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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Evaluate the ...

Question

Evaluate the following:
(i) $2 x^{3} + 2 x^{2} - 7 x + 72, when x = \frac{3 - 5 i}{2}$
(ii) $x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i$
(iii) $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, when x = - 1 + i \sqrt{2}$
(iv) $x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}} .$

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Solution

$(i) x = \frac{3 - 5 i}{2} \Rightarrow x^{2} = {(\frac{3 - 5 i}{2})}^{2} = \frac{9 + 25 i^{2} - 30 i}{4} = \frac{- 16 - 30 i}{4} \Rightarrow x^{3} = \frac{- 16 - 30 i}{4} \times \frac{3 - 5 i}{2} = \frac{- 48 + 80 i - 90 i + 150 i^{2}}{8} = \frac{- 198 - 10 i}{8} ∴ 2 x^{3} + 2 x^{2} - 7 x + 72 = 2 (\frac{- 198 - 10 i}{8}) + 2 (\frac{- 16 - 30 i}{4}) - 7 (\frac{3 - 5 i}{2}) + 72 = \frac{- 198 - 10 i - 32 - 60 i - 42 + 70 i + 288}{4} = \frac{16}{4} = 4 (ii) x = 3 + 2 i \Rightarrow x^{2} = {(3 + 2 i)}^{2} = 9 + 4 i^{2} + 12 i = 5 + 12 i \Rightarrow x^{3} = x^{2} \times x = (5 + 12 i) \times (3 + 2 i) = 15 + 10 i + 36 i - 24 = - 9 + 46 i \Rightarrow x^{4} = {(x^{2})}^{2} = {(5 + 12 i)}^{2} = 25 + 144 i^{2} + 120 i = - 119 + 120 i \Rightarrow x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44 = - 119 + 120 i - 4 (- 9 + 46 i) + 4 (5 + 12 i) + 8 (3 + 2 i) + 44 = - 119 + 120 i + 36 - 184 i + 20 + 48 i + 24 + 16 i + 44 = 5 (iii) x = - 1 + \sqrt{2} i \Rightarrow x^{2} = {(- 1 + \sqrt{2} i)}^{2} = 1 + 2 i^{2} - 2 \sqrt{2} i = - 1 - 2 \sqrt{2} i \Rightarrow x^{3} = (- 1 - 2 \sqrt{2} i) \times (- 1 + \sqrt{2} i) = 1 - \sqrt{2} i + 2 \sqrt{2} i - 4 i^{2} = 5 + \sqrt{2} i \Rightarrow x^{4} = {(- 1 - 2 \sqrt{2} i)}^{2} = 1 + 8 i^{2} + 4 \sqrt{2} i = - 7 + 4 \sqrt{2} i \Rightarrow x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9 = - 7 + 4 \sqrt{2} i + 4 (5 + \sqrt{2} i) + 6 (- 1 - 2 \sqrt{2} i) + 4 (- 1 + \sqrt{2} i) + 9 = - 7 + 4 \sqrt{2} i + 20 + 4 \sqrt{2} i - 6 - 12 \sqrt{2} i - 4 + 4 \sqrt{2} i + 9 = 12 (iv) x = \frac{1 + i}{\sqrt{2}} \Rightarrow x^{2} = {(\frac{1 + i}{\sqrt{2}})}^{2} = (\frac{1 + i^{2} + 2 i}{2}) = \frac{2 i}{2} = i \Rightarrow x^{6} = {(x^{2})}^{3} = i^{3} = - i \Rightarrow x^{2} = i \Rightarrow x^{4} = {(x^{2})}^{2} = i^{2} = - 1 Now, x^{6} + x^{4} + x^{2} + 1 = - i - 1 + i + 1 = 0$

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Similar questions

Q. Evaluate the following:
(i)

2 x^{3} + 2 x^{2} - 7 x + 72, when x = \frac{3 - 5 i}{2}

(ii)

x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i

(iii)

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, when x = - 1 + i \sqrt{2}

(iv)

x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}}

(v)

2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, when x = - 2 - \sqrt{3} i

Evaluate the following :

$(i) 2 x^{3} + 2 x^{2} - 7 x + 72, w h e n x = \frac{3 - 5 i}{2} (i i) x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, w h e n x = 3 + 2 i (i i i) x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, w h e n x = - 1 + i \sqrt{2} (i v) x^{6} + x^{4} + x^{2} + 1, w h e n x = \frac{1 + i}{\sqrt{2}} (v) 2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, w h e n x = - 2 - \sqrt{3} i$

Q. The remainder obtained when (

x^{4} + 4 x^{3} + 2

) is divided by (

x + 1

) = .

Q. Write the quotient and remainder when we divide:

(x^{4} - 2 x^{3} + 2 x^{2} + x + 4)

(x^{2} + x + 1)

Q. Find the remainder when

x^{4} + 4 x^{3} - 5 x^{2} - 6 x + 7

is divided by

x - 1

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Evaluate the following: (i) 2x3+2x2-7x+72, when x=3-5i2 (ii) x4-4x3+4x2+8x+44, when x=3+2i (iii) x4+4x3+6x2+4x+9, when x=-1+i2 (iv) x6+x4+x2+1, when x=1+i2.

Evaluate the following:
(i) $2 x^{3} + 2 x^{2} - 7 x + 72, when x = \frac{3 - 5 i}{2}$
(ii) $x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i$
(iii) $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, when x = - 1 + i \sqrt{2}$
(iv) $x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}} .$