Find **Solved Practice Paper for IIT JEE Main** in this article. This practice paper consists of 10 questions. These questions are from the chapter complex numbers of Mathematics. We have tried to include all important concepts in these questions. Also, detailed solution of each question has been provided. All questions are very important from examination point of view.

**Q1.** If *z*_{1}, z_{2},* z*_{3} ,.* z*_{4} are the four complex numbers represented by* *the vertices of a quadrilateral taken in order such that *z*_{1 }– z_{4} = * z*_{2} − *z*_{3}

(a) rhombus

(b) square

(c) rectangle

(d) cyclic quadrilateral

Since two angle of the quadrilateral is 90^{° }and the sum of opposite angles of a quadrilateral is 180^{°} so, the quadrilateral ABCD is cyclic quadrilateral.

**UPSEE 2017 Solved Sample Paper Set-1**

Since all the elements of the second row of the determinant is 0, so the value of determinant is zero.

(a) 3ω

(b) 3ω (ω − 1)

(c) 3ω^{2}

(d) 3ω (1 − ω)

**Correct Option: (b)**

**Sol.**

**Applying C_{1}→ C_{1} + C_{2} + C_{3} , then**

**JEE Main Mathematics Solved Sample Paper Set-VII**

(a) Roots are purely imaginary for all negative realvalues of *m*

(b) Roots are not real for all complex numbers *m*

(c) Roots are purely irnaginary for all positive real values of *m*

(d) None of the above

**Correct Option: (a)**

**Sol.**

Given:

10z^{2} -4*iz* -*m *= 0

By using quadratic formula, we get

If *m *is negative, then *D* will also be negative.

So, the roots of *z* will be purely imaginary.

Hence, the correct option is (a).

**JEE Main Physics Solved Sample Paper Set-VII**

**Q6.** Common roots of the equations *z*^{3}* +*2*z*^{2}* + *2*z + *1 = 0 and *z*^{1985} + *z*^{100}* *+ 1 = 0 are

(a) ω, ω^{2}

(b) ω, ω^{3}

(c) ω^{2}, ω^{3}

(d) None of these

**Sol.**

**Correct Option: (a)**

We have equation,

*z*^{3}* +*2*z*^{2}* + *2*z + *1 = 0

(*z* + 1) (*z*^{2} + *z *+ 1) = 0

Roots of the above equations are −1, ω and ω^{2}

Let *f*(*z*) = *z*^{1985} + *z*^{100} + 1

Putting z = −1, ω and ω^{2} respectively, we get

*f*(−1) = (−1)^{1985} + (−1)^{100} + 1 ≠ 0

Therefore, −1 is not a root of the equation

Again, *f*(ω) = ω^{1985} + ω^{100}+ 1

= (ω^{3})^{661 }ω^{2} + (ω^{3})^{33 }ω+1

= ω^{2} +ω + 1 = 0

Therefore, ω is a root of the equation *f*(*z) = *0.

Similarly, *f*(ω^{2}) = 0

Hence, ω and ω^{2} are the common roots.

**Q7. **The value of the expression 1×(2 - ω) (2 - ω^{2}) + 2(3 - ω) (3 - ω^{2})+........+ (*n* - 1) (*n* - ω) (*n* - ω^{2}), where ω is an imaginary cube root of unity is

According to the properties of cube root of unity:

**Q8.** Let *z* be a complex number such that the imaginary part of *z* is non zero and

*a* = *z*^{2} + *z* + 1 is real. Then *a* cannot take the value

(a) - 1

(b) 1/3

(c) 1/2

(d) 3/4

**Correct option: (c)**

**Sol.**

We have,

(a) is equal to 5/2

(b) lies in the interval (1, 2)

(c) is strictly greater than 5/2

(d) is strictly greater than 3/2 but less than 5/2

**Sol. **

**Correct option: (b)**

**JEE Main Chemistry Solved Sample Paper Set-VII**

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