Dear Readers,
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 equation of the base of an equilateral triangle is x+y=2 and the vertex is (2,-1),then the length of the side of the triangle is

Solution

(c) Let p be the length of the perpendicular from the vertex (2,-1) to the base x+y=2, then p=|(2-1-2)/√(1

(c) Let p be the length of the perpendicular from the vertex (2,-1) to the base x+y=2, then p=|(2-1-2)/√(1

^{2}+1^{2})| =1/√2 If a be the length of the side of triangle then, p=a sin〖60°〗 ⇒ 1/√2=(a√3)/2 ⇒ a=√(2/3)**Q2.**The equation y

^{2}-x

^{2}+2x-1=0 represents

Solution

c) Given equation of curve is y

c) Given equation of curve is y

^{2}-x^{2}+2x-1=0 Here, a=-1,b=1,c=-1,h=0,g=1,f=0 ∴ ∆=abc+2fgh-af^{2}-bg^{2}-ch^{2}=(-1)(1)(-1)+2(0)1(0)-0-1-0 =1-1=0 ∴ Given equation is equation of pair of straight lines.**Q3.**If the diagonals of a parallelogram ABCD are along the lines x+5y=7 and 10x-2y=9, then ABCD must be a

Solution

d) Clearly, diagonals are perpendicular So, ABCD must be a rhombus

d) Clearly, diagonals are perpendicular So, ABCD must be a rhombus

**Q4.**The pairs of straight lines ax

^{2}+2hxy-ay

^{2}=0

and hx

^{2}-2axy-hy

^{2}are such that

Solution

d) The two pairs of lines are ax

d) The two pairs of lines are ax

^{2}+2hxy-ay^{2}=0 …(i) hx^{2}-2axy-hy^{2}=0 …(ii) Clearly, these two equations represent two pairs of lines such that the lines in each pair are mutually perpendicular. The combined equation of the bisectors of the angles between the lines given in (i) is (x^{2}-y^{2})/(a+a)=xy/h⇒hx^{2}-2axy-hy^{2}=0 Clearly it is same as (ii). Thus, each pair bisects the angle between the other pair.**Q5.**The equation of straight line equally inclined to the axes and equidistance from the points (1,-2) and (3,4) is ax+by+c=0

, where

Solution

(c) The equation of straight line equally inclined to the axes is x/a+y/a=1 ⇒ x+y=a. Since, it is equidistant from the points (1,-2) and (3, 4), so perpendicular distances from these points on the line will be equal. ⇒ |(1-2-a)/√(1

(c) The equation of straight line equally inclined to the axes is x/a+y/a=1 ⇒ x+y=a. Since, it is equidistant from the points (1,-2) and (3, 4), so perpendicular distances from these points on the line will be equal. ⇒ |(1-2-a)/√(1

^{2}+1^{2})|=|(3+4-a)/√(1^{2}+1^{2})| ⇒ (1+a)/√2=(7-a)/√2 ⇒ 2a=6 ⇒ a=3 ∴ Equation is x+y-3=0 But, given equation is ax+by+c=0 ∴a=1,b=1,c=-3**Q6.**The equation y=±√3 x,y=1 are the sides of

Solution

(a) Equation of OA is y=√3 x. Equation of OB is y=-√3 x and equation of AB is y=1 Its an equilateral triangle.

(a) Equation of OA is y=√3 x. Equation of OB is y=-√3 x and equation of AB is y=1 Its an equilateral triangle.

**Q7.**If A(2,-1) and B(6,5) are two points, then the ratio in which the foot of the perpendicular from (4,1) to AB divided

it, is

Solution

(b) Let P(4,1)and PD⊥AB. Equation of AB is 3x-2y-8=0 ∴ Equation of PD is 2x+3y-11=0 Let line AB is divided by PD in the ratio Î»:1, then intersecting point D((6Î»+2)/(Î»+1),(5Î»-1)/(Î»+1)) lies on 2x+3y-11=0 ⇒ 2((6Î»+2)/(Î»+1))+3((5Î»-1)/(Î»+1))-11=0 ⇒ 16Î»-10=0 ⇒ Î»:1=5:8

(b) Let P(4,1)and PD⊥AB. Equation of AB is 3x-2y-8=0 ∴ Equation of PD is 2x+3y-11=0 Let line AB is divided by PD in the ratio Î»:1, then intersecting point D((6Î»+2)/(Î»+1),(5Î»-1)/(Î»+1)) lies on 2x+3y-11=0 ⇒ 2((6Î»+2)/(Î»+1))+3((5Î»-1)/(Î»+1))-11=0 ⇒ 16Î»-10=0 ⇒ Î»:1=5:8

**Q8.**Origin containing angle bisector of two lines L

_{1}≡a

_{1}x+b

_{1}y+c

_{1}=0 and L

_{2}≡a

_{2}x+b

_{2}y+c

_{2}=0 (where c

_{1}c

_{2}<0) is

Solution

b)

b)

**Q9.**f the distance of any point (x,y) from the origin is defined as d(x,y)=max {|x|,|y|},d(x,y)=a, non-zero constant, then the locus is

Solution

b) d(x,y)=max{|x|,|y|} …(i) but d(x,y)=a …(ii) From, Eqs. (i) and (ii), a=〖max 〗{|x|,|y|} If |x|>|y|, then a=|x| ∴ x=±a and if |y|>|x|, then a=|y| ∴y=±a Therefore, locus represents a straight line

b) d(x,y)=max{|x|,|y|} …(i) but d(x,y)=a …(ii) From, Eqs. (i) and (ii), a=〖max 〗{|x|,|y|} If |x|>|y|, then a=|x| ∴ x=±a and if |y|>|x|, then a=|y| ∴y=±a Therefore, locus represents a straight line

**Q10.**Which of the following pair of straight lines intersect at right angle?

Solution

(a)

(a)