datavorous/entrance-exam-dataset huggingface dataset:
Let \( k \) be sum of all \( x \) in the interval \( [0,2 \pi] \) such that \( 3 \cot ^{2} x+8 \cot x+3=0 \), then the value of \( \frac{k}{\pi} \) is

Options:

The correct answer is:
1.50

Solution:
We have,

3cot2x+8cotx+3=0

∴cotx=−4±73

Both roots are negative.

∴π2<x1,x2<π

∴π<x1+x2<2π.

Product of roots =cotx1cotx2

=1

⇒cotx1=1cotx2

⇒cotx1=tanx2

⇒tan3π2-x1=tanx2

⇒3π2-x1=x2

⇒x1+x2=3π2

⇒k=3π2

⇒kπ=32=1.50

Topic: Mathematics, Trigonometric Equations, JEE Main
Let K be the sum of the coefficients of the odd powers of x in the expansion of 1+x99. Let a be the middle term in the expansion of 2+12200. If C99200Ka=2lmn, where m and n are odd numbers, then the ordered pair l,n is equal to:

Options:
A 50,51
B 51,99
C 50,101
D 51,101

The correct answer is:
50,101

Solution:
In the binomial expansion of
1+x99, the sum of odd coefficients of x is given by

K=C1+C3+….+C99=299-02=298

Also, a=Middle term in the binomial expansion of 2+12200 is

T2002+1=T100+1=C100200210012100

⇒T100+1=C100200·250

So,

C99200Ka

=C99200×298C100200×250=100101×248

=25101×250=mn2l

On comparing, we get

l=50, m=25 and n=101

Therefore, l, n≡50,101.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Let l1 and l2 be the two lines which are normal to y2=4x and tangent to x2=-12y respectively (where, l1 and l2 are not the x-axis). Then, the product of the slopes of l1 and l2 is

Options:
A 3
B 2
C 1
D 12

The correct answer is:
2

Solution:
Let the slope of the tangent be k, then

the equation of the tangent to x2=-12y is kx=y+-3k2

Let the slope of the normal be m, then
the equation of the normal to y2=4x is y=mx-2m-m3
⇒k=m and 2m+m3=-3k2
⇒mm2+3m+2=0⇒m=-1,-2 (m=0 is rejected)

Topic: Mathematics, Parabola, JEE Main
Let $l_1$ and $l_2$ be two lines intersecting at $P$. If $A_1, B_1, C_1$ are points on $l_1$, and $A_2, B_2, C_2, D_2, E_2$ are points on $l_2$ and if none of these coincides with $P$, then the number of triangles formed by these eight points, is :

Options:
A 56
B 55
C 46
D 45

The correct answer is:
45

Solution:
If triangle is formed including point ' $P$ ' the other points must be one from $l_1$ and other point from $l_2$. Number of triangles formed with $P$,
$n\left(E_1\right)={ }^3 C_1 \times{ }^5 C_1=15 \text { ways }$
When $P$ is not included, Number of triangles formed
$n\left(E_2\right)={ }^3 C_2 \times{ }^5 C_1+{ }^3 C_1 \times{ }^5 C_2$
$=15+15=30$
Total number of triangle $=n\left(E_1\right)+n\left(E_2\right)$ $=15+30=45$

Topic: Mathematics, Permutation Combination, JEE Main
Let L1 be a tangent to the parabola y2=4(x+1) and L2 be a tangent to the parabola y2=8(x+2) such that L1 and L2 intersect at right angles. Then L1 and L2 meet on the straight line:

Options:
A x + 3 = 0
B 2x + 1 = 0
C x + 2 = 0
D x + 2y = 0

The correct answer is:
x + 3 = 0

Solution:
Let tangent of y2=4(x+1) be L1:t1y=(x+1)+t12
and tangent of y2=8(x+2) be L2:t2y=(x+2)+2t22
L1⊥L2
1t1·1t2=-1
t1t2=-1
t2×(i)-t1×(ii) ⇒ x+3=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Let $\mathrm{L}_1$ be the length of the common chord of the curves $x^2+y^2=9$ and $y^2=8 x$, and $L_2$ be the length of the latus rectum of $y^2=8 x$, then:

Options:
A
$\mathrm{L}_1>\mathrm{L}_2$

B
$\mathrm{L}_1=\mathrm{L}_2$

C
$\mathrm{L}_1 < \mathrm{L}_2$
D
$\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\sqrt{2}$

The correct answer is:
$\mathrm{L}_1 < \mathrm{L}_2$

Solution:
We have
$$
\begin{aligned}
&x^2+(8 x)=9 \\
&x^2+9 x-x-9=0 \\
&x(x+9)-1(x+9)=0 \\
&(x+9)(x-1)=0 \\
&x=-9,1 \\
&\text { for } x=1, y=\pm 2 \sqrt{2 x}=\pm 2 \sqrt{2} \\
&\mathrm{~L}_1=\text { Length of } \mathrm{AB} \\
&\quad=\sqrt{(2 \sqrt{2}+2 \sqrt{2})^2+(1-1)^2}=4 \sqrt{2} \\
&\mathrm{~L}_2=\text { Length of latus rectum } \\
&=4 a=4 \times 2=8 \\
&\mathrm{~L}_1 < \mathrm{L}_2
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2014 (11 Apr Online)
Let l1 be the line in xy-plane with x and y intercepts 18 and 142 respectively, and l2 be the line in zx-plane with x and z intercepts -18 and -163 respectively. If d is the shortest distance between the line l1 and l2, then d-2 is equal to _____.

Options:

The correct answer is:
51

Solution:
Given,

Line L1:x-1818=y-142=z0

or x-181=y-2=z0 ...1

Equation of L2:x+18-63=y0=z8

Shortest distance between two lines is given by,

d=c→-a→·b→×d→b→×d→=14i^·42i^+4j^+36k^422+42+362

=232+16+54=151

∴ d-2=51

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let L1 be the projection and L2 be the image of z-axis in the plane 3x-4y+z+1=0. The distance of the point 1,2,3 from the plane containing the lines L1 and L2 is

Options:
A 2 units
B 4 units
C 5 units
D 6 units

The correct answer is:
2 units

Solution:
The required plane must contain z-axis and parallel to normal vector of the given plane
A parallel vector to z-axis is <0,0,1>
Normal vector to the given plane is <3,-4,1>
A normal vector to the required plane =i^j^k^0013-41
=4i^+3j^
∵ Required plane contain z axis, so it must pass through 0,0,0
Equation of the required plane is 4x-0+3y-0=0⇒4x+3y=0
So, the perpendicular distance from 1,2,3 to 4x+3y=0 is 4+65=2 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let L1 (respectively L2 ) be the line passing through 2i^-k^ (respectively 2i^+j^-3k^) and parallel to 3i^-j^+2k^ (respectively i^-2j^+k^). Then the shortest distance between the lines L1 and L2 is equal to

Options:
A 1035
B 835
C 1135
D 935

The correct answer is:
935

Solution:
Given that line L1 passes through 2i^-k^ and parallel to 3i^-j^+2k^

and line L2 passes through 2i^+j^-3k^ and parallel to i^-2j^+k^.

Then, a1=2i^-k^, b1=3i^-j^+2k^ and a2=2i^+j^-3k^, b2=i^-2j^+k^

Shortest distance between lines is given by, d=b→1×b→2·a→2-a→1b→1×b→2.

Now, b→1×b→2=i^j^k^3-121-21=3i^-j^-5k^

⇒b→1×b→2=32+-12+-52=35

And, a→2-a→1=j^-2k^

Putting these values, we get d=3i^-j^-5k^·j^-2k^35=-1+1035=935

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let L1:x-22=y-3-1=z-13, L2:x-2-1=y-33=z-153 and L3:x-2-32=y-3-19=z-115 are three lines intersecting each other at the point P and a given plane at A,B,C respectively, such that PA=2,PB=3,PC=6.The volume (in cubic units) of the tetrahedron PABC is

Options:
A 2
B 18
C 6
D 10

The correct answer is:
6

Solution:
âµ All three lines are perpendicular
â Volume of the tetrahedron =16PAÃPBÃPC=16Ã2Ã3Ã6=6 cubic units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let L1: x=y=z and L2: x-1=y-2=z-3 be two lines. The foot of perpendicular drawn from the origin O0,0,0 on L1 to L2 is A. If the equation of a plane containing the line L1 and perpendicular to OA is 10x+by+cz=d, then the value of b+c+d is equal to

Options:
A 10
B -10
C 12
D -7

The correct answer is:
-10

Solution:
From diagram,
OA is perpendicular to line L2=0
â1Î»+1+1Î»+2+1Î»+3=0âÎ»=-2
So, coordinates of point A are -1,0,1
OAâ=-i^+k^
âµ A normal vector to required plane is OAâ
Equation of the required plane is -1x-0+1z-0=0
â-x+z=0
â10x-10z=0
b=0,c=-10,d=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let L be a common tangent line to the curves 4x2+9y2=36 and 2x2+2y2=31. Then the square of the slope of the line L is ______.

Options:

The correct answer is:
3

Solution:
Given curves are x29+y24=1

x2+y2=314

let slope of common tangent be m

so tangents are y=mx±9m2+4

y=mx±3121+m2

hence 9m2+4=3141+m2

⇒36m2+16=31+31m2⇒m2=3

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let \( L \) be a common tangent line to the curves \( 4 x^{2}+9 y^{2}=36 \) and \( (2 x)^{2}+(2 y)^{2}=31 . \) Then the square of the slope of the line \( L \) is

Options:

The correct answer is:
3

Solution:
Given curves are x29+y24=1

x2+y2=314

let slope of common tangent be m

so tangents are y=mx±9m2+4

y=mx±3121+m2

hence 9m2+4=3141+m2

⇒36m2+16=31+31m2⇒m2=3

Topic: Mathematics, Ellipse, JEE Main
Let L be a line obtained from the intersection of two planes x+2y+z=6 and y+2z=4. If point Pα,β,γ is the foot of perpendicular from 3,2,1 on L, then the value of 21α+β+γ equals:

Options:
A 102
B 142
C 68
D 136

The correct answer is:
102

Solution:
Given, x+2y+z-6=0 ...1y+2z-4=0 ...2

On solving, equations 1 and 2, we get

x-3z=-2⇒x=3z-2⇒y=4-2z

Hence, Equation of the line is

x+23=y-4-2=z1

⇒ line of intersection of two planes is

x+23=y-4-2=z1=λ (Let)

It is clear from the diagram, AP⊥ar to line

∴ AP¯.3i^-2j^+k^=0

3λ-5.3+-2λ+2-2+λ-1.1=0

9λ-15+4λ-4+λ-1=0

14λ=20

λ=107⇒P167,87,107

⇒α+β+γ=16+8+107=347

21α+β+γ=102

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let \( L \) be a line obtained from the intersection of two planes \( x+2 y+z=6 \) and \( y+2 z=4 \). If point \( P(\alpha, \beta, \gamma) \) is the foot of perpendicular from \( (3,2,1) \) on \( L \), then the value of \( 21(\alpha+\beta+\gamma) \) equals:

Options:
A \( 102 \)
B \( 142 \)
C \( 68 \)
D \( 136 \)

The correct answer is:
\( 102 \)

Solution:
x+2y+z=6

y+2z=4×2

x-3z=-2⇒x=3z-2⇒y=4-2z

x+23=z y-4-2=z

⇒ line of intersection of two planes is

x+23=y-4-2=z=λ (Let)

∵AP⊥ar to line

∴ AP¯.3i^-2j^+k^=0

3λ-5.3+-2λ+2-2+λ-1.1=0

9λ-15+4λ-4+λ-1=0

14λ=20

λ=107⇒P167,87,107

⇒α+β+γ=16+8+107=347

21α+β+γ=102

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let l be a line which is normal to the curve y=2x2+x+2 at a point P on the curve. If the point Q6,4 lies on the line l and O is origin, then the area of the triangle OPQ is equal to _____.

Options:

The correct answer is:
13

Solution:
The given curve is y=2x2+x+2

dydx=4x+1

Let P be h,k, then normal at P is

y-k=-14h+1x-h

Since this normal passes through Q6,4

∴ 4-k=-14h+16-h

⇒4h+14-k+6-h=0

Also k=2h2+h+2

∴ 4h+14-2h2-h-2+6+h=0

⇒4h3-3h2+3h-8=0

⇒h=1,k=5

So P≡1,5

Now area of ∆OPQ will be =12100115164=13

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let $L$ be the length and d be the diameter of cross-section of a wire. Wires of the same material with different $L$ and $d$ are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

Options:
A $L=200 \mathrm{cm}, \mathrm{d}=0.5 \mathrm{mm}$
B $L=300 \mathrm{cm}, \mathrm{d}=1.0 \mathrm{mm}$
C $L=50 \mathrm{cm}, \quad d=0.05 \mathrm{mm}$
D $L=100 \mathrm{cm}, \mathrm{d}=0.2 \mathrm{mm}$

The correct answer is:
$L=50 \mathrm{cm}, \quad d=0.05 \mathrm{mm}$

Solution:
We know
$$
\Delta l=\frac{F \times l}{A \times Y}=\frac{4 L}{\pi d^{2}}
$$
Here $\frac{4}{\pi}$ is a constant.
Hence, $\Delta l \propto \frac{L}{d^{2}}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Let L be the line of intersection of planes r→·(i^-j^+2k^)=2 and r→·(2i^+j^-k^)=2. If P(α, β, γ) is the foot of perpendicular on L from the point (1, 2, 0), then the value of 35(α+β+γ) is equal to:

Options:
A 101
B 119
C 143
D 134

The correct answer is:
119

Solution:
The given planes are r→·(i^-j^+2k^)=2 and r→·(2i^+j^-k^)=2.

We can convert these planes into cartesian form by taking r→=xi^+yj^+zk^

Thus, we have xi^+yj^+zk^·(i^-j^+2k^)=2 and xi^+yj^+zk^·(2i^+j^-k^)=2

⇒x-y+2z=2 and 2x+y-z=2

So, let P1≡x-y+2z=2 and P2≡2x+y-z=2

Let, the line of Intersection L of planes P1 and P2 cuts xy plane in point Q and it is given that the foot of perpendicular P on L from a point F is F1, 2, 0, and this point is also on the xy plane.

From the diagram it is clear that, PQ is a line in the xy plane and hence at every point on it the z-coordinate is zero.

⇒z-coordinate of point Q, is zero and it is on both the planes,

⇒ x-y=2 ...i

And, 2x+y=2 ...ii

On solving the above two equation, we get x=43, y=-23.

⇒Q43, -23, 0

A vector parallel to the line of intersection can be obtained by finding a vector which is perpendicular to both the planes.

And, we know that a vector perpendicular to two vectors can be obtained by finding their cross product, i.e. the vector parallel to the line of intersection is

a→=i^j^k^1-1221-1=-i^+5j^+3k^

Now, we know that the equation of a line passing through a point x1, y1 and parallel to a vector ai^+bj^+ck^ is x-x1a=y-y1b=z-z1c.

Thus, the equation of the line of intersection of the planes is

x-43-1=y+235=z-03

x-43-1=y+235=z3=λ (let)

⇒x=-λ+43, y=5λ-23 & z=3λ.

Let, coordinates of foot of perpendicular be

P≡-λ+43, 5λ-23, 3λ

Now, PF→=-λ+43-1i^+5λ-23-2j^+3λ-0k^

⇒PF→=-λ+13i^+5λ-83j^+3λk^

Now, since PF→ is perpendicular to the line, hence the dot product of their direction ratios is zero.

⇒-1-λ+13+55λ-83+33λ=0

⇒λ-13+25λ-403+9λ=0

⇒35λ=413

⇒λ=41105.

Now, α=-λ+43, β=5λ-23, γ=3λ

⇒α+β+γ=7λ+23

⇒α+β+γ=741105+23

⇒α+β+γ=5115

⇒35(α+β+γ)=5115×35=119.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let $L$ be the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$. If $L$ makes an angles $\alpha$ with the positive $x$-axis, then $\cos \alpha$ equals

Options:
A
$\frac{1}{\sqrt{3}}$

B
$\frac{1}{2}$

C
$1$

D
$\frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{1}{\sqrt{3}}$

Solution:
If direction cosines of $L$ be $l$, $m, n$, then
$\begin{aligned}
& 2 l+3 m+n=0 \\
& l+3 m+2 n=0
\end{aligned}$
Solving, we get, $\frac{l}{3}=\frac{m}{-3}=\frac{n}{3}$
$\therefore l: m: n=\frac{1}{\sqrt{3}}:-\frac{1}{\sqrt{3}}: \frac{1}{\sqrt{3}} \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}} \text {. }$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2007
Let L be the line passing through the point P1,2 such that its intercepted segment between the co-ordinate axes is bisected at P. If L1 is the line perpendicular to L and passing through the point -2, 1, then the point of intersection of L and L1 is

Options:
A 35,2310
B 45,125
C 1120,2910
D 310,175

The correct answer is:
45,125

Solution:
If P is the midpoint of the segment between the axes, them point A2, 0 and B0, 4.

The equation of the line is, x2+y4=1

or, 2x+y=4 …1

The line perpendicular to it is of form, x-2y=k

Since it passes through (-2, 1),

-2-2=k⇒k=-4

∴ Line is, x-2y=-4 …2

Solving 1 and 2 we get 45,125.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2015 (10 Apr Online)
Let L be the line through the intersection of the planes 3x-y+2z+1=0 and 3x-2y+z=3. Then, the equation of the plane passing through 2,1,4 and perpendicular to the line L is

Options:
A x+y-z=2
B x+y-z+1=0
C x+y+z-7=0
D 2x-3y+4z=17

The correct answer is:
x+y-z+1=0

Solution:
A vector parallel to the line of intersection is i^j^k^3-123-21
=i^3-j^-3+k^-3
=3i^+3j^-3k^
Also, the line of intersection is perpendicular to the required plane.
Hence, equation of the plane is 1x-2+1y-1-1z-4=0
⇒x+y-z+1=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let $L$ be the line $y=2 x$, in the two dimensional plane.
Statement 1: The image of the point $(0,1)$ in $L$ is the point $\left(\frac{4}{5}, \frac{3}{5}\right)$
Statement 2: The points $(0,1)$ and $\left(\frac{4}{5}, \frac{3}{5}\right)$ lie on opposite sides of the line $\mathrm{L}$ and are at equal distance from it.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
Statement - 1
Let $P^{\prime}\left(x_1, y_1\right)$ be the image of $(0,1)$ with respect to the line $2 x-y=0$ then
$$
\begin{aligned}
& \frac{x_1}{2}=\frac{y_1-1}{-1}=\frac{-4(0)+2(1)}{5} \\
& \Rightarrow x_1=\frac{4}{5}, y_1=\frac{3}{5}
\end{aligned}
$$
Thus, statement- 1 is true.
Also, statement-2 is true and correct explanation for statement-1.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (19 May Online)
Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point-1, -4 in the line is :

Options:
A 115, 285
B 295, 85
C 85, 295
D 295, 115

The correct answer is:
115, 285

Solution:
Equation of line is

x3+y1=1⇒x+3y-3=0

if image isx1,y1

Then,

x1+11=y1+43=-2-1-12-310

x1+1=y1+43=165

⇒x1=115,y1=285

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Let l, m and n are three distinct numbers in arithmetic progression. Also l2, m2 and n2 are in geometric progression and l+m+n=3. If l<m<n, then n is equal to

Options:
A 1
B 1-2
C 1+2
D 2+5

The correct answer is:
1+2

Solution:
l,m,n are in A.P.

⇒2m=l+n

Given, l+m+n=3⇒m=1

Let, l=1-d, n=1+d, d>0
Now, m4=l2n2⇒m2=-ln or ln
So 1+d1-d=±1

⇒d=0 or 2 (d=0 is rejected)
Hence, n=1+2

Topic: Mathematics, Sequences and Series, JEE Main
Let $\int_\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^\alpha$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :

Options:
A $x^2+2 x-8=0$
B $x^2-2 x-8=0$
C $2 x^2-5 x+2=0$
D $2 x^2-5 x-2=0$

The correct answer is:
$2 x^2-5 x+2=0$

Solution:
$\begin{aligned} & \int_\alpha^{\log 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{x}}-1}}=\frac{\pi}{6} \\ & \text { Let } \mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^2 \\ & \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{t} \mathrm{dt} \\ & =\int \frac{2 \mathrm{dt}}{\mathrm{t}^2+1} \\ & =2 \tan ^{-1} \mathrm{t} \\ & =\left.2 \tan ^{-1}\left(\sqrt{\mathrm{e}^{\mathrm{x}}-1}\right)\right|_\alpha ^{\log _{\mathrm{e}}^4} \\ & =2\left[\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}\right]=\frac{\pi}{6} \\ & =\frac{\pi}{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{12} \\ & \Rightarrow \tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{4} \\ & \mathrm{e}^\alpha=2 \quad \mathrm{e}^{-\alpha}=\frac{1}{2} \\ & \mathrm{x}^2-\left(2+\frac{1}{2}\right) \mathrm{x}+1=0 \\ & 2 \mathrm{x}^2-5 \mathrm{x}+2=0\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let lx-2y=1 intersects the parabola y2=4ax at points P and Q. If PS and QS meet the parabola again at R and T respectively (where S is the focus of y2=4ax ), then the slope of RT is equal to

Options:

The correct answer is:
0.5

Solution:
Let, P=t12,2t1 and Q=t22,2t2
Equation of PS is t1+t2y=2x+2t1t2
⇒ lines -2x+t1+t2y=2t1t2 and lx-2y=1 are same
⇒-2l=t1+t2-2=2t1t21
⇒t1+t2=4l & t1t2=-1l
Now, R=1t12,-2t1 and T=1t22,-2t2
The slope of RT is -2t2+2t11t22-1t12=-2t1-t2t1t2t1-t2t1+t2t12t22
=-2t1t2t1+t2=2l4l=12=0.5

Topic: Mathematics, Parabola, JEE Main
Let \( l_{1}=\int_{0}^{\infty} \frac{x^{2} \sqrt{x}}{(1+x)^{6}} d x, l_{2}=\int_{0}^{\infty} \frac{x \sqrt{x}}{(1+x)^{6}} d x \), then

Options:
A \( l_{1}=2 l_{2} \)
B \( l_{2}=2 l_{1} \)
C \( l_{1}=l_{2}^{2} \)
D \( l_{1}=l_{2} \)

The correct answer is:
\( l_{1}=l_{2} \)

Solution:
We have, l1=∫0∞x2x1+x6dx

Let x=1t⇒dx=-1t2dt then,

⇒l1=∫∞01t2t1+1t6−1t2dt

⇒l1=∫0∞tt1+t6dt=l2

Topic: Mathematics, Definite Integration, JEE Main
Let limx→0⁡sin⁡2xtan⁡xk=L1 and limx→0⁡e2x-1x=L2, and the value of L1L2 is 8, then k is

Options:
A 4
B 8
C 6
D 2

The correct answer is:
2

Solution:
L1=limxâ0â¡sinâ¡2x2x⋅2xtanâ¡xkxk⋅xk=2k
L2=limxâ0â¡e2x-1x=limxâ0â¡e2x-12x⋅2=2
âL1L2=8â4k=8âk=2

Topic: Mathematics, Limits, JEE Main
Let lnx denote the logarithm of x with respect to the base e Let S⊂R be the set of all points where the function lnx2-1 is well-defined. Then, the number of functions f:S→R that are differentiable, satisfy f'x=lnx2-1 for all x∈S and f2=0, is

Options:
A 0
B 1
C 2
D infinite

The correct answer is:
infinite

Solution:
We have,

f'x=lnx2+1

fx=∫lnx2-1dx

fx=xlnx2-1-∫2x2x2-1dx

fx=xlnx2-1-2∫x2-1x2-1+1x2-1dx

fx=xlnx2-1-2x-lnx-1x+1+C

f2=2ln3-4-ln13+C=0 ∵f2=0

⇒C=4-3ln3

∴fx=xlnx2-1-2nlnx-1x+1+4-3ln3

defined for S infinite C values possible in set S such that f'x=lnx2-1

Topic: Mathematics, Indefinite Integration, JEE Main
Let M=0-αα0, where α is a non-zero real number and N=∑k=149M2k. If I-M2N=-2I, then the positive integral value of α is ______.

Options:

The correct answer is:
1

Solution:
Given, M=0-αα0

Now, M2=-α200-α2=-α2I

Also given, N=∑k=149M2k

⇒N=M2+M4+…….+M98=-α2+α4-α6+…..I

=-α21--α2491+α2.I

⇒N1+α2=-α2α98+1

⇒NI-M2=-α2α98+1

Now comparing with I-M2N=-2I

We get, α=1.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let M=abcdef111 and N=M22. If a-b2+d-e2=36, b-c2+e-f2=64, a-c2+d-f2=100, then value of N is equal to

Options:
A 1152
B 48
C 144
D 288

The correct answer is:
288

Solution:
M=2ÃArea of the triangle with vertices a,d, b,e & c,f with the length of sides as 6, 8, 10
âµ Triangle is a right angle triangle, hence,
Area=12Ã6Ã8=29
M=48
N=M22=18M2=4828=288

Topic: Mathematics, Determinants, JEE Main
Let M and m be respectively the absolute maximum and the absolute minimum values of the function, fx=2x3-9x2+12x+5 in the interval [0,3] . Then M-m is equal to

Options:
A 9
B 4
C 1
D 5

The correct answer is:
9

Solution:
Given fx=2x3-9x2+12x+5, x∈0,3

Differentiating w.r.t x,

⇒f′x=6x2-18x+12

⇒f′x=6x-1x-2

For maxima/minima f'x=0

⇒6x-1x-2=0⇒x=1,2

Now, f"x=12x-18

So at x=1, f"x<0

and at x=2, f"x>0

So, for absolute maxima/minima

⇒f1=10

⇒f2=9

⇒f0=5

⇒f3=14

⇒M=14

⇒m=5

∴ M-m=9

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (16 Apr Online)
Let m and M be respectively the minimum and maximum value values of cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x Then the ordered pair (m, M) is equal to:

Options:
A 3,3
B -3,-1
C 4,1
D 1,3

The correct answer is:
-3,-1

Solution:
D=cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x

C1→C1+C2

D=21+sin2xsin2x2sin2xsin2x1sin2x1+sin2x

R1→R1-2R3 and R2→R2-2R3

D=0cos2x-(2+sin2x)0-sin2x-(2+sin2x)1sin2x1+sin2x⇒D=-2-sin2x

As range of sin2x∈-1,1, thus

m=-3,M=-1

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Let M and m respectively be the maximum and minimum values of the function f(x)=tan-1(sinx+cosx) in 0,π2. Then the value of tan(M-m) is equal to:

Options:
A 2-3
B 3-22
C 3+22
D 2+3

The correct answer is:
3-22

Solution:
Given, the function, f(x)=tan-1(sinx+cosx) in 0,π2

f'x=cosx-sinx1+sinx+cosx2

=cosx-sinx1+sin2x

f'x=0

cosx-sinx=0

tanx=1

x=π4
Then, 1≤sinx+cosx≤2 x∈π4,tan-12
M=tan-1(2)

m=tan-1(1)

M-m=tan-12-tan-11

M-m=tan-12-11+2

M-m=tan-1(3-22)

tan(M-m)=3-22

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Let M and N are two non singular matrices of order 3 with real entries such that adjM=2N and adjN=M. If MN=Î»Î, then the value of Î» is equal to (where, adjX represents the adjoint matrix of matrix X and I represents an identity matrix)

Options:

The correct answer is:
2

Solution:
adjM=2N⇒adjM=2N⇒M2=8N …. 1
adjN=M⇒adjN=M⇒N2=M …. 2
From 1 and 2, we get,

N=2,M=4
Now, MN=adjNN=NΙ=2Ι⇒λ=2

Topic: Mathematics, Matrices, JEE Main
Let M and N be the number of points on the curve y5-9xy+2x=0, where the tangents to the curve are parallel to x-axis and y-axis, respectively. Then the value of M+N equals _______.

Options:

The correct answer is:
2

Solution:
Given,

y5-9xy+2x=0 .......1

Now differentiating both side we get,

5y4dyx-9xdydx-9y+2=0

⇒dydx5y4-9x=9y-2

⇒dydx=9y-25y4-9x=0 .....2

⇒9y-2=0 (for horizontal tangent)

⇒y=29, which does not satisfy the original equation so M=0.

As 295-9x×29+2x=0

⇒295-2x+2x=0

⇒295=0 which is a false statement

Now 5y4-9x=0 (from equation 2 for vertical tangent)

⇒5y49y-2-9y5=0 (putting the value of x from equation 1)

⇒y445y-10-9y=0

⇒y=0 or 36y=10⇒y=518

So at y=0⇒x=0 and y=518⇒x=x1

0,0x1,518 (where x1=51859×518-2)

So, N=2

M+N=0+2=2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (27 Jul Shift 1)