As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

**Q1.**The general value of Î¸ which satisfies the equation (cosÎ¸+i sinÎ¸ )(cos3Î¸+i sin3Î¸ )(cos5Î¸+i sin5Î¸ )… (cos(2n-1)Î¸+i sin(2n-1)Î¸=1) is

(d) We have, (cosÎ¸+ i sinÎ¸ )(cos3 Î¸ + i sin3 Î¸ ) …[cos(2 n-1)Î¸+i sin(2n-1)Î¸ ]=1+i 0

^{2}Î¸)+i sin(n

^{2}Î¸)=1+i0

^{2}Î¸=1 and sin n

^{2}Î¸=0

^{2}Î¸=2 r Ï€

^{2}

**Q2.**If n is a positive integer, then (1+i)

^{n}+(1-i)

^{n}is equal to

(c) (1+i)

^{n}+(1-i)

^{n}=(√2 (1/√2+i/√2))

^{n}+(√2 (1/√2-i/√2))

^{n}

^{n/2}(cosÏ€/4+i sinÏ€/4 )

^{n}+2

^{n/2}(cosÏ€/4-i sinÏ€/4)

^{n}

^{n/2}(cosn Ï€/4+sin nÏ€/4+cos nÏ€/4-i sin nÏ€/4)

^{n/2+1}cos(nÏ€/4)

^{n+2}cos(nÏ€/4)

**Q3.**Let a=e

^{(i 2Ï€/3 )}. Then, the equation whose roots are a+a

^{(-2)}and a

^{2}+a

^{(-4)}is

(d) Here, a=e

^{(i 2Ï€/3)}=Ï‰

^{2}=Ï‰+1/Ï‰

^{2}=Ï‰+Ï‰=2Ï‰

^{2}+1/a

^{2}=Ï‰

^{2}+1/Ï‰

^{4}=2Ï‰

^{2}

^{2}+a

^{2}+1/a

^{4}=2Ï‰+2Ï‰

^{2}=-2

^{2})(a

^{2}+1/a

^{4})=2Ï‰.2Ï‰

^{2}=4

^{2}+2x+4=0

**Q4.**If z

^{2}+z|z|+|z|

^{2}=0, then the locus of z is

(c) We have, z

^{2}+z|z|+|z|

^{2}=0

^{2}+z/|z| +1=0

^{2}

^{2}|z|

^{2}-3x

^{2}=0

**Q5.**The points representing cube roots of unity

(c) The cube roots or unity are 1,Ï‰,Ï‰

^{2}.

^{2}respectively.

^{2}+(√3/2)

^{2})=√3 QR=|Ï‰-Ï‰

^{2}|=√3, and RP=|1-Ï‰

^{2}|=√3

^{2}form an equilateral triangle.

^{i}=1,z_2=Ï‰ and z_3=Ï‰

^{2}.

_{1}

^{2}+z

_{2}

^{2}+z

_{2}

^{3}=z

_{1}z

_{2}+z

_{2}z

_{3}+z

_{3}z

_{1}

^{2}form an equilateral triangle

**Q6.**If (x-2) is a common factor of the expressions x

^{2}+ax+b and x

^{2}+cx+d, then (b-d)/(c-a) is equal to

(d) Since, (x-2) is a common factor of the expressions x

^{2}+ax+b and x

^{2}+cx+d ⇒4+2a+b=0 ...(i)

**Q7.**The centre and the radius of the circle z.z

^{c}+(2+3i) z

^{c}+(2-3i)z+12=0 are respectively

(a) Given equation of circle is zz

^{c}+(2+3i) z

^{c}+(2-3i)z+12=0

^{2}-12)=√(13-12)=1

**Q8.**For any complex number z, the minimum value of |z|+|z-1| is

(b) Since, |-z|=|z| And |z

_{1}+z

_{2}|≤|z

_{1}|+|z

_{2}|

**Q10.**Number of solutions of the equation z

^{2}+|z|

^{2}=0, where z∈C is

(d) Let z=x+i y.

^{2}+|z|

^{2}=0

^{2}-y

^{2}+2 ixy+x

^{2}+y

^{2}=0

^{2}+2ixy=0

^{2}=0,2xy=0