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प्रश्न
Match the statements of column A and B.
| Column A | Column B |
| (a) The value of 1 + i2 + i4 + i6 + ... i20 is | (i) purely imaginary complex number |
| (b) The value of `i^(-1097)` is | (ii) purely real complex number |
| (c) Conjugate of 1 + i lies in | (iii) second quadrant |
| (d) `(1 + 2i)/(1 - i)` lies in | (iv) Fourth quadrant |
| (e) If a, b, c ∈ R and b2 – 4ac < 0, then the roots of the equation ax2 + bx + c = 0 are non real (complex) and |
(v) may not occur in conjugate pairs |
| (f) If a, b, c ∈ R and b2 – 4ac > 0, and b2 – 4ac is a perfect square, then the roots of the equation ax2 + bx + c = 0 |
(vi) may occur in conjugate pairs |
उत्तर
| Column A | Answers |
| (a) The value of 1+ i2 + i4 + i6 + ... i20 is | (ii) purely real complex number |
| (b) The value of `i^(-1097)` is | (i) purely imaginary complex number |
| (c) Conjugate of 1 + i lies in | (iv) Fourth quadrant |
| (d) `(1 + 2i)/(1 - i)` lies in | (iii) second quadrant |
| (e) If a, b, c ∈ R and b2 – 4ac < 0, then the roots of the equation ax2 + bx + c = 0 are non real (complex) and |
(vi) may occur in conjugate pairs |
| (f) If a, b, c ∈ R and b2 – 4ac > 0, and b2 – 4ac is a perfect square, then the roots of the equation ax2 + bx + c = 0 |
(v) may not occur in conjugate pairs |
Explanation:
(a) Because 1 + i2 + i4 + i6 + ... i20
= 1 – 1 + 1 – 1 + ... + 1 = 1 ......(Which is purely a real complex number.)
(b) Because `i^(-1097)` = `1/((i)^1097)`
= `1/(i^(4 xx 274 + 1)`
= `1/((i^4)^274i)`
= `1/i`
= `i/i^2`
= –i
Which is purely imaginary complex number.
(c) Conjugate of 1 + i is 1 – i which is represented by the point (1, –1) in the fourth quadrant.
(d) Because `(1 + 2i)/(1 - i) = (1 + 2i)/(1 - i) xx (1 + i)/(1 + i)`
= `(-1 + 3i)/2`
= `-1/2 + 3/2 i`
Which is represented by the point `(- 1/2, 3/2)` in the second quadrant.
(e) If b2 – 4ac < 0 = D < 0 i.e., square root of D is a imaginary number.
Therefore, roots are x = `(-b +- "Imaginary Number")/(2a)`
i.e., roots are in conjugate pairs.
(f) Consider the equation `x^2 - (5 + sqrt(2)) x + 5 sqrt(2)` = 0, Where a = 1, b = `-(5 + sqrt(2))`, c = `5 sqrt(2)`, Clearly a, b, c ∈ R.
Now D = b2 – 4ac = `{- (5 + sqrt(2))}^2 - 4.1.5 sqrt(2) = (5 - sqrt(2))^2`.
Therefore x = `(5 + sqrt(2) +- 5 - sqrt(2))/2` = `5sqrt(2)` which do not form a conjugate pair.
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