datavorous/entrance-exam-dataset huggingface dataset:
Let he sum of the coefficient of first three terms in the expansion of x-3x2n; x=0, n∈N be 376. Then, the coefficient of x4 is equal to:

Options:

The correct answer is:
405

Solution:
We know that general term in the expansion of x-3x2n is given by,

Tr+1=Crnxn-r-3x2r

=-1r×Crn3r xn-r-2r

Tr+1=-1r×Crn3r xn-3r...............1

So, T1=T0+1= C0n30 xn=xn

T2=-1× C1n31 xn-3

T3=C2n32 xn-6

Now given sum is

1- C1n·3+ C2n·32=376

⇒1-3n+nn-12·9=376

⇒1-3n+n2-n2·9=376

⇒2-6n+9n2-9n=752

⇒9n2-15n-750=0

⇒3n2-5n-250=0

⇒n=5±25+30006

⇒n=5±556

⇒n=10 ignoring negative sign

Now, Tr+1=-1r Cr10 3r x10-3r

So, for coefficient of x4 we take

10-3r=4

⇒3r=6

⇒r=2

So, coefficient of x4 is given by,

T2+1=-12·C210 32=45×9=405.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let I=â«01sinâ¡xx dx and J=â«01cosâ¡xx dx. Then, which one of the following is true?

Options:
A I>23Â andÂ J<2
B I>23Â andÂ J>2
C I<23 and J<2
D I<23Â andÂ Â J>2

The correct answer is:
I<23 and J<2

Solution:
Since, I=∫01sin⁡xxdx<∫01xxdx
as in x∈0,1, x>sin⁡x
I<∫01xdx=23x3/201⇒I<23
For, x∈0,1, cos⁡xx<1x
Hence, J=∫01cos⁡xxdx<∫01x-12dx=2⇒J<2

Topic: Mathematics, Definite Integration, JEE Main
Let $\mathrm{I}=\int_0^1 \frac{\sin \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$ and $\mathrm{J}=\int_0^1 \frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$. Then which one of the following is true?

Options:
A
$\mathrm{I}>\frac{2}{3}$ and $\mathrm{J}>2$

B
$I < \frac{2}{3}$ and $J < 2$
C
$\mathrm{I} < \frac{2}{3}$ and $\mathrm{J}>2$
D
$I>\frac{2}{3}$ and $\mathrm{J} < 2$

The correct answer is:
$I < \frac{2}{3}$ and $J < 2$

Solution:
$$
\begin{aligned}
& \mathrm{I}=\int_0^1 \frac{\sin x}{\sqrt{x}} \mathrm{dx} < \int_0^1 \frac{\mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}=\int_0^1 \sqrt{\mathrm{x}} \mathrm{dx}=\left.\frac{2}{3} \mathrm{x}^{3 / 2}\right|_0 ^1=\frac{2}{3} \\
& \Rightarrow \mathrm{I} < \frac{2}{3} \\
& \mathrm{~J}=\int_0^1 \frac{\cos x}{\sqrt{\mathrm{x}}} \mathrm{dx} < \int_0^1 \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}=\left.2 \sqrt{\mathrm{x}}\right|_0 ^1=2 \\
& \therefore \mathrm{J} \leq 2 .
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2008
Let I1=∫0α1+2cos⁡x1+exdx and I2=∫0α1+ex1+2cos⁡xdx, where α is the root of the equation 2cos⁡x-ex=0. Then,

Options:
A I1=I2
B I1>I2
C I1+I2=0
D I1<I2

The correct answer is:
I1>I2

Solution:
As, x∈0,α ⇒2cos⁡x>ex
∴1+2cos⁡x1+ex>1+ex1+2cos⁡x
⇒∫0α1+2cos⁡x1+exdx>∫0α1+ex1+2cos⁡xdx
⇒I1>I2
Also, I1,I2>0

Topic: Mathematics, Definite Integration, JEE Main
Let $I_1=\int_1^2 \frac{d x}{\sqrt{1+x^2}}$ and $I_2=\int_1^2 \frac{d x}{x}$ then

Options:
A $I_1 \gt I_2$
B $I_2 \gt I_1$
C $I_1=I_2$
D $I_1 \gt 2 I_2$

The correct answer is:
$I_2 \gt I_1$

Solution:
We have, $\left(1+x^2\right) \gt x^2, \forall x ; \sqrt{1+x^2} \gt x, \forall x \in(1,2)$
$\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{1+x^2}} \lt \frac{1}{x}, \forall x \in(1,2) \Rightarrow \int_1^2 \frac{d x}{\sqrt{1+x^2}} \lt \int_1^2 \frac{d x}{x} \\
& \Rightarrow I_1 \lt I_2 \Rightarrow I_2 \gt I_1 .
\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
Let ∪i=150Xi=∪i=1nYi=T, where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of sets Yi's then n is equal to :

Options:
A 15
B 50
C 45
D 30

The correct answer is:
30

Solution:
∪i=150Xi=∪i=1nYi=T

From question's conditions

10×5020=5n6

⇒n=30

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Let i=-1. If -1+i3211-i24+1+i3211+i24=k, and n=k be the greatest integral part of k.

Then ∑j=0n+5j+52-∑j=0n+5j+5 is equal to ________.

Options:

The correct answer is:
310

Solution:
k=129-12+i322112-12i24+12+i322112+12i24

k=1512ei2π321e-iπ424+eiπ321eiπ424]

k=1512ei14π+6π+ei7π-6π

k=1512e20πi+eπi

k=1512[1+-1]=0

n=k=0

∑j=05j+52-∑j=05j+5

=∑j=05j2+25+10j-j-5

=∑j=05j2+9j+20

=∑j=05j2+9∑j=05j+20∑j=051

=5×6×116+95×62+20×6

=55+135+120

=310

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let $I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x$. Then

Options:
A $\frac{1}{2} \leq 1 \leq 1$
B $4 \leq 1 \leq 2 \sqrt{30}$
C $\frac{\sqrt{3}}{8} \leq 1 \leq \frac{\sqrt{2}}{6}$
D $1 \leq 1 \leq \frac{2 \sqrt{3}}{\sqrt{2}}$

The correct answer is:
$\frac{\sqrt{3}}{8} \leq 1 \leq \frac{\sqrt{2}}{6}$

Solution:
We have,
$$
I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x
$$
since, $\frac{\sin x}{x}$ is a decreasing function.
$$
\begin{array}{c}
\frac{\pi}{12} \times \frac{\sin \pi / 3}{\pi / 3} \leq 1 \leq \frac{\pi}{12} \times \frac{\sin \pi / 4}{\pi / 4} \\
\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main
Let I=∫abx4-2x2dx. If I is minimum then the ordered pair (a, b) is

Options:
A 0, 2
B 2,Â -2
C -Â 2,0
D -2,Â 2

The correct answer is:
-2, 2

Solution:
Given that, I=∫abx4-2x2dx

Let fx=x4-2x2

To draw it's graph, finding the points where it intersects the x-axis, put fx=0

⇒x4-2x2=0

⇒x2x2-2=0

So, x=0 and x=±2.

The graph of fx is,
From the figure, we get minimum area when -2≤x≤2.

Hence, the ordered pair can be written as a, b=-2, 2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let $I$ be a unit matrix of order 6 . Let $A=\left(a_{i j}\right)$ be a square matrix of order 6 such that $a_{i j}=\left\{\begin{array}{l}1, \text { if } i+j=7 \\ 0, \text { if } i+j \neq 7\end{array}\right.$ then $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=$

Options:
A 1
B $A$
C $-A$
D -1

The correct answer is:
$-A$

Solution:
Given that,
$a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases}$
Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2=\left(A(\operatorname{Adj} A) A^{-1}\right) A^2$
$=\left((|A| I) A^{-1}\right) A^2 \quad[\because A$ adj $A=|A| I]$
$=-1\left(A^{-1} A\right) A \quad[\because|A|=-1]$
$=-I A \quad\left[\because A^{-1} A=I\right]$
$=-A$

Topic: Mathematics, Matrices, JEE Main
Let I be an identity matrix of order 2×2 and P=2-15-3. Then the value of n∈N for which Pn=5I-8P is equal to ___ .

Options:

The correct answer is:
6

Solution:
P=2-15-3

5I-8P=5005-16-840-24=-118-4029

P2=2-15-32-15-3=-11-54

P3=2-15-3-11-54=3-210-7

⇒P6=3-210-73-210-7=-118-4029=Pn

⇒n=6

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let I be the purchase value of an equipment and $\mathrm{V}(\mathrm{t})$ be the value after it has been used for t years. The value $\mathrm{V}(\mathrm{t})$ depreciates at a rate given by differential equation $\frac{\mathrm{dV}(\mathrm{t})}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$, where $\mathrm{k}>0$ is a constant and $\mathrm{T}$ is the total life in years of the equipment. Then the scrap value $\mathrm{V}(\mathrm{T})$ of the equipment is

Options:
A
$\mathrm{I}-\frac{\mathrm{kT}}{2}$

B
$1-\frac{k(T-t)^2}{2}$

C
$e^{-k T}$

D
$\mathrm{T}^2-\frac{1}{\mathrm{k}}$

The correct answer is:
$\mathrm{I}-\frac{\mathrm{kT}}{2}$

Solution:
$$
\frac{d V}{d t}=-k(T-t) \Rightarrow d V=-k(T-t) d t
$$
Integrate
$$
V=\frac{-k(T-t)^2}{(-2)}+c \Rightarrow V=\frac{k(T-t)^2}{2}+c
$$
at $\mathrm{t}=\mathrm{0} \Rightarrow \mathrm{V}=\mathrm{I}$
$$
\mathrm{I}=\frac{\mathrm{kT}}{2}+\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2} \Rightarrow \mathrm{c}=\mathrm{V}(\mathrm{T})=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2011
Let I=â«dx1+3sin2x=12tan-12fx+C (where, C is the constant of integration). If fÏ4=1, then the fundamental period of y=fx is

Options:
A π4
B π
C 2π
D π6

The correct answer is:
π

Solution:
Dividing the numerator and denominator by cos2x, we get
I=â«sec2xsec2x+3tan2xdx
=â«sec2x1+4tan2xdx
Let, tanâ¡x=t
âsec2xdx=dt
â´I=â«dt1+4t2
=12tan-12t+C
I=12tan-12tanâ¡x+Câfx=tanâ¡x
Hence, the fundamental period is âÏâ

Topic: Mathematics, Indefinite Integration, JEE Main
Let I=∫dxcos⁡x-sin⁡x2=1fx+C (where, C is the constant of integration). If fπ3=1-3, then the number of solution(s) of fx=2020 in x∈π2,π is/are

Options:

The correct answer is:
1

Solution:
I=∫sec2⁡x1-tan⁡x2dx
Let tan⁡x=t
⇒sec2⁡xdx=dt
⇒I=∫dt1-t2=∫1-t-2dt
⇒I=11-t+C
=11-tan⁡x+C
∴fx=1-tan⁡x
If fx=2020⇒1-tan⁡x=2020
⇒tan⁡x=-2019
Which happens once in x∈π2,π

Topic: Mathematics, Indefinite Integration, JEE Main
Let $I_{n}=\int_{0}^{1} e^{-y} y^{n} d y$, where $n$ is a non-negative integer. Then $\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{I}_{\mathrm{n}}}{\mathrm{n} !}$ is

Options:
A 1
B $1-\frac{1}{\mathrm{e}}$
C $\frac{1}{\mathrm{e}}$
D $1+\frac{1}{\mathrm{e}}$

The correct answer is:
$\frac{1}{\mathrm{e}}$

Solution:
$I_{n}=\int_{0}^{1} e^{-y} y^{n} d y$
$I_{n}=-\frac{1}{e}+n I_{n-1}$ (by reduction formula)
$I_{n}-n I_{n-1}=-\frac{1}{e}$
$\begin{array}{l}
\frac{I_{n}}{n !}-\frac{I_{n-1}}{n-1 !}=-\frac{I}{e(n !)} \\
n=1 \quad \frac{I_{1}}{1 !}-\frac{I_{0}}{0 !}=-\frac{1}{e} \\
n=2 \quad \frac{I_{2}}{2 !}-\frac{I_{1}}{1 !}=-\frac{1}{e(2 !)} \\
n=3 \quad \frac{I_{3}}{3 !}-\frac{I_{2}}{2 !}=-\frac{1}{e(3 !)} \\
n=4 \quad \frac{I_{4}}{4 !}-\frac{I_{3}}{3 !}=-\frac{1}{e(4 !)} \\
\text { eq. }(1)+e q \cdot(2) \times 2+\operatorname{eq} \cdot(3) \times 3+(\text { eq. } 4) \times 4+\ldots \ldots \\
-\left\{\frac{I_{1}}{1 !}+\frac{I_{2}}{2 !}+\frac{I_{3}}{3 !}+\frac{I_{4}}{4 !}\right\}-\mathrm{I}_{0} \\
=-\frac{1}{e}\left\{1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right\}
\end{array}$
Let $\mathrm{S}=\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{I}_{\mathrm{n}}}{\mathrm{n} !}$
$\begin{array}{l}
-\mathrm{S}-\mathrm{I}_{0}=-\frac{1}{\mathrm{e}} \times \mathrm{e} \\
\mathrm{S}=1-\mathrm{I}_{0} \\
\mathrm{~S}=1-\int_{0}^{1} \mathrm{e}^{-\mathrm{y}} \mathrm{dy} \\
\mathrm{S}=1+\left[\mathrm{e}^{-\mathrm{y}}\right]_{0}^{1} \\
\mathrm{~S}=1+\left[\frac{1}{\mathrm{e}}-1\right] \\
\mathrm{S}=\frac{1}{\mathrm{e}}
\end{array}$

Topic: Mathematics, Definite Integration, JEE Main
Let In=∫1ex19(log|x|)ndx, where n∈N. If 20I10=αI9+βI8, for natural numbers α and β, then α-β equal to _______.

Options:

The correct answer is:
1

Solution:
Given In=∫1ex19(log|x|)ndx, n∈N

⇒In=∫1ex19logxndx, ∵ for 1<x<e, x>0

Integrating by using by parts rule, ∫abu·vdx=u∫vdxab-∫abddxu∫vdxdx, taking u=logx & v=x19, we get

In=logxn∫x19dx1e-∫1eddxlogxn∫x19dxdx

⇒In=logxnx20201e-∫1enlogxn-1×1x×x2020dx

⇒In=logene2020-0-∫1enlogxn-1×x1920dx

⇒In=e2020-n20∫1ex19logxn-1dx

⇒20In=e20-nIn-1

Hence, we get 20I10=e20-10I9 ...i and

20I9=e20-9I8 ...ii

On subtracting the above two equations, we get

20I10-20I9=-10I9+9I8

⇒20I10=10I9+9I8

Given 20I10=αI9+βI8

⇒α=10, β=9 and hence α-β=1.

Topic: Mathematics, Definite Integration, JEE Main
Let In=∫1ex19(log|x|)ndx, where n∈N. If 20I10=αI9+βI8, for natural numbers α and β, then α-β equal to _______.

Options:

The correct answer is:
1

Solution:
Given In=∫1ex19(log|x|)ndx, n∈N

⇒In=∫1ex19logxndx, ∵ for 1<x<e, x>0

Integrating by using by parts rule, ∫abu·vdx=u∫vdxab-∫abddxu∫vdxdx, taking u=logx & v=x19, we get

In=logxn∫x19dx1e-∫1eddxlogxn∫x19dxdx

⇒In=logxnx20201e-∫1enlogxn-1×1x×x2020dx

⇒In=logene2020-0-∫1enlogxn-1×x1920dx

⇒In=e2020-n20∫1ex19logxn-1dx

⇒20In=e20-nIn-1

Hence, we get 20I10=e20-10I9 ...i and

20I9=e20-9I8 ...ii

On subtracting the above two equations, we get

20I10-20I9=-10I9+9I8

⇒20I10=10I9+9I8

Given 20I10=αI9+βI8

⇒α=10, β=9 and hence α-β=1.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let, In=∫tannxdxn>1 . If I4+I6=atan5x+bx5+c, then the ordered pair a,b, is equal to

Options:
A -15,1
B 15,0
C 15,-1
D -15,0

The correct answer is:
15,0

Solution:
Given, In=∫tannxdx

⇒I4+I6=∫tan4xdx+∫tan6xdx

⇒I4+I6=∫tan4xdx1+tan2xdx

⇒I4+I6=∫tan4xsec2xdx

Put, tanx=t⇒sec2xdx=dt

⇒I4+I6=∫t4dt

⇒I4+I6=t55+c, where c is the constant of integration.

⇒atan5x+bx5+c=tan5x5+c

∴ a=15& b=0

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2017 (02 Apr)
Let Inx=∫0x1t2+5ndt,n=1,2,3,…. Then

Options:
A 50I6-9I5=xI5'
B 50I6-11I5=xI5'
C 50I6-9I5=I5'
D 50I6-11I5=I5'

The correct answer is:
50I6-9I5=xI5'

Solution:
Given,

Inx=∫0xdtt2+5n

Applying integration by parts we get,

⇒Inx=tt2+5n0x-∫0xnt2+5-n-1·2t2

⇒Inx=xx2+5n+2n∫0xt2t2+5n+1dt

⇒Inx=xx2+5n+2n∫0xt2+5-5t2+5n+1dt

⇒Inx=xx2+5n+2n∫0xdtt2+5n-10n∫0xdtt2+5n+1

⇒Inx=xx2+5n+2nInx-10nIn+1x

⇒10nIn+1x+1-2nInx=xx2+5n

⇒10nIn+1x+1-2nInx=xI'n {where I'n=1x2+5n we get by differentiating Inx=∫0xdtt2+5n }

Now put n=5

We get, 50I6-9I5=xI5'

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let Ix=∫x+1x1+xex2dx, x>0. If limx→∞Ix=0 then I1 is equal to

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

The correct answer is:
e+2e+1-logee+1

Solution:
Given,

Ix=∫x+1x1+xex2dx

Now let 1+xex=t

⇒exx+1dx=dt

So, Ix=∫1t-1t2dt

⇒Ix=∫1-t2+t2t-1t2dt

⇒Ix=∫-t+1t2+1t-1dt

⇒Ix=∫-1t-1t2+1t-1dt

⇒Ix=-lnt+1t+lnt-1+C

⇒Ix=lnxexxex+1+1xex+1+C

Also given,limx→∞Ix=0

⇒limx→∞Ix=limx→∞ln1-1xex+1+1xex+1+C

⇒limx→∞Ix=ln1-0+0+C

⇒C=0

Now finding,

I1=lnee+1+1e+1=1+1e+1-ln(e+1)

⇒I1=e+2e+1-lne+1

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let Ix=∫x+7x dx and I9=12+7loge7. If I1=α+7loge1+22, then α4 is equal to _____.

Options:

The correct answer is:
64

Solution:
Given,

Ix=∫x+7x dx

Now let, x+7x=t2⇒-7x2dx=2t dt

⇒dx=-14tt2-12dt

So, Ix=-14∫t2t2-12dt

⇒Ix=-14∫dtt2+1t2-2

⇒Ix=-142∫1-1t2t+1t2-4+1+1t2t-1t2dt

⇒Ix=-714lnt+1t-2t+1t+2-1t-1t+c

Now when x=9, t=43

⇒I9= 12+7×ln7=-74ln172+7×127+c

⇒ c=72ln7

Now when x=1, t=22

⇒ I1=+74ln22+122-12+7×227+72ln7

⇒I1=72ln22+127+22+72ln7

⇒I1=7 ln22+1-72ln7+22+72ln7

⇒ α=22⇒α4=64

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let in a âABC, vertex A, the circumcentre S and the orthocentre H are 1,10, -13,23 and 113,43 respectively, then mid-point of the side BC is

Options:
A 1,-113
B 1,-223
C 2,-113
D -1,113

The correct answer is:
1,-113

Solution:
We know that centroid divides orthocentre and cirumcentre in the ratio 2:1
Hence, centroid G is -23+1133,43+433
âG=1,89
Let, D is the mid-point of side BC then G also divides A and D in the ratio 2:1
2h+13=1 and 2k+103=89
âh=1 and k=-113âD=1,-113

Topic: Mathematics, Straight Lines, JEE Main
Let in ΔABC coordinates of vertex A is (0, 0). Equation of the internal angle bisector of ∠ABC is x+y-1=0 and mid-point of BC is (1, 3). The ordinate of vertex C is

Options:

The correct answer is:
6

Solution:
Reflection of A (0, 0) about the line x+y-1=0 is (1, 1) which lies on side BC
Equation of BC is x=1 {since, line is passing through (1,1) and (1, 3)}
⇒ B is the point of intersection of x=1 and x+y-1=0

⇒ coordinates of B is (1, 0)
⇒coordinates of C is (1, 6) { ∵ mid point of B and C is (1, 3)}

Topic: Mathematics, Straight Lines, JEE Main
Let in ∆ABC the coordinates of A are 0,0. Internal angle bisector of ∠ABC is x-y+1=0 and mid-point of BC is -1,3. Then, the ordinate of C is

Options:

The correct answer is:
6

Solution:
Reflection of A0,0 about the line x-y+1=0 is -1,1 which lies on BC

The equation of BC is x=-1 (line passing through -1,1 & -1,3)
⇒B is the point of intersection of x=-1 and x-y+1=0

The point B is -1,0
Hence, the coordinates of C are-1,6 (since, the mid point of B & C is -1,3)

Topic: Mathematics, Straight Lines, JEE Main
Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :

Options:
A 32625
B 80243
C 40243
D 128625

The correct answer is:
32625

Solution:
PX=1=C15·p.q4=0.4096

PX=2=C25·p2·q3=0.2048

⇒q2p=2

⇒q=4p and p+q=1

⇒p=15 and q=45

Now

PX=3=C35·153·452=10×16125×25=32625

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let in a right angled triangle, the smallest angle be θ. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sinθ is equal to:

Options:
A 5+14
B 5-12
C 2-12
D 5-14

The correct answer is:
5-12

Solution:
Let A=θ, then B=90°-θ.

Hence, a is the smallest side.

Then, in ∆ACB

c2=a2+b2

On taking reciprocal, longest side become shortest and vice versa.

⇒1a2=1b2+1c2

⇒b2c2a2=b2+c2 ...i

Use a=2RsinA=2Rsinθ

b=2RsinB=2Rsin(90°-θ)=2Rcosθ

c=2RsinC=2sin90°=2R

Hence, by i

4R2cos2θ·4R24R2sin2θ=4R2cos2θ+4R2

⇒cos2θ=sin2θcos2θ+sin2θ

⇒1-sin2θ=sin2θ1-sin2θ+sin2θ

⇒1=3sin2θ-sin4θ

⇒sin4θ-3sin2θ+1=0

⇒sin2θ=3-52=5+1-254=5-122

⇒sinθ=5-12

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Let in a series of 2n observations, half of them are equal to a and remaining half are equal to -a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2+b2 is equal to :

Options:
A 425
B 650
C 250
D 925

The correct answer is:
425

Solution:
Let observations are denoted by xi for 1≤i<2n

x¯=∑xi2n=(a+a+…+a)-(a+a+…+a)2n

⇒x¯=0

and σx2=∑xi22n-x¯2=a2+a2+…+a22n-0=a2

⇒σx=a

Now, adding a constant b then y¯=x¯+b=5

⇒b=5

and σy=σx (No change in S.D.) ⇒a=20

⇒a2+b2=425

Topic: Mathematics, Statistics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let Ï is the angular velocity of the earth's rotation about its own axis and acceleration due to gravity on the earth's surface at the poles is g. An object weighed by a spring balance gives the same reading at the equator, as well as at a height h above the poles (h âª R). What is the value of h?

Options:
A gRω
B ω2R2g
C ω2R22g
D 2ω2R2g

The correct answer is:
ω2R22g

Solution:
The value of acceleration due to gravity at the equator is gE=g-Rω2 and the acceleration due to gravity at a height h above the pole is gh=1-2hRg . It is given that
∵mgE=mgh
∴gE=gh
∴g-Rω2=1-2hRg=g-2hgR
∴Rω2=2hgR ∴h=R2ω22g

Topic: Physics, Gravitation, JEE Main
Let J=∫01x1+x8dx Consider the following assertions:

I. J>14

II. J<π8

Then,

Options:
A only I is true
B only II is true
C both I and II are true
D neither I nor II is true

The correct answer is:
only I is true

Solution:
As we know, 1+x8<2,∀x∈0,1

⇒12<11+x8,∀x∈0,1

⇒x2<x1+x8,∀x∈0,1

⇒∫01x2dx<∫01x1+x8dx

⇒∫01x1+x8dx>14

⇒J>14

Similarly, as we know, 1+x8<1+x4,∀x∈0,1

⇒∫01x1+x4dx<∫01x1+x8dx

⇒J>12tan-1x201

⇒J>π8

Statement J=∫01x1+x8dx>14 is true, while the statement J=∫01x1+x8dx<π8 is not true.

Topic: Mathematics, Definite Integration, JEE Main
Let Jn,m=∫01/2xnxm-1dx,∀n>m and n, m∈N. Consider a matrix A=aij3×3 where aij=J6+i,3-Ji+3,3, i≤j0, i>j. Then adjA-1 is :

Options:
A (15)2×234
B (15)2×242
C (105)2×236
D (105)2×238

The correct answer is:
(105)2×238

Solution:
∵A=a11a12a130a22a2300a33⇒A=a11a22a33

⇒|A|=J7,3-J4,3·J8,3-J5,3·J9,3-J6,3

=∫01/2x7-x4x3-1dx·∫01/2x8-x5x3-1dx·∫01/2x9-x6x3-1dx

=∫01/2x4dx·∫01/2x5dx·∫01/2x6dx

=1(210)·(2)18

Now

adjA-1=A-12=1|A|2=(210)·2182=(105)2·(2)38

Topic: Mathematics, Definite Integration, JEE Main
Let Jn,m=∫01/2xnxm-1dx,∀n>m and n, m∈N. Consider a matrix A=aij3×3 where aij=J6+i,3-Ji+3,3, i≤j0, i>j. Then adjA-1 is :

Options:
A (15)2×234
B (15)2×242
C (105)2×236
D (105)2×238

The correct answer is:
(105)2×238

Solution:
∵A=a11a12a130a22a2300a33⇒A=a11a22a33

⇒|A|=J7,3-J4,3·J8,3-J5,3·J9,3-J6,3

=∫01/2x7-x4x3-1dx·∫01/2x8-x5x3-1dx·∫01/2x9-x6x3-1dx

=∫01/2x4dx·∫01/2x5dx·∫01/2x6dx

=1(210)·(2)18

Now

adjA-1=A-12=1|A|2=(210)·2182=(105)2·(2)38

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let ∑k=110fa+k=16210-1, where the function f satisfies fx+y=fxfy for all natural numbers x, y and f1=2. Then the natural number 'a' is:

Options:
A 3
B 16
C 4
D 2

The correct answer is:
3

Solution:
Given f1=2 and fx+y=fx·fy

Put x=y=1

⇒f2=f1·f1=22

Now, put x=1, y=2

⇒f3=f1·f2=23

Now, put x=1, y=3

⇒f4=f1·f3=24

Similarly, f10=210

Now ∑k=110 fa+k=16210-1

⇒∑k=110fafk=16210-1

⇒fa∑k=110fk=16210-1

⇒faf1+f2…+f10=16210-1

⇒fa21+22+23+…+210=16210-1

Using the sum of n terms of a geometric progression, i.e. a+ar+ar2+ar3+...+arn-1=arn-1r-1, we get

fa2210-12-1=16210-1

⇒fa=8=23

Also, we know that f3=8

⇒fa=f3

⇒a=3.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ1=3λ2 then :

Options:
A K1>K23
B K1<K23
C K1=K23
D K2=K13

The correct answer is:
K1<K23

Solution:
According to Einstein's photoelectric equation
h⁡cλ=ϕ0+KEmax, where ϕ0 is a work function of a metal
∴ K⁡ 1 = h⁡ c λ 1 - ϕ 0 … i
K⁡ 2 = h⁡ c λ 2 - ϕ 0 … i i
orK1-K2=hc1λ1-1λ2
=hc13λ2-1λ2=-2hc3λ2Givenλ1=3λ2
= - 2 3 K⁡ 2 + ϕ 0 Using (ii)
or⁡ K⁡ 1 = K⁡ 2 - 2 3 K⁡ 2 - 2 3 ϕ 0 = K⁡ 2 3 - 2 3 ϕ 0
or K⁡ 1 < K⁡ 2 3

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let k and K be the minimum and the maximum values of the function fx=1+x0.61+x0.6 in 0, 1, respectively, then the ordered pair (k, K) is equal to:

Options:
A 2-0.4,1
B 2-0.6, 1
C 2-0.4,20.6
D 1,20.6

The correct answer is:
2-0.4,1

Solution:
fx= 1+x0.61+x0.6
f′x= 0.61+x-0.4 1+x0.6-0.6 x0.41+x0.61+x0.62
=0.6 1+x-0.4 1+x0.6-x-0.4 1+x21+x0.62=0.6x0.4-11+x0.41+x0.62
Clearly it is always negative in [0, 1]
minimum value would be at x=1.
(1+1)0.61+10.6= 20.62=2-0.4=k
& maximum values would be at x=0
(1+0)0.61+00.6=11=K
Hence 2-0.4,1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (11 Apr Online)
Let k and m be positive real numbers such that the function fx=3x2+kx+1,0<x<1mx2+k2,x≥1 is differentiable for all x>0. Then 8f'(8)f'18 is equal to

Options:

The correct answer is:
309

Solution:
Since, fx=3x2+kx+1; 0<x<1mx2+k2; x≥1 is differentiable at x=1, so function must be continuous at x=1, hence

LHL=RHL

3+k2=m+k2

⇒k2-k2+m-3=0 ....1

And,

f'x=6x+k2x+1; 0<x<12mx; x>1

So,

f'1-=f'1+

⇒6+k22=2m

⇒m=3+k42 ...2

Putting in 1, we get

k2-k2+3+k42-3=0

⇒k2-2-142k=0

⇒k2-728k=0

⇒kk-728=0

⇒k=728

So,

m=1+k122=1+72962=10396

Hence,

f'x=6x+7216x+1; 0<x<1103x16; x>1

So,

f'8f'18=103×816×168+721698

⇒f'8f'18=1032×34

⇒8f'8f'18=309

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let k be a non - zero real number. If fx=(ex−1)2sin xklog 1+x4,x ≠ 012                                       ,x = 0  is a continuous function at x=0, then the value of k is

Options:
A 2
B 4
C 3
D 1

The correct answer is:
3

Solution:
If the function is continuous then
limx→0fx=f0
limx→0(ex−1)2sinxk.log1+x4=12
⇒ limx→0ex-1x2sinxkk . xk . log1+x44. x4 = 12
⇒ 11k .14=12
⇒ 4k=12
⇒ k=3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2015 (11 Apr Online)
Let $k$ be a positive real number and let $\begin{aligned} A & =\left[\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1\end{array}\right] \text { and } \\ B & =\left[\begin{array}{ccc}0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0\end{array}\right]\end{aligned}$
If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then $[k]$ is equal to
[Note : adj $M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$ ].

Options:

The correct answer is:
4

Solution:
$|A|=(2 k+1)^3,|B|=0$
But $\operatorname{det}(\operatorname{adj} A)=\operatorname{det}(\operatorname{adj} B)=10^6$
$$
\begin{aligned}
& \Rightarrow(2 k+1)^6=10^6 \\
& \Rightarrow k=\frac{9}{2} \Rightarrow[k]=4
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
Let k be an integer such that the triangle with vertices k,-3k, 5, k and -k, 2 has area 28 sq. units. Then the orthocenter of this triangle is at the point:

Options:
A 2, -12
B 1,34
C 1, -34
D 2,12

The correct answer is:
2,12

Solution:
âââââââ

Given, Area of the triangle âABC=28⇒12kk-2+52+3k-k-4k=28

⇒kk-2+52+3k-k-4k=56
⇒ 5k2+13k−46=0

⇒ k=2 and k=-235 (Not possible)
Altitude from A : x=2
Altitude from B : y-2=12x-5
So, their point of intersection is H2,12

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2017 (02 Apr)