Complex Numbers & Quadratic Equations Practice Questions - DPP for JEE

 Page 2


 
= 
Product of the roots
 
? The required equation is
or  
2. (a)
which is represented by a point in first quadrant.
3. (a) Roots =
=  = 
 and 
Page 3


 
= 
Product of the roots
 
? The required equation is
or  
2. (a)
which is represented by a point in first quadrant.
3. (a) Roots =
=  = 
 and 
4. (b) = (cos? + i sin?)
4
 (cos? – i sin?)
–3
= (cos 4 ? + i sin 4?) {cos (–?) + i sin (– ?)}
–3
= (cos 4 ? + i sin 4?) {cos(– 3) (– ?) + i sin (–3) (– ?)}
= (cos 4? + i sin 4?) {cos3? + i sin 3?}
= cos 4? cos 3? – sin 4? sin 3?
+ i (sin 4? cos 3? + sin3? cos 4?)
= cos (4? + 3?) + i sin (4? + 3?)  = cos 7? + i sin 7?
5. (b) Let a > 0, b > 0, c > 0
Given equation ax
2
 + bx + c = 0
we know that D = b
2
 – 4ac and 
Let 
If a > 0, c > 0 then 
? Roots are negative
Let , then  
? roots are imaginary and have negative real part. .
6. (c)
Let the point A represents z and B represents ?z, then ?AOB = . If C
represents z + ?z then OACB is a parallelogram.
? Area of ? ABC = Area of ? OAB
Page 4


 
= 
Product of the roots
 
? The required equation is
or  
2. (a)
which is represented by a point in first quadrant.
3. (a) Roots =
=  = 
 and 
4. (b) = (cos? + i sin?)
4
 (cos? – i sin?)
–3
= (cos 4 ? + i sin 4?) {cos (–?) + i sin (– ?)}
–3
= (cos 4 ? + i sin 4?) {cos(– 3) (– ?) + i sin (–3) (– ?)}
= (cos 4? + i sin 4?) {cos3? + i sin 3?}
= cos 4? cos 3? – sin 4? sin 3?
+ i (sin 4? cos 3? + sin3? cos 4?)
= cos (4? + 3?) + i sin (4? + 3?)  = cos 7? + i sin 7?
5. (b) Let a > 0, b > 0, c > 0
Given equation ax
2
 + bx + c = 0
we know that D = b
2
 – 4ac and 
Let 
If a > 0, c > 0 then 
? Roots are negative
Let , then  
? roots are imaginary and have negative real part. .
6. (c)
Let the point A represents z and B represents ?z, then ?AOB = . If C
represents z + ?z then OACB is a parallelogram.
? Area of ? ABC = Area of ? OAB
=  |z| |?z| sin   =  |z| |z| [|?z| = |z|]   = 
Now given that area of ? ABC = .
? 
. That is z lies on a circle or radius 4
7. (b) We have two equations
 and 
Putting  these equations become
 i.e. ...(1)
and 
or     i.e    y = – 2 ...(2)
Hence the complex numbers z satisfying the given
equations are  and 
that is, 
8. (b)
Squaring both side,
i.e. 
?
9. (a) Since  = 1,
? The given equation becomes
Page 5


 
= 
Product of the roots
 
? The required equation is
or  
2. (a)
which is represented by a point in first quadrant.
3. (a) Roots =
=  = 
 and 
4. (b) = (cos? + i sin?)
4
 (cos? – i sin?)
–3
= (cos 4 ? + i sin 4?) {cos (–?) + i sin (– ?)}
–3
= (cos 4 ? + i sin 4?) {cos(– 3) (– ?) + i sin (–3) (– ?)}
= (cos 4? + i sin 4?) {cos3? + i sin 3?}
= cos 4? cos 3? – sin 4? sin 3?
+ i (sin 4? cos 3? + sin3? cos 4?)
= cos (4? + 3?) + i sin (4? + 3?)  = cos 7? + i sin 7?
5. (b) Let a > 0, b > 0, c > 0
Given equation ax
2
 + bx + c = 0
we know that D = b
2
 – 4ac and 
Let 
If a > 0, c > 0 then 
? Roots are negative
Let , then  
? roots are imaginary and have negative real part. .
6. (c)
Let the point A represents z and B represents ?z, then ?AOB = . If C
represents z + ?z then OACB is a parallelogram.
? Area of ? ABC = Area of ? OAB
=  |z| |?z| sin   =  |z| |z| [|?z| = |z|]   = 
Now given that area of ? ABC = .
? 
. That is z lies on a circle or radius 4
7. (b) We have two equations
 and 
Putting  these equations become
 i.e. ...(1)
and 
or     i.e    y = – 2 ...(2)
Hence the complex numbers z satisfying the given
equations are  and 
that is, 
8. (b)
Squaring both side,
i.e. 
?
9. (a) Since  = 1,
? The given equation becomes
 where y = 
? y
2
 – 14y + 1 = 0 ? y = 7 ± 
Now y = 7 + 4 ? x
2
 – 4x + 3 = –1 ? x = 2, 2
Also y = 7 – 4 ? x
2
 – 4x + 3 = 1 ? x = 2 ± 
10. (c) Let  then 
where ? = cube root of unity
Consider 
( )
11. (d) Let z = 
= 
=  = 
=  = 
=  = 
12. (a) Discriminant of the equation
 is  and that of
the equation