If i -1- then the value of - t -12 t -1- i -32 t -1- i -3-4 isA- t3-t2-t-1B- t3-t2-5C- t3-5 t2-5 t-2D- t3-1

The correct option is D t3 1Assume t= 2 The question changes to [(5 i√(3))(5+i√(3))]/4 [25 (i√(3))^2]/4since ,(a b)(a+b)=a^2 b^2 (25+3)/4, since i^2= 1 =7 Only ...

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If i =√-1, th...

Question

If i = $\sqrt{- 1}$ , then the value of $\frac{[(t - 1) (2 t + 1 - i \sqrt{3}) (2 t + 1 + i \sqrt{3})]}{4}$ is

t³+t²+t+1

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t³+t²-5

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t³+ √5t² +√5t -2

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t³-1

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Solution

The correct option is D t³-1
Assume t= 2

The question changes to

$\frac{[(5 - i \sqrt{3}) (5 + i \sqrt{3})]}{4}$

$\frac{[25 - (i \sqrt{3})^{2}]}{4}$ since , $(a - b) (a + b) = a^{2} - b^{2}$

$\frac{(25 + 3)}{4}$ , since $i^{2} = - 1$

=7

Only answer option (d) gives 7 on substitution of t=2

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If i = √−1, then the value of [(t−1)(2t+1−i√3)(2t+1+i√3)]4 is

The correct option is D t3-1Assume t= 2 The question changes to [(5−i√3)(5+i√3)]4 [25−(i√3)2]4since ,(a−b)(a+b)=a2−b2 (25+3)4, since i2=−1 =7 Only answer option (d) gives 7 on substitution of t=2

If i = $\sqrt{- 1}$ , then the value of $\frac{[(t - 1) (2 t + 1 - i \sqrt{3}) (2 t + 1 + i \sqrt{3})]}{4}$ is

The correct option is D t³-1
Assume t= 2

The question changes to

$\frac{[(5 - i \sqrt{3}) (5 + i \sqrt{3})]}{4}$

$\frac{[25 - (i \sqrt{3})^{2}]}{4}$ since , $(a - b) (a + b) = a^{2} - b^{2}$

$\frac{(25 + 3)}{4}$ , since $i^{2} = - 1$

=7

Only answer option (d) gives 7 on substitution of t=2