datavorous/entrance-exam-dataset huggingface dataset:
Let A be a square matrix of order 3 such that A+AT= 10 4 6a21+a12 6 a23+a32a31+a13 8 4, where a12,a23,a31 are positive roots of the equation x3-6x2+px-8=0, ∀p∈R, then the absolute value of A is equal to

Options:

The correct answer is:
37

Solution:
Clearly, A+AT is a symmetric matrix
Also, a12+a23+a31=6, a12a23a31=8
AM=GM⇒a12=a23=a31=2
a21+a12=4⇒a21=a12=2
a31+a13=6⇒a31=a13=3
a23+a32=8⇒a23=a32=4
For diagonal elements
2a11=10⇒a11=5
2a22=6⇒a22=3
2a33=4⇒a33=2
A=523234342
A=56-16-24-12+38-9
=-50+16-3
=-37

Topic: Mathematics, Determinants, JEE Main
Let A be a square matrix such that AAT=I. Then 12 A A+AT2+A-AT2 is equal to

Options:
A A2+I
B A3+I
C A2+AT
D A3+AT

The correct answer is:
A3+AT

Solution:
Given: AAT=1

⇒AT=A-1

Now, solving

⇒12AA+AT2+A-AT2=12A×2A2+2AT2+2AAT-2AAT⇒12AA+AT2+A-AT2=12A×2A2+AT2

⇒12AA+AT2+A-AT2=A3+AAT2

Now using, AAT=1,

⇒12AA+AT2+A-AT2=A3+AT

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is:

Options:
A 12
B 4
C 1
D 6

The correct answer is:
4

Solution:
A=abbc, a,b,c∈I

A2=abbcabbc=a2+b2ba+cba+cb2+c2

Sum of the diagonal entries of

A2=a2+2b2+c2

Given a2+2b2+c2=1,a,b,c∈I

b=0 & a2+c2=1

Case-1: a=0⇒c=±1 (2-matrices)

Case-2: c=0⇒a=±1 (2-matrices)

Total =4 matrices

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Let A be a symmetric matrix such that A=2 and 21332A=12αβ If the sum of the diagonal elements of A is s, then βsα2 is equal to _________.

Options:

The correct answer is:
5

Solution:
Let

A=abbc

And,

21332A=12αβ

⇒21332abbc=12αβ

⇒2a+b2b+c3a+3b23b+3c2=12αβ

So,

2a+b=12b+c=2

So, 4a-c=0 ....i

Also,

A=2⇒ac-b2=2

⇒a4a-1-2a2=2

⇒4a2-4a2-1+4a=2

⇒a=34

⇒b=1-2a=1-64=-12

c=4a=3

And,

3a+3b2=α⇒α=94-34=323b+3c2=β⇒β=-32+92=3

So,

A=34-12-123

So, TrA=s=34+3=154

β×sα2=3×15494=5

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Let a→ be a unit vector coplanar with i^-j^+2k^ and 2i^-j^+k^ such that a→ is perpendicular to i^-2j^+k^. If the projection of a→ along i^-j^+2k^ is λ, then the value of 1λ2 is equal to

Options:

The correct answer is:
12

Solution:
Let, α→=i^-j^+2k^, β→=2i^-j^+k^, γ→=i^-2j^+k^
a→=γ→×α→×β→γ→×α→×β→
γ→×α→×β→=γ→⋅β→α→-γ→⋅α→β→
=5α→-5β→
=5-i^+k^
a→=-i^+k^2
Projection of a→ on α→ is a→⋅α^=a→⋅α→α→=-1+226=112

Topic: Mathematics, Vector Algebra, JEE Main
Let aâ be a vector in the xy- plane making an angle of 60o with the positive x-axis and aâ-i^ is the geometric mean of aâ and aâ-2i^, then the value of aâ is equal to

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

The correct answer is:
2-1

Solution:
Let, a→=xi^+3xj^
Given, a→-i^2=a→a→-2i^
⇒x-12+3x2=2xx-22+3x2
On simplifying, we get,

x=122-1⇒a→=2-1

Topic: Mathematics, Vector Algebra, JEE Main
Let Aâ be a vector parallel to the line of intersection of the planes P1 and P2. The plane P1 is parallel to the vectors 2j^+3k^ and 4j^-3k^ while plane P2 is parallel to the vectors j^-k^ and i^+j^. The acute angle between Aâ and 2i^+j^-2k^ is

Options:
A π6
B π4
C π3
D 5π12

The correct answer is:
π4

Solution:
Let, a→=2j^+3k^
b→=4j^-3k^
c→=j^-k^
d→=i^+j^
e→=2i^+j^-2k^
A normal vector perpendicular to the plane P1 is a→×b→
A normal vector perpendicular to the plane P2 is c→×d→
A vector parallel to the line of intersection is a→×b→×c→×d→=k^-j^
Let, θ be the acute angle between them then,
cos⁡θ=A→·e→A→e→=-32×3=12
⇒θ=π4

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let a→ be a vector which is perpendicular to the vector 3i^+12j^+2k^. If a→×2i^+k^=2i^-13j^-4k^, then the projection of the vector a→ on the vector 2i^+2j^+k^ is

Options:
A 13
B 1
C 53
D 73

The correct answer is:
53

Solution:
Given, a→×2i^+k^=2i^-13j^-4k^

Now multiplying both side by 3i^+12j^+2k^ we get,

⇒a→×2i^+k^×3i^+12j^+2k^

=2i^-13j^-4k^×3i^+12j^+2k^

⇒3i^+12j^+2k^.a→2i^+k^-6+2a→=i^j^k^2-13-43122

⇒0-6+2a→=i^j^k^2-13-43122

a→=3i^+2j^-5k^

Now, projection of a→ on vector 2i^+2j^+k^ is a→·2i^+2j^+k^3=53

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let a be an integer such that all the real roots of the polynomial 2x5+5x4+10x3+10x2+10x+10 lie in the interval a,a+1. Then, |a| is equal to ______.

Options:

The correct answer is:
2

Solution:
Let 2x5+5x4+10x3+10x2+10x+10=fx

Now f-2=-34 and f-1=3

Hence fx has a root in -2,-1

Further f'x=10x4+20x3+30x2+20x+10

=10x2x2+2x+3+2x+1x2

=10x2x2+1x2+2x+1x+3

=10x2x+1x2-2+2x+1x+3

=10x2x+1x+12>0 for all x belongs to R.

fx is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

Hence, fx has only one real root, so a=2.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let a be an integer such that limx→718-1-xx-3a exists, where t is greatest integer ≤t. Then a is equal to

Options:
A -2
B 6
C -6
D -7

The correct answer is:
-6

Solution:
Since limx→718-1-xx-3a exists a∈𝕀

∴L.H.L = R.H.L

⇒limh→018--67-h-3a=limh→018--77+h-3a

⇒limh→0247-3a+-h=limh→0257-3a+h

⇒246-3a=257-3a ⇒a=-6

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let a,Î² be an ordered pair of real numbers satisfying the equation x2-4x+4y2+3=0. If the maximum and minimum values of a2+Î²2 are l and s respectively, then the value of l-sl+s is equal to ______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
1

Solution:
a2+Î²2 represents the distance of a,Î² from the origin 0,0. Now, let the line joining O0,0 and center of circle C2,0 cuts the circle at points 1,0=A and 3,0=B,
then, l=OB=3 and s=OA=1
âl-sl+s=24=0.5

Topic: Mathematics, Circle, JEE Main
Let a,Î² be an ordered pair of real numbers satisfying the equation x2-4x+4y2+3=0. If the maximum and minimum values of a2+Î²2 are l and s respectively, then the value of l-sl+s is equal to

Options:

The correct answer is:
0.5

Solution:
a2+Î²2 represents the distance of a,Î² from the origin 0,0. Now, let the line joining O0,0 and center of circle C2,0 cuts the circle at points 1,0=A and 3,0=B,
then, l=OB=3 and s=OA=1
âl-sl+s=24=0.5

Topic: Mathematics, Circle, JEE Main
Let A be any 3×3 invertible matrix. Then which one of the following is not always true?

Options:
A adj adj (A)=A2. adj (A)-1
B adj adj (A)=A.adj (A)-1
C adj adj (A)=A . A
D adj (A)=A.A-1

The correct answer is:
adj adj (A)=A.adj (A)-1

Solution:
âµA-1=1AadjA
â´adjadjA=adj Aadj A-1=A2adj A-1â¦â¦i
âµAadj A=AI
â´adjA=AA-1
â´adj A-1=1AA
â´ from (i) adjadj A=A.A

Topic: Mathematics, Matrices, JEE Main, JEE Main 2017 (08 Apr Online)
Let A be the area bounded by the curve y=xx-3, the x-axis and the ordinates x=-1 and x=2. Then 12 A is equal to _____ .

Options:

The correct answer is:
62

Solution:
Plotting the y=xx-3 and x-axis between x=-1 and x=2 we get,

Now area of the above region is given by A=∫-103x-x2dx+∫023x-x2dx

⇒A=3×x22-x33-10+3×x22-x3302

⇒A=116+103=11+206=316

Hence, the value of 12A=62

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let A be the area of the region (x,y):y≥x2, y≥(1-x)2, y≤2x(1-x). Then 540A is equal to

Options:

The correct answer is:
25

Solution:
Given,

A be the area of the region

(x,y):y≥x2,y≥(1-x)2,y≤2x(1-x)

Now solving y=x2 & y=2x1-x we get, x=0, 23

And solving y=1-x2 & y=2x1-x we get,

⇒1+x2-2x=2x-2x2

⇒3x2-4x+1=0

⇒x=1, 13

Now on plotting the diagram of the above region we get,

Now from the above diagram area of the shaded region will be,

A=∫13232x-2x2dx-∫13121-x2dx+∫1223x2dx

⇒A=x2-2x331323-x-1331312+x331223

⇒A=5108

∴A=5108⇒540 A=5108×540=25

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let A be the centre of the circle x2+y2-2x-4y-20=0. If the tangents at the points B1,7 and D4,-2 on the circle meet at the point C, then the perimeter of the quadrilateral ABCD is

Options:
A 60 units
B 20 units
C 40 units
D 50 units

The correct answer is:
40 units

Solution:
The radius of the given circle is 5=AB=AD
Equation of the tangent at B1,7 is y=7 and

equation of the tangent at D4,-2 is 3x-4y=20
Point of intersection of y=7 and 3x-4y=20 is 16,7=C
Now, AB=AD=5 & CB=CD=15
⇒ The perimeter of the quadrilateral ABCD is equal to 5+5+15+15=40 units

Topic: Mathematics, Circle, JEE Main
Let A be the foot of the perpendicular from the origin to the plane x-2y+2z+6=0 and B0,-1,-4 be a point on the plane. Then, the length of AB is

Options:
A 13 units
B 15 units
C 4 units
D 17 units

The correct answer is:
13 units

Solution:
From the diagram
OB=0+1+16=17
OA=0-0+0+61+4+4=2
∵∆OAB is right angled triangle
AB=OB2-OA2=13 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let A be the point 1,2,3 and B be a point on the line x-1-2=y+13=z-54=k. The value of k such that line AB is perpendicular to the plane 4x+9y-18z=6 is

Options:
A -25
B 15
C 25
D no such value of k is possible

The correct answer is:
25

Solution:
Coordinates of the point B are 2k+1,3k−1,4k+5
A vector parallel to AB is −2k i ^ + 3k−3 j ^ + 4k+2 k ^
∵ the line AB is perpendicular to the given plane
Hence, -2k4=3k-39=4k+2-18
-18k=12k-12
k=25

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let A be the point 1,2 and B be any point on the curve x2+y2=16. If the centre of the locus of the point P, which divides the line segment A B in the ratio 3:2 is the point Cα,β, then the length of the line segment AC is

Options:
A 355
B 455
C 255
D 655

The correct answer is:
355

Solution:
Given,

A be the point 1,2 and B be any point on the curve x2+y2=16,

And the centre of the locus of the point P, which divides the line segment AB in the ratio 3:2 is the point Cα,β,

Now let the point on the circle x2+y2=16 be B4cosθ,4sinθ

Now using section formula in A1,2 and B4cosθ,4sinθ we get,

P12cosθ+25,12sinθ+45≡(h,k)

⇒cosθ=5h-212 & sinθ=5k-412

Now squaring and adding we get,

5h-2122+5k-4122=1

⇒h-252+k-452=1252

So, the centre of the locus is C25,45

Hence, by distance formula we get,AC=352+652=355

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let A be the point of intersection of the lines 3x+2 y=14,5 x-y=6 and B be the point of intersection of the lines 4 x+3 y=8, 6 x+y=5. The distance of the point P(5,-2) from the line AB is

Options:
A 132
B 8
C 52
D 6

The correct answer is:
6

Solution:
Solving lines L1(3x+2y=14) and L2(5x-y=6).

⇒x=14-2y3=6+y5

⇒70-10y=18+3y

⇒52=13y

⇒y=4

⇒x=14-83=2

⇒A≡2,4

Now, solving lines L3(4x+3y=8) and L4(6x+y=5)

⇒x=8-3y4=5-y6

⇒48-18y=20-4y

⇒28=14y

⇒y=2

⇒x=8-64=12

⇒B≡12,2

So, equation of AB is given by,

AB: y-4=2-412-2x-2

⇒AB: y-4=43x-2

⇒AB: y-4=43x-2

⇒3y-12=4x-8

⇒4x-3y+4=0

The distance of P5,-2 from the line AB is given by,

D=4(5)-3(-2)+45

⇒D=6 units.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let $A$ be the set of all $3 \times 3$ determinants with entries 0 or 1 only and $B$ be the subset of $A$ consisting of all determinants with value 1 . If $C$ is the subset of $A$ consisting of all determinants with value -1 , then

Options:
A $n(C)=0$
B $n(B)=n(C)$
C $A=B \cup C$
D $n(B)=2 n(A)$

The correct answer is:
$n(B)=n(C)$

Solution:
We know that the interchange of two adjacent rows (columns) changes the value of a determinant only in sign but not in magnitude.
Hence, corresponding to every element $\Delta$ at $B$ there is an element $\Delta^{\prime}$ in $C$ obtained by inter changing two adjacent rows (columns) in $\Delta$. It follows that $n(B) \leq n(c)$, i.e. the number of elements in $B$ is less than an equal to the number of elements in $C$.
Similarly, $\quad n(C) \leq n(B)$
Hence, $\quad n(B)=n(C)$
i.e. $B$ has an many elements as $C$.

Topic: Mathematics, Determinants, JEE Main
Let $A$ be the set of all $3 \times 3$ scalar matrices with real entries. If $f: A \rightarrow R$ is defined by $f(m)=\operatorname{det}(m) \forall ; m \in A$, then $f$ is

Options:
A one-one but not onto
B onto but not one-one
C bijective
D neither one-one nor onto

The correct answer is:
bijective

Solution:
$A$ be a scalar matrix
$$
\begin{aligned}
A & =\left[\begin{array}{ccc}
m & 0 & 0 \\
0 & m & 0 \\
0 & 0 & m
\end{array}\right] \\
|A| & =m^3 \Rightarrow f(m)=m^3
\end{aligned}
$$
$f(m)$ is increasing function for all values of $m$.
$\therefore f(m)$ is injective.
Range of $f(m)=R=$ codomain
$\therefore f(m)$ is surjective.
Hence, $f(m)$ is bijective.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 1)
Let A be the set of all points α, β such that the area of triangle formed by the points 5, 6, 3, 2 and α, β is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :

Options:
A 85
B 125
C 165
D 45

The correct answer is:
85

Solution:
Given:Area of triangle formed by 5,6,3,2 & α,β=12 square units

∴12αβ1561321=12

⇒αβ1561321=24

⇒α6-2-β5-3+110-18=±24

⇒4α-2β-8=±24

⇒4α-2β=32, 4α-2β+16=0

⇒2α-β-16=0, 2α-β+8=0

So point will be either α,2α+8 , α,2α-16

For the point α,2α+8,

Distance from origin
D=α2+(2α+8)2=5α2+32α+64 ∵β=2α+8

⇒D2=5α2+32α+64

For maximum or minimum length,

dD2dα=10α+32=0

⇒α=-165

∴β=-325+8=85

∴D=-1652+852=855=85

Similarly if β=2α-16, D=α2+2α-162=5α2-64α+256

∴D2=5α2-64α+256

For maximum or minimum length,

dD2dα=0

⇒10α-64=0

∴α=325

∴D=3252+645-162=165 at α=325
So, least possible length of line segment is 85

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let A be the set of values of k for which 2 lies between the roots of the quadratic equation x2+k+2x-k+3=0, then A is given by

Options:
A -∞,-5
B 5,∞
C -∞,-5
D 5,∞

The correct answer is:
-∞,-5

Solution:
Opening upward parabola as the coefficient of x2 is +ve
2 lies between the roots ⇒f2<0
f2=4+2k+2-k+3<0
4+2k+4-k-3<0
k+5<0⇒k<-5

Topic: Mathematics, Quadratic Equation, JEE Main
Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

12+2·22+32+2·42+52+2·62+…

If B-2A=100λ, then λ is equal to :

Options:
A 496
B 232
C 248
D 464

The correct answer is:
248

Solution:
Given series, 12+2·22+32+2·42+52+2·62+…

Here, A=12+2·22+32+2·42+...20 terms
=12+ 22+32+42+...20 terms+22+42+ 62+...10 terms
=20·21·416 +410·11·216
⇒A=4410
Similarly for B=12+22+33+...40 terms+22+42+62+...20 terms
⇒B=40×41×816+4×20×21×416
⇒B=33620
Now, B-2A=24800=100λ (Given)
⇒λ=248

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (08 Apr)
Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is

Options:
A 4664
B 27565
C 4155
D 3654

The correct answer is:
4664

Solution:
P4 heads=P5 heads

5C4p4q= 5C5p5⇒p=5q ...i

Also, p+q=1⇒q=16,p=56 (from i)

So, P (at most 2 heads) =Px=0+Px=1+Px=2

=q5+5C1pq4+5C2p2q3

=165+556164+10562163 =4664

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Let a circle C touch the lines L1:4x-3y+K1=0 and L2:4x-3y+K2=0, K1, K2∈R. If a line passing through the centre of the circle C intersects L1 at -1,2 and L2 at 3,-6, then the equation of the circle C is

Options:
A x-12+y-22=4
B x-12+y+22=16
C x+12+y-22=4
D x-12+y-22=16

The correct answer is:
x-12+y+22=16

Solution:
Given,

L1=4x-3y+K1=0; L2=4x-3y+K2=0

Here line L1 and L2 are parallel.

So,

Now point A-1,2 will satisfy L1 so, -4-3×2+K2=0 ⇒ K2=10

Also point B3,-6 will satisfy L2=4x-3y+K2

So 4×3-3×-6+K2=0 ⇒K2=-30

Now distance between the parallel line will be the diameter

Diameter =K1-K242+32

=10+305=8

So radius 82=4

Now mid-point of AB will give us centre of circle by symmetry, so by midpoint formula in AB we get centre 1,-2,

Now equation of circle will be x-12+y+22=42

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let a circle C:x-h2+y-k2=r2,k>0, touch the x-axis at 1,0. If the line x+y=0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h+k+r is equal to _____.

Options:

The correct answer is:
7

Solution:
Since the circle is touching x-axis so, k=r and h=1

Now by diagram, OP=r,PR=1

OR=r+12

r2=1+r+122

2r2=2+r2+1+2r

r2-2r-3=0

r-3r+1=0

r=3,-1

h+k+r=1+3+3

=7

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let a circle passing through $(2,0)$ have its centre at the point $(h, k)$. Let $\left(x_c, y_c\right)$ be the point of intersection of the lines $3 x+5 y=1$ and $(2+c) x+5 c^2 y=1$. If $\mathrm{h}=\lim _{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$ and $\mathrm{k}=\lim _{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$, then the equation of the circle is :

Options:
A $25 x^2+25 y^2-2 x+2 y-60=0$
B $5 x^2+5 y^2-4 x+2 y-12=0$
C $5 x^2+5 y^2-4 x-2 y-12=0$
D $25 x^2+25 y^2-20 x+2 y-60=0$

The correct answer is:
$25 x^2+25 y^2-20 x+2 y-60=0$

Solution:
$\begin{aligned} & (2+c) x+5 c^2\left(\frac{1-3 x}{5}\right)=1 \\ & x=\frac{1-c^2}{2+c-3 c^2}, y=\frac{1-3 x}{5}=\frac{c-1}{5\left(2+c-3 c^2\right)} \\ & h=\lim _{c \rightarrow 1} \frac{(1-c)(1+c)}{(1-c)(2+3 c)}=\frac{2}{5} \\ & K=\lim _{c \rightarrow 1} \frac{c-1}{-5(c-1)(3 c+2)}=-\frac{1}{25} \\ & \operatorname{Centre}\left(\frac{2}{25},-\frac{1}{25}\right)\end{aligned}$
$\begin{aligned} & r=\sqrt{\left(2-\frac{2}{5}\right)^2+\left(0-\frac{1}{25}\right)^2}=\sqrt{\frac{64}{25}+\frac{1}{625}} \\ & r=\frac{\sqrt{161}}{25} \\ & \left(x-\frac{2}{5}\right)^2+\left(y+\frac{1}{25}\right)^2=\frac{161}{125} \\ & \Rightarrow 25 x^2+25 y^2-20 x+2 y-60=0\end{aligned}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let a common tangent to the curves y2=4x and x-42+y2=16 touch the curves at the points P and Q. Then PQ2 is equal to ________.

Options:

The correct answer is:
32

Solution:
Given,

Common tangent to the curves y2=4x and x-42+y2=16 touch the curves at the points P and Q

Let, the equation of tangent be,

y=mx+am, where a=1 is focus and m is slope,

So, equation of tangent to parabola will be, y=mx+1m

Also line is tangent to circle,

So, using perpendicular distance formula from centre 4, 0to tangent which will be radius r=4, we get,

4m+1mm2+1=4

⇒16m2+1m2+8=16m2+16

⇒m2=18

⇒m=±122

Now if m=122 then P=am2, 2am=8, 42

Now equation of tangent will be, x-22y+8=0

And Q will be foot of perpendicular from 4, 0 on tangentx-22y+8=0 ,

So, using foot of perpendicular formula we get,

x-41=y-0-22=-4-0+812+222

⇒x,y=83, 823

Hence, by distance formula we get,

PQ2=1632+4232

⇒PQ2=256+329=32

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let a complex number z, z≠1, satisfy log12z+11z-12≤2. Then, the largest value of z is equal to _________.

Options:
A 8
B 7
C 6
D 5

The correct answer is:
7

Solution:
log12z+11z-12≤2

⇒z+11z-12≥12, ∵logab≤c⇒b≥ac if 0<a<1

⇒2z+22≥z-12, ∵z≠1 & z-12>0

⇒2z+22≥z2+1-2z

⇒z2-4z-21≤0

⇒z-7z+3≤0

⇒z≤7

∴ Largest value of z is 7

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be 12 and probability of occurrence of 0 at the odd place be 13. Then the probability that 10 is followed by 01 is equal to :

Options:
A 118
B 13
C 16
D 19

The correct answer is:
19

Solution:
Given the probability that the digit 0 appear at the even place is 12, hence the probability that the digit 1 appear at the even place is 1-12=12.

And the probability that the digit 0 appear at the odd place is 13, hence the probability that the digit 1 appear at the odd place is 1-13=23.

For the digits '10' and '01' we have the following possibilities

1001odd placeeven placeodd placeeven place

Or

1001even placeodd placeeven placeodd place

Thus, the required probability is 12·13·12·23+23·12·13·12

=118+118=19.

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let a continuous and differentiable function fx is such that fx and ddxfx have opposite signs everywhere. Then,

Options:
A f ′ x is always increasing
B fx is always increasing
C fx is non-decreasing
D fx is decreasing

The correct answer is:
fx is decreasing

Solution:
fx=fx:fx≥0-fx:fx<0
ddxfx=f'x:fx≥0-f'x:fx<0
As fx and f ′ x have opposite signs, then ddxfx is negative everywhere, hence fx is a decreasing function.

Topic: Mathematics, Application of Derivatives, JEE Main
Let Aα=cos⁡α0sin⁡α010sin⁡α0cos⁡α and Axyz=010. If the system of equations has infinite solutions and sum of all the possible value of α in 0,2π is kπ, then the value of k is equal to

Options:
A 0
B 2
C 4
D 8

The correct answer is:
4

Solution:
All the three equations obtained are
cos⁡αx+sin⁡αz=0
y=1
sin⁡αx+cos⁡αz=0
So, the equations cos⁡αx+sin⁡αz=0 and sin⁡αx+cos⁡αz=0 has infinite solutions
Then, ∆=cos⁡αsin⁡αsin⁡αcos⁡α=0

⇒cos2⁡α-sin2⁡α=0⇒cos⁡2α=0
⇒α=π4,3π4,5π4,7π4
Sum of all values of α =4π

Hence, k=4

Topic: Mathematics, Determinants, JEE Main
Let a→=cos⁡θi^-sinθj^, b→=sin⁡θi^+cos⁡θj^, c→=k^ and r→=7ı^+j^+10k^. If r→=xa→+yb→+zc→, then the value of x2+y2z is equal to

Options:
A 3
B 5
C 50
D 7

The correct answer is:
5

Solution:
r→=xcos⁡θ+ysin⁡θi^+-xsin⁡θ+ycos⁡θj^+zk^
∵r→=7ı^+j^+10k^
on comparison
xcos⁡θ+ysin⁡θ=7…….i
-xsin⁡θ+ycos⁡θ=1…….ii
z=10
Squaring and adding equation i and ii
x2+y2=72+12=50
⇒x2+y2z=5010=5

Topic: Mathematics, Vector Algebra, JEE Main
Let A=cosα-sinαsinαcosα, a∈R such that A32=0-110. Then, a value of α is:

Options:
A 0
B π16
C π64
D π32

The correct answer is:
π64

Solution:
∵ A=cosα-sinαsinαcosα

⇒A2=cos2α-sin2αsin2αcos2α

⇒A3 =cos3α-sin3αsin3αcos3α

Similarly, A32 =cos32α-sin32αsin32αcos32α
Given,

A32 =0-110

⇒ cos32α-sin32αsin32αcos32α=0-110

On comparing cos32α=0 & sin32α=1, we get,
α=π64

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Let A=cos⁡αsin⁡α-sin⁡αcos⁡α and matrix B is defined such that B=A+3A2+3A3+A4 . If B=8 , then the number of values of α in 0, 10π is

Options:
A 10
B 12
C 5
D 3

The correct answer is:
10

Solution:
B=AI+3A+3A2+A3=AI+A3
B=A  I+A3=8…i
A=cos⁡αsin⁡α-sin⁡αcos⁡α=cos2α+sin2α=1
I+A=1+cos⁡αsin⁡α-sin⁡α1+cos⁡α=1+cos⁡α2+sin2α =2+2cos⁡α
From i
12+2cos⁡α3=8⇒2+2cos⁡α=2
⇒cos⁡α=0
α=π2, 3π2, 5π2, 7π2, 9π2, 11π2, 13π2,15π2,17π2, 19π2
Total number of values =10

Topic: Mathematics, Determinants, JEE Main
Let a curve y=fx pass through the point 2,loge22 and have slope 2yxlogex for all positive real values of x. Then the value of fe is equal to _____.

Options:

The correct answer is:
1

Solution:
Let a curve y=fx pass through the point 2,loge22 and have slope 2yxlogex for all positive real values of x.

Therefore, y'=2yxlnx

⇒dyy=2dxxℓnx

⇒ℓny=2ℓnℓnx+C

Put x=2,y=ℓn22

⇒c=0

⇒y=ℓnx2

⇒fe=1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Let a curve y=yx be given by the solution of the differential equation cos12cos-1e-xdx=e2x-1dy. If it intersects y-axis at y=-1, and the intersection point of the curve with x-axis is α,0, then eα is equal to

Options:

The correct answer is:
2

Solution:
We have,

cos12cos-1e-xdx=e2x-1dy ...i

Put cos-1e-x=θ, θ∈0,π

⇒cosθ=e-x

⇒2cos2θ2-1=e-x

⇒cosθ2=e-x+12=ex+12ex

Hence, by i, we have

cosθ2dx=e2x-1dy

⇒ex+12exdx=e2x-1dy

⇒ex+12exdx=ex-1ex+1dy

⇒12∫dxexex-1=∫dy

Put ex=t⇒dt=exdx

12∫dtexexex-1=∫dy

∫dttt2-t=2∫dy

Put t=1z⇒dtdz=-1z2

∫-dzz21z1z2-1z=2∫dy

⇒-∫dz1-z=2∫dy

⇒-2(1-z)1/2-1=2y+c

⇒21-1t1/2=2y+c

⇒21-e-x1/2=2y+c

Now, it meets y-axis at 0,-1, hence

0=-2+c⇒c=2

Hence,

21-e-x1/2=2y+1

It passes through α, 0

21-e-α1/2=2

⇒1-e-α=12

⇒1-e-α=12

⇒e-α=12⇒eα=2

Topic: Mathematics, Differential Equations, JEE Main
Let a curve y=yx be given by the solution of the differential equation cos12cos-1e-xdx=e2x-1dy. If it intersects y-axis at y=-1, and the intersection point of the curve with x-axis is α,0, then eα is equal to

Options:

The correct answer is:
2

Solution:
We have,

cos12cos-1e-xdx=e2x-1dy ...i

Put cos-1e-x=θ, θ∈0,π

⇒cosθ=e-x

⇒2cos2θ2-1=e-x

⇒cosθ2=e-x+12=ex+12ex

Hence, by i, we have

cosθ2dx=e2x-1dy

⇒ex+12exdx=e2x-1dy

⇒ex+12exdx=ex-1ex+1dy

⇒12∫dxexex-1=∫dy

Put ex=t⇒dt=exdx

12∫dtexexex-1=∫dy

∫dttt2-t=2∫dy

Put t=1z⇒dtdz=-1z2

∫-dzz21z1z2-1z=2∫dy

⇒-∫dz1-z=2∫dy

⇒-2(1-z)1/2-1=2y+c

⇒21-1t1/2=2y+c

⇒21-e-x1/2=2y+c

Now, it meets y-axis at 0,-1, hence

0=-2+c⇒c=2

Hence,

21-e-x1/2=2y+1

It passes through α, 0

21-e-α1/2=2

⇒1-e-α=12

⇒1-e-α=12

⇒e-α=12⇒eα=2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (20 Jul Shift 2)