datavorous/entrance-exam-dataset huggingface dataset:
Let a line y=mxm>0, intersect the parabola, y2=x, at a point P, other than the origin. Let the tangent to it a P, meet the x-axis at the point Q. If area ΔOPQ=4 square unit, then m is equal to

Options:

The correct answer is:
0.5

Solution:
Let the equation of tangent at P is 2ty=x+t2
So, Q-t2,0

As given, area ΔOPQ=4 square unit, so
12001t2t1-t201=±4
t3=±8
t=2 t>0

So, m=tt2=12.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let an=â«0Ï21-cosâ¡2nx1-cosâ¡2xdx, then a1,a2,a3,...... are in

Options:
A Arithmetic Progression
B Geometric Progression
C Harmonic Progression
D Arithmetic Geometric Progression

The correct answer is:
Arithmetic Progression

Solution:
an+2+an-2an+1=

∫0π21-cos⁡2n+2x1-cos⁡2xdx+∫0π21-cos⁡2nx1-cos⁡2xdx-2∫0π21-cos⁡2n+1x1-cos⁡2xdx
=∫0π/21-cos⁡2n+4x+1-cos⁡2nx-2+2cos⁡2n+2x1-cos⁡2xdx
=∫0π22cos⁡2n+2x-cos⁡2n+4x+cos⁡2nx1-cos⁡2xdx
=∫0π22cos⁡2n+2x-2cos⁡2n+2xcos⁡2x1-cos⁡2xdx
=∫0π/22cos⁡2n+2x1-cos⁡2x1-cos⁡2xdx
=2sin⁡2n+2x2n+10π2=1n+1sin⁡n+1π-sin⁡0=0
⇒an+2+an=2an+1
⇒ They are in A.P.

Topic: Mathematics, Definite Integration, JEE Main
Let $A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $B_n=1-A_n$. Find a least odd natural number $n_0$, so that $B_n>A_n, \forall n \geq n_0$.

Options:

The correct answer is:
7

Solution:
$B_n=1-A_n>A_n \Rightarrow A_n < \frac{1}{2}$ Now, $A_n=\frac{\frac{3}{4}\left(1-\left(-\frac{3}{4}\right)^n\right)}{1+\frac{3}{4}} < \frac{1}{2} \Rightarrow\left(-\frac{3}{4}\right)>-\frac{1}{6}$
Obviously, it is true for all even values of $n$. But, for
$$
\begin{aligned}
n & =1, & \frac{3}{4} & < \frac{1}{6} \\
n & =3, & \frac{27}{64} & < \frac{1}{6} \\
n & =5, & \frac{243}{1024} & >\frac{1}{6}
\end{aligned}
$$
which is true for $n=7$.
Obviously, $n_0=7$

Topic: Mathematics, Sequences and Series, JEE Main
Let $A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $B_n=1-A_n$. Then, the least odd natural number $p$, so that $B_n>A_n$, for all $n \geq p$ is

Options:
A
5

B
7

C
11

D
9

The correct answer is:
7

Solution:
$$
A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n
$$
Which is a G.P. with $a=\frac{3}{4}, r=\frac{-3}{4}$ and number of terms $=n$
$$
\begin{aligned}
\therefore A_n &=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{1-\left(\frac{-3}{4}\right)}=\frac{\frac{3}{4} \times\left(1-\left(\frac{-3}{4}\right)^n\right)}{\frac{7}{4}} \\
\Rightarrow A_n=\frac{3}{7}\left[1-\left(\frac{-3}{4}\right)^n\right] \\
\text { As, } B_n=1-A_n
\end{aligned}
$$
For least odd natural number $\mathrm{p}$, such that $B_n>A_n$
$$
\Rightarrow 1-A_n>A_n \Rightarrow 1>2 \times A_n \Rightarrow A_n < \frac{1}{2}
$$
From eqn. (1), we get
$$
\begin{aligned}
&\frac{3}{7} \times\left[1-\left(\frac{-3}{4}\right)^n\right] < \frac{1}{2} \Rightarrow 1-\left(\frac{-3}{4}\right)^n < \frac{7}{6} \\
&\Rightarrow 1-\frac{7}{6} < \left(\frac{-3}{4}\right)^n \Rightarrow \frac{-1}{6} < \left(\frac{-3}{4}\right)^n
\end{aligned}
$$
As $n$ is odd, then $\left(\frac{-3}{4}\right)^n=-\frac{3^n}{4}$
So $\frac{-1}{6} < -\left(\frac{3}{4}\right)^n \Rightarrow \frac{1}{6}>\left(\frac{3}{4}\right)^n$
$$
\log \left(\frac{1}{6}\right)=n \log \left(\frac{3}{4}\right) \Rightarrow 6.228 < n
$$
Hence, $n$ should be 7 .

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Let an=16,4,1,…… be a geometric sequence. The value of ∑n=1∞Pnn, where Pn is the product of the first n terms, is equal to.

Options:
A 8
B 16
C 32
D 64

The correct answer is:
32

Solution:
For the G.P. a,ar,ar2,………
Pn=aarar2…..arn-1
=anrnn-12
⇒Pnn=arn-12
⇒∑n=1∞Pnn=a1+r+r+rr+.……
=a1-r
Given, a=16 and r=14
=161-14=32

Topic: Mathematics, Sequences and Series, JEE Main
Let $A=\{n \in[100,700] \cap \mathbb{N}: n$ is neither a multiple of 3 nor a multiple of 4$\}$. Then the number of elements in $A$ is

Options:
A 290
B 280
C 300
D 310

The correct answer is:
300

Solution:
$\begin{aligned} & \mathrm{n}(3) \Rightarrow \text { multiple of } 3 \\ & 102,105,108, \ldots . ., 699 \\ & \mathrm{~T}_{\mathrm{n}}=699=102+(\mathrm{n}-1)(3) \\ & \mathrm{n}=200 \\ & \mathrm{n}(3)=200 \\ & \because \mathrm{n}(4) \Rightarrow \text { multiple of } 4\end{aligned}$
$\begin{aligned} & 100,104,108, \ldots ., 700 \\ & T_n=700=100+(n-1)(4) \\ & n=151 \\ & n(4)=151 \\ & n(3 \cap 4) \Rightarrow \text { multiple of } 3 \& 4 \text { both } \\ & 108,120,132, \ldots ., 696 \\ & T_n=696=108+(n-1)(12) \\ & n=50 \\ & n(3 \cap 4)=50 \\ & n(3 \cup 4)=n(3)+n(4)-n(3 \cap 4) \\ & \quad=200+151-50 \\ & \quad=301 \\ & n(\overline{3 \cup 4})=\text { Total }-n(3 \cup 4)=\text { neither a multiple } \\ & \text { of } 3 \text { nor a multiple of } 4 \\ & =601-301=300\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let <an> be a sequence such that a1+a2+...+an=n2+3nn+1n+2. If 28∑k=1101ak=p1 p2 p3 ... pm, where p1, p2, ... pm are the first m prime numbers, then m is equal to

Options:
A 5
B 8
C 6
D 7

The correct answer is:
6

Solution:
Given,

Sn=∑i=1nai=n2+3nn+1n+2

Now we know that,

an=Sn-Sn-1

⇒an=n2+3nn+1n+2-n-12+3n-1nn+1

⇒an=4nn+1n+2

Now solving, ∑k=1101ak=14∑k=110kk+1k+2

⇒∑k=1101ak=116∑k=110kk+1k+2k+3-k-1kk+1k+2

⇒∑k=1101ak=1161·2·3·4.-0+2·3·4·5-1·2·3·4....+10·11·12·13-9·10·11·12

⇒∑k=1101ak=11610·11·12·13-0

⇒∑k=1101ak=125·11·3·13

⇒ 28∑k=1101ak=28×5×11×3×132

⇒28∑k=1101ak=2·3·5·7·11·13

Hence, there are six prime numbers in multiplications,

So, m=6

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let an be nth term of the series 5+8+14+23+35+50+.......and Sn=∑k=1nak. Then S30-a40 is equal to

Options:
A 11310
B 11260
C 11290
D 11280

The correct answer is:
11290

Solution:
The given series is an, then we can write as

an=5+8+14+23+............+anan= 5+8+14+23+......+an-1+an

Subtracting above equations, we get

0=5+3+6+9+....+an-an-1-an

⇒an=5+n-122×3+n-1-13

⇒an=5+n-123n

⇒an=123n2-3n+10

So,

⇒a40=123×402-3×40+10

⇒a40=124800-120+10

⇒a40=2345

⇒S30=∑n=130an

⇒S30=123∑n=130n2-3∑n=130n+∑n=13010

=1233031616-330312+10×30

⇒S30=13635

⇒S30-a40=13635-2345=11290

Therefore, the required value is 11290.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let an be the nth term of a G.P. of positive terms. If ∑n=1100a2n+1=200 and ∑n=1100a2n=100, then ∑n=1200an is equal to:

Options:
A 300
B 225
C 175
D 150

The correct answer is:
150

Solution:
Let GP is a,ar,ar2…

∑n=1100a2n+1=a3+a5+…a201=200⇒ar2(r200-1)r2-1=200 …1

∑n=1100a2n=a2+a4+…a200=100=arr200-1r2-1=100 …2

Form 1 and 2, r=2

adding both,

⇒a2+a3+…a200+a201=300⇒ra1+…a200=300

∑n=1200300r=150

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 .

Question: The value of $b_{6}$ is

Options:
A 7
B 8
C 9
D 11

The correct answer is:
8

Solution:
$\because a_{n}=$ number of all $n$ digit +ve integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 .

and $b_{n}=$ number of such $n$ digit integers ending with 1 $c_{n}=$ number of such $n$ digit integers ending with 0

Clearly, $a_{n}=b_{n}+c_{n}\left(\because a_{n}\right.$ can end with 0 or 1)

Also $b_{n}=a_{n-1}$ and $c_{n}=a_{n-2}[\because$ if last digit is 0 , second last has to be 1]

$\therefore$ We get $a_{n}=a_{n-1}+a_{n-2}, n \geq 3$

Also $a_{1}=1, a_{2}=2$,

Now by this recurring formula, we get

$\begin{aligned} a_{3} &=a_{2}+a_{1}=3 \\ a_{4} &=a_{3}+a_{2}=3+2=5 \\ a_{5} &=a_{4}+a_{3}=5+3=8 \\ \text { Also } b_{6} &=a_{5}=8 \end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 .

Question: Which of the following is correct?

Options:
A $a_{17}=a_{16}+a_{15}$
B $c_{17} \neq c_{16}+c_{15}$
C $b_{17} \neq b_{16}+c_{16}$
D $a_{17}=c_{17}+b_{16}$

The correct answer is:
$a_{17}=a_{16}+a_{15}$

Solution:
By recurring formula, $a_{17}=a_{16}+a_{15}$ is correct

Also $c_{17} \neq c_{16}+c_{15}$

$\Rightarrow a_{15} \neq a_{14}+a_{13}\left(\because c_{n}=a_{n-2}\right)$

$\therefore$ Incorrect

Similarly, other parts are also incorrect.

Topic: Mathematics, Permutation Combination, JEE Main
Let ann=0∞ be a sequence such that a0=a1=0 and an+2=2an+1-an+1 for all n≥0. Then, ∑n=2∞an7n is equal to

Options:
A 6343
B 7216
C 8343
D 49216

The correct answer is:
7216

Solution:
Given an+2=2an+1-an+1

Dividing by 7n+2, we get

an+27n+2=27·an+17n+1-149·an7n+17n+2

So, ∑n=2∞an+27n+2=27∑n=2∞an+17n+1-149∑n=2∞an7n+∑n=2∞17n+2

Let
∑n=2∞an7n=P

⇒P-a373-a272=27P-a272-149P+17411-17

∵a2=1 & a3=3

⇒P1-27+149=373+172-273+16·73

⇒36P49=142 ⇒P=7216

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Let $a_n=\frac{10^n}{n !}$ for $n=1,2,3, \ldots$ then the greatest value of $n$ for which $a_n$ is the greatest is

Options:
A 11
B 20
C 10
D 8

The correct answer is:
10

Solution:
Given, $a_n=\frac{10^n}{n !}, n=1,2,3, \ldots$, here we see that when we increase the value of $n$ like as $1,2,3, \ldots$ the value of $a_n$ increases but when we reach at $n=9$ or 10 the value of $a_n$ remain unchanged, $i e$, minor difference in after decimal places and when we cross the value $n=10 \mathrm{ie}$, $n=11$, then we see that the value of $a_n$ is monotonically decreasing.
Hence, $a_n$ have its maximum value at $n=9$ or 10.

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2010
Let $a_n=\frac{10^n}{n !}$ for $n=1,2,3, \ldots$ then the greatest value of $n$ for which $a_n$ is the greatest is

Options:
A 11
B 20
C 10
D 8

The correct answer is:
10

Solution:
Given, $a_n=\frac{10^n}{n !}, n=1,2,3, \ldots$, here we see that when we increase the value of $n$ like as $1,2,3, \ldots$ the value of $a_n$ increases but when we reach at $n=9$ or 10 the value of $a_n$ remain unchanged, $i e$, minor difference in after decimal places and when we cross the value $n=10 \mathrm{ie}$, $n=11$, then we see that the value of $a_n$ is monotonically decreasing.
Hence, $a_n$ have its maximum value at $n=9$ or 10.

Topic: Mathematics, Permutation Combination, JEE Main
Let $a_n=\frac{10^n}{n !}$ for $n=1,2,3, \ldots$ then the greatest value of $n$ for which $a_n$ is the greatest is

Options:
A 11
B 20
C 10
D 8

The correct answer is:
10

Solution:
Given, $a_n=\frac{10^n}{n !}, n=1,2,3, \ldots$, here we see that when we increase the value of $n$ like as $1,2,3, \ldots$ the value of $a_n$ increases but when we reach at $n=9$ or 10 the value of $a_n$ remain unchanged, $i e$, minor difference in after decimal places and when we cross the value $n=10 \mathrm{ie}$, $n=11$, then we see that the value of $a_n$ is monotonically decreasing.
Hence, $a_n$ have its maximum value at $n=9$ or 10.

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2010
Let an,n≥1, be an arithmetic progression with first term 2 and common difference 4 . Let Mn be the average of the first n terms. Then the sum ∑n=110Mn is

Options:
A 110
B 335
C 770
D 1100

The correct answer is:
110

Solution:
The sum of first n,n≥1 terms of arithmetic progression with first term 2 and common difference 4, is
Sn=n2[4+(n-1)4]=2n2
So, the average of the first n terms

Mn=Snn=2n

Now, ∑n=110 Mn=2∑n=110 n

=2×10×112=110

Topic: Mathematics, Sequences and Series, JEE Main
Let $A=N \times N$ and ${ }^*$ be the binary operation on A defined by $(a, b) *(c, d)=(a+c, b+d)$.
Show that ${ }^*$ is commutative and associative. Find the identity element for ${ }^*$ on $A$, if any.

Options:

None

Solution:
We are given $\mathrm{f}: \mathrm{W} \rightarrow \mathrm{W}$, defined as $\mathrm{f}(\mathrm{n})$
$= \begin{cases}n-1, & n \text { is odd } \\ n+1, & n \text { is even }\end{cases}$
Let $f\left(n_1\right)=f\left(n_2\right)$. If $n$, is odd and $n_2$ is even then
$\mathrm{n}_1-1=\mathrm{n}_2+1$, i.e. $\mathrm{n}_1-\mathrm{n}_2=2$
Which is not possible as $\mathrm{n}_1$ is odd and $\mathrm{n}_2$ is even. When $\mathrm{n}_1$ and $\mathrm{n}_2$ both are odd.
then $\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right) \Rightarrow \mathrm{n}_1-1=\mathrm{n}_2-1 \Rightarrow \mathrm{n}_1=\mathrm{n}_2$
When $\mathrm{n}_1$ and $\mathrm{n}_2$ both are even $\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)=\mathrm{n}_1+1$
$=\mathrm{n}_2+1=\mathrm{n}_1=\mathrm{n}_2$
This shows $f$ is one-one. Any odd number $2 r+1$ in co-domain is the image of $2 r$ and any even number $2 r$ in the codomain $W$ in the image of $2 r+1$.

Topic: Mathematics, Relations and Functions (Class 12)
Let A=n∈N∣n2≤n+10,000, B=3k+1∣k∈N and C=2k∣k∈N, then the sum of all the elements of the set A∩B-C is equal to ________.

Options:

The correct answer is:
832

Solution:
We have, A=n∈N∣n2≤n+10,000,B=3k+1∣k∈N and C=2k∣k∈N,

B≡4,7,10,13,16,19,……

C≡2,4,6,8,10,12,14,16,20,……

B-C≡7,13,19,…97,…

Now, n2-n≤100×100

⇒nn-1≤100×100

⇒A=1,2,…,100

So, A∩B-C=7,13,19,…,97

Let Sn=7+13+19+…+97

Clearly, the terms are in arithmetic progression. Here, a=7 and d=13-7=6 and l=97

⇒l=a+n-1d=97 where n is the number of terms.

⇒7+n-16=97

⇒n-16=90

⇒n-1=15

⇒n=16

Hence, sum Sn=7+13+19+…+97

Sn=n2a+l

Sn=1627+97

=832.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let An=∫tann⁡xdx, ∀n∈N. If A10+A12=tanm⁡xm+λ (where λ is an arbitrary constant), then the value of m is equal to

Options:
A 10
B 11
C 12
D 13

The correct answer is:
11

Solution:
A10+A12=â«tan10â¡xdx+â«tan12â¡xdx
=â«tan10â¡x+tan12â¡xdx
=â«tan10â¡x1+tan2â¡xdx
=â«tan10â¡xâsec2â¡xdx
Let, tanâ¡x=tâsec2â¡xdx=dt
â´A10+A12=â«t10dt=t1111+Î»
=tan11â¡x+Î»11

Topic: Mathematics, Indefinite Integration, JEE Main
Let $\mathrm{A}$, other than $\mathrm{I}$ or $-\mathrm{I}$, be a $2 \times 2$ real matrix such that $\mathrm{A}^2=\mathrm{I}$, I being the unit matrix. Let $\operatorname{Tr}(\mathrm{A})$ be the sum of diagonal elements of A.
Statement-1: $\operatorname{Tr}(\mathrm{A})=0$
Statement-2: $\operatorname{det}(\mathrm{A})=-1$

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 not a correct explanation for Statement-1.

Solution:
$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$$
\begin{aligned}
& {\left[\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& b(a+d)=0, b=0 \text { or } a=-d \\
& c(a+d)=0, c=0 \text { or } a=-d \\
& a^2+b c=1, b c+d^2=1
\end{aligned}
$$
' $a$ ' and ' $d$ ' are diagonal elements $a+d=0$ statement- 1 is correct.
Now, $\operatorname{det}(A)=a d-b c$
Now, from (3) $a^2+b c=1$ and $d^2+b c=1$
So, $a^2-d^2=0$
Adding $a^2+d^2+2 b c=2$
$$
=(a+d)^2-2 a d+2 b c=2
$$
or $0-2(a d-b c)=2$
So, $a d-b c=1 \Rightarrow \operatorname{det}(A)=-1$
So, statement $-2$ is also true.
But statement $-2$ is not the correct explanation of statement-I

Topic: Mathematics, Matrices, JEE Main, JEE Main 2013 (23 Apr Online)
Let a parabola P be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0,0) to the parabola P which meet P at S and R, then the area (in sq. units) of ΔSOR is equal to :

Options:
A 162
B 16
C 32
D 82

The correct answer is:
16

Solution:
From the diagram, clearly RS is latus-rectum

∵ VF=4-22+0-02=2=a

∴ RS=4a=8

Now OF=2a=4

So, the area of triangle ORS=12×4×8=16

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let a plane P contain two lines r→=i^+λi^+j^, λ∈R and r→=-j^+μj^-k^, μ∈R. If Qα, β, γ is the foot of the perpendicular drawn from the point M1, 0, 1 to P, then 3α+β+γ equals .......

Options:

The correct answer is:
5

Solution:
Normal vector to the plane containing to given lines n→=i^j^k^11001-1=-i^+j^+k^

Direction ratios of normal vector =-1, 1, 1.

Equation of plane will be -1x-1+1y-0+1z-0=0

⇒ x-y-z-1=0

If x, y, z is foot of perpendicular drawn from M1, 0, 1 on the plane, then

⇒x-11=y-0-1=z-1-1=-1-0-1-13

⇒x=43, y=-13, z=23

⇒α+β+γ=43-13+23=53

⇒3α+β+γ=5.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let a plane P contain two lines r→=i^+λi^+j^, λ∈R and r→=-j^+μj^-k^, μ∈R. If Qα, β, γ is the foot of the perpendicular drawn from the point M1, 0, 1 to P, then 3α+β+γ equals .......

Options:

The correct answer is:
5

Solution:
Normal vector to the plane containing to given lines n→=i^j^k^11001-1=-i^+j^+k^

Direction ratios of normal vector =-1, 1, 1.

Equation of plane will be -1x-1+1y-0+1z-0=0

⇒ x-y-z-1=0

If x, y, z is foot of perpendicular drawn from M1, 0, 1 on the plane, then

⇒x-11=y-0-1=z-1-1=-1-0-1-13

⇒x=43, y=-13, z=23

⇒α+β+γ=43-13+23=53

⇒3α+β+γ=5.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let a plane P pass through the point 3,7,-7 and contain the line, x-2-3=y-32=z+21. If distance of the plane P from the origin is d, then d2 is equal to

Options:

The correct answer is:
3

Solution:
The required plane passes through the point 3,7,-7.

Assume, the equation of the plane is ax-3+by-7+cz+7=0

The line x-2-3=y-32=z+21 lies on the required plane.

So, we can say that

a2-3+b3-7+c-2+7=0

⇒-a-4b+5c=0 ...1

And -3×a+2×b+1×c=0 ...2

Use cross-multiplication to solve the equations 1 & 2

a-4×1-2×5=b5×-3+1×1=c-1×2--4×-3

⇒a-14=b-14=c-14=k (Assume)

So, a=k, b=k, c=k

Put the values of a,b,c in the required equation of plane we get x-3+y-7+z+7=0

⇒x+y+z=3

So, the distance of the plane from the origin is 312+12+12=3

Hence, d2=32=3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let a point P be such that its distance from the point 5,0 is thrice the distance of P from the point -5,0. If the locus of the point P is a circle of radius r, then 4r2 (in the nearest integer) is equal to __________.

Options:

The correct answer is:
56

Solution:
Let point is h,k

So, h-52+k2=3h+52+k2

8x2+8y2+100x+200=0

x2+y2+252x+25=0

r2=25242-25

4r2=2524-100

4r2=156.25-100

4r2=56.25

After round off 4r2=56

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let A=θ∈R:13sinθ+23cosθ2=13sin2θ+23cos2θ. Then

Options:
A A∩0,π is an empty set
B A∩0,π has exactly one point
C A∩0,π has exactly two points
D A∩0,π has more than two points

The correct answer is:
A∩0,π has exactly one point

Solution:
Given trigonometric relation is

13sinθ+23cosθ2=13sin2θ+23cos2θ

⇒ 19sin2θ+49cos2θ+49sinθcosθ

=13sin2θ+23cos2θ

⇒ 29sin2θ+29cos2θ-49sinθcosθ=0

⇒ sin2θ+cos2θ-2sinθcosθ=0

⇒ sin2θ=1

⇒ 2θ=2nπ+π2, n∈I

⇒ θ=nπ+π4, n∈I

∴ A=θ∈R:θ=nπ+π4,n∈I

∴ A∩0,π=π4

∴ A∩0,π has exactly one point.

Topic: Mathematics, Trigonometric Equations, JEE Main
Let $A=R-\{3\}$ and $B=R-\{1\}$. Let $f: A \rightarrow B$ defined as $f(x)$ $=\frac{\mathrm{x}-2}{\mathrm{x}-3} \forall \mathrm{x} \in \mathrm{A}$. Then show that $\mathrm{f}$ is bijective.

Options:

None

Solution:
One-one/many-one : Let $\mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R}-\{3\}$ are the elements such that
$f\left(x_1\right)=f\left(x_2\right)$ : then $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \frac{\mathrm{x}_1-2}{\mathrm{x}_1-3}=\frac{\mathrm{x}_2-2}{\mathrm{x}_2-3}$ $\Rightarrow\left(\mathrm{x}_1-2\right)\left(\mathrm{x}_2-3\right)=\left(\mathrm{x}_2-2\right)\left(\mathrm{x}_1-3\right)$ $\Rightarrow \mathrm{x}_1 \mathrm{x}_2-2 \mathrm{x}_2-3 \mathrm{x}_1+6=\mathrm{x}_2 \mathrm{x}_1-2 \mathrm{x}_1-3 \mathrm{x}_2+6$ $\Rightarrow-2 \mathrm{x}_2-3 \mathrm{x}_1=-2 \mathrm{x}_1-3 \mathrm{x}_2$ $\Rightarrow \mathrm{x}_2=\mathrm{x}_1, \therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2$ $\Rightarrow \mathrm{f}$ is one-one function
Onto/Into : Let $\mathrm{y} \in \mathrm{R}-\{1\}$ (co-domain)
Then one element $\mathrm{x} \in \mathrm{R}-\{3\}$ in domain is such that
$$
f(x)=y \Rightarrow \frac{x-2}{x-3}=y \Rightarrow x-2=x y-3 y
$$
$$
\Rightarrow x=\left(\frac{3 y-2}{y-1}\right)
$$
$\therefore$ The pre-image of each element of co-domain $\mathrm{R}-\{1\}$ exists in domain $\mathrm{R}-\{3\}$.
$\Rightarrow \mathrm{f}$ is onto

Topic: Mathematics, Relations and Functions (Class 11)
Let a∈R and let α,β be the roots of the equation x2+6014x+a=0. If α4+β4=-30, then the product of all possible values of a is _____ .

Options:

The correct answer is:
45

Solution:
Since, α,β be the roots of the equation x2+6014x+a=0, therefore

α+β=-6014

αβ=a

Given:

α4+β4=-30

⇒α2+β22-2α2β2=-30

⇒α+β2-2αβ2-2a2=-30

⇒6012-2a2-2a2=-30

⇒ 60+4a2-4a×6012-2a2=-30

⇒ 2a2-4·6012a+90=0

⇒a2-2·6012a+45=0

Product of roots=45

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let a,Î»,Î¼âR. Consider the system of linear equations ax+2y=Î» and 3x-2y=Î¼. Which of the following statement is INCORRECT?

Options:
A If a=-3, then the system has infinitely many solutions for all values of λ and μ
B If a≠-3, then the system has a unique solution for all values of λ and μ
C If λ+μ=0, then the system has infinitely many solutions for a=-3
D If λ+μ≠0, then the system has no solution for a=-3

The correct answer is:
If a=-3, then the system has infinitely many solutions for all values of λ and μ

Solution:
ax+2y=λ
3x-2y=μ
For a=-3, above lines will be parallel or coincident
Also parallel for λ+μ≠0 and coincident λ+μ=0
and if a≠-3, lines are intersecting ⇒ unique solution.

Topic: Mathematics, Determinants, JEE Main
Let ar=r 4Cr , br=4-r 4Cr, Ar=ar23br and A=âr=04Ar, then the value of A is equal to

Options:

The correct answer is:
874

Solution:
A=∑r=04Ar=∑r=04r 4Cr∑r=042∑r=043∑r=044-r 4Cr
Now, ∑r=04r 4Cr=∑r=044-r 4Cr=4×23=25
A=32101532
A=1024-150=874

Topic: Mathematics, Matrices, JEE Main
Let a random variable X has a binomial distribution with mean 8 and variance 4. If PXâ¤2=k216, then the value of k is equal to

Options:
A 121
B 1
C 17
D 137

The correct answer is:
137

Solution:
Let n be the number of times the experiment is done, p is the probability of the success and q is the probability of the failure.

∴ np=8 (mean), npq=4 (variance)

⇒q=12⇒p=12

⇒n=16

⇒PX≤2=pX=0+pX=1+pX=2

⇒PX≤2=16C0p0q16+16C1p1q15+16C2p2q14

⇒PX≤2=1216+161216+120×1216

⇒PX≤2=1216137=137216=k216

∴ k=137

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Let a random variable X has a binomial distribution with mean 8 and variance 4. If PXâ¤2=k216, then the value of k is equal to

Options:
A 121
B 1
C 17
D 137

The correct answer is:
137

Solution:
Let n be the number of times the experiment is done, p is the probability of success and q is the probability of failure
∴ np=8 (mean), npq=4 (variance)
⇒q=12⇒p=12
⇒n=16
PX≤2=pX=0+pX=1+pX=2
= 16C0p0q16 + 16C1p1q15 + 16C2p2q14
=1216+161216+ 120×1216
=1216137=137216=k216
∴ k=137

Topic: Mathematics, Probability, JEE Main