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.**The distance between the pair of parallel lines given by x

^{2}-1005x+2006=0 is

Solution

a) Given equation is x

a) Given equation is x

^{2}-1005x+2006=0 ⇒ (x-2)(x-1003)=0 ⇒ x=2,x=1003 ∴ Required distance between the lines =1003-2=1001**Q2.**The area of the triangle formed by the axes and the line (coshÎ±-sinhÎ± )+(coshÎ±+sin hÎ±)y=2 in square units, is

Solution

c) Given line meets the coordinate axes at A(2/cosh Î±-sinh Î± ,0)and B(0,2/cosh Î±+sinh Î±) ∴Area of ∆ OAB =1/2×OA×OB =1/2×2/cosh Î±-sinh Î± ×2/cosh Î±+sinhÎ± =2/(cosh

c) Given line meets the coordinate axes at A(2/cosh Î±-sinh Î± ,0)and B(0,2/cosh Î±+sinh Î±) ∴Area of ∆ OAB =1/2×OA×OB =1/2×2/cosh Î±-sinh Î± ×2/cosh Î±+sinhÎ± =2/(cosh

^{2}Î±-sinh^{2}Î±)=2 sq.units**Q3.**The area enclosed within the curve |x|+|y|=1, is

Solution

(b) The given curve is |x|+|y|=1 ⇒x+y=1 for x≥0,y≥0 -x-y=1 for x<0,y<0 x-y=1 for x≥0,y<0 -x+y=1 for x≤0,y≥0 These lines represent a square as shown in Fig.S.5 such that the length of each side is √2 units ∴ Area enclosed =√2×√2=2 sq.units

(b) The given curve is |x|+|y|=1 ⇒x+y=1 for x≥0,y≥0 -x-y=1 for x<0,y<0 x-y=1 for x≥0,y<0 -x+y=1 for x≤0,y≥0 These lines represent a square as shown in Fig.S.5 such that the length of each side is √2 units ∴ Area enclosed =√2×√2=2 sq.units

**Q4.**The vertices of ∆OBC are (0, 0), (-3,-1) and (-1,-3), then the equation of the line parallel to BC which is a distance 1/2 from the origin and cut OB and OC intercept, is

Solution

(a) Since, the angle is right angle. ∴Homogenising,x

(a) Since, the angle is right angle. ∴Homogenising,x

^{2}+y^{2}=4((y-3x)/c)^{2}⇒ c^{2}(x^{2}+y^{2})=4(y^{2}+9x^{2}-6xy) Since, lines are at right angle. ∴ Coefficient of x^{2}+coefficient of y^{2}=0 ⇒ c^{2}-36+c^{2}-4=0 ⇒ c^{2}=20**Q5.**A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve

x

^{2}+y

^{2}=4 with x+y=a. The set containing the value of ‘a’ is

Solution

a) To make the given curves x

a) To make the given curves x

^{2}+y^{2}=4 and x+y=a homogeneous. x^{2}+y^{2}-4((x+y)/a)^{2}=0 ⇒ a^{2}(x^{2}+y^2 )-4(x^{2}+y^{2}+2xy)=0 ⇒ x^{2}(a^{2}-4)+y^{2}(a^{2}-4)-8xy=0 Since, this is a perpendicular pair of straight lines. ∴ a^{2}-4+a^{2}-4=0 ⇒ a^{2}=4 ⇒ a=±2 Hence, required set of a is {-2,2}.**Q6.**The equation of a straight line which passes through the point acos

^{3}Î¸,asin

^{3}Î¸ and perpendicular to xsecÎ¸ + ycosec Î¸=a is

Solution

b) Required equation of line is y-a sin

b) Required equation of line is y-a sin

^{3}Î¸=cosecÎ¸/secÎ¸ (x-acos^{3}Î¸) ⇒y-a sin^{3}Î¸=cosÎ¸/sinÎ¸ (x-a cos^{3}Î¸) ⇒y sinÎ¸-asin^{4}Î¸=x cosÎ¸-a cos^{4}Î¸ ⇒x cosÎ¸-y sinÎ¸+asin^{4}Î¸-a cos^{4}Î¸=0 ⇒x cosÎ¸- y sinÎ¸+a cos2 Î¸=0**Q7.**The image of the point (1,3) in the line x+y-6=0, is

Solution

a) Image of (1,3) in the line x+y-6=0 is given by (x-1)/1=(y-3)/1=-2((1+3-6)/(1

a) Image of (1,3) in the line x+y-6=0 is given by (x-1)/1=(y-3)/1=-2((1+3-6)/(1

^{2}+1^{2}))⇒x=3,y=5 Hence, the image of the given point has coordinates (3,5)**Q8.**The angle between the lines √3 x-y-2=0 and x-√3 y+1=0 is

Solution

(d) Slopes of given lines are m_1=√3 and m_2=1/√3 ∴ tanÎ¸= |(√3-1/√3)/(1+1)|=|(3-1)/(2√3)|=1/√3 ⇒ Î¸=30°

(d) Slopes of given lines are m_1=√3 and m_2=1/√3 ∴ tanÎ¸= |(√3-1/√3)/(1+1)|=|(3-1)/(2√3)|=1/√3 ⇒ Î¸=30°

**Q9.**The lines bisecting the angle between the bisectors of the angles between the lines ax

^{2}+2hxy+by

^{2}are given by

Solution

(c) The equation representing the bisectors of the angles between the lines given by ax

(c) The equation representing the bisectors of the angles between the lines given by ax

^{2}+2hxy+by^{2}=0 is (x^{2}-y^{2})/(a-b)=xy/h ⇒hx^{2}-(a-b)xy-hy^{2}=0 …(i) The combined equation of the bisectors of the angles between these lines is (x^{2}-y^{2})/(h+h)=xy/(-((a-b))/2)⇒(a-b)(x^{2}-y^{2})+4hxy=0**Q10.**The equation of the straight line which is perpendicular to y=x and passes through (3, 2) is

Solution

(b) ∵ Slope of given line y=x is 1 ∴ Slope of required line which is perpendicular to given line is -1 Thus, the equation of required line passing through (3, 2) and slope -1, is y-2=-1(x-3) ⇒ x+y=5

(b) ∵ Slope of given line y=x is 1 ∴ Slope of required line which is perpendicular to given line is -1 Thus, the equation of required line passing through (3, 2) and slope -1, is y-2=-1(x-3) ⇒ x+y=5