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.

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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.

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**Q1.**If the lines kx-2y-1=0 and 6x-4y-m=2 are identical (coincident ) lines, then the values of k and m are

Solution

(a) Given lines are kx-2y-1=0 and 6x-4y-m=0 Since, these lines are coincident. ∴ k/6=(-2)/(-4)=(-1)/(-m) ⇒k/6=1/2 and 1/m=1/2 ⇒k=3 and m=2

(a) Given lines are kx-2y-1=0 and 6x-4y-m=0 Since, these lines are coincident. ∴ k/6=(-2)/(-4)=(-1)/(-m) ⇒k/6=1/2 and 1/m=1/2 ⇒k=3 and m=2

**Q2.**The ratio in which the line 3 x+4 y+2=0 divides the distance between 3 x+4 y+5=0, and 3 x+4 y-5=0, is

Solution

(b) Lines 3 x+4 y+2=0 and 3 x+4 y+5=0 are on the same side of the origin. The distance d

d

Lines 3 x+4 y+2=0 and 3 x+4 y-5=0 are on the opposite sides of the origin. The distance d

d

Thus, 3 x+4 y=0 divides the distance between 3x+4y+5=0 and 3 x+4 y-5=0 in the ratio d

(b) Lines 3 x+4 y+2=0 and 3 x+4 y+5=0 are on the same side of the origin. The distance d

_{1}between these lines is given byd

_{1}=|(2-5)/√(3^{2}+4^{2})|=3/5Lines 3 x+4 y+2=0 and 3 x+4 y-5=0 are on the opposite sides of the origin. The distance d

_{2}between these lines is given byd

_{2}=|(2+5)/√(3^{2}+4^{2})|=7/5Thus, 3 x+4 y=0 divides the distance between 3x+4y+5=0 and 3 x+4 y-5=0 in the ratio d

_{1}∶d_{2}i.e. 3∶7**Q3.**The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0,(1+q)x-qy+q(1+q)=0 and y=0, where

p≠q is

Solution

d)

d)

**Q4.**The locus of the point of intersection of lines x cos〖Î±+y sinÎ±=a and x sinÎ±- ycosÎ±=b is (Î± is a variable)

Solution

(c) Let (h,k) be the point of intersection of the line x cosÎ±+y sinÎ±=a and x sinÎ±-y cosÎ±=b. Then,

h cosÎ±+k sinÎ±=a …(i)

h sinÎ±-k cosÎ±=b …(ii)

Squaring and adding (i) and (ii), we get

(h cosÎ±+k sinÎ±)

⇒h

Hence, locus of (h,k) is x

(c) Let (h,k) be the point of intersection of the line x cosÎ±+y sinÎ±=a and x sinÎ±-y cosÎ±=b. Then,

h cosÎ±+k sinÎ±=a …(i)

h sinÎ±-k cosÎ±=b …(ii)

Squaring and adding (i) and (ii), we get

(h cosÎ±+k sinÎ±)

^{2}+(h sinÎ±-k cosÎ±)^{2}=a^{2}+b^{2}⇒h

^{2}+k^{2}=a^{2}+b^{2}Hence, locus of (h,k) is x

^{2}+y^{2}=a^{2}+b^{2}**Q5.**If the equation 3x

^{2}+xy-y

^{2}-3x+6y+k=0 represents a pair of straight lines, then the values of k is

Solution

(c)

We have, 3x

Comparing this equation with

ax

a=3,b=-1,h=1/2,c=k,f=3 and g=-3/2

Equation (i) will represent a pair of straight lines if

abc+2fgh-af

⇒-3k-9/2-27+9/4-k/2=0

⇒-13k/3-117/4=0⇒k=-9

(c)

We have, 3x

^{2}+xy-y^{2}-3x+6y+k=0 …(i)Comparing this equation with

ax

^{2}+2hxy+by^{2}+2gx+2fy+c=0, we havea=3,b=-1,h=1/2,c=k,f=3 and g=-3/2

Equation (i) will represent a pair of straight lines if

abc+2fgh-af

^{2}-bg^{2}-ch^{2}=0⇒-3k-9/2-27+9/4-k/2=0

⇒-13k/3-117/4=0⇒k=-9

**Q6.**The equation 2x

^{2}-24xy+11y

^{2}=0 represents

Solution

a) The required lines are obtaining by shifting the origin at (4,0). So, the required equation is y=|x-4|

a) The required lines are obtaining by shifting the origin at (4,0). So, the required equation is y=|x-4|

**Q7.**If A(cosÎ±, sinÎ±), B(sinÎ±,-cosÎ±), C(1,2) are the vertices of a ∆ ABC, then as Î± varies the locus of its centroid is

Solution

b) Let (h,k) be the centroid of the triangle having vertices A(cosÎ±,-cosÎ±) and C(1,2). Then,

h=cosÎ±+sinÎ±+1/3 and k=sinÎ±-cosÎ±+2/3

⇒3 h-1=cosÎ±+sinÎ± and 3 k-2=sinÎ±-cosÎ±

⇒(3 h-1)

⇒9(h

⇒3(h

Hence, the locus of (h, k) is 3(x

b) Let (h,k) be the centroid of the triangle having vertices A(cosÎ±,-cosÎ±) and C(1,2). Then,

h=cosÎ±+sinÎ±+1/3 and k=sinÎ±-cosÎ±+2/3

⇒3 h-1=cosÎ±+sinÎ± and 3 k-2=sinÎ±-cosÎ±

⇒(3 h-1)

^{2}+(3 k-2)^{2}=2 [Squaring and adding]⇒9(h

^{2}+k^{2})-6 h-12 k+3=0⇒3(h

^{2}+k^{2})-2 h-4k+1=0Hence, the locus of (h, k) is 3(x

^{2}+y^{2})-2 x-4 y+1=0**Q8.**The number of points on the line 3x+4y=5, which are at a distance of sec

^{2}Î¸ + 2cossec

^{2}Î¸,Î¸∈R, from the point (1, 3) is

Solution

b) The perpendicular distance of (1, 3) from the line 3x+4y=5 is 2 units while, sec

b) The perpendicular distance of (1, 3) from the line 3x+4y=5 is 2 units while, sec

^{2}Î¸+2 cosec^{2}Î¸≥3 [as sec^{2}Î¸,cosec^{2}Î¸≥1] So, there will be two such points on the line**Q9.**If the pair of straight lines given by Ax

^{2}+2Hxy+By

^{2}=0(H

^{2}>AB) forms an equilateral triangle with line ax+by+c=0, then (A+3B)(3A+B) is equal to

Solution

(d) Given, Ax

(d) Given, Ax

^{2}+2Hxy+By^{2}=0 …(i) and ax+by+c=0 …(ii) Since, triangle is equilateral, then angle between the two lines is 60° Angle between pair of lines is given by cos60°=(A+B)/√((A-B)^{2}+4H^{2}) ⇒ (A+B)/√((A-B)^{2}+4H^{2})=1/2 ⇒ (A-B)^{2}+4H^{2}=4(A+B)^{2}⇒ 4(A^{2}+B^{2}+2AB)-(A^{2}+B^{2}-2AB)=4H^{2}⇒ 3A^{2}+10AB+3B^{2}=4H^{2}⇒ (3A+B)(A+3B)=4H^{2}**Q10.**The equation of pair of lines joining origin to the points of intersection of x

^{2}+y

^{2}=9 and x+y=3 is

Solution

(b) Given, x

and x+y=3 …(ii)

From Eqs. (i) and (ii), we make a homogeneous equation.

⇒ x

⇒ x

⇒ xy=0

(b) Given, x

^{2}+y^{2}=9 …(i)and x+y=3 …(ii)

From Eqs. (i) and (ii), we make a homogeneous equation.

⇒ x

^{2}+y^{2}=(x+y)^{2}⇒ x

^{2}+y^{2}=x^{2}+y^{2}+2xy⇒ xy=0