MHT CET · Maths · Complex Number
If \(x=\frac{5}{1-2 \mathrm{i}}, \mathrm{i}=\sqrt{-1}\), then the value of \(x^3+x^2-x+22\) is
- A \(7\)
- B \(9\)
- C \(17\)
- D \(39\)
Answer & Solution
Correct Answer
(A) \(7\)
Step-by-step Solution
Detailed explanation
\(x=\frac{5}{1-2 \mathrm{i}}=\frac{5(1+2 \mathrm{i})}{1+4}=1+2 \mathrm{i} \)
\( \therefore x^2=(1+2 \mathrm{i})^2=1-4+4 \mathrm{i}=-3+4 \mathrm{i} \)
\( \therefore x^3=(-3+4 \mathrm{i})(1+2 \mathrm{i}) \)
\( =-3-6 i+4 i-8 \)
\( =-11-2 \mathrm{i} \)
\( \therefore x^3+x^2-x+22 \)
\( =(-11-2 \mathrm{i})+(-3+4 \mathrm{i})-(1+2 \mathrm{i})+22 \)
\( =-11-2 \mathrm{i}-3+4 \mathrm{i}-1-2 \mathrm{i}+22 \)
\( =7\)
\( \therefore x^2=(1+2 \mathrm{i})^2=1-4+4 \mathrm{i}=-3+4 \mathrm{i} \)
\( \therefore x^3=(-3+4 \mathrm{i})(1+2 \mathrm{i}) \)
\( =-3-6 i+4 i-8 \)
\( =-11-2 \mathrm{i} \)
\( \therefore x^3+x^2-x+22 \)
\( =(-11-2 \mathrm{i})+(-3+4 \mathrm{i})-(1+2 \mathrm{i})+22 \)
\( =-11-2 \mathrm{i}-3+4 \mathrm{i}-1-2 \mathrm{i}+22 \)
\( =7\)
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