – 22) (y² + 4y + y + 4) 

= (y² + 5y – 22) [y(y + 4) + 1(y + 4)] 

= (y² + 5y – 22) (y + 4) (y + 1)

Correct option is (C) 2

\(x^2 – 7x + 12=0\)

\(\Rightarrow\) (x - 3) (x - 4) = 0

\(\Rightarrow\) x = 3 or x = 4

\(\therefore\) Number of zeros of \(x^2 – 7x + 12\) is 2.

Correct option is C) 2

Correct option is (A) -8, -7

Let consecutive numbers are x & (x+1).

\(\therefore\) x (x+1) = 56

\(\Rightarrow\) \(x^2+x-56=0\)

\(\Rightarrow\) \(x^2+8x-7x-56=0\)

\(\Rightarrow\) x (x+8) - 7 (x+8) = 0

\(\Rightarrow\) (x-7) (x+8) = 0

\(\Rightarrow\) x = 7 or x = -8

\(\therefore\) Consecutive numbers either -8, -7 or 7, 8.

Correct option is A) -8, -7

Correct option is B) 6x – 1 – 7y

(b) \(\frac{8}{4x^2-1}\)

A - B = \(\frac{2x+1}{2x-1}\) - \(\frac{2x-1}{2x+1}\)

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

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

= \(\frac{8}{4x^2-1}\)

(c) 2

\(\frac{1}{x+1}-\frac{1}{x-1}-\frac{x^2}{x+1}+\frac{x^2}{x-1}\)

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

= \(\frac{x-1-x-1-x^3+x^2+x^3+x^2}{x^2-1}\)

= \(\frac{2x^2-2}{x^2-1}=\frac{2(x^2-1)}{(x^2-1)}\) = 2

(d) 1

\(\bigg(\frac{2x+y}{x+y}-1\bigg)÷\bigg(1-\frac{y}{x+y}\bigg)\)

= \(\bigg[\frac{2x+y-(x+y)}{x+y}\bigg]÷\bigg[\frac{x+y-y}{x+y}\bigg]\)

= \(\frac{x}{x+y}\times\frac{x+y}{x}=1\)

(c) 12x²(x² - 3x +2)(x² - 4)

LCM = \(\frac{\text{product of the polynomials}}{HCF}\)

= \(\frac{4x^2(x^2-3x+2)\times12x(x-2)(x^2-4)}{4x(x-2)}\)

= \(\frac{4x^2(x-2)(x-1)\times12x(x-2)(x-2)(x+2)}{4x(x-2)}\)

= 12x²(x - 1)(x - 2)(x - 2)(x + 2)

= 12x²(x² - 3x +2)(x² - 4)

(a) (x² - 1)(x + 1)³

\(x^3 + 3x^2 +3x +1\) = (x + 1)³

\(x^2 + 2x + 1\) = (x + 1)²

\(x^2 - 1\) = (x + 1)(x - 1)

∴ Reqd. LCM = (x - 1)(x + 1)³

= (x² - 1)(x + 1)³

(b) \(\frac{y}{x}\)

Reqd. product = \(\frac{x^2 - y^2}{x^2+2xy+y^2}\) x \(\frac{xy+y^2}{x^2 - xy}\)

= \(\frac{(x + y)(x + y)}{(x+y)^2}\times\frac{y(x+y)}{x(x-y)}\)

= \(\frac{y}{x}\)

x + y + z = 0 

x + y = -z 

by cubing on both sides, 

(x + y)³ = (-z)³ 

x³+ y³ + 3xy(x + y) = -z³ 

x³ + y³ + 3xy (-z) 

= -z (∵ x + y =-z data given) 

x³ + y³ – 3xyz = -z³ 

∴ x³ + y³ + z³ = 3xyz.

(c) \(\frac{x+y+z}{a+b+c}\)

Reqd. exp.

= \(\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}\times\frac{a^2+b^2+c^2-ab-bc-ca}{x^2+y^2+z^2-xy-yz-zx}\)

= \(\frac{x+y+z}{a+b+c}\)

Given x³ + y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)²] 

R-H.S = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)² ] 

= 1/2 (x + y + z) [x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2xz] 

= 1/2 (x + y + z) [2x² + 2y² + 2z² – 2xy – 2yz – 2zx] 

= 1/2 (x + y + z) (2) [x² + y² + z² – xy – yz – zx] 

= (x + y + z) (x² + y² + z² – xy – yz – zx) 

= L.H.S 

Hence proved.

Given x + y + z = 0 

To prove x³ + y³ + z³ = 3xyz 

We have an identity 

(x + y + z) (x² + y² + z² – xy – yz – zx) 

= x³ + y³ + z³ – 3xyz 

Substituting x + y + z = 0in the above equation, we get 

0 x (x² + y² + z² -xy-yz-zx) 

= x³ + y³ + z³ – 3xyz 

⇒ x³ + y³ + z³ – 3xyz = 0 

⇒ x³ + y³ + z³ = 3xyz

Let f(x) = 8x² – 4

= 4 ((√2x)² – (1)²)

= 4(√2x + 1)(√2x – 1)

To find the zeroes, set f(x) = 0

(√2x + 1)(√2x – 1) = 0

(√2x + 1) = 0 or (√2x – 1) = 0

x = (-1)/√2 or x = 1/√2

So, the zeroes of f(x) are (-1)/√2 and x = 1/√2

Again,

Sum of zeroes = -1/√2 + 1/√2 = (-1+1)/√2 = 0

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = -1/√2 x 1/√2 = -1/2 = -4/8

= c/a

= Constant term / Coefficient of x²

Correct option is (B) 0

\(\frac{1}{1+x+x^2}-\frac{1}{1-x+x^2}+\frac{2x}{1+x^2+x^4}\)

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

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

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

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

Correct option is B) 0