datavorous/entrance-exam-dataset huggingface dataset:
Let \( S \) be the set of points where the function, \( f(x)=|2-| x-3 \mid, x \in R \), is not differentiable. Then \( \sum_{x \in S} f(f(x)) \) is equal to

Options:

The correct answer is:
3

Solution:
Check non-differentiability of function, we get,

∵fx is non differentiable at x=1,3,5
∑ffx=ff1+ff3+ff5
=1+1+1
=3

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let S be the set of positive integral values of a for which ax2+2a+1x+9a+4x2-8x+32<0, ∀x∈ℝ. Then, the number of elements in S is:

Options:
A 1
B 0
C ∞
D 3

The correct answer is:
0

Solution:
Given:

ax2+2(a+1)x+9a+4x2-8x+32<0 ∀x∈R

For quadratic x2-8x+32=0, D1=-82-432=-64

Since the discriminant is less than zero and the leading coefficient is positive, this quadratic will always be positive.

Now, solving ax2+2(a+1)x+9a+4<0

We know that, for a quadratic to be always negative, the coefficient of x2<0, D<0.

⇒a<0

But we want positive values.

So, no positive integral value exist.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let S be the set of values of λ, for which the system of equations

6λx-3y+3z=4λ2, 2x+6λy+4z=1 and 3x+2y+3λz=λ has no solution. Then,12∑λ∈Sλ is equal to _______.

Options:

The correct answer is:
24

Solution:
Given,

System of equations,

6λx-3y+3z=4λ2

2x+6λy+4z=1

3x+2y+3λz=λ

For no solution △=0,

So, 6λ-3326λ4323λ=0

⇒9λ3-7λ-2=0

⇒(λ-1)(3λ+1)(3λ+2)=0

Hence, λ∈1,-13,-23

Now putting the value of λ in 12∑λ∈Sλ we get,

12∑λ∈Sλ=12×1+13+23=24

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let S be the sum of the first 9 term of the series :

x+ka+x2+k+2a+x3+k+4a+x4+k+6a+… where a≠0 and x ≠ 1. If S=x10-x+45ax-1x-1 , then k is equal to

Options:
A -5
B 1
C -3
D 3

The correct answer is:
-3

Solution:
S=x+ka+x2+k+2a+x3+k+4a+x4+k+6a+…

S=x+x2+x3+x4+...+ka+k+2a+k+4a+k+6a+…

S=xx9-1x-1+a·922k+82

xx9-1x-1+a·922k+82=x10-x+9·ak+8x-1x-1

So 45=9k+8⇒k=-3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let S denote the set of all real values of λ such that the system of equations

λx+y+z=1

x+λy+z=1

x+y+λz=1
is inconsistent, then ∑λ∈Sλ2+λ is equal to

Options:
A 2
B 12
C 4
D 6

The correct answer is:
6

Solution:
System of equations

λx+y+z=1x+λy+z=1x+y+λz=1

has no solution, so

λ111λ111λ=0

⇒λ+22+λ2+λ1λ111λ=0

⇒2+λ1111λ111λ=0

⇒2+λ1001λ-1010λ-1=0

⇒2+λλ-12=0

⇒λ=1, -2

For λ=1, we get infinite solutions, since equation will be identical.

So,

∑λ2+λ=4+2=6

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let S=E,E2…E8 be a sample space of raddom experiment such that PEn=n36 for every n=1,2….8. Then the number of elements in the set A⊂S:PA≥45 is _____.

Options:

The correct answer is:
19

Solution:
Given,

PA≥45

A is subset of S hence

A can have elements;

type 1:

type 2: E1,E2,……E8

type 3: E1,E2,E1,E3……E1,E8

⋮

⋮

type 6: E1,E2,…….E5,……E4,E5,E6,E7,E8

type 7: E1,E2,…….E6,……E3,E4,……….E8

type 8: E1,E2,…….E7E2,E3,…….E8

type 9: E1,E2,……E8

As PA≥45;

Note: Type 1 to Type 4 elements can not be in set A as maximum probability of type 4 elements.

E5,E6,E7,E8 is 536+636+736+836=1318<45

Now for Type 5 acceptable elements let's call probability as P5

P5=n1+n2+n3+n4+n536≥45

⇒n1+n2+n3+n4+n3≥28.8

Hence, 2 possible ways E5,E6,E7,E8,E3 or E4

P6=n1+n1+n3+n4+n3+n6≥28.8

⇒ 9 possible ways

P7⇒n1+n1+………+n7≥28.8

⇒ 7 possible ways

P8⇒n1+n1+……+nn≥28.8

⇒ 1 possible way

Total =19

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let SK=1+2+...+KK and ∑j=1nS2j=nABn2+Cn+D where A, B, C, D ∈ N and A Has least value then

Options:
A A+C+D is not divisible by D
B A+B=5(D-C)
C A+B+C+D is divisible by 5
D A+B is divisible by D

The correct answer is:
A+B is divisible by D

Solution:
Given that SK=1+2+...+KK

We know that 1+2+3....+n=nn+12

⇒SK=KK+12K=K+12

Now,

∑j=1nSj2=∑j=1n1422+32+42...n+12

=∑j=1n1412+22+32+42...n+12-12

⇒14n+1n+22n+36-1

=14n2+3n+22n+3-66

=142n3+6n2+4n+3n2+9n+6-66

=142n3+9n2+13n6

=n242n2+9n+13

On comparing this with ∑j=1nS2j=nABn2+Cn+D we get,

A=24, B=2, C=9, D=13

⇒A+BD=2613=2

That means, A+B is divisible by D.

Hence this is the correct option.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let $S(K)=1+3+5+\ldots+(2 K-1)=3+K^2$. Then which of the following is true?

Options:
A
$S(1)$ is correct

B
Principle of mathematical induction can be used to prove the formula

C
$S(K) \neq S(K+1)$

D
$\mathrm{S}(\mathrm{K}) \Rightarrow \mathrm{S}(\mathrm{K}+1)$

The correct answer is:
$\mathrm{S}(\mathrm{K}) \Rightarrow \mathrm{S}(\mathrm{K}+1)$

Solution:
$$
\begin{aligned}
& S(k)=1+3+5+\ldots \ldots . .+(2 k-1)=3+k^2 \\
& S(k+1)=1+3+5+\ldots \ldots \ldots+(2 k-1)+(2 k+1) \\
& =\left(3+k^2\right)+2 k+1=k^2+2 k+4\left[\text { from } S(k)=3+k^2\right] \\
& =3+\left(k^2+2 k+1\right)=3+(k+1)^2=S(k+1)
\end{aligned}
$$
Although $S(k)$ in itself is not true but it considered true will always imply towards $S(k+1)$.

Topic: Mathematics, Mathematical Induction, JEE Main, JEE Main 2004
Let $S_k, k=1,2, \ldots, 100$, denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k !}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|$ is

Options:

The correct answer is:
4

Solution:
We have, $S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}$
Now,
$$
\begin{aligned}
& \left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \\
& =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \\
& \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \\
& =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4
$$

Topic: Mathematics, Sequences and Series, JEE Main
Let Sk=∑r=1ktan-16r22r+1+32r+1, then limk→∞Sk is equal to :

Options:
A tan-132
B π2
C cot-132
D tan-13

The correct answer is:
cot-132

Solution:
Given Sk=∑r=1ktan-16r22r+1+32r+1

⇒Sk=∑r=1ktan-12r·3r22r·2+32r·3

Divide by 32r we get,

Sk=∑r=1ktan-123r232r.2+3

Sk=∑r=1ktan-123r3232r+1+1

Let 23r=t

Sk=∑r=1ktan-1t31+23t2

Sk=∑r=1ktan-1t-2t31+t.2t3

Sk=∑r=1ktan-1t-tan-12t3

Sk=∑r=1ktan-123r-tan-123r+1

Sk=tan-123-tan-1232+tan-1232-tan-1233+...+tan-123k-tan-123k+1

Sk=tan-123-tan-123k+1

Then, S∞=limk→∞tan-123-tan-123k+1

=tan-123-tan-10

∴ S∞=tan-123=cot-132, ∵tan-1x=cot-11x

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let
S=α:log292α-4+13-log252·32α-4+1=2. Then the maximum value of β for which the equation x2-2∑α∈sα2x+∑a∈sα+12β=0 has real roots, is _____ .

Options:

The correct answer is:
25

Solution:
Given,

log292α-4+13-log232α-4·52+1=2

Now let 32α-4=t, so the equation becomes,

log2t2+13-log25t2+1=2

⇒log2t2+135t2+1=2

⇒t2+135t2+1=22

⇒t2+13=10t+4

⇒t2-10t+9=0

⇒t=1 or 9

So,

32α-4=1 or 9

⇒32α-4=30 or 32

⇒2α-4=0 or 2

⇒α=2, 3

Now,

x2-2∑α∈sα2x+∑a∈sα+12β=0

⇒x2-22+32x+32+42β=0

⇒x2-50x+25β=0

Now for real roots

D≥0

⇒502-4×25β≥0

⇒50-2β≥0

⇒β≤25

So, maximum value of β is 25.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let Sn=113+1+213+23+1+2+313+23+33+…+1+2+…,+n13+23+…n3 . If 100 Sn=n, then n is equal to:

Options:
A 200
B 199
C 99
D 19

The correct answer is:
199

Solution:
Given, Sn=113+1+213+23+1+2+313+23+33+…+1+2+…+n13+23+…+n3

Tn=nn+12nn+122
Tn=2nn+1=21n-1n+1
Sn=2∑n=1n1n-1n+1
Sn=21-1n+1=2nn+1
Given, 100Sn=n⇒100×2nn+1=n
n+1=200
n=199

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2017 (09 Apr Online)
Let S={n:1⩽n⩽50 and n is odd}. Let a∈S and A=10a-110-a01. If Σa∈SdetadjA=100λ, then λ is equal to

Options:
A 218
B 221
C 663
D 1717

The correct answer is:
221

Solution:
We know that

Adj A=An-1

Here n is 3×3 matrix

so Adj A=A3-1=A2

Now find A=10a-110-a01

=1+a2

Now Adj A=A2=1+a22

Now

Σa∈SdetadjA=Σ1+a22=100λ

⇒22+42...+482+502=100λ

⇒412+22...+252=100λ

⇒4×25×26×516=100λ

⇒221=λ

Topic: Mathematics, Matrices, JEE Main
Let S={n:1⩽n⩽50 and n is odd}. Let a∈S and A=10a-110-a01. If Σa∈SdetadjA=100λ, then λ is equal to

Options:
A 218
B 221
C 663
D 1717

The correct answer is:
221

Solution:
We know that

Adj A=An-1

Here n is 3×3 matrix

so Adj A=A3-1=A2

Now find A=10a-110-a01

=1+a2

Now Adj A=A2=1+a22

Now

Σa∈SdetadjA=Σ1+a22=100λ

⇒22+42...+482+502=100λ

⇒412+22...+252=100λ

⇒4×25×26×516=100λ

⇒221=λ

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Let $S_{n}=1+q+q^{2}+\ldots .+q^{n}$ and

$\mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^{2}+\ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}$

where $q$ is a real number and $q \neq 1$. If ${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \cdot \mathrm{S}_{1}+\ldots .+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{S}_{100}=\alpha \mathrm{T}_{100},$ then $\alpha$ is

equal to :

Options:
A $2^{99}$
B 202
C 200
D $2^{100}$

The correct answer is:
$2^{100}$

Solution:
$S_{n}=\left(\frac{1-q^{n+1}}{1-q}\right), T_{n}=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)}$

$\Rightarrow T_{100}=\frac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\left(\frac{q+1}{2}\right)}$

$S n=\frac{1}{1-q}-\frac{q^{n+1}}{1-q}, T_{100}=\frac{2^{101}-(\mathrm{q}+1)^{101}}{2^{100}(1-\mathrm{q})}$

Now, ${ }^{101} C_{1}+{ }^{101} C_{2} S_{1}+{ }^{101} C_{3} S_{2}+\ldots+{ }^{101} C_{101} S_{100}$

$=\left(\frac{1}{1-q}\right)\left({ }^{101} C_{2}+\ldots+{ }^{101} C_{101}\right)$

$$

\begin{aligned}

& \quad-\frac{1}{1-q}\left({ }^{101} C_{2} q^{2}+{ }^{101} C_{3} q^{3}+\ldots+{ }^{101} C_{101} q^{101}\right)+101 \\

=& \frac{1}{1-q}\left(2^{101}-1-101\right)-\left(\frac{1}{1-q}\right)\left((1+q)^{101}-1\right.\\

\left.-{ }^{101} C_{1} q\right)+101 \\

=& \frac{1}{1-q}\left[2^{101}-102-(1+q)^{101}+1+101 q\right]+101 \\

=& \frac{1}{1-q}\left[2^{101}-101+101 q-(1+q)^{101}\right]+101 \\

=&\left(\frac{1}{1-q}\right)\left[2^{101}-(1+q)^{101}\right]=2^{100} T_{100}

\end{aligned}

$$

Hence, by comparison $\alpha=2^{100}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let Sn be the sum of the first n terms of an arithmetic progression. If S3n=3S2n, then the value of S4nS2n is :

Options:
A 6
B 4
C 2
D 8

The correct answer is:
6

Solution:
Let a be first term and d be common difference of this A.P.

Given S3n=3S2n

⇒3n22a+3n-1d=3×2n22a+2n-1d

⇒2a+(3n-1)d=4a+(4n-2)d

⇒2a+(n-1)d=0

Now S4nS2n=4n22a+4n-1d2n22a+2n-1d=22a+n-1d+3nd2a+n-1d+nd

=6ndnd=6

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let Sn be the sum to n-terms of an arithmetic progression 3,7,11,……, if 40<6n(n+1)∑k=1nSk<42, then n equals ____________.

Options:

The correct answer is:
9

Solution:
Given, AP: 3,7,11,15,...

⇒an=3+n-14

⇒an=4n-1

⇒Sn=n26+4n-4

⇒Sn=n×22n+12

⇒Sn=2n2+n

Now, solving

⇒∑k=1nSk=2∑n=1nn2+∑n=1nn

⇒∑k=1nSk=nn+12n+13+nn+12

⇒∑k=1nSk=nn+121+4n+23

Now, using 40<6nn+1∑k=1nSk<42 we get,

⇒40<6nn+1×nn+125+4n3<42

⇒40<4n+5<42

⇒35<4n<37

⇒n=9

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $S_{n}=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots .$ to $n^{\text {th }}$ term. Then $\lim _{n \rightarrow \infty} S_{n}$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D $\frac{\pi}{8}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
$\begin{aligned} & t_{n}=\cot ^{-1} 2 n^{2} \\=& \tan ^{-1} \frac{1}{2 n^{2}}=\tan ^{-1} \frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)} \\=& \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1) \\ \therefore & S_{n}=\tan ^{-1}(2 n+1)-\tan ^{-1} 1 \\ \therefore & \operatorname{Lim}_{n \rightarrow \infty} S_{n}=\pi / 2-\pi / 4=\pi / 4 \end{aligned}$

Topic: Mathematics, Limits, JEE Main
Let Sn denote the sum of first n-terms of an arithmetic progression. If S10=530, S5=140, then S20-S6 is equal to:

Options:
A 1862
B 1842
C 1852
D 1872

The correct answer is:
1862

Solution:
We know that the sum of n terms of an arithmetic progression with its first term as a and common difference d is Sn=n22a+n-1d.

Given, S10=530

⇒1022a+9d=530

⇒2a+9d=106 …1

And S5=140

⇒522a+4d=140

⇒2a+4d=56 …2

Subtracting the two equation, we get, 5d=50

⇒d=10

On putting the value of d in the equation 2, we get

2a+4×10=56

⇒2a=16

⇒a=8.

Now, S20-S6=2022a+19d-622a+5d

=14a+175 d

=(14×8)+(175×10)

=1862.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let Sn denote the sum of the first n terms of an A.P.. If S4=16 and S6=-48 , then S10 is equal to:

Options:
A -320
B -380
C -260
D -410

The correct answer is:
-320

Solution:
Let the first term of A.P. is ‘a’ and the common difference is ‘d’.
∴ S4=422a+4-1d=16
⇒2a+3d=8 ...i
and S6=622a+6-1d=-48
⇒2a+5d=-16 ...ii
∴ From (i) & (ii), a=22 & d=-12
∴ S10=102(2a+9d)=52×22+9-12
=-320

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Let Sn denote the sum of the first n terms of an arithmetic progression. If S10=390 and the ratio of the tenth and the fifth terms is 15 : 7, then S15−S5 is equal to:

Options:
A 800
B 890
C 790
D 690

The correct answer is:
790

Solution:
Given: a10a5=157

⇒a+9da+4d=157

⇒7a+63d=15a+60d

⇒3d=8a

Also, S10=390

⇒1022a+9d=390

⇒2a+24a=78

⇒a=3

⇒d=8

⇒S15-S5=1526+14×8-526+4×8

⇒S15-S5=152×118-52×38

⇒S15-S5=885-95

⇒S15-S5=790

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Let S=n∈N, 0i10nabcd=abcd∀a,b,c,d∈R, where i=-1. Then the number of 2- digit numbers in the set S is

Options:

The correct answer is:
11

Solution:
Let X=abcd & A=0i10n

⇒AX=IX

⇒A=I

⇒0i10n=I

⇒A8=1001

⇒n is multiple of 8.

So number of 2 digit numbers in the set

S=16,24,32,…,96

Clearly, 16,24,32,…,96 are in arithmetic progression.

Here, a=16, d=24-16=8 and l=a+n-1d=96

Where, a is the first term, d is the common difference, l is the last term and n is the number of terms.

Consider, l=a+n-1d=96

⇒16+n-18=96

⇒n-18=80

⇒n-1=10

⇒n=11

Hence, number of two-digit numbers in set S is 11.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let Sn(x)=loga1/2x+loga1/3x+loga1/6x+loga1/11x+loga1/18x+loga1/27x+… up to n-terms, where a>1. If S24(x)=1093 and S12(2x)=265, then value of a is equal to _____ .

Options:

The correct answer is:
16

Solution:
Given Sn(x)=loga1/2x+loga1/3x+loga1/6x+loga1/11x+loga1/18x+loga1/27x+…+n-terms

⇒Sn(x)=(2+3+6+11+18+27+…+n-terms )logax

Let S1=2+3+6+11+18+27+…+Tn

S1=2+3+6+11+18+...+Tn ...i

S1=0+2+3+6+11+18+...+Tn-1+Tn ...ii

Subtract i from ii we get,

Tn=2+1+3+5+...+upto n-1-term

Tn=2+(n-1)2

S1=ΣTn=2n+(n-1)n(2n-1)6

⇒Snx=2n+n(n-1)(2n-1)6logax

S24(x)=1093 (Given)

logax48+23.24.476=1093

logax=14 ......1

S12(2x)=265

S12(2x)=265

loga(2x)24+11.12.236=265

loga2x=12 ......2

(2)-(1)

loga2x-logax=14

loga2=14⇒a=16

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let $S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^t-\mu\right) \cdot \sin (2|t|)\right.$, $t \in R$, is a differentiable function $\}$. Then $S$ is a subest of?

Options:
A
$R \times[0, \infty)$

B
$(-\infty, 0) \times R$

C
$[0, \infty) \times R$

D
$R \times(-\infty, 0)$

The correct answer is:
$R \times[0, \infty)$

Solution:
$$
\begin{aligned}
&\quad S=\left\{(\lambda, \mu) \in \mathrm{R} \times \mathrm{R}: \mathrm{f}(\mathrm{t})=\left(|\lambda| \mathrm{e}^{\mid t}-\mu\right) \sin (2 \mathrm{t} \mid),\right. \\
&t \in R \\
&f(t)=\left(|\lambda| \mathrm{e}^{|t|}-\mu\right) \sin (2|t|) \\
&\left\{\begin{array}{l}
\left(|\lambda| e^t-\mu\right) \sin 2 t \quad t>0 \\
\left(|\lambda| e^{-t}-\mu\right)(-\sin 2 t) t < 0
\end{array}\right. \\
&f^{\prime}(t) \\
&\left\{\begin{array}{l}
\left(|\lambda| \mathrm{e}^t\right) \sin 2 t+\left(|\lambda| \mathrm{e}^t-\mu\right)(2 \cos 2 t) t>0 \\
|\lambda| \mathrm{e}^{-t} \sin 2 t+\left(|\lambda| e^{-t}-\mu\right)(-2 \cos 2 t) < t < 0
\end{array}\right. \\
&\text { As, } f(t) \text { is differentiable } \\
&\therefore \text { LHD }=\mathrm{RHD} \text { at } t=0 \\
&|\lambda| \cdot \sin 2(0)+\left(|\lambda| \mathrm{e}^0-\mu\right) 2 \cos (0) \\
&=|\lambda| \mathrm{e}^{-0} \cdot \sin 2(0)-2 \cos (0)\left(|\lambda| \mathrm{e}^{-0}-\mu\right) \\
&\Rightarrow 0+(|\lambda|-\mu) 2=0-2(|\lambda|-\mu) \\
&\quad 4(|\lambda|-\mu)=0 \\
&\quad|\lambda|=\mu
\end{aligned}
$$
So, $S \equiv(\lambda, \mu)=\{\lambda \in R \& \mu \in[0, \infty)\}$
Therefore set $\mathrm{S}$ is subset of $R \times[0, \infty)$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^t-\mu\right) \cdot \sin (2|t|)\right.$, $t \in R$, is a differentiable function $\}$. Then $S$ is a subest of?

Options:
A
$R \times[0, \infty)$

B
$(-\infty, 0) \times R$

C
$[0, \infty) \times R$

D
$R \times(-\infty, 0)$

The correct answer is:
$R \times[0, \infty)$

Solution:
$$
\begin{aligned}
&\quad S=\left\{(\lambda, \mu) \in \mathrm{R} \times \mathrm{R}: \mathrm{f}(\mathrm{t})=\left(|\lambda| \mathrm{e}^{\mid t}-\mu\right) \sin (2 \mathrm{t} \mid),\right. \\
&t \in R \\
&f(t)=\left(|\lambda| \mathrm{e}^{|t|}-\mu\right) \sin (2|t|) \\
&\left\{\begin{array}{l}
\left(|\lambda| e^t-\mu\right) \sin 2 t \quad t>0 \\
\left(|\lambda| e^{-t}-\mu\right)(-\sin 2 t) t < 0
\end{array}\right. \\
&f^{\prime}(t) \\
&\left\{\begin{array}{l}
\left(|\lambda| \mathrm{e}^t\right) \sin 2 t+\left(|\lambda| \mathrm{e}^t-\mu\right)(2 \cos 2 t) t>0 \\
|\lambda| \mathrm{e}^{-t} \sin 2 t+\left(|\lambda| e^{-t}-\mu\right)(-2 \cos 2 t) < t < 0
\end{array}\right. \\
&\text { As, } f(t) \text { is differentiable } \\
&\therefore \text { LHD }=\mathrm{RHD} \text { at } t=0 \\
&|\lambda| \cdot \sin 2(0)+\left(|\lambda| \mathrm{e}^0-\mu\right) 2 \cos (0) \\
&=|\lambda| \mathrm{e}^{-0} \cdot \sin 2(0)-2 \cos (0)\left(|\lambda| \mathrm{e}^{-0}-\mu\right) \\
&\Rightarrow 0+(|\lambda|-\mu) 2=0-2(|\lambda|-\mu) \\
&\quad 4(|\lambda|-\mu)=0 \\
&\quad|\lambda|=\mu
\end{aligned}
$$
So, $S \equiv(\lambda, \mu)=\{\lambda \in R \& \mu \in[0, \infty)\}$
Therefore set $\mathrm{S}$ is subset of $R \times[0, \infty)$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Let S={λ, μ∈R×R : ft=λ et-μsin⁡2t, t∈R is a differential function}. Then, S is a subset of :

Options:
A -∞, 0×R
B R×0, ∞)
C 0, â)ÃR
D RÃ-â, 0

The correct answer is:
R×0, ∞)

Solution:
Let S=λ, μ∈R×R and

ft=λet-μ. sin⁡2t

RHD=limh→0⁡f0+h-0h=limh→0⁡λeh-μsin2⁡hh=2(λ-μ)

LHD=limh→0⁡f0-h-0-h=limh→0⁡λeh-μsin2⁡h-h=-2(λ-μ)

Now, LHD=RHD if λ=μ

⇒μ≥0 and λ∈R

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2018 (15 Apr)
Let S=t∈R:fx=x-π ⋅ex-1sinx is not differentiable at t. Then, the set S is equal to:

Options:
A 0, π
B ϕ (an empty set)
C 0
D π

The correct answer is:
ϕ (an empty set)

Solution:
fx=x-π.ex-1.sinx

In the neighbourhood of x=π, fx=x-π.ex-1.sinx
Rf′π=limh→0⁡fπ+h-fπh=limh→0hsin⁡π+heπ+h-1h=0
Lf′π=limh→0fπ-fπ-hh=-limh→0-hsin⁡π-heπ-h-1h=0
Differentiable at x=π
Rf′0=limh→0fh-f0h=limh→0h-π×sinheh-1h=0
Lf′0=limh→0⁡f0-f-hh=-limh→0-h-π×sin-he-h-1h=0
Differentiable at x=0
Hence, fx is differentiable everywhere.
⇒S=ϕ

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2018 (08 Apr)
Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be so that $g o f: s \rightarrow U$ is surjective. Then,

Options:
A $g$ and $f$ are both surjective
B $g$ is surjective, $f$ may not be so
C $f$ is surjective, $g$ may not be so
D $f$ and $g$ both may not be surjective

The correct answer is:
$g$ is surjective, $f$ may not be so

Solution:
We have,
gof : $S \rightarrow U$ is an onto function.
Let $Z$ be any arbitrary element such that $Z \in U$. Now, $g o f: S \rightarrow U$ is onto $\Rightarrow g_{0} f(x)=Z,$ for $x \in S$
$\Rightarrow g(f(x))=Z$
$\Rightarrow g(y)=Z,$ where $y=f(x) \in T$ for all $Z \in U$
$\therefore$ For all $Z \in U,$ there exists $y=f(x) \in T$ such that $g(y)=z$
$\therefore g: T \rightarrow U$ is an onto function.

Topic: Mathematics, Functions, JEE Main
Let S=x∈-π2,π2:91-tan2x+9tan2x=10 and β=∑x∈Stan2x3, then 16(β-14)2 is equal to

Options:
A 16
B 8
C 64
D 32

The correct answer is:
32

Solution:
Given,

91-tan2x+9tan2x=10

Now let 9tan2x=t, then above equation will be,

9t+t=10

⇒t2-10t+9=0

⇒t=9 or t=1

So, when 9tan2x=9⇒tan2x=1

⇒tanx=±1⇒x=±π4, as given x∈-π2,π2

Now when 9tan2x=1

⇒tan2x=0⇒x=0

Hence, β=∑x∈Sx3=tan203+tan2π12+tan2-π12

=0+22-32

=14-83

So, the value of 16β-142=64×36=32

Hence this is the correct option.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let S=x∈R:0<x<1 and 2tan-11-x1+x=cos-11-x21+x2. If n(S) denotes the number of elements in S then :

Options:
A n(S)=2 and only one element in S is less then 12
B n(S)=1 and the element in S is more than 12
C n(S)=1 and the element in S is less then 12
D n(S)=0

The correct answer is:
n(S)=1 and the element in S is less then 12

Solution:
Given, 0<x<1

2tan-11-x1+x=cos-11-x21+x2

Put x=tanθ ; 0<θ<π4

2tan-11-tan θ1+tan θ=cos-11-tan2θ1+tan2θ

⇒2tan-1tanπ4-θ=cos-1(cos2θ)

2π4-θ=2θ⇒θ=π8

Put θ=tan-1x we get,

x=tanπ8∴x=2-1≃0.414<12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let S=x∈R:3+2x+3−2x=10. Then the number of elements in S is:

Options:
A 4
B 0
C 2
D 1

The correct answer is:
2

Solution:
Given: 3+2x+3-2x=10

Now, let 3+2x=t⇒3-2x=1t

So, 3+2x+3-2x=10

⇒t+1t=10

⇒t2-10t+1=0

⇒t=5±26

⇒t=3-22, 3+22

Now, taking 3+2x=3-22=3+2-2 we get, x=-2

And equating 3+2x=3+22 we get, x=2

Hence, the number of elements in S is 2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let S=x∈R :x≥0 & 2x-3+x x-6+6=0 . Then S:

Options:
A Contains exactly four elements
B Is an empty set
C Contains exactly one element
D Contains exactly two elements

The correct answer is:
Contains exactly two elements

Solution:
2x-3+xx-6+6=0
Let x=t>0
2t-3+tt-6+6=0
⇒2t-3=-t2-6t+6

⇒2t-3=-t2-6t+9+3
So the equation 2 solutions.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2018 (08 Apr)
Let
S=x:x∈ℝ and 3+2x2-4+3-2x2-4=10. Then nS is equal to

Options:
A 2
B 4
C 6
D 0

The correct answer is:
4

Solution:
Given:

S=x:3+2x2-4+3-2x2-4=10

Now,

3-2=3+23-23+2=13+2

So,

3+2x2-4+3-2x2-4=10

⇒3+2x2-4+13+2x2-4=10

Put 3+2x2-4=u, then

u+1u=10

⇒u2-10u+1=0

⇒u=10±100-42

⇒u=5±24

⇒u=5±232

⇒u=3±22

⇒3+2x2-4=3±22

So, 3+2x2-4=3+22 and 3+2x2-4=13+22

Therefore,

x2-4=2 & x2-4=-2

⇒x=±6 & x=±2

So, S=6, 2, -6, -2

Hence, nS=4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let S=x,y∈R2:y21+r-x21-r=1, where r≠±1. Then S represents:

Options:
A An ellipse whose eccentricity is 1r+1, when r>1.
B A hyperbola whose eccentricity is 2r+1, when 0<r<1.
C An ellipse whose eccentricity is 2r+1, when r>1
D A hyperbola whose eccentricity is 21-r, when 0<r<1

The correct answer is:
An ellipse whose eccentricity is 2r+1, when r>1

Solution:
Given y21+r-x21-r=1
If r∈0,1 then this curve is a hyperbola whose eccentricity is 1+1-r1+r=2r+1.
If r>1 , then this curve is an ellipse whose
eccentricity is 1-r-1r+1=2r+1.

Topic: Mathematics, Ellipse, JEE Main
Let S=x,y∈R2:y21+r-x21-r=1, where r≠±1. Then S represents:

Options:
A An ellipse whose eccentricity is 1r+1, when r>1.
B A hyperbola whose eccentricity is 2r+1, when 0<r<1.
C An ellipse whose eccentricity is 2r+1, when r>1
D A hyperbola whose eccentricity is 21-r, when 0<r<1

The correct answer is:
An ellipse whose eccentricity is 2r+1, when r>1

Solution:
Given y21+r-x21-r=1
If r∈0,1 then this curve is a hyperbola whose eccentricity is 1+1-r1+r=2r+1.
If r>1 , then this curve is an ellipse whose
eccentricity is 1-r-1r+1=2r+1.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let Sα=x,y:y2≤x, 0≤x≤α and Aα is area of the region Sα. If for a λ,0<λ<4, Aλ:A4=2:5, then λ equals:

Options:
A 4 2513
B 2 42513
C 4 42513
D 2 2513

The correct answer is:
4 42513

Solution:
Sλ=2∫0λx dx=2x32320λ
=43λ32
∴S4=43432
Given SλS4=25⇒λ32432=25
⇒λ=442513

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (08 Apr Shift 2)