datavorous/entrance-exam-dataset huggingface dataset:
Let C be the centre of the circle x2+y2-x+2y=114 and P be a point on the circle. A line passes through the point C, makes an angle of π4 with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2) is

Options:
A 2
B 22
C 8sinπ8
D 8cosπ8

The correct answer is:
22

Solution:
Given,

x2+y2-x+2y=114

On rearranging terms we get,

⇒x-122+y+12=22

Now given in ∆PQR, ∠PCR=π4 so ∠PQR=2212 by circle property, radius is 2 so RC=QC=2

Now, PR=QRsin2212=4sinπ8

And PQ=QRcos2212=4cos2212

Now area of ΔPQR=12PR×PQ

=124sinπ84cosπ8

=4×2sinπ8cosπ8

=4sinπ4=42=22

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then, which one of the following points lies on the circle $C$ ?

Options:
A $(1,2)$
B $(1,1)$
C $(2,2)$
D $(2,4)$

The correct answer is:
$(2,4)$

Solution:
Equation of circle
$\mathrm{x}^2+(\mathrm{y}-(6-\mathrm{r}))^2=\mathrm{r}^2$
touches $\sqrt{3} \mathrm{x}-\mathrm{y}=0$
$\begin{aligned}
& \mathrm{p}=\mathrm{r} \\
& \frac{|0-(6-\mathrm{r})|}{2}=\mathrm{r} \\
& |\mathrm{r}-6|=2 \mathrm{r} \\
& \mathrm{r}=2
\end{aligned}$
$\therefore$ Circle $\mathrm{x}^2+(\mathrm{y}-4)^2=4$
$(2,4)$ Satisfies this equation

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let C be the circle with center at 1,1 and radius =1. If T is the circle centered at 0,y, passing through the origin and touching the circle C externally, then the radius of T is equal to

Options:
A 1 2
B 1 4
C 3 2
D 3 2

The correct answer is:
1 4

Solution:
Let r be the required radius
As origin lies on the circle, the distance of 0,0 from 0,y=r

∴ y=r
As the two circles touch externally, distance between centres is sum of the radii
⇒0-12+y-12=r+1
⇒1+y2-2y+1=r2+2r+1
⇒1+r2-2r=r2+2r, ∵ y=r
∴1=4r
∴ r=14

Topic: Mathematics, Circle, JEE Main, JEE Main 2014 (06 Apr)
Let $C$ be the circle with centre $(0,0)$ and radius 3 units. The equation of the locus of the mid points of the chords of the circle $C$ that subtend an angle of $\frac{2 \pi}{3}$ at its centre is

Options:
A
$x^2+y^2=\frac{3}{-}$

B
$x^2+y^2=1$

C
$x^2+y^2=\frac{27}{4}$

D
$x^2+y^2=\frac{9}{4}$

The correct answer is:
$x^2+y^2=\frac{9}{4}$

Solution:
$\cos \frac{\pi}{3}=\frac{\sqrt{h^2+k^2}}{3} \Rightarrow h^2+k^2=\frac{9}{4}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2006
Let C(α, β) be the circumcentre of the triangle formed by the lines 4x+3y=69, 4y-3x=17, and x+7y=61. Then (α-β)2+α+β is equal to

Options:
A 18
B 17
C 15
D 16

The correct answer is:
17

Solution:
Given,

C(α, β) be the circumcentre of the triangle formed by the lines 4x+3y=69, 4y-3x=17, and x+7y=61,

Now plotting the diagram of the above equations and finding the intersection point we get,

Now from diagram we can see that it is a right-angled triangle, so the circumcentre is given by midpoint of the hypotenuse,

Hence, circumcentre will be, C12+52,7+82

⇒C172,152=(α,β)

Hence, (α-β)2+α+β

=172-1522+172+152

=1+16=17

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let C be the largest circle centred at 2,0 and inscribed in the ellipse x236+y216=1. If 1,α lies on C, then 10α2 is equal to ______

Options:

The correct answer is:
118

Solution:
Given:

x236+y216=1

Let any point on the ellipse be P6cosθ, 4sinθ.

Equation of normal of ellipse at any point P is

3secθx–2cosecθy=10

This normal is also the normal of the circle passing through the point 2,0.

So, 6secθ=10 or sinθ=0 (Not possible)

cosθ=35 and sinθ=45

So, point P≡185,165.

So, the largest radius of circle r=3205=645

So the equation of circle x-22+y2=645

This circle is passing through 1,α, hence

1-22+α2=645

⇒α2=595

⇒10α2=118.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let C be the locus of the mirror image of a point on the parabola y2=4x with respect to the line y=x. Then the equation of tangent to C at P2,1 is :

Options:
A x-y=1
B 2x+y=5
C x+3y=5
D x+2y=4

The correct answer is:
x-y=1

Solution:
Given y2=4x

Mirror image on y=x⇒C : x2=4y

2x=4·dydx⇒dydx=x2

dydxP2,1=22=1

Equation of tangent at 2,1

⇒y-1=1×x-2

⇒x-y=1

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let CNaCl and CBaSO4, be the conductances (in S) measured for saturated aqueous solutions of NaCl and BaSO4 respectively, at a temperature T. Which of the following is false?

Options:
A lonic mobilities of ions from both salts increase with T.
B CBaSO4T2>CBaSO4T1 for T2>T1
C CNaClT2>CNaClT1 for T2>T1
D CNaCl≫CBaSO4 at given T

The correct answer is:
lonic mobilities of ions from both salts increase with T.

Solution:
(i) Ionic mobilities decreases with increase in temperature due to increase in random motion and hence decrease in relaxation time so derease in drift speed.

(i) NaCl is completely soluble salt while BaSO4 is sparingly soluble salt so C1>>C2.

(ii) On increasse in temperature conductance increase.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Let Ι=∫cos3x1+sin2xdx, then Ι is equal to (where c is the constant of integration)

Options:
A 2tan-1x+sin⁡x+c
B 2tan-1sin⁡x-sin⁡x+c
C 2tan-1x-x+c
D 2tan-1sin⁡x+sin⁡x+c

The correct answer is:
2tan-1sin⁡x-sin⁡x+c

Solution:
Let, sinâ¡x=tâcosâ¡x dx=dt
Î=â«cos2x 1+t2dt=â«1-sin2x 1+t2dt=â«1-t2 1+t2dt
or Î=â«2-1+t2 1+t2dt
=â«2dt 1+t2-â«dt
=2tan-1t-t+c
=2tan-1sinâ¡x-sinâ¡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
Let Cr denote the binomial coefficient of xr in the expansion of 1+x10. If for α,β∈R, C1+3·2C2+5·3C3+… upto 10 terms =α×2112β-1 (C0+C12+C23+… upto 10 terms) then the value of α+β is equal to _____.

Options:

The correct answer is:
286

Solution:
1+x10=C0+C1x+C2x2+……+C10x10

Differentiating, we get

101+x9=C1+2C2x+3C3x2+…+10C10x9

Now replacing x→x2

101+x29=C1+2C2x2+3C3x4+….+10C10x18

Multiplying by x on both the sides

10·x1+x29=C1x+2C2x3+3C3x5+…+10C10x19

Differentiating again, we get

101+x29·1+91+x282x

=C1+2C2·3x2+3·5·C3x4+…+10·19C10x18

putting x=1

1029+18·28=C1+3·2·C2+5·3·C3+….+19·10·C10

C1+3·2·C2+……+19·10·C10

=10·29·10=100·29

Now 1+x10=C0+C1x+C2x2+……+C10x10

Integrating, we get

1+x111101=C0x+C1x22+C2x33+……+C10x111101

C0+C12+C23+….+C910+C1011=211-111

C0+C12+C23+….+C910=211-211

Now, 100·29=α·2112β-1211-211

⇒α=275, β=11⇒α+β=286

Topic: Mathematics, Binomial Theorem, JEE Main
Let Cr denote the binomial coefficient of xr in the expansion of 1+x10. If for α,β∈R, C1+3·2C2+5·3C3+… upto 10 terms =α×2112β-1 (C0+C12+C23+… upto 10 terms) then the value of α+β is equal to _____.

Options:

The correct answer is:
286

Solution:
1+x10=C0+C1x+C2x2+……+C10x10

Differentiating, we get

101+x9=C1+2C2x+3C3x2+…+10C10x9

Now replacing x→x2

101+x29=C1+2C2x2+3C3x4+….+10C10x18

Multiplying by x on both the sides

10·x1+x29=C1x+2C2x3+3C3x5+…+10C10x19

Differentiating again, we get

101+x29·1+91+x282x

=C1+2C2·3x2+3·5·C3x4+…+10·19C10x18

putting x=1

1029+18·28=C1+3·2·C2+5·3·C3+….+19·10·C10

C1+3·2·C2+……+19·10·C10

=10·29·10=100·29

Now 1+x10=C0+C1x+C2x2+……+C10x10

Integrating, we get

1+x111101=C0x+C1x22+C2x33+……+C10x111101

C0+C12+C23+….+C910+C1011=211-111

C0+C12+C23+….+C910=211-211

Now, 100·29=α·2112β-1211-211

⇒α=275, β=11⇒α+β=286

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let Crn denote the binomial coefficient of xr in the expansion of (1+x)n. If ∑k=01022+3kCkn=α.310+β·210, α, β∈R, then α+β is equal to ___ .

Options:

The correct answer is:
19

Solution:
Instead of Ckn it must be Ck10 i.e.

∑k=01022+3kCk10=α·310+β·210

LHS=4∑k=010Ck10+3∑k=010k·10k·Ck-19

=4.210+3.10.29

=19.210=α.310+β.210

⇒α=0,β=19⇒α+β=19

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let C:x2+y2=4 and C':x2+y2−4λx+9=0 be two circles. If the set of all values of λ so that the circles C and C' intersect at two distinct points, is R−a,b, then the point 8a+12,16b−20 lies on the curve:

Options:
A x2+2y2−5x+6y=3
B 5x2−y=−11
C x2−4y2=7
D 6x2+y2=42

The correct answer is:
6x2+y2=42

Solution:
Given: x2+y2=4

So, centre and radius of C are 0,0 and r1=2 respectively.

Also, C':x2+y2−4λx+9=0

So, centre and radius of C' are 2λ,0 and r2=4λ2−9 respectively.

⇒r1−r2<CC'<r1+r2

⇒2−4λ2−9<2λ<2+4λ2−9

⇒4+4λ2−9−44λ2−9<4λ2

⇒λ∈R ...i

Also, 4λ2<4+4λ2−9+44λ2−9

⇒5<44λ2−9 λ2≥94⇒λ∈−∞,−32∪32,∞

⇒2516<4λ2−9

⇒16964<λ2

⇒16964<λ2

⇒λ∈−∞,−138∪138,∞ ...ii

Using i and ii,

⇒λ∈−∞,−138∪138,∞

⇒λ∈R−−138,138

As per question a=−138 and b=138

∴ required point is 8a+12, 16b-20≡-1, 6 with satisfies option (d).

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let circles C1, C2 and C3 with centres O1, O2 and O3 respectively touch each other externally, where O1=-36,7, O2=20,7 and O3=0,-8. The coordinates of the centre of a circle passing through the points of contact of circles C1,C2 and C2,C3 and C3,C1 are

Options:
A -1,0
B 1,0
C 0,1
D 0,-1

The correct answer is:
-1,0

Solution:
The circle which passes through the points of contact of circles C1,C2 & C2,C3 & C3,C1 is the incircle of ΔO1O2O3.

Its centre lies on the angle bisectors of ΔO1O2O3.
Hence, required point is incentre of ΔO1O2O3.
Now, O1O2=56, O2O3=25 & O3O1=39
⇒Ι=56×0+25×-36+39×2056+25+39,25×7+56×-8+39725+56+39
=-1,0

Topic: Mathematics, Circle, JEE Main
Let $\cos (\alpha+\beta)=\frac{4}{5}$ and let $\sin (\alpha-\beta)=\frac{5}{13}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$, then $\tan 2 \alpha=$

Options:
A
$\frac{56}{33}$

B
$\frac{19}{12}$

C
$\frac{20}{7}$

D
$\frac{25}{16}$

The correct answer is:
$\frac{56}{33}$

Solution:
$\begin{array}{ll}\cos (\alpha+\beta)=\frac{4}{5} & \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\ \sin (\alpha-\beta)=\frac{5}{13} & \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\ \tan 2 \alpha=\tan (\alpha+\beta+\alpha-\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \frac{5}{12}}=\frac{56}{33}\end{array}$

Topic: Mathematics, Trigonometric Equations, JEE Main
Let $\cos (\alpha+\beta)=\frac{4}{5}$ and let $\sin (\alpha-\beta)=\frac{5}{13}$, where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$, then $\tan 2 \alpha=$

Options:
A
$\frac{56}{33}$

B
$\frac{19}{12}$

C
$\frac{20}{7}$

D
$\frac{25}{16}$

The correct answer is:
$\frac{56}{33}$

Solution:
$\begin{array}{ll}\cos (\alpha+\beta)=\frac{4}{5} & \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\ \sin (\alpha-\beta)=\frac{5}{13} & \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\ \tan 2 \alpha=\tan (\alpha+\beta+\alpha-\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \frac{5}{12}}=\frac{56}{33}\end{array}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2010
Let D and E be the midpoints of the sides AC and BC of a triangle ABC respectively. If O is an interior point of the triangle ABC such that OA→+2OB→+3OC→=0→, then the area (in sq. units) of the triangle ODE is

Options:
A 6
B 5
C 34
D 0

The correct answer is:
0

Solution:
It is given that,

OA→+2OB→+3OC→=0

The figure below represents the triangle ABC and midpoints.

Ar△ODE=12|OD→×OE→|

=12a→+c→2×b→+c→2

=18(a→×b→+a→×c→+c→×b→) ...i

The given equation can be written as,

a→+2b→+3c→=0→ ...ii

Multiply b→ in equation ii we get,

a→×b→+2(b→×b→)+3(c→×b→)=0→

⇒a→×b→+3(c→×b→)=0→ ...iii

Multiplying c→ in equation ii we get,

a→×c→+2(b→×c→)+3(c→×c→)=0→

⇒a→×c→+2(b→×c→)=0→ ...iv

Adding equations iii & iv we get,

a→×b→+3(c→×b→)+a→×c→+2(b→×c→)=0→

⇒a→×b→+a→×c→+c→×b→=0→

Substituting the above value in equation i we get,

Ar△ODE=0

Topic: Mathematics, Vector Algebra, JEE Main
Let D be the centroid of the triangle with vertices 3,-1 , 1,3 and 2,4 . Let P be the point of intersection of the lines x+3y-1=10 and 3x-y+1=0 . Then, the line passing through the points D and P also passes through the point:

Options:
A -9,-6
B 9,7
C 7,6
D -9,-7

The correct answer is:
-9,-6

Solution:
Centroid x1+x2+x33,y1+y2+y33

D2,2
Point of intersection P-15,25
Equation of line DP 8x-11y+6=0.

Hence, -9,-6 satisfy above equation.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Let d be the distance between the foot of perpendiculars of the points P1,2-1 and Q2,-1,3 on the plane -x+y+z=1. Then d2 is equal to ______.

Options:

The correct answer is:
26

Solution:
Since -1+2-1-1<0 & -2-1+3-1<0

so points P1,2,-1 and Q2,-1,3 lie on same side of the given plane.

Perpendicular distance of point P from the plane is

-1+2-1-112+12+12=13

Perpendicular distance of point Q from the plane is

=-2-1+3-112+12+12=13

⇒ PQ→ is parallel to given plane. So, distance between P and Q= distance between their foot of perpendiculars.

⇒PQ→=1-22+2+12+-1-32

=26

PQ→2=26=d2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Let $\mathrm{d}$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $\mathrm{d}^2+6$ is equal to :

Options:
A 69
B 78
C 72
D 75

The correct answer is:
75

Solution:
$\begin{aligned} & \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda....(1)\\ & x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \\ & \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu....(2)\end{aligned}$
$\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4$
$3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13$ $\ldots(3) \times 2$
$2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9$ $\ldots(4) \times 3$
$\begin{array}{c}
6 \lambda-8 \mu=26 \\
6 \lambda-9 \mu=27 \\
-\quad+\quad- \\
\hline\mu=-1
\end{array}$
$\begin{gathered}
\Rightarrow 3 \lambda-4(-1)=13 \\
3 \lambda=9 \\
\lambda=3
\end{gathered}$
int. point $(3,6,2) ;(7,8,9)$
$\mathrm{d}^2=16+4+49=69$
Ans. $\mathrm{d}^2+6=69+6=75$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Let $D$ be the domain of a twice differentiable function $f$. For all $x \in D, f^{\prime \prime}(x)+f(x)=0$ and $f(x)=\int g(x) d x+$ constant. If $h(x)=\{\mathrm{f}(x)\}^2+\{g(x)\}^2$ and $h(0)=5$, then $h(2015)-h(2014)$ is equal to

Options:
A $5$
B $3$
C $0$
D $1$

The correct answer is:
$0$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2015
Let D be the domain of the function fx=sin-1log3x6+2log3x-5x. If the range of the function g:D→ℝ defined by gx=x-x, (x is the greatest integer function), is α, β, then α2+5β is equal to

Options:
A 135
B 45
C 46
D 136

The correct answer is:
135

Solution:
We have been given

fx=sin-1log3x6+2log3x-5x

We know that the domain of sin-1x is -1≤x≤1.

⇒log3x6+2log3x-5x∈-1 1-1, 3x>0-2, 3x≠1-3, 6+2log3x-5x-4>0

⇒ From 2, x>0, from 3, x≠13

From 4, 6+2log3x<0 ∵ x>0

⇒log3x<-3

⇒x<3-3

⇒x<127

From 2, 3, 4, 0<x<127 ...5

From 1, -1≤log3x6+2log3x-5x≤1

∵ 3x∈0, 19

⇒13x≥6+2log3x-5x≥3x

⇒-53≤6+2log3x≤-15x2

⇒-236≤log3x≤-15x2-62

⇒3-236≤x≤3-15x2-62

⇒13236≤x≤1315x2+62

⇒3-236≤x<1315x2+62 ......6, here 3-236 is very small positive quantity,

From 5 and 6, 3-236<x<127, x=0

⇒gx=x

⇒ Range gx is domain of fx

⇒gx∈3-236,127≡α,β.

⇒α2+5β=3-2362+5127=3-2362+135≈135

Note: This question was bonus in Jee Main April 2023 session.

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let $\mathrm{D}$ be the domain of the real valued function $f$ defined by $f(x)=\sqrt{25-x^2}$. Then, write $D$.

Options:

None

Solution:
Given function is, $f(x)=\sqrt{25-x^2}$
For real value of $f(x) 25-x^2 \geq 0$
$$
\begin{aligned}
x^2 & \leq 25 \\
-5 & \leq x \leq+5 \\
D &=[-5,5]
\end{aligned}
$$

Topic: Mathematics, Relations and Functions (Class 11)
Let D is a point on the line l1:x+y-2=0 and S3,3 is a fixed point. The line l2 is perpendicular to DS and passes through S. If M is another point on the line l1 (other than D ), then the locus of the point of intersection of l2 and the angle bisector of the angle MDS is

Options:
A x+y-22=2x-32+2y-32
B x+y-22=x-32+y-32
C x+y-22=x-32+y-322
D None of these

The correct answer is:
x+y-22=2x-32+2y-32

Solution:
Let the point of intersection of l2 and the angle bisector of angle MDS is Ph,k.

Draw perpendicular PF to l1 from P.

Now, ∆PFD & ∆PSD are congruent triangles.

Hence, PF=PS
h+k-22=h-32+k-32
⇒x+y-22=2x-32+y-32

Topic: Mathematics, Straight Lines, JEE Main
Let D is a point on the line l1:x+y-2=0, S3,3 is a fixed point and line l2 is the perpendicular to DS and passing through S. If MK is another point on the line l1 (other than D), then the locus of the point of intersection of l2 and angle bisector of the angle MDS is a conic whose length of latus rectus rectum is equal to

Options:
A 42
B 4
C 8
D 22

The correct answer is:
42

Solution:
Let the point of intersection of l2 and angle bisector of the angle MDS is Ph,k.

Draw perpendicular PF to l1 from P.

Now, ÎPFD and ÎPSD are congruent triangles.

Hence, PF=PS
â locus of P is a parabola having S as focus and l1 as the directrix
Hence, the length of the latus rectum is 23+3-21+1=42 units

Topic: Mathematics, Parabola, JEE Main
Let d∈R, and A=-24+dsinθ-21sinθ+2d52sinθ-d-sinθ+2+2d, θ∈0, 2π. If the minimum value of detA is 8, then a value of d is:

Options:
A 22+2
B 22+1
C -5
D -7

The correct answer is:
-5

Solution:
A=-24+dsinθ-21sinθ+2d52sinθ-d-sinθ+2+2d

R3→R3-2R2+R1

=-24+dsinθ-21sinθ+2d100

=14+dd-sinθ+2sinθ-2

=4d+d2-sin2θ+4=d+22-sin2θ

We know that 0≤sin2θ≤1

Given, the minimum value of A=8, which is possible, when sin2θ=1

⇒d+22=9

⇒d+2=±3

⇒d=1 or d=-5.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let $\frac{d}{d x} F(x)=\left(\frac{e^{\text {sinx }}}{x}\right) x>0$. If $\int_1^4 \frac{3}{x} e^{\text {sin }^3} d x=F(k)-F(1)$ then one of the possible values of $k$, is

Options:
A
64

B
15

C
16

D
63

The correct answer is:
64

Solution:
$$
\frac{d}{d x} F(x)=\frac{e^{\sin x}}{x} \text { or } \int_1^4 \frac{3}{x} e^{\sin x^3} d x=\int_1^4 \frac{3 x^2}{x^3} e^{\sin x^3} d x
$$
Let $\mathrm{x}^3=\mathrm{t}, 3 \mathrm{x}^2 \mathrm{dx}=\mathrm{dt}$
when $x=1, t=1 \& x-4, t=64$
$$
\begin{aligned}
& \mathrm{F}(\mathrm{t})=\int_1^{64} \frac{\mathrm{e}^{\sin t}}{\mathrm{t}} \mathrm{dt}=\int_1^{64} \mathrm{~F}(\mathrm{t}) \mathrm{dt}=\mathrm{F}(64)-\mathrm{F}(1) \\
& \mathrm{K}=64
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
Let ∫dxx2+1-x=fx+C such that f0=0 and C is the constant of integration, then the value of f1 is

Options:
A 12+12ln⁡1+2
B 12+12ln⁡1+2
C 12+12ln⁡2+1
D 12+121+ln⁡1+2

The correct answer is:
12+121+ln⁡1+2

Solution:
∫x2+1+x1dx
∫x2+1dx+∫xdx
=x2x2+1+12ln⁡x+x2+1+x22+C
fx=x2x2+1+12ln⁡x+1x2+1+x22
f0=0
f1=22+12ln⁡1+2+12=12+121+ln⁡1+2

Topic: Mathematics, Indefinite Integration, JEE Main
Let dydx=ax-by+abx+cy+a, where a,b,c are constants. represent a circle passing through the point 2,5. Then the shortest distance of the point 11,6 from this circle is

Options:
A 10
B 8
C 7
D 5

The correct answer is:
8

Solution:
Let equation of circle is

x2+y2+2gx+2fy+c=0

Differentiating above equation w.r.t x we get,

⇒dydx=-2x+2g2y+2f

Now, comparing with dydx=ax-by+abx+cy+a

We get, b=0,a=-2,c=2

⇒-2g=-2⇒g=1

Also, 2f=-2

So, f=-1

Now circle will be

x2+y2+2x-2y+c=0

its passes through 2,5

which will give c=-23

so circle will be x2+y2+2x-2y-23=0

centre C=-1,1

and radius 5

Now P is 11,6

So minimum distance of P from circle will be =11+12+6-12-5

=13-5=8

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let $\alpha, \beta$ denote the cube roots of unity other than 1 and $\alpha \neq \beta .$ Let $S=\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{\alpha}{\beta}\right)^{n} .$ Then the value of $S$ is

Options:
A either $-2 \omega$ or $-2 \omega^{2}$
B either $-2 \omega$ or $2 \omega^{2}$
C either $2 \omega$ or $-2 \omega^{2}$
D either $2 \omega$ or $2 \omega^{2}$

The correct answer is:
either $-2 \omega$ or $-2 \omega^{2}$

Solution:
Case I Let $\alpha=\omega$ and $\beta=\omega^{2}$
$S=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega}{\omega^{2}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\omega^{2}\right)^{n}$
$=1-\omega^{2}+\omega^{4}-\omega^{6}+\omega^{8}-\omega^{10}+\omega^{12}$
$\quad+\ldots+\omega^{600}-\omega^{602}+\omega^{604}$
$=1-\omega^{2}+\omega-1+\omega^{2}-\omega+1+\ldots$
$=0+\ldots+1-\omega^{2}+\omega$
$=-\omega^{2}-\omega^{2}=-2 \omega^{2} \quad\left[\because 1+\omega+\omega^{2}=0\right]$
Case II Let $\alpha=\omega^{2}$ and $\beta=\omega$
$\therefore S=\sum_{n=0}^{\infty_{2}}(-1)^{n}\left(\frac{\omega^{2}}{\omega}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega^{4}}{\omega^{3}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}(\omega)$
$=1-\omega+\omega^{2}-\omega^{3}+\omega^{4}-\omega^{5}+\omega^{6}-\ldots$
$+\omega^{300}-\omega^{301}+\omega^{302}$
$=1-\omega+\omega^{2}-1+\omega-\omega^{2}+1$
$=0+\ldots+1+\omega^{2}-\omega$
$=-\omega-\omega=-2 \omega$

Topic: Mathematics, Complex Number, JEE Main
Let α denote the greatest integer ≤α. Then 1+2+3+.............+120 is equal to

Options:

The correct answer is:
825

Solution:
Given,

1+2+3+.............+120

Now we now that,

x=1, when x∈[1,4), x=2, when x∈[4,9) and similarly x=10, when x∈[100,120)

Now using the above formula we get,

E=1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+4+4+…

⇒E=3×1+5×2+7×3+…..+19×9+10×21

⇒E=∑r=1102r+1r

⇒E=2∑r=110r2+r

⇒E=210×11×216+10×112

⇒E=770+55

⇒E=825

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let λ denote the number of terms in the expansion of 1+5x+10x2+10x3+5x4+x520. If unit’s place and ten’s place digits in 3λ are O and T, then O+T is equal to

Options:

The correct answer is:
3

Solution:
1+5x+10x2+10x3+5x4+x520
=1+x520=1+x100
Number of terms =101=λ
3101=3⋅3100=3⋅950=310-150
=3C500⋅1050-C501⋅1049+.……..+C5048⋅102-C5049⋅10+C5050
=3100K+50×492×100-50×10+1
=3K×100+3×25×49×100-500×3+3
= Multiple of 100+3
⇒ one’s place is 3 and ten’s place is 0
Hence, O+T=3

Topic: Mathematics, Binomial Theorem, JEE Main
Let e1 and e2 be the eccentricities of the ellipse x225+y2b2=1 b<5 and the hyperbola x216-y2b2=1 respectively satisfying e1e2=1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair α, β is equal to:

Options:
A 8, 10
B 203, 12
C 8, 12
D 245, 10

The correct answer is:
8, 10

Solution:
e1=1-b225 and e2=1+b216

Given e1e2=1

⇒e1e22=1⇒1-b2251+b216=1

⇒b216-b225-b425×16=0

⇒916×25b2-b425×16=0⇒b2=9

Thus, e1=1-925=45 and e2=1+916=54

Therefore, α=25e1=8 and β=24e2=10

∴α, β=8, 10

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let E1 and E2 be two events such that the conditional probabilities PE1∣E2=12, PE2∣E1=34 and PE1∩E2=18. Then

Options:
A PE1∩E2=PE1·PE2
B PE1'∩E2'=PE1'·PE2
C PE1∩E2'=PE1·PE2
D PE1∪E2=PE1PE2

The correct answer is:
PE1∩E2'=PE1·PE2

Solution:
PE1∩E2PE2=12, PE2∩E1PE1=34, PE1∩E2=18

⇒ PE2=14, PE1=16, PE1∪E2=16+14-18=724

PE'∩E2'=PE1∪E2'=1-724=1724

PE2'=34 ⇒PE1∩E2=PE1PE2'

PE1∩E2'=PE1-PE1∩E2=16-18=124

PE1∩E2'=PE1·PE2

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let E1 and E2 denotes gravitational field at distance r1 and r2 from the axis of an infinitely long solid cylinder of the radius R. Which of the following must hold true?

Options:
A E1<E2 if r1<r2<R
B E1 > E2 if R<r1<r2
C E1 > E2, if r1=R- E, r2=R+E E is positive constant <R
D All of the above

The correct answer is:
All of the above

Solution:
Graph of gravitational field versus distance from axis for a solid cylinder

Topic: Physics, Gravitation, JEE Main