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.**For real values of x, the expression ((x-b)(x-c))/((x-a)) will assume all real values provided

(b) Let m=((x-b)(x-c))/(x-a)

^{2}-(b+c+m)x+(bc+am)=0

^{2}-4(bc+am)≥0

^{2}+2(b+c-2a)m+(b-c)

^{2}≥0 for all m

^{2}-4(b-c)

^{2}≤0

^{2}-(b-c)

^{2}≤0

**Q3.**If x-c is a factor of order m of the polynomial f(x) of degree n(1<m<n), then x=c is a root of the polynomial

(b) Since x-c is a factor of order m of the polynomial f(x)

^{m}Ï•(x), where Ï•(x) is a polynomial of degree n-m

^{m-1}(x) are all zero for x=c but f

^{m}(x)≠0 at x=c

^{m-1}(x)

**Q4.**Let x

_{1}, x

_{2}be the roots of the equation x

^{2}-3x+p=0 and let x

_{3}, x

_{4}be the roots of the equation x

^{2}-12x+q=0. If the numbers x

_{1}, x

_{2}, x

_{3}, x

_{4}(in order) form an increasing G.P., then

(b) It is given that x

_{1},x

_{2}are roots of x

^{2}-3x+p=0

_{1}+x

_{2}=3,x

_{1}x

_{2}=p x

_{3},x

_{4}are roots of x

^{2}-12x+q=0 ⇒x

_{3}+x

_{4}=12 and x

_{3}x

_{4}=q

_{1},x

_{2},x

_{3},x

_{4}form an increasing G.P.

_{1}=a,x

_{2}=ar,x

_{3}=ar

^{2},x

_{4}=ar

^{3}, where r>1

_{1}+x

_{2}=3

_{3}+x

_{4}=12

^{2}(1+r)=12)}

_{1}=1,x

_{2}=2,x

_{3}=4,x

_{4}=8

_{1}x

_{2}=2 and q=x

_{3}x

_{4}=32

**Q5.**If (√3+i)

^{10}=a+ib, then a and b are respectively

(c) We have, (√3+i)

^{10}=a+ib ⇒i

^{10}(1-i√3)

^{10}=a+i b

^{10}=a+i b [∵Ï‰=-1/2+i √3/2]

^{10}Ï‰

^{10}=a+i b

^{10}Ï‰=a+i b

^{10}((-1)/2+i √3/2)=a+i b

^{9}-2

^{9}√3 i=a+i b

^{9}and b=-2

^{9}√3

**Q6.**If Ï‰ is an imaginary cube root of unity and x=a+b, y=aÏ‰+bÏ‰

^{2},z=aÏ‰

^{2}+bÏ‰, then x

^{2}+y

^{2}+z

^{2}is equal to

(a) x

^{2}+y

^{2}+z

^{2}=(a+b)

^{2}+Ï‰

^{2}(a+bÏ‰)

^{2}+(aÏ‰

^{2}+bÏ‰)

^{2}

^{2}+b

^{2}+2ab+a

^{2}Ï‰

^{2}+b

^{2}Ï‰

^{4}+2abÏ‰

^{3}+a

^{2}Ï‰

^{4}+b

^{2}Ï‰

^{2}+2abÏ‰

^{3}=a

^{2}(1+Ï‰+Ï‰

^{2})+b

^{2}(1+Ï‰+Ï‰

^{2})+6ab [∵Ï‰

^{4}=Ï‰] =6ab [∵1+Ï‰+Ï‰

^{2}=0]

**Q7.**The conjugate of the complex number (1+i)

^{2}/(1-i) is

(d) Given complex number is (1+i)

^{2}/(1-i)=((1+i

^{2}+2i))/(1-i)×(1+i)/(1+i)=(2i+2i

^{2})/(1+1)=i-1

**Q8.**Number of non-zero integral solutions of the equation (1-i)

^{n}=2

^{n}is

(d) We have, (1-i)

^{n}=2

^{n}

^{n}=|2|

^{n}

^{n}=2

^{n}

^{n/2}=2

^{n}

^{n/2}=1

**Q9.**If the roots of the equation qx

^{2}+px+q=0 are complex, where p,q are real, then the roots of the equation x

^{2}-4qx+p

^{2}=0 are

(a) The given equations are qx

^{2}+px+q=0 …(i)

^{2}-4qx+p

^{2}=0 …(ii)

^{2}-4q

^{2}<0

^{2}-4p

^{2}=-4(p

^{2}-4q

^{2})>0

**Q10.**If z

_{1},z

_{2},z

_{3},z

_{4}represent the vertices of a rhombus taken in the anticlockwise order, then

(c) Since the diagonals of a rhombus bisect each other at right-angle

_{1}+z

_{3})/2=(z

_{2}+z

_{4})/2

_{1}+z

_{3}=z

_{2}+z

_{4}

_{2}-z

_{4})/(z

_{1}-z

_{3})=Ï€/2