CBSE 11th Standard Maths Subject Complex Numbers and Quadratic Equations Ncert Exemplar 4 Marks Questions With Solution 2021
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CBSE 11th Standard Maths Subject Complex Numbers and Quadratic Equations Ncert Exemplar 4 Marks Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Mathematics
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What is the value of \(\frac { { i }^{ 4x+1 }-{ i }^{ 4x-1 } }{ 2 } ?\)
(a) -
Find the real value of 'a' for which 3i3- 2ai2+(1-a)i + 5 is real.
(a) -
If f(z) = \(\frac { 7-z }{ 1-{ z }^{ 2 } } \) where z= 1 + 2i , then find \(|f(z)|\)
(a) -
If \(\frac { z-1 }{ z+1 } \) is a purely imaginary number \((z\neq -1)\) then find the value of \(|z|\)
(a) -
Find \(\left| (1+i)\frac { (2+i) }{ (3+i) } \right| \)
(a)
4 Marks
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CBSE 11th Standard Maths Subject Complex Numbers and Quadratic Equations Ncert Exemplar 4 Marks Questions With Solution 2021 Answer Keys
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Consider, \(\frac { { i }^{ 4x+1 }-{ i }^{ 4x-1 } }{ 2 } =\frac { { i }^{ 4x }.i-{ i }^{ 4x }.{ i }^{ -1 } }{ 2 } =\frac { i-\frac { 1 }{ i } }{ 2 } \quad [\because { i }^{ 4x }=1]\)
\(=\frac { { i }^{ 2 }-1 }{ 2i } =\frac { -2 }{ 2i } \quad [\because { i }^{ 2=-1 }]\)
\(=\frac { -1 }{ i } =\frac { -i }{ { i }^{ 2 } } =\frac { -i }{ -1 } =i\quad [\because { i }^{ 2 }=-1]\)
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3i3- 2ai2+(1-a)i + 5
= 3( - i) + 2a + (1 - a)i+5 [i3 = -i and i2 = -1]
= (2a + 5) + i(1-a-3), which will be real,
if 1 - a - 3 =0,
i.e, a = - 2 -
\(\frac { 2-i }{ 2 } \)
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Let z = x + iy, then
\(\frac { z-1 }{ z+1 } =\frac { ({ x }^{ 2 }-1)+{ y }^{ 2 }+i[y(x+1)-y(x-1)] }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } \\ \)
\(\because \frac { z-1 }{ z+1 } \) is purely imaginary.
\(\therefore Re(\frac { z-1 }{ z+1 } ) =0\ i.e.\ \frac { ({ x }^{ 2 }-1)+{ y }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } =0\)
\({ x }^{ 2 }-1-{ y }^{ 2 }=0\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1= |z|=1\) -
Let,
\(z=\frac { (1+i)(2+i) }{ (3+i) } =\frac { 2+i+2i+{ i }^{ 2 } }{ 3+i } =\frac { 2+3i-1 }{ 3+i } \)
\( \Rightarrow z=\frac { 1+3i }{ 3+i } \quad [\because { i }^{ 2 }=-1]\)
Now,
\(\left| z \right| =\left| \frac { 1+3i }{ 3+i } \right| =\frac { \left| 1+3i \right| }{ \left| 3+i \right| } \quad \left[ \left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } \right] \)
\(=\frac { \sqrt { { 1 }^{ 2 }+{ 3 }^{ 2 } } }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } } =1\)
Hence, \(\left| (1+i)\frac { (2+i) }{ (3+i) } \right| \)
4 Marks
