Let the equation of the plane, that passes through the point and contains the line of intersection of the planes and , be , then is equal to :
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept: A plane passing through the line of intersection of two planes and is given by the family of planes . Step 2: Detailed Explanation: The family of planes is:
Since the plane passes through the point :
Substitute back into the family equation:
To make the constant term , multiply the entire equation by :
Comparing this with : , , . Calculate :
Step 3: Final Answer: The sum is , which is option (D).
02
PYQ 2021
medium
mathematicsID: jee-main
If the variable line lies between the two circles and , without intercepting a chord on either circle, then the sum of all the integral values of is _____________
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: For a line to lie between two circles without intersecting them, its distance from the centers must be greater than or equal to the respective radii, and the centers must lie on opposite sides of the line. Step 2: Detailed Explanation: Circle 1: Center , Radius . Circle 2: Center , Radius . Distance of line from centers: 1. or . 2. or . 3. Since the line is between the circles, the centers must be on opposite sides:
Combining all conditions: - From and ( or ): . - From and ( or ): . Integral values of . Sum = . Step 3: Final Answer: The sum of integral values is 165.
03
PYQ 2023
medium
mathematicsID: jee-main
From the top A of a vertical wall AB of height 30 m, the angles of depression of the top P and bottom Q of a vertical tower PQ are 15 and 60 respectively. B and Q are on the same horizontal level. If C is a point on AB such that CB = 15 m, then the area (in m ) of the quadrilateral BCPQ is equal to :
1
2
3
4
Official Solution
Correct Option: (4)
Tower and Wall Problem
Let AB be the wall and PQ be the tower. Let x be the height of the tower PQ and y be the distance BQ.
In , we have and .
.
Let C be a point on AB such that CB = 15. Then AC = AB - CB = 30 - 15 = 15.
In , we have and . CP = BQ = y = .
Here there is an error! it should be
Let's assume that in ACP we have AP not x so it becomes and and such that
We need to calculate so x is incorrect!
Let's try another option we use where CQ is (AC to B which is 15 ) + (CQ that we need to find). PC = AB - AP = x Here, x = AP, so PQ = AB - AP = 30 - x . The problem is that tan(15) cannot be derived like this. This would work: x the tower is then which we don't have a proper answer for. But then what is BCPQ Area of quadrilateral BCPQ = Area of rectangle - Area of PCQ BC = 15 and PQ = 30 - x BCPQ is a trapezium, not a rectangle.
Area of trapezium BCPQ
Area Area = x.y where x = and y =
This is assuming we calculated X as but it is still incorrect!
04
PYQ 2025
medium
mathematicsID: jee-main
Let be a tetrahedron such that the edges , , and are mutually perpendicular. Let the areas of the triangles , , and be 5, 6, and 7 square units respectively. Then the area (in square units) of the is equal to:
1
2
12
3
4
7
Official Solution
Correct Option: (3)
We are asked to find the area of the face BCD of a tetrahedron ABCD. We are given that the edges AB, AC, and AD are mutually perpendicular, and the areas of the faces ABC, ACD, and ADB are 5, 6, and 7 square units, respectively.
Concept Used:
This problem can be solved using De Gua's Theorem, which is a generalization of the Pythagorean theorem to three dimensions. For a tetrahedron that has a right-angled vertex (like the corner of a cube), the square of the area of the face opposite the right-angled vertex is the sum of the squares of the areas of the other three faces.
Let A be the right-angled vertex. Let , , and be the areas of the three mutually orthogonal faces. Let be the area of the face opposite the right-angled vertex. The theorem states:
Step-by-Step Solution:
Step 1: Identify the given information based on the problem statement.
The edges AB, AC, and AD are mutually perpendicular, which means the vertex A is a right-angled vertex. The faces , , and are the three orthogonal faces. Their areas are given as:
We need to find the area of the face .
Step 2: Apply De Gua's Theorem to the given tetrahedron.
Using the formula from the concept, we can write:
Step 3: Substitute the given area values into the equation.
Final Computation & Result:
Step 4: Perform the calculation to find the squared area.
Step 5: Take the square root to find the final area.
Therefore, the area of the is square units.
05
PYQ 2025
easy
mathematicsID: jee-main
Let a circle with radius passes through four distinct points , such that , then is equal to:
Official Solution
Correct Option: (1)
06
PYQ 2025
medium
mathematicsID: jee-main
Let for two distinct values of , the lines touch the ellipse at the points and . Let the line intersect at the points and . Then the area of the quadrilateral is equal to:
1
36
2
24
3
48
4
20
Official Solution
Correct Option: (2)
Step 1: Given points of contact:
and the point is:
Step 2: Area Calculation of Triangle :
The area of triangle is given by:
Step 3: Area of Quadrilateral :
The area of quadrilateral is:
07
PYQ 2025
medium
mathematicsID: jee-main
The center of a circle is at the center of the ellipse , where . Let pass through the foci and of such that the circle and the ellipse intersect at four points. Let be one of these four points. If the area of the triangle is 30 and the length of the major axis of is 17, then the distance between the foci of is:
1
8
2
10
3
12
4
14
Official Solution
Correct Option: (2)
We are given the equation of the ellipse: - The foci and of the ellipse are located at , where is given by: - The length of the major axis of the ellipse is , so: - The area of the triangle is given as 30. The area of a triangle with base and height (since the height of the triangle is the distance from the point to the major axis, which is the semi-minor axis of the ellipse) is given by: Given that the area is 30: - Now, we can substitute the value of (the semi-minor axis) using the relation . From the earlier equation for , we have: Thus, we can solve for the distance between the foci, . The solution is given by: Thus, the correct answer is .
08
PYQ 2025
medium
mathematicsID: jee-main
Let the sum of the focal distances of the point on the hyperbola be . If for , the length of the latus rectum is and the product of the focal distances of the point is , then is equal to:
1
184
2
186
3
185
4
187
Official Solution
Correct Option: (3)
We are given the hyperbola equation , and the sum of the focal distances of the point is . Step 1: Find the distance using the given conditions:
Step 2: Use the relationship for the focal distance in a hyperbola:
Step 3: Use the relationship between and : Substitute into this equation: Step 4: Now, calculate the length of the latus rectum : Substitute , we get , so: Step 5: Calculate (the product of the focal distances): Substitute and : Step 6: Finally, calculate :
Thus, the correct answer is .
09
PYQ 2025
medium
mathematicsID: jee-main
The axis of a parabola is the line and its vertex and focus are in the first quadrant at distances and units from the origin, respectively. If the point lies on the parabola, then a possible value of is:
1
4
2
9
3
3
4
8
Official Solution
Correct Option: (2)
The vertex of the parabola is at the origin , and the axis of the parabola is along the line . The focus is at , and the directrix is the line . Using the definition of a parabola, the distance from any point on the parabola to the focus equals the distance from that point to the directrix. Let the point be on the parabola. Let be the distance from to the focus and be the distance from to the directrix. First, calculate the distance : Next, calculate the distance from the point to the directrix . The formula for the distance from a point to a line is: For the directrix , , , and , so: Now, equate and (since the point lies on the parabola): After solving this equation, we find that . Thus, the correct answer is .
10
PYQ 2025
medium
mathematicsID: jee-main
Let the values of , for which the shortest distance between the lines and is , be , where . Then the length of the latus rectum of the ellipse is:
1
9
2
3
4
18
Official Solution
Correct Option: (3)
The shortest distance between two skew lines is given by the formula: where , , , and . We compute as: Now, we compute the cross product : Now, we calculate the magnitude of the cross product: Using the formula for the shortest distance: This yields the condition for the distance, and solving for the length of the latus rectum of the ellipse: Thus, the correct answer is .
11
PYQ 2025
easy
mathematicsID: jee-main
Let the three sides of a triangle ABC be given by the vectors $ $ _______
Official Solution
Correct Option: (1)
We are given the sides of the triangle as vectors: Step 1: Centroid Calculation The centroid of a triangle is given by the average of the position vectors of the three vertices: Since and , we can solve for the position vectors of and then calculate . Step 2: Compute , , and From the centroid formula: Thus, . Now, we find the squared distances from to each point: - : The distance from to is: - : The distance from to is: - : The distance from to is: Step 3: Calculate the Final Expression Now, we calculate: Thus, the final value is .
12
PYQ 2025
medium
mathematicsID: jee-main
Let be the radius of the circle, which touches the -axis at point , and the parabola at the point . Then is equal to:
Official Solution
Correct Option: (1)
Given the equation of the tangent to at the point :
The equation is:
Also, the equation of the circle is:
Expanding the equation of the circle:
Now, comparing the coefficients of , , and constant terms:
From the equation:
Now solve for :
Solving the quadratic equation:
Now, the radius is given by:
Since the center lies in the second quadrant:
Thus,
13
PYQ 2025
hard
mathematicsID: jee-main
Let and be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.
Official Solution
Correct Option: (1)
The area of triangle is:
The maximum area of parallelogram is given by:
14
PYQ 2025
hard
mathematicsID: jee-main
The distance of the point from the line along the line is:
1
2
3
4
Official Solution
Correct Option: (2)
The given line can be written in parametric form as:
where is the parameter. Let the point be the point from which we want to find the distance. The direction ratios of the line are , and the coordinates of the point on the line are . The distance of the point from the line can be calculated using the formula:
where is the vector from a point on the line to the point , and is the direction vector of the line. Substitute the values:
Now, calculate the cross product :
Thus, the magnitude of the cross product is:
Finally, the distance is:
Thus, the required distance is .
15
PYQ 2026
easy
mathematicsID: jee-main
Identify the total number of surfaces in the given 3D object.
1
9
2
10
3
8
4
12
Official Solution
Correct Option: (2)
Step 1: Analyze the given 3D object. The given object is a 3D shape composed of several surfaces, including the top, bottom, and sides. We need to count all the visible surfaces. Step 2: Counting the surfaces. The 3D object has 10 surfaces, including:
- The top and bottom surfaces.
- The front, back, left, and right sides.
- Additional angled or slanted surfaces formed due to the shape. Step 3: Conclusion. By counting the surfaces, we determine that the total number of surfaces in this 3D object is 10. Final Answer: 10.
16
PYQ 2026
hard
mathematicsID: jee-main
Let the midpoints of the sides of a triangle be and . If its incentre is , then is equal to:
1
2
3
4
Official Solution
Correct Option: (3)
Concept: The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle. The centroid of the original triangle and the medial triangle coincide. If the vertices of the medial triangle are known, we can first find the vertices of the original triangle and then determine its incentre using the side–length weighted formula. Step 1:Let the given midpoints be} These correspond to midpoints of and . Step 2:Recover the vertices of triangle .} Using midpoint relations, Thus, Step 3:Compute side lengths.} Step 4:Use the incentre formula.} For triangle vertices , where . Substituting, Thus, Step 5:Compute the required value.}
17
PYQ 2026
medium
mathematicsID: jee-main
The question figure shows the 3D view of the object. Choose the correct 2D view (Elevation) of the object looking in the direction of the arrow
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (3)
The 3D view shows a specific object, and the correct 2D projection (elevation) is the one that matches the viewpoint looking in the direction of the arrow. By analyzing the 3D figure and understanding how the surfaces appear from that perspective, we conclude that the correct 2D view is option (C). Final Answer: (C).
18
PYQ 2026
medium
mathematicsID: jee-main
Identify the Mirror image of the given figure along X – X' axis.
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
The given figure is reflected along the X – X' axis. When a figure is mirrored along the X – X' axis, the left and right sides are swapped. By analyzing the options, we find that option (B) shows the correct mirror image of the given figure along the X – X' axis. Final Answer: (B).
19
PYQ 2026
medium
mathematicsID: jee-main
Find out the odd one.
1
D
2
C
3
B
4
A
Official Solution
Correct Option: (1)
Let’s analyze the given options one by one: - Option A: The shape inside the box in option A has a triangle and a circle, which is different from other shapes that involve a combination of different geometric shapes (square, triangle, and circle). - Option B: The shapes in the box involve a circle, square, and triangle, and follow a similar pattern as the other options. - Option C: Similarly, option C involves different shapes (circle, square, and triangle) in different orientations. - Option D: This option is the most different as it involves a circle and square, with the other shapes (triangle) placed differently. Thus, option A is the odd one out as it doesn't follow the same pattern of shapes as the others.
Final Answer: (D) A
20
PYQ 2026
medium
mathematicsID: jee-main
If you draw an object measuring 1 cm on the paper, what would it be equivalent to at a scale of 1:100?
1
10 cm
2
1 meter
3
10 meters
4
1 cm
Official Solution
Correct Option: (3)
Step 1: Understand the scale. At a scale of 1:100, 1 cm on the paper represents 100 cm in real life. Thus, to find the actual size of the object, we multiply the drawing size by 100. Step 2: Calculate the real-life size. Since 1 cm on the paper is equivalent to cm in real life, we convert this to meters: [
100 cm = 1 \textmeter
] Step 3: Conclusion. Therefore, the equivalent size in real life is 1 meter. Final Answer: 10 meters.
21
PYQ 2026
medium
mathematicsID: jee-main
If and , then the distance of the centre of to the farthest point on is:
1
10
2
11
3
12
4
13
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept:
The farthest point on a circle from an external point lies on the line passing through the centre of the circle and that external point. The distance is calculated as the distance between the two centres plus the radius of the target circle. Step 2: Key Formula or Approach:
1. Centre of , and Radius .
2. Centre of , .
3. Farthest distance . Step 3: Detailed Explanation:
1. For : .
Centre .
Radius .
2. For : .
Centre .
3. Distance between centres and :
.
4. Farthest distance = . Step 4: Final Answer:
The distance to the farthest point is 12.
22
PYQ 2026
medium
mathematicsID: jee-main
P is the point of intersection of the two half-lines Points and lie on these lines at a distance from .
If the length of perpendicular from on is and the radius of the circumcircle of is , then find
Official Solution
Correct Option: (1)
Concept:
The lines represent two symmetric half-lines forming an angle between them. Step 1: Point of intersection For the lines intersection occurs at Step 2: Points and Since and lie on the two rays and are at distance from , Thus triangle is symmetric with vertex angle Step 3: Altitude The perpendicular from to is given as In an equilateral triangle of side , altitude From the figure geometry, Step 4: Circumradius For an equilateral triangle: Thus Step 5: Required value
23
PYQ 2026
medium
mathematicsID: jee-main
Let A, B be points on the two half-lines at a distance of from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If and R is the radius of the circumcircle of , then is equal to ________
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
By graphing the half-lines, we can establish the exact geometric nature of the triangle PAB. Observing the slopes of the lines determines the angle between them, which in turn identifies special properties (like an equilateral triangle) allowing swift computation of the circumradius.
Step 2: Key Formula or Approach:
Line 1: (Slope ).
Line 2: (Slope ).
Radius of circumcircle of an equilateral triangle of side : .
Step 3: Detailed Explanation:
The intersection P of the two half-lines occurs where .
.
The two lines extend outward from P. Their slopes are and , meaning they make angles of and with the positive x-axis.
The angle between the two half-lines is .
Since A and B are at distance from P, .
In , the sides and are equal, and the included angle is . This forces to be an equilateral triangle with side length . The angle bisector of the two half-lines lies directly on the x-axis. Q is the intersection of the angle bisector and segment AB. Because is equilateral, the angle bisector is also the median and the altitude.
Thus, PQ is the altitude of the equilateral triangle PAB.
Length of altitude .
We are given :
. Let R be the radius of the circumcircle of equilateral .
. We need the value of :
.
Step 4: Final Answer:
The value is 9.
24
PYQ 2026
medium
mathematicsID: jee-main
The value of is equal to:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
This is a standard improper integral that can be solved using substitution or complex analysis (residues). A common substitution for integrals of the form is . Step 2: Key Formula or Approach:
Let .
Using the property . Step 3: Detailed Explanation:
1. Here , so .
2. Substitute , then .
3. Limits: ; .
4. Split the integral:
5. We know .
6. So, . Step 4: Final Answer:
The value of the integral is .
25
PYQ 2026
hard
mathematicsID: jee-main
If and are two parallel lines and is an equilateral triangle as shown in the figure, then the area of triangle is:
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: From the figure, point lies on the median of the equilateral triangle.
The distances are:
Step 2: In an equilateral triangle, the centroid divides the median in the ratio .
Hence, total height of the triangle is:
Step 3: Height of an equilateral triangle is given by:
Step 4: Area of an equilateral triangle:
26
PYQ 2026
easy
mathematicsID: jee-main
A rectangle is formed by the lines . A line perpendicular to divides the rectangle into two equal parts. Then the distance of the line from the point is:
1
2
3
4
Official Solution
Correct Option: (2)
Concept:
A line dividing a plane figure into two equal areas must pass through the centroid of the figure. Centroid of a rectangle Distance of a point from a line is Step 1: Find the centroid of the rectangle. The rectangle is bounded by:
Hence, centroid:
Step 2: Find the slope of the required line. Given line:
Slope Slope of a perpendicular line:
Step 3: Equation of the line passing through the centroid. Using point–slope form:
Simplifying:
Thus, the line is:
Step 4: Find the distance of the point from the line.
27
PYQ 2026
medium
mathematicsID: jee-main
If Then the value of is:
1
8
2
9
3
10
4
11
Official Solution
Correct Option: (3)
Step 1: Use series expansions. To solve this, we expand the terms , , and into their Taylor series around . Step 2: Analyze the denominator. Similarly, expand using the series for and . Step 3: Apply the limit. After simplifying the expression and equating the limit to 2, we solve for . Final Answer:
28
PYQ 2026
medium
mathematicsID: jee-main
If lies on the hyperbola and the length of the latus rectum is 8, then the square of the area of is [where and are the foci of the hyperbola].
1
2700
2
2400
3
1750
4
3600
Official Solution
Correct Option: (1)
Step 1: Use the properties of hyperbolas. The length of the latus rectum of a hyperbola is given by: We are given that the length of the latus rectum is 8, so: This gives the relation between and . Step 2: Use the coordinates of point . Since the point lies on the hyperbola, substitute and into the equation of the hyperbola: This gives another equation involving and . Step 3: Solve the system of equations. By solving these equations, we find the values for and , and then calculate the square of the area of . Final Answer:
29
PYQ 2026
medium
mathematicsID: jee-main
If is the vertex of the parabola , is a point on the parabola.
If is the locus of a point which divides in the ratio ,
then the equation of the chord of which is bisected at the point is:
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Parametric coordinates of a point on the parabola are:
Vertex . Step 2: Point divides internally in the ratio :
Step 3: Eliminate to find the locus of : Hence, the locus is:
Step 4: Let the chord of this parabola be bisected at .
Let the corresponding parameters of the endpoints be . Using midpoint formula:
Step 5: Slope of the chord:
Equation of chord through :
30
PYQ 2026
easy
mathematicsID: jee-main
The locus of the point of intersection of tangents drawn to the circle
which subtends an angle of , is:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: The given circle has centre
and radius
Step 2: If the angle between tangents drawn from a point to a circle is , then:
Here,
Step 3: Distance of point from the centre:
Step 4: Hence, the locus of point is a circle with centre and radius . Multiplying by 3:
31
PYQ 2026
easy
mathematicsID: jee-main
The line intersects the ellipseat points and . Find the angle subtended by the segment at the centre of the ellipse.
1
2
3
4
Official Solution
Correct Option: (2)
Concept:
The centre of the ellipse is at the origin . The angle subtended by chord at the centre is the angle between vectors and . If slopes of lines and are and , then Step 1: Find the points of intersection. Substitute into the ellipse:
Corresponding points:
Step 2: Slopes of and . Step 3: Angle between the two lines. Angle between vertical line and a line of slope :
32
PYQ 2026
medium
mathematicsID: jee-main
If and , where and and , then is:
Official Solution
Correct Option: (1)
Step 1: Use trigonometric identities. We are given and , which we can use to set up equations for and . Using these values, we find relations between and . Step 2: Solve for . Using the given equation for and the condition , we solve for the values of and , which leads to . Final Answer:
