datavorous/entrance-exam-dataset huggingface dataset:
Let $x \in(0,1)$. The set of all $x$ such that $\sin ^{-1} x>\cos ^{-1} x$, is the interval:

Options:
A
$\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$

B
$\left(\frac{1}{\sqrt{2}}, 1\right)$

C
$(0,1)$

D
$\left(0, \frac{\sqrt{3}}{2}\right)$

The correct answer is:
$\left(\frac{1}{\sqrt{2}}, 1\right)$

Solution:
Given $\sin ^{-1} x>\cos ^{-1} x$ where $x \in(0,1)$
$$
\begin{aligned}
&\Rightarrow \sin ^{-1} x>\frac{\pi}{2}-\sin ^{-1} x \\
&\Rightarrow 2 \sin ^{-1} x>\frac{\pi}{2} \Rightarrow \sin ^{-1} x>\frac{\pi}{4} \\
&\Rightarrow x>\sin \frac{\pi}{4} \Rightarrow x>\frac{1}{\sqrt{2}}
\end{aligned}
$$
Maximum value of $\sin ^{-1} x$ is $\frac{\pi}{2}$
So, maximum value of $x$ is 1 . So, $x \in\left(\frac{1}{\sqrt{2}}, 1\right)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (25 Apr Online)
Let x0 be the point of local maxima of fx=a→·b→×c→, wherea→=xi^-2j^+3k^, b→=-2i^+xj^-k^ and c→=7i^-2j^+xk^. Then the value of a→·b→+b→·c→+c→·a→ at x=x0 is:

Options:
A -4
B -30
C 14
D -22

The correct answer is:
-22

Solution:
fx=a→  b→  c→=x      −2     3−2     x     −17      −2     x

=xx2−2+2−2x+7+34−7x

=x3−2x−4x+14+12−21x

fx=x3−27x+26

f'x=3x2−27=3x−3x+3

so local maxima point is x0=-3

Now a→.b→+b→.c→+c→.a→=−2x−2x−3-14−2x−x+7x+4+3x=3x−13

at x=x0=−3

a→.b→+b→.c→+c→.a→=−9−13=−22

Topic: Mathematics, Vector Algebra, JEE Main
Let x0 be the point of local maxima of fx=a→·b→×c→, wherea→=xi^-2j^+3k^, b→=-2i^+xj^-k^ and c→=7i^-2j^+xk^. Then the value of a→·b→+b→·c→+c→·a→ at x=x0 is:

Options:
A -4
B -30
C 14
D -22

The correct answer is:
-22

Solution:
fx=a→  b→  c→=x      −2     3−2     x     −17      −2     x

=xx2−2+2−2x+7+34−7x

=x3−2x−4x+14+12−21x

fx=x3−27x+26

f'x=3x2−27=3x−3x+3

so local maxima point is x0=-3

Now a→.b→+b→.c→+c→.a→=−2x−2x−3-14−2x−x+7x+4+3x=3x−13

at x=x0=−3

a→.b→+b→.c→+c→.a→=−9−13=−22

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let $x \neq 0,|x| < \frac{1}{2}$ and $f(x)=1+2 x+4 x^2+8 x^3+\ldots$ Then, $f^{-1}(x)=$

Options:
A $\frac{x-1}{2 x}$
B $\frac{x-1}{2}$
C $\frac{x-1}{x}$
D $1-2 x$

The correct answer is:
$\frac{x-1}{2 x}$

Solution:
We have,
$$
f(x)=1+2 x+4 x^2+8 x^3+\ldots \text { and }|x| < \frac{1}{2}
$$
Let $f(x)=y$
$$
\begin{aligned}
y & =1+2 x+4 x^2+8 x^3+\ldots \\
& =1+(2 x)+(2 x)^2+(2 x)^3+\ldots \\
y & =\frac{1}{1-2 x}\left[\because \text { sum of infinite GP, } S_{\infty}=\frac{a}{1-r}\right]
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad y-2 x y=1 \quad \Rightarrow \quad 2 x y=y-1 \\
& \Rightarrow \quad x=\frac{y-1}{2 y} \quad \Rightarrow f^{-1}(y)=\frac{y-1}{2 y} \\
&
\end{aligned}
$$
Hence, $f^{\prime}(x)=\frac{x-1}{2 x}$.

Topic: Mathematics, Functions, JEE Main
Let x0,y0 be fixed real numbers such that x02+y02>1. If x, y are arbitrary real numbers such that x2+y2≤1, then the minimum value of x-x02+y-y02 is

Options:
A x02+y02-12
B x02+y02-1
C x0+y0-12
D x0+y02-1

The correct answer is:
x02+y02-12

Solution:
LetPx0,y0

Given x2+y2≤11$

Let any arbitrary point 8 (x, y).

PQ2=x-x02+y-y02PQ2=(OP-OQ)2
PQ2=(OP-OQ)2PQ2=x02+x02-12 [∵0Q=1]

∴ Mininimum value of PQ2 is

x02+y02-12

Topic: Mathematics, Circle, JEE Main
Let $X=\{1,2,3,4,5\}$. The number of different ordered pairs $(Y, Z)$ that can be formed such that $Y \subseteq X, Z$ $\subseteq \mathrm{X}$ and $\mathrm{Y} \cap \mathrm{Z}$ is empty, is

Options:
A
$5^2$

B
$3^5$

C
$2^5$

D
$5^3$

The correct answer is:
$3^5$

Solution:
$\mathrm{Y} \subseteq \mathrm{X}, \mathrm{Z} \subseteq \mathrm{X}$
Let $a \in X$, then we have following chances that
(1) $a \in Y, \quad a \in Z$
(2) $a \notin Y, \quad a \in Z$
(3) $a \in Y, \quad a \notin Z$
(4) $a \notin Y, \quad a \notin Z$
We require $Y \cap Z=\phi$
Hence (2), (3), (4) are chances for '$a$' to satisfy $Y \cap Z=\phi$.
$\therefore Y \cap Z=\phi$ has $3$ chances for $a$.
Hence for five elements of $\mathrm{X}$, the number of required chances is $3 \times 3 \times 3 \times 3 \times 3=3^5$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2012 (Offline)
Let $X=\{1,2,3\}$ and $Y=\{4,5\}$. Find whether the following subsets of $\mathrm{X} \times \mathrm{Y}$ are functions from $\mathrm{X}$ to $\mathrm{Y}$ or not.
(i) $\mathrm{f}=\{(1,4),(1,5),(2,4),(3,5)\}$
(ii) $g=\{(1,4),(2,4),(3,4)\}$
(iii) $\mathrm{h}=\{(1,4),(2,5),(3,5)\}$
(iv) $\mathrm{k}=\{(1,4),(2,5)\}$

Options:

None

Solution:
Given that, $X=\{1,2,3\}$ and $Y=\{4,5\}$
$$
X \times Y=\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}
$$
(i) $\mathrm{f}=\{(1,4),(1,5),(2,4),(3,5)\}$
$\mathrm{f}$ is not a function because $\mathrm{f}$ does not have a unique image.
(ii) Since $g=\{(1,4),(2,4),(3,4)\}$
$\mathrm{g}$ is a function as each element of the domain has unique image.
(iii) $\mathrm{h}=\{(1,4),(2,5),(3,5)\}$
It is clear that $h$ is a function.
(iv) $\mathrm{k}=\{(1,4),(2,5)\}$
$\mathrm{k}$ is not a function as 3 has not any image under the mapping.

Topic: Mathematics, Relations and Functions (Class 11)
Let x=12 and x=1 are the extreme points of fx=alogx+bxx+2cosπxπ, then

Options:
A a-b=2
B a+b=1
C ab=12
D ab=2

The correct answer is:
a-b=2

Solution:
f'x=ax+2bx-2sinπx
Now f'12=0⇒2a+b-2=0
& f'1=0⇒a+2b=0
⇒b=-23 and a=43

Topic: Mathematics, Application of Derivatives, JEE Main
Let X1, X2, X3.... are in arithmetic progression with a common difference equal to d which is a two digit natural number. y1, y2, y3.... are in geometric progression with common ratio equal to 16 . Arithmetic mean of X1,X2....Xn is equal to the arithmetic mean of y1,y2....yn which is equal to 5. If the arithmetic mean of X6,X7....Xn+5 is equal to the arithmetic mean of yP+1,yP+2....yP+n, then d is equal to

Options:

The correct answer is:
15

Solution:
Mean X1,X2…Xn=n22X1+n-1dn=5
Mean of y1,y2....yn=y116n-115n=5
2X1+n-1d=10 … 1
y116n-1=75n … 2
Mean of X6,X7....Xn+5= Mean of yP+1,yP+2,yP+n
n22X6+n-1dn=yP+116n-115n=y116P16n-115n
X6+n-12d=16P75n15n=5×16P
X1+5d+n-12d=5×16P
5-n-12d+5d+n-12d=5×16P⇒d=16P-1
∵d is 2 digit natural number ⇒P=1,d=15

Topic: Mathematics, Sequences and Series, JEE Main
Let x1,x2,x3,…..,x20 be in geometric progression with x1=3 and the common ration 12. A new data is constructed replacing each xi by xi-i2. If x is the mean of new data, then the greatest integer less than or equal to x is

Options:

The correct answer is:
142

Solution:
Given,

x1,x2,x3,…..,x20 are in G.P with common difference 12 and first term as 3

Now using the formula for sum of G.P we get,

∑i=120xi=31-12201-12=61-1220

Now new sum of G.P with new data will be=∑i=120xi-i2

=∑i=120xi2+i2-2xii

Now ∑i=120xi2=91-14201-14=121-1240

And ∑i=120i2=16×20×21×41=2870

And ∑i=120xii=3+2.312+3.3122+4.3123+…….AGP

=64-11218

So, mean of the new data will be,

x=12-12240+2870-64-1121820

⇒x=285820+-12240+66218×120

Now ⇒x=142.4+0.00001=142

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Let x1, x2, x3....xk be k observations and wi=axi+b for i=1, 2, 3.....k, where a and b are constants. If mean of xi is 52 and their standard deviation is 12 and mean of wi is 60 and their standard deviation is 15, then the value of a and b should be

Options:
A a=1.25,b=-5
B a=-1.25,b=5
C a=2.5,b=-5
D a=2.5,b=5

The correct answer is:
a=1.25,b=-5

Solution:
wi=axi+b
⇒w¯=ax¯+b
⇒60=a52+b …. (i)
S.D of w=aS.D of xi
15=a12
⇒a=±1.25 …. (ii)

From (i) and (ii), we get,
If a=1.25⇒60=65+b⇒b=-5
If a=-1.25⇒60=-65+b⇒b=125

Topic: Mathematics, Statistics, JEE Main
Let x1,x2,…,x11 be 11 distinct positive integers. If we replace the largest of these integers by the median of the other 10 integers, then

Options:
A the median remains the same
B the mean increases
C the median decreases

D the mean remains the same

The correct answer is:
the mean remains the same

Solution:
Let the given 11 distinct positive integers are in increasing order x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11, so x11
is largest of these integers and the median is x6. Now, median of first 10 numbers is
x6+x62=m( Let )
Now, we have to replace largest number x11 by m and then increasing order will be
x1,x2,x3,x4,x5,m,x6,x7,x8,x9,x10
∵ m<x6 as x5<x5+x62<x6
So, median decreases.

Topic: Mathematics, Statistics, KVPY, KVPY 2019 (SA)
Let x1,x2,…,x100 be in an arithmetic progression, with x1=2 and their mean equal to 200 . If yi=ixi-i,1≤i≤100, then the mean of y1,y2,…, y100 is

Options:
A 10100
B 10101.50
C 10049.50
D 10051.50

The correct answer is:
10049.50

Solution:
Given,

Mean of x1, x2..........,x100 is 200

So, the sum of observation will be,

∑xi=100×200

Now using the sum of A.P formula in above equation as all terms are in arithmetic progression we get,

1002x1+x100=100×200

⇒502+x100=100×200

⇒x100=398

⇒x1+99d=398

⇒d=4

Now, xi=2+(i-1)4=4i-2

So, yi=ixi-i=3i2-2i

Now finding mean we get,

y¯=1100∑yi

⇒y¯=1100∑3i2-2i

⇒y¯=11003×100×101×2016-2100×1012

⇒y¯=101×2012-101

⇒y¯=10049.5

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let $\mathrm{x}_1, \mathrm{x}_2, \ldots \ldots, \mathrm{x}_{\mathrm{n}}$ be $\mathrm{n}$ observations, and let $\overline{\mathrm{x}}$ be their arithematic mean and $\sigma^2$ be their variance.
Statement $1$: Variance of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n$ is $4 \sigma^2$.
Statement $2$: Arithmetic mean of $2 \mathrm{x}_1, 2 \mathrm{x}_2, \ldots . ., 2 \mathrm{x}_{\mathrm{n}}$ is $4 \overline{\mathrm{x}}$.

Options:
A
Statement $1$ is false, statement $2$ is true

B
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

C
Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$

D
Statement $1$ is true, statement $2$ is false

The correct answer is:
Statement $1$ is true, statement $2$ is false

Solution:
$\sigma^2=\sum \frac{x_i^2}{n}-\left(\sum \frac{x_i}{n}\right)^2$
Variance of $2 x_1, 2 x_2, \ldots . ., 2 x_n=\sum \frac{\left(2 x_i\right)^2}{n}-\left(\sum \frac{2 x_i}{n}\right)^2=4\left[\sum \frac{x_i^2}{n}-\left(\sum \frac{x_i}{n}\right)^2\right]=4 \sigma^2$ Statement $1$ is true.
A.M. of $2 x_1, 2 x_2, \ldots \ldots, 2 x_n=\frac{2 x_1+2 x_2+\cdots+2 x_n}{n}=2\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)=2 \bar{x}$
Statement $2$ is false.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2012 (Offline)
Let x1,x2…,xn be n observations such that ∑xi2=300 and ∑xi=90. Then a possible value of n among the following is

Options:
A 25
B 18
C 29
D 22

The correct answer is:
29

Solution:
Variance =Σxi2n-Σxin2
⇒Σxi2n≥Σxin2
⇒n≥Σxi2Σxi2⇒n≥8100300⇒n≥27

Topic: Mathematics, Statistics, JEE Main
Let x1, x2,...., xn be n observations such that âxi2=400 and âxi=40, then a possible value of n among the following is

Options:
A 5
B 1
C 2
D 3

The correct answer is:
5

Solution:
Since, variance is always greater than or equal to zero hence x 1 2 + x 2 2 + ... + x n 2 n â¥ xâ¡ 1 + x 2 + ... + x n n 2
â 4 0 0 n â¥ 1 6 0 0 n 2
â n â¥ 4

Topic: Mathematics, Statistics, JEE Main
Let $x_1, x_2, \ldots, x_n$ be $n$ observations such that $\sum x_i^2=400$ and $\sum x_i=80$. Then a possible value of $n$ among the following is

Options:
A $9$
B $12$
C $15$
D $18$

The correct answer is:
$18$

Solution:
Since, root mean square $\geq$ arithmetic mean
$\therefore \sqrt{\frac{\sum_{i=1}^n x_i^2}{n}} \geq \frac{\sum_{i=1}^n x_i}{n}=\sqrt{\frac{400}{n}} \geq \frac{80}{n} \Rightarrow n \geq 16$
Hence, possible value of $n=18$.

Topic: Mathematics, Statistics, JEE Main
Let $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{\mathrm{n}}$ be $\mathrm{n}$ observations such that $\sum \mathrm{x}_{\mathrm{i}}^2=400$ and $\sum \mathrm{x}_{\mathrm{i}}=80$. Then a possible value of $\mathrm{n}$ among the following is

Options:
A
15

B
18

C
9

D
12

The correct answer is:
18

Solution:
$$
\begin{aligned}
& \frac{\sum x_i^2}{n} \geq\left(\frac{\sum x_i}{n}\right)^2 \\
& \Rightarrow n \geq 16
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2005
Let x1,x2,…xn be n observations such that ∑xi2=500 and ∑xi=100. Then, an impossible value of n among the following is

Options:
A 24
B 18
C 29
D 22

The correct answer is:
18

Solution:
Variance =∑xi2n-∑xin2≥0
⇒∑xi2n≥∑xi2n2
⇒n≥∑xi2∑xi2⇒n≥10000500⇒n≥20

Topic: Mathematics, Statistics, JEE Main
Let x+1x=2, y+1y=-2 and sin −1 x+ cos −1 y=mπ, then the value of m is ______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
x+1x=2⇒x=1 ….[x+1x≥2 as equality occurs only when x=1]
⇒sin-1x=π2

Similarly, y=-1 ....[y+1y≤-2 as equality occurs only when y=-1]
⇒cos-1y=π
sin-1x+cos-1y=π2+π
=3π2
⇒m=32=1.5

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Let x+1x=2, y+1y=-2 and sin −1 x+ cos −1 y=mπ, then the value of m is

Options:

The correct answer is:
1.5

Solution:
x+1x=2⇒x=1 ….[x+1x≥2 as equality occurs only when x=1]
⇒sin-1x=π2

Similarly, y=-1 ....[y+1y≤-2 as equality occurs only when y=-1]
⇒cos-1y=π
sin-1x+cos-1y=π2+π
=3π2
⇒m=32=1.5

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Let x1,y1, x2,y2, x3,y3 and x4,y4 are four points which are at unit distance from the lines 3x-4y+1=0 and 8x+6y+1=0, then the value of ∑i=14xi∑i=14yi is equal to

Options:
A 2
B –2
C 1
D –1

The correct answer is:
–2

Solution:
Point of intersection of lines 3x-4y+1=0 and 8x+6y+1=0 is -15,110
Now, âi=14xi=4-15=-45
and âi=14yi=4110=25
ââi=14xiâi=14yi=-4525=-2

Topic: Mathematics, Straight Lines, JEE Main
Let X=2103 be a matrix. If X6=abcd, then the number of divisors of a+b+2020c+d is equal to

Options:
A 7
B 14
C 21
D 28

The correct answer is:
14

Solution:
X2=21032103=4509
X3=X2X=45092103=819027
X 6 = X 3 â X 3 = 8 19 0 27 8 19 0 27 = 64 665 0 729
âa=64,b=665,c=0,d=729
âa+b+2020c+d=1458=2Ã36
Number of divisors =1+16+1=14

Topic: Mathematics, Matrices, JEE Main
Let x-23=y+1-2=z+3-1 lie on the plane px-qy+z=5, for some p, q∈R. The shortest distance of the plane from the origin is:

Options:
A 3109
B 5142
C 571
D 1142

The correct answer is:
5142

Solution:
Given, line x-23=y+1-2=z+3-1

So, the point 2, -1, -3 will satisfy the given plane px-qy+z=5

So, 2p+q-3=5 or 2p+q=8 ... (i)

Also given line is perpendicular to normal plane so

3p+2q-1=0 ... (ii)

⇒p=15, q=-22

So,equation of plane 15x-22y+z-5=0

Now its distance from origin =6710=5142

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let x2=4ky,k>0 be a parabola with vertex A. Let BC be its latusrectum. An ellipse with centre on BC touches the parabola at A, and cuts BC at points D and E such that BD=DE=EC (B, D, E, C in that order). The eccentricity of the ellipse is

Options:
A 12
B 13
C 53
D 32

The correct answer is:
53

Solution:
Given, x2=4ky

BC is latusrectum.

∴BC=4k

BD=DE=EC

∴DE=BC3=4k3

P is centre of ellipse.

∴PE=2k3

OP=k

∵ Eccentricity of ellipse

1-PE2OP2=1-4k29k2

⇒e=9-49=53

Topic: Mathematics, Ellipse, JEE Main
Let x2a2+y2b2=1a>b be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕt=512+t-t2, then a2+b2 is equal to :

Options:
A 145
B 116
C 126
D 135

The correct answer is:
126

Solution:
LR=2b2a=10 ⇒   b2=5a

ϕt=512−t2-t+14−14=512+14−t−122

=23−t−122

max ϕt=23=e

b2=a21−e2

5a=a21−49

⇒  5=59a

⇒   a2=81,  b2=45

a2+b2=126.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let x2a2+y2b2=1,a>b be an ellipse, whose eccentricity is 12 and the length of the latus rectum is 14. Then the square of the eccentricity of x2a2−y2b2=1 is:

Options:
A 3
B 72
C 32
D 52

The correct answer is:
32

Solution:
Given: Length of latus rectum of ellipse is 14.

⇒2b2a=14 ...i

Also, eccentricity is given as 12

⇒12=1−b2a2

⇒12=1−b2a2

⇒12=1−a142a2

⇒-12=−142a

⇒a=14

⇒a2=14, b2=7

Now, eH=1+b2a2

⇒eH=1+12

⇒eH=32

⇒eH2=32

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let x2a2+y2b2=1,a>b be an ellipse with foci F1 and F2 Let AO be its semi-minor axis, where O is the centre of the ellipse. The lines AF1 and AF2, when extended, cut the ellipse again at points B and C respectively. Suppose that the ΔABC is equilateral. Then, the eccentricity of the ellipse is

Options:
A 12
B 13
C 13
D 12

The correct answer is:
12

Solution:
Given, x2a2+y2b2=1

ABC is an equilateral traingles.

In ΔAOF2,

tan30°=OF2OA

13=aeb

⇒b=3ae

⇒b2=3a2e2

⇒b2a2=3e2

⇒1-e2=3e2 ∵e=1-b2a2

⇒4e2=1

⇒e2=14⇒e=12

Topic: Mathematics, Ellipse, JEE Main
Let x=2 be a local minima of the function fx=2x4-18x2+8x+12,x∈-4,4. If M is local maximum value of the function f in -4,4, then M=

Options:
A 126-332
B 126-312
C 186-332
D 186-312

The correct answer is:
126-332

Solution:
Given,

x=2 be a local minima of the function fx=2x4-18x2+8x+12,x∈-4,4

Now differentiating the above function we get,

f'x=8x3-36x+8

⇒f'x=42x3-9x+2

Now equating it to zero we get, f'x=0

⇒2x3-9x+2=0

⇒x-22x2+4x-1=0

⇒x=-4±244

⇒x=-2±62

∴ x=6-22 {by checking sign change of f'x for maxima}

Now rewriting fx=2x4-18x2+8x+12 we get,

fx=x2-2x-922x2+4x-1+24x+7.5

Hence, f6-22=M=126-332

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let x=2 be a root of the equation x2+px+q=0 and fx=1-cosx2-4px+q2+8q+16x-2p4,x≠2p0,x=2p. Then limx→2p+fx
where · denotes greatest integer function, is

Options:
A 2
B 1
C 0
D -1

The correct answer is:
0

Solution:
Given,

x=2 be the root of the given equation x2+px+q=0,

Putting x=2 in given equation we get,

⇒4+2p+q=0⇒q+4=-2p

∵ x2-4px+q2+8q+16

=x2-4px+q+42

=x2-4px+4p2 (∵q+4=-2p)

=(x-2p)2

Now, solving the limit limx→2p+fx=limx→2p+1-cosx-2p2x-2p4

Let x-2p=θ

⇒limθ→0+fx=limθ→0+1-cosθ2θ4

∵limθ→01-cosθ2θ4=12 (Using L'Hospital's)

∴limθ→0+fx=limθ→0+12

∴limθ→0+fx=0

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let x2+y2=4r2 and xy=1 intersects at A and B in first quadrant. If AB=14 units, then the value of r is

Options:

The correct answer is:
1.5

Solution:
Let A=t,1t & B=1t,t âµBoth curves are symmetric about line y=x
then, AB2=t-1t2+1t-t2=14
â2t2+1t2=14+4ât2+1t2=9
Now, OA2=t2+1t2=9â2r2=9
â2r=3 âr=1.5

Topic: Mathematics, Hyperbola, JEE Main
Let x2+y2=r2 and xy=1 intersect at A and B in the first quadrant. If AB=14 units, then the square of the distance of AB from the origin is equal to

Options:

The correct answer is:
5.5

Solution:
Let, A=t,1t and B=1t,t { âµ both curves are symmetric about line y=x }
Then, AB2=t-1t2+1t-t2=14
â2t2+1t2=14+4ât2+1t2=9
Now, OA2=t2+1t2=9âr2=9
âr=3
âOF2=9-144=5.5

Topic: Mathematics, Straight Lines, JEE Main
Let λx-2y=μ be a tangent to the hyperbola a2x2-y2=b2. Then λa2-μb2 is equal to

Options:
A -2
B -4
C 2
D 4

The correct answer is:
4

Solution:
Given that λx-2y=μ is a tangent to the curve a2x2-y2=b2

So a2x2-λx-μ22=b2

⇒4a2-λ2x2+2λμx-μ2-4b2=0

So for the line to be tangent, the condition is discriminant =0

i.e. 4λ2μ2+44a2-λ2μ2+4b2=0

4λ2b2-4a2μ2=16a2b2

Hence, λ2a2-μ2b2=4

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Let ∫x3+x2+x12x3+15x2+20xdx=fx where f1=4730. If f22 is equal to p225, then the value of p is equal to

Options:

The correct answer is:
196

Solution:
In the given integral, multiplying numerator and denominator by x, we get,

fx=∫x4+x3+x212x5+15x4+20x3dx
Now substitute 12x5+15x4+20x3=t2
⇒60x4+x3+x2dx=2tdt
⇒fx=130∫tdtt=13012x5+15x4+20x3+C
As f1=4730⇒C=0
So, f2=78430⇒f22=784900=196225

Hence, p=196

Topic: Mathematics, Indefinite Integration, JEE Main
Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12. If P(1,β),β>0 is a point on this ellipse, then the equation of the normal to it at P is

Options:
A 4x–3y=2
B 8x–2y=5
C 7x–4y=1
D 4x–2y=1

The correct answer is:
4x–2y=1

Solution:
ae=4⇒a=4e⇒a=2

b2=a21-e2=3

(1,β) lies on x24+y23=1⇒14+β23=1⇒ β2=94⇒β=32(∵β>0)

Normal at (1,β)⇒ a2x1-b2yβ=a2-b2 ⇒4x-3yβ=1

so equation of normal is 4x-2y=1

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Let x=83+1313 and y=72+99. If t denotes the greatest integer ≤t, then

Options:
A x+y is even
B x is odd but y is even
C x is even but y is odd
D x and y are both odd

The correct answer is:
x+y is even

Solution:
We know the binomial expansion of a+bn is,

a+bn=C0nanb0+C1nan-1b1+C2nan-2b2+......+Cnna0bn

Using the above expansion,

x=83+1313=C0138313+C1138312131+…

Let x'=83-1313=C0138313-C1138312131+…

So, x-x'=2C113·8312131+C3138310·133…

Therefore, x-x' is an even integer, hence x is even

Now, y=72+99=C09729+C1972891+C2972792+....Now let,y'=(72-9)9=C09(72)9-C19(72)8(9)1+C2972792

So, y-y'=2C1972891+C3972693+…

y-y'= Even integer hence y is even

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let $(x+10)^{50}+(x-10)^{50}=\mathrm{a}_{0}+\mathrm{a}_{1} x+\mathrm{a}_{2} x^{2}+\ldots .+\mathrm{a}_{50} x^{50}$,

for all $x \in \mathbf{R} ;$ then $\frac{\mathrm{a}_{2}}{\mathrm{a}_{0}}$ is equal to :

Options:
A 12.5
B 12
C 12.25
D 12.75

The correct answer is:
12.25

Solution:
$(x+10)^{50}+(x-10)^{50}$

$=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{50} x^{50}$

$\therefore \quad a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{50} x^{50}$

$=2\left({ }^{50} C_{0} x^{50}+{ }^{50} C_{2} x^{48} \cdot 10^{2}+{ }^{50} C_{4} x^{46} \cdot 10^{4}+\ldots\right)$

$\therefore \quad a_{0}=2 \cdot{ }^{50} C_{50} 10^{50}$

$\quad a_{2}=2 \cdot{ }^{50} C_{2} \cdot 10^{48}$

$\therefore \quad \frac{a_{2}}{a_{0}}=\frac{{ }^{50} C_{2} \times 10^{48}}{{ }^{50} C_{50} 10^{50}}$

$=\frac{50 \times 49}{2 \times 100}=\frac{49}{4}=12.25$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let X=11,12,13,….,40,41 and Y=61,62,63,...,90,91 be the two sets of observations. If x¯ and y¯ are their respective means and σ2 is the variance of all the observations in X∪Y, then x¯+y¯-σ2 is equal to ________

Options:

The correct answer is:
603

Solution:
We have,

X=11,12,13,….,40,41

⇒nX=31

Y=61,62,63,...,90,91

⇒nY=31

So,

x¯=∑i=1141i31=13141×422-10×112=26

y¯=∑i=6191i31=13191×922-60×612=73

Combined mean,

μ=31×26+31×7631+31

⇒μ=26+762=51

Also,

σ2=162×∑i=131xi-μ2+∑i=131yi-μ2

Since, xi∈X are in A.P. with 31 elements common difference 1 , same is yi∈y, when written in increasing order.

∴∑i=131xi-μ2=∑i=131yi-μ2=102+112+….+402

=102+112+….+402

=40×41×816-9×10×196=21855

Hence, σ2=2185562+2185562=705

∴x¯+y¯-σ2=|26+76-705|=603

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Let x=a2b2c be the physical quantity. If the percentage error in the measurement of physical quantities a, b and c are 2%, 3% and 4% respectively then percentage error in the measurement of x is

Options:

The correct answer is:
14

Solution:
Îx x =2 Îa a +2 Îb b + Îc c
Îx y =2Ã2+2Ã3+4
=4+6+4
Îx x =14%

Topic: Physics, Units and Dimensions, JEE Main
Let $\bar{X}$ and M.D. be the mean and the mean deviation about $\bar{X}$ of $n$ observations $\mathrm{x}_{\mathrm{i}}, \mathrm{i}=1,2$,n. If each of the observations is increased by 5 , then the new mean and the mean deviation about the new mean, respectively, are :

Options:
A
$\bar{X}$, M.D.

B
$\overline{\mathrm{X}}+5$, M.D.

C
$\bar{X}$, M.D. $+5$

D
$\overline{\mathrm{X}}+5$, M.D. $+5$

The correct answer is:
$\overline{\mathrm{X}}+5$, M.D.

Solution:
Let $x_i$ be $n$ observations, $i=1,2, \ldots n$ Let $\bar{X}$ be the mean and M.D be the mean deviation about $\bar{X}$. If each observation is increased by 5 then new mean will be $\bar{X}+5$ and new M.D. about new mean will be M.D.
$$
\left(\because \text { Mean }=\sum_{i=1}^n \frac{x_i}{n}\right)
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2014 (12 Apr Online)
Let x and x denote the fractional part of x and the greatest integer ≤x respectively of a real number x. if ∫0nxdx, ∫0nxdx and 10n2-n,(n∈N,n>1) are three consecutive terms of a G.P. then n is equal to__

Options:

The correct answer is:
21

Solution:
∫0nxdx=n∫01xdx=nx2201=n2
and ∫0nxdx=∫0nx-xdx=x220n-∫0nxdx=n22-n2
Now, n2,n2-n2 and 10n2-n are in Geometric progression

So,n2-n22=n2·10n2-n
⇒n2(n-1)24=5.n2(n-1)
⇒n-1=20⇒n=21 ∵n≠1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (04 Sep Shift 2)