datavorous/entrance-exam-dataset huggingface dataset:
Let $f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ for $\mathrm{k}=1,2,3, \ldots$ Then for all $\mathrm{x} \in \mathrm{R},$ the value of $f_{4}(x)-f_{6}(x)$ is equal to :

Options:
A $\frac{1}{12}$
B $\frac{1}{4}$
C $\frac{-1}{12}$
D $\frac{5}{12}$

The correct answer is:
$\frac{1}{12}$

Solution:
$f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$

$f_{4}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]$

$=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\frac{(\sin 2 x)^{2}}{2}\right]$

$=\frac{1}{4}\left[1-\frac{(\sin 2 x)^{2}}{2}\right]$

$f_{6}(x)=\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right]$

$=\frac{1}{6}\left[\left(\sin ^{2} x+\left(\cos ^{2} x\right)-\frac{3}{4}\left(\sin ^{2} x\right)^{2}\right]\right.$

$=\frac{1}{6}\left[1-\frac{3}{4}(\sin 2 x)^{2}\right]$

Now $f_{4}(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2 x)^{2}}{8}+\frac{1}{8}(\sin 2 x)^{2}$

$=\frac{1}{12}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let f⁡n=13+3n100n, where [n] denotes the greatest integer less than or equal to n. Then ∑n=156fn is equal to

Options:
A 56
B 1287
C 1399
D 689

The correct answer is:
1399

Solution:
Given fn=13+3n100n=9n+100300n

If 9n+100300<1

⇒9n+100<300

⇒n<2009

⇒n=22

We know that for a number 0⩽x<1, x=0
⇒f(n)=0, ∀ 1⩽n⩽22

Again, if 9n+100300<2

⇒9n+100<600

⇒n=5009

⇒n=55

We know that for a number 1⩽x<2, x=1

⇒f(n)=1, 23⩽n⩽55

And, f(n)=2 if n=56

Thus, ∑n=156fn=∑n=1220×n+∑n=23551×n+2×56

=0+23+24+...+55+56×2

=33223+55+112

=33×39+112

=1399.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (19 Apr Online)
Let f:N→N be a function such that fm+n=fm+fn for every m, n∈N. If f6=18 then f2·f3 is equal to :

Options:
A 54
B 6
C 36
D 18

The correct answer is:
54

Solution:
Given functional equation is

fm+n=fm+fn;m,n∈N ....1

Put m=n=3

f(3+3)=f(3)+f(3)

⇒f(3)=9∵f6=18

Put m=2,n=1 in equation 1

∴f(3)=f(2+1)=f(2)+f(1)

=f(1+1)+f(1)

=f(1)+f(1)+f(1)

9=3f(1)

⇒f(1)=3

∴f(2)=f(1+1)=f(1)+f(1)=6

Now,

f(2)·f(3)=(6)(9)=54

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ be defined by $\mathrm{f}(\mathrm{n})$ $=\left\{\begin{array}{l}\frac{\mathrm{n}+1}{2}, \text { if } \mathrm{n} \text { isodd } \\ \frac{\mathrm{n}}{2}, \text { if } \mathrm{n} \text { iseven }\end{array}\right.$ for all $\mathrm{n} \in$
State whether the function $f$ is bijective. Justify your answer.

Options:

None

Solution:
$\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$, defined by $\mathrm{f}(\mathrm{n})= \begin{cases}\frac{\mathrm{n}+1}{2}, & \text { if } \mathrm{n} \text { is odd } \\ \frac{\mathrm{n}}{2}, & \text { if } \mathrm{n} \text { is even }\end{cases}$
(a) $\mathrm{f}(1)=\frac{\mathrm{n}+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1, \mathrm{f}(2)=\frac{\mathrm{n}}{2}=\frac{2}{2}=1$ The elements 1,2 belonging to domain of $f$ have the same image 1 in its codomain $\Rightarrow \mathrm{f}$ is not one-one. $\quad \therefore$ it is not injective.
(b) Every member of codomain has pre-image in its domain e.g. 1 has two pre-images 1 and 2
$\Rightarrow f$ is onto. Thus $f$ is not one-one but it is onto $\Rightarrow \mathrm{f}$ is not bijective.

Topic: Mathematics, Relations and Functions (Class 12)
Let $f: N \rightarrow N$ defined by $f(x)=x^2+x+1, x \in N$, then $f$ is

Options:
A One-one onto
B Many one onto
C One-one but not onto
D None of these

The correct answer is:
One-one onto

Solution:
Let $x, y \in N$ such that $f(x)=f(y)$
Then $f(x)=f(y) \Rightarrow x^2+x+1=y^2+y+1$
$\Rightarrow(x-y)(x+y+1)=0 \Rightarrow x=y$ or $x=(-y-1) \notin N$
$\therefore f$ is one-one.
Again, since for each $y \in N$, there exist $x \in N$
$\therefore f$ is onto.

Topic: Mathematics, Functions, JEE Main
Let fn=∑r=110n6+rd and gn=∑r=1n6+rd, where n∈N, d≠0. If fngn is independent of n, then d is equal to

Options:
A 12
B -6
C 6
D -12

The correct answer is:
-12

Solution:
fngn=60n+5n10n+1d6n+nn+12d
=512+dn+50n2d6+d2n+n2d2
If d=-12, fngn=100

Topic: Mathematics, Sequences and Series, JEE Main
Let f:N→R be a function such that fx+y=2 fx fy for natural numbers x and y. If f1=2, then the value of α for which ∑k=110fα+k=5123220-1 holds, is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
4

Solution:
Given fx+y=2fx·fy & f1=2

Now putting x=1 & y=1 in fx+y=2fx·fy we get

f1+1=2f1·f1=2×22=23

So f2=23, Similarly

f3=25, f4=27.....

Now

∑k=110fα+k=∑k=1102fα·fk=5123220-1

⇒ 2fα∑k=110fk=5123220-1

⇒ 2fαf1+f2⋯f10=5123220-1

⇒ 2fα2+23+25…=5123220-1

⇒2fα22210−122−1=5123220−1

⇒2×2fα×220-13=5123220-1

Now comparing both side we get

4fα=512⇒fα=128

⇒fα=27⇒α=4.

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let fn=sinâ¡1Ãsinâ¡2Ã.....sinn, ânâN, then the number of elements in the set A=f1, f2,.........., f6 that are positive are

Options:
A 3
B 4
C 5
D 2

The correct answer is:
4

Solution:
sin⁡1, sin⁡2, sin⁡3 are positive and sin⁡4, sin⁡5, sin⁡6 are negative
Hence f1, f2, f3, f5 are positive while in f4 there are exactly one negative and three positive terms. Similarly, f(6) is negative

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let fn,x=∫ncosnxdx, with fn,0=0. If the expression ∑x=189f1,x simplifies to sin⁡asin⁡bsin⁡c, then the value of bac is (where a>b)

Options:
A 45
B 89
C 8945
D 4589

The correct answer is:
8945

Solution:
fn,x=nsinnxn+C
As fn,0=C=0
âfn,x=sinnx
Thus, âx=189f1,x=sinâ¡1+sinâ¡2+.....+sinâ¡89
=sin1+892sin89Ã12sin12
=sin45sin892sin12

Topic: Mathematics, Indefinite Integration, JEE Main
Let $f: N \rightarrow Y$ be a function defined as $f(x)=4 x+3$, where $Y=\{y \in N: y=4 x+3$ for some $x \in N\}$. Show that $\mathrm{f}$ is invertible and its inverse is

Options:
A
$g(y)=\frac{3 y+4}{3}$

B
$g(y)=4+\frac{y+3}{4}$

C
$g(y)=\frac{y+3}{4}$

D
$g(y)=\frac{y-3}{4}$

The correct answer is:
$g(y)=\frac{y-3}{4}$

Solution:
Function is increasing
$$
x=\frac{y-3}{4}=g(y)
$$

Topic: Mathematics, Functions, JEE Main, JEE Main 2008
Let f:R-0,1→R be a function such that fx+f11-x=1+x. Then f2 is equal to :

Options:
A 92
B 94
C 74
D 73

The correct answer is:
94

Solution:
Given,

fx+f11-x=1+x

x=2⇒f2+f-1=3 …1

x=-1⇒f-1+f12=0 …2

x=12⇒f12+f2=32 …3

On subtracting equation 3 and equation 2 we get,

f(2)-f(-1)=34 ........(4)

On adding equation 1 and equation 4 we get,

2f2=92

⇒ f2=94

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let f:→R→0,∞ be strictly increasing function such that limx→∞f7xfx=1. Then, the value of limx→∞f5xfx−1 is equal to

Options:
A 4
B 0
C 75
D 1

The correct answer is:
0

Solution:
Let, x be any poistive real number.

⇒x<5x<7x

⇒fx<f5x<f7x

⇒fxfx<f5xfx<f7xfx

⇒1<f5xfx<f7xfx

⇒limx→∞1<limx→∞f5xfx<limx→∞f7xfx

It is given that, limx→∞f7xfx=1

So, by using Sandwich theorem,

⇒limx→∞f5xfx=1

⇒limx→∞f5xfx-1=0

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let f:R-2,6→R be real valued function defined as fx=x+2x+1x2-8x+12. Then range of f is

Options:
A -∞,-214∪214,∞
B -∞,-214∪[0,∞)
C -∞,-214∪0,∞
D -∞,-214∪[1,∞)

The correct answer is:
-∞,-214∪[0,∞)

Solution:
Given,

f:R-2,6→R be real valued function defined as fx=x+2x+1x2-8x+12

Now let, y=x2+2x+1x2-8x+12=x+12x-2x-6 …1

Now differentiating the above function we get,

⇒dydx=-2x+15x-16x-22x-62

Now by wavy curve method we get,

So Graph of y=x+12x-2x-6 for given domain will be,

So, from graph range is y∈-∞,214∪[0,∞)

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let f:Râ2,â be a function defined as fx=x2-12ax+15-2a+36a2. If fx is surjective on R, then the value of a is equal to

Options:
A 92
B 112
C 132
D 152

The correct answer is:
132

Solution:
Rewriting the given function, we get,fx=x-6a2+15-2a
âµ fx is surjective on R
â15â2a=2â2a=13âa=132

Topic: Mathematics, Functions, JEE Main
Let f:R-{3}→R-{1} be defined by fx=x-2x-3. Let g : R→R be given as g(x)=2x-3. Then, the sum of all the values of x for which f-1x+g-1x=132 is equal to

Options:
A 7
B 2
C 5
D 3

The correct answer is:
5

Solution:
fx=y=x-2x-3

∴x=3y-2y-1

∴f-1x=3x-2x-1

& g(x)=y=2 x-3

∴x=y+32

∴g-1x=x+32

∵f-1x+g-1x=132

3x-2x-1+x+32=132

∴ x2-5x+6=0 has roots x1,x2

∴ sum of roots x1+x2=5

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let f:R-α6→R be defined by fx=5x+36x-α. Then the value of α for which fofx=x, for all x∈R-α6, is

Options:
A No such α exists
B 5
C 8
D 6

The correct answer is:
5

Solution:
We have,

fx=y=5x+36x-α …i

⇒5x+3=6xy-αy

⇒x(6y-5)=αy+3

⇒x=αy+36y-5

Hence,

f-1x=αx+36x-5 …ii

Now,

fofx=x

f(x)=f-1(x)

From equation i and ii, we have

5x+36x-α=αx+36x-5

Clearly, α=5

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Let f:RâB be a function defined by fx=tan-12x1+x2, then f is both one-one and onto when B is in the interval

Options:
A 0,π4
B 0,π3
C -Ï4,Ï4
D -Ï4,Ï4

The correct answer is:
-π4,π4

Solution:
Let, y=2x1+x2âyx2-2x+y=0
âµxâRâDâ¥0â4-4y2â¥0âyâ-1,1
-1â¤2x1+x2â¤1âtan-1-1â¤tan-12x1+x2â¤tan-11
-Ï4â¤fxâ¤Ï4

Topic: Mathematics, Functions, JEE Main
Let f:R→B, be a function defined by fx=tan-12x31+x2, then f is both one-one and onto when B, is the interval

Options:
A 0,x6
B 0,π6
C -π6,π6
D -π6,π6

The correct answer is:
-π6,π6

Solution:
Let y=2x31+x2⇒3yx2-2x+3y=0
∵x∈R⇒D≥0⇒4-12y2≥0⇒y∈-13,13
-13≤2x31+x2≤13⇒tan-1-13≤tan-12x31+x2≤tan-113
-π6≤fx≤π6

Topic: Mathematics, Functions, JEE Main
Let f:R→R and g:R→R be defined as fx=x+a,x<0|x-1|,x≥0 and gx=x+1,x<0(x-1)2+b,x≥0, where a, b are non-negative real numbers. If gof(x) is continuous for all x∈R, then a+b is equal to ______ .

Options:

The correct answer is:
1

Solution:
gfx=f(x)+1f(x)<0(f(x)-1)2+bf(x)≥0

gfx=x+a+1x+a<0; x<0|x-1|+1|x-1|<0; x≥0(x+a-1)2+bx+a≥0; x<0(|x-1|-1)2+b|x-1|≥0; x≥0

gfx=x+a+1x∈(-∞,-a); x∈(-∞,0)|x-1|+1x∈ϕ(x+a-1)2+bx∈[-a,∞); x∈(-∞,0)(|x-1|-1)2+bx∈R; x∈[0,∞)

gfx=x+a+1x∈(-∞,-a)(x+a-1)2+bx∈[-a,0)(|x-1|-1)2+bx∈[0,∞)

g(fx) is continuous

at x=-a & at x=0

1=b+1 & (a-1)2+b=b

b=0 & a=1

⇒a+b=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:R→R and g:R→R be defined as fx=x+a,x<0|x-1|,x≥0 and gx=x+1,x<0(x-1)2+b,x≥0, where a, b are non-negative real numbers. If gof(x) is continuous for all x∈R, then a+b is equal to ______ .

Options:

The correct answer is:
1

Solution:
gfx=f(x)+1f(x)<0(f(x)-1)2+bf(x)≥0

gfx=x+a+1x+a<0; x<0|x-1|+1|x-1|<0; x≥0(x+a-1)2+bx+a≥0; x<0(|x-1|-1)2+b|x-1|≥0; x≥0

gfx=x+a+1x∈(-∞,-a); x∈(-∞,0)|x-1|+1x∈ϕ(x+a-1)2+bx∈[-a,∞); x∈(-∞,0)(|x-1|-1)2+bx∈R; x∈[0,∞)

gfx=x+a+1x∈(-∞,-a)(x+a-1)2+bx∈[-a,0)(|x-1|-1)2+bx∈[0,∞)

g(fx) is continuous

at x=-a & at x=0

1=b+1 & (a-1)2+b=b

b=0 & a=1

⇒a+b=1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let f:R→R be a continuous and differentiable function such that f2=6 and f'2=148. If ∫6f(x)4t3dt=x-2gx, then limx→2gx is equal to

Options:
A 24
B 18
C 12
D 36

The correct answer is:
18

Solution:
limx→2gx=limx→2∫6fx4t3dtx-2 00 form.

By using L'Hospital rule
=limx→24.f3(x)⋅f'x1
=4f32f'2=4×6×6×6×148=18.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Let $f: R \rightarrow R$ be a continuous function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$.
Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$.
Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, for all $x \in R$

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is false

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Solution:
$f(x)=\frac{1}{e^x+2 e^{-x}}=\frac{e^x}{e^{2 x}+2}$
$f^{\prime}(x)=\frac{\left(e^{2 x}+2\right) e^x-2 e^{2 x} \cdot e^x}{\left(e^{2 x+2}\right)^2}$
$f^{\prime}(x)=0 \quad \Rightarrow e^{2 x}+2=2 e^{2 x}$
$\mathrm{e}^{2 \mathrm{x}}=2 \quad \Rightarrow \mathrm{e}^{\mathrm{x}}=\sqrt{2}$
maximum $f(x)=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}}$
$0 < f(x) \leq \frac{1}{2 \sqrt{2}} \quad \forall x \in R$
Since $0 < \frac{1}{3} < \frac{1}{2 \sqrt{2}} \quad \Rightarrow$ for some $c \in R$
$f(c)=\frac{1}{3}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2010
Let f:R→R be a continuous function such that fx+fx+1=2 for all x∈R. If I1=∫08fxdx and I2=∫-13fxdx, then the value of I1+2I2 is equal to ________.

Options:

The correct answer is:
16

Solution:
fx+fx+1=2

Replace x by x+1

fx+1+fx+2=2

By subtracting the equations, we get

fx=fx+2

⇒fx is periodic with period =2

I1=∫08fxdx=4∫02fxdx

=4∫01fx+f1+xdx=8

Similarly, I2=2×2=4

Hence, I1+2I2=16

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let f : R→R be a continuous function. Then limx→π/4π4∫2sec2xf(x)dxx2-π216 is equal to:

Options:
A f(2)
B 2f(2)
C 2f(2)
D 4f(2)

The correct answer is:
2f(2)

Solution:
limx→π4π4∫2sec2xfxdxx2-π216

Using L-Hospital's rule, we get

=limx→π4π4fsec2x·2·secx·secx·tanx-02x

=π4·2(2)2·(1)·f(2)2·π4

=2f2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3}{x-2} d t$ equals

Options:
A
24

B
36

C
12

D
18

The correct answer is:
18

Solution:
$$
\lim _{x \rightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t
$$
Applying $L$ Hospital rule
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left[4 f(x)^2 f^{\prime}(x)\right]=4 f(2)^3 f^{\prime}(2) \\
& =4 \times 6^3 \times \frac{1}{48}=18 .
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2005
Let f:R→R be a differentiable function satisfying f'3+f'2=0. Then limx→01+f3+x-f31+f2-x-f21x is equal to

Options:
A 1
B e
C e2
D e-1

The correct answer is:
1

Solution:
This limit is 1∞ form

∴limx→01+f3+x-f31+f2-x-f21x=elimx→01x1+f3+x-f31+f2-x-f2-1

=elimx→01xf3+x-f3-f2-x+f21+f2-x-f200 form

=elimx→0 f'3+x+f'2-xx-f'2-x+1+f2-x-f2

=e0=1.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let f:R→R be a differentiable function such that fπ4=2,fπ2=0 and f'π2=1 and let gx=∫xπ4f'tsect+tantsectftdt for x∈π4,π2. Then limx→π2-gx is equal to

Options:
A 2
B 3
C 4
D -3

The correct answer is:
3

Solution:
Given,

gx=∫xπ4f'tsect+tantsectftdt

⇒gx=∫xπ4dft·sect⇒gx=ftsectxπ4

gx=fπ4secπ4-fx·secx

gx=2-fxsecx=2-fxcosx

Now taking limit both side, we get

limx→π2-gx=2-limx→π2-fxcosx

Using L'Hospital Rule

=2-limx→π2-f'x-sinx

=2+f'π2sinπ2=2+11=3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let f:R→R be a differentiable function with f0=1 and satisfying the equation

fx+y=fxf'y+f'xfy for all x, y∈R.

Then, the value of loge⁡f4 is

Options:

The correct answer is:
2

Solution:
Px,y: fx+y=fxf'y+f'xfy ∀ x, y∈R
P0,0: f0=f0f'0+f'0f0
⇒1=2f'0
⇒f'0=12
Px,0: fx=fx.f'0+f'x.f0
⇒ fx=12fx+f'x
⇒ f'x=12fx
⇒ fx=e12x
⇒ ln⁡f4=2

Topic: Mathematics, Differential Equations, JEE Main
Let f:R→R be a function defined as
fx= 5, if x≤1a+bx, if 1<x<3b+5x, if 3≤x<5 30, if x≥5
Then f is:

Options:
A continuous if a=-5 and b=10
B continuous if a=0 and b=5
C not continuous for any values of a and b
D continuous if a=5 and b=5

The correct answer is:
not continuous for any values of a and b

Solution:
fx=5:x≤1a+bx:1<x<3b+5x:3≤x<530:x≥5
For f(x) to be continuous at x=5
limx→5-fx=limx→5+fx=f5
⇒limx→5-b+5x=30
⇒b+25=30⇒b=5
For f(x) to be continuous at x=1
limx→1+fx=limx→1-fx=f1
⇒limx→1+a+5x=5
⇒a=0
Now at x=3,
LHL=limx→3-0+5x=15
RHL=limx→3+5+5x=20
Hence function f(x) is discontinuous at x=3 for all a, b∈R .

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Let f:RâR be a function defined as fx=6ifxâ¤1a+bx if1<x<3b+5xif3â¤x<515ifxâ¥5. Then, f is

Options:
A continuous in R if a=-5 and b=10
B continuous in R if a=5 and b=5
C continuous in R if a=0 and b=5
D not continuous in R for any values of a and b

The correct answer is:
not continuous in R for any values of a and b

Solution:
Given, fx=6ifx≤1a+bxif1<x<3b+5xif3≤x<515ifx≥5
Now, f1=6,f1-=6,f1+=a+b…..i
Also, f3=b+15,f3-=a+3b,f3+=b+15…..ii
Also, f5=15,f5-=b+25,f5+=15…….iii
From i,ii and iii , clearly, f is not continuous for any values of a and b .

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f : R→R be a function defined as

fx=sin(a+1)x+sin2x2x,if x<0b,if x=0x+bx3-xbx5/2,if x>0

If f is continuous at x=0, then the value of a+b is equal to :

Options:
A -52
B -2
C -3
D -32

The correct answer is:
-32

Solution:
f(x) is continuous at x=0

limx→0+fx=f0=limx→0-fx ...1

f(0)=b ...2

limx→0-fx=limx→0-sin(a+1)x2x+sin2x2x

=a+12+1 ...3

limx→0+fx=limx→0+x+bx3-xbx5/2

=limx→0+x+bx3-xbx5/2x+bx3+x

=limx→0+xx1+bx2+1=12 ...4

Use (2),(3) & (4) in (1)

12=b=a+12+1

⇒b=12,a=-2
a+b=-32

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let f:RâR be a function defined as fx=x2-6x2+2, then f is

Options:
A one-one but not onto
B one-one and onto
C onto but not one-one
D neither one-one nor onto

The correct answer is:
neither one-one nor onto

Solution:
Since, fx=f-x
∴f is not one-one.
Let, y∈R. Then, fx=y⇒y=x2-6x2+2
⇒x2=6+2y1-y
For x to be real, 6+2y1-y≥0
and 1-y≠0⇒y+3y-1≤0 and y≠1
⇒-3≤y<1
∴ Range of f=-3,1⊂R
∴ f is not onto

Topic: Mathematics, Functions, JEE Main
Let $f: R \rightarrow R$ be a function defined as $f(x)=x|x|$; for each $x \in R, R$ being the set of real numbers. Which one of the following is correct?

Options:
A $f$ is one-one but not onto
B $f$ is onto but not one-one
C $f$ is both one-one and onto
D $f$ is neither one-one nor onto

The correct answer is:
$f$ is both one-one and onto

Solution:
Given, $f(x)=x|x|$
If $\quad f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|$
$\Rightarrow x_{1}=x_{2}$
$\therefore f(x)$ is one-one.
Also, range of $f(x)=$ co-domain of $f(x)$
$\therefore f(x)$ is onto.
Hence, $f(x)$ is both one-one and onto.

Topic: Mathematics, Functions, NDA, NDA 2009 (Phase 1)
Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $\mathrm{f}(\mathrm{x})=\frac{2 \mathrm{x}+1}{3}$. If $\alpha$ is an element in the domain of $\mathrm{f}$ whose image is $\frac{1}{\alpha}$, then the sum of all possible values of such $\alpha$ is

Options:
A $\frac{-1}{2}$
B $\frac{1}{2}$
C $\frac{5}{2}$
D 0

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

Solution:
Given that : $f(x)=\frac{2 x+1}{3}$ ...(i)
Also, $f(\alpha)=\frac{1}{\alpha}$
$\begin{aligned} & \Rightarrow \frac{2 \alpha+1}{3}=\frac{1}{\alpha} \\ & \Rightarrow 2 \alpha^2+\alpha=3 \Rightarrow 2 \alpha^2+\alpha-3=0 \\ & \Rightarrow(\alpha-1)(2 \alpha+3)=0 \\ & \Rightarrow \alpha=-\frac{3}{2}, 1\end{aligned}$
Then sum of all possible value of $\alpha=-\frac{3}{2}+1=-\frac{1}{2}$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2023 (16 May Shift 1)
Let f:R→R be a function defined by fx=max x,x2.Let S denote the set of all points in R,where f is not differentiable.Then :

Options:
A 0,1
B 0
C ϕ (an empty set)
D 1

The correct answer is:
0,1

Solution:
From the graph we can say maxx,x2 has 2 sharp corners at x=0,1.

⇒ Function is non-derivable at x=0 & 1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Let $f: R \rightarrow R$ be a function defined by $f(x)=\operatorname{Min}\{x+1,|x|+1\}$. Then which of the following is true?

Options:
A
$f(x) \geq 1$ for all $x \in R$

B
$f(x)$ is not differentiable at $x=1$

C
$f(x)$ is differentiable everywhere

D
$f(x)$ is not differentiable at $x=0$

The correct answer is:
$f(x)$ is differentiable everywhere

Solution:
$f(x)=\min \{x+1,|x|+1\}$
$f(x)=x+1 \forall x \in R$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2007
Let f: R→R be a function defined by fx=x-3n1x-5n2, n1, n2∈N. The, which of the following is NOT true?

Options:
A For n1=3, n2=4, there exists α∈3, 5 where f attains local maxima.
B For n1=4, n2=3, there exists α∈3, 5 where f attains local maxima.
C For n1=3, n2=5, there exists α∈3, 5 where f attains local maxima.
D For n1=4, n2=6, there exists α∈3, 5 where f attains local maxima.

The correct answer is:
For n1=3, n2=5, there exists α∈3, 5 where f attains local maxima.

Solution:
Given,

fx=x-3n1x-5n2

Differentiating the function to get maxima and minima, we get

f'x=x-3n1-1x-5n2-1n1+n2x-5n1+3n2n1+n2

Now checking the option, we get

Option (3) is incorrect since

for n1=3, n2=5

f'x=8x-32x-54x-308

minima at x=308.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{x-m}{x-n}$, where $m \neq n$. Then

Options:
A $f$ is one-one onto
B $f$ is one-one into
C $f$ is many one onto
D $f$ is many one into

The correct answer is:
$f$ is one-one into

Solution:
For any $x, y \in R$, we have
$f(x)=f(y) \Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n} \Rightarrow x=y$
$\therefore f$ is one-one.
Let $\alpha \in R$ such that $f(x)=\alpha \Rightarrow \frac{x-m}{x-n}=\alpha \Rightarrow x=\frac{m-n \alpha}{1-\alpha}$ Clearly $x \notin R$ for $\alpha=1$. So, $f$ is not onto.

Topic: Mathematics, Functions, JEE Main
Let f:R→R be a function defined fx=2e2xe2x+e. Then f1100+f2100+f3100+…+f99100 is equal to ______.

Options:

The correct answer is:
99

Solution:
fx=2e2xe2x+e

f1−x=2e21−xe21−x+e=2ee+e2x

Now, fx+f1−x=2

i.e. f1100+f99100=2

⇒f1100+f2100+...f99100=2×49+f50100

=98+1=99

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Let f:R→R be a function defined fx=x1+x41/4 and gx=ffffx then 18∫025x2gxdx

Options:
A 33
B 36
C 42
D 39

The correct answer is:
39

Solution:
Given,

fx=x1+x41/4

Now, finding

fofx=fx1+f4x1/4=x1+x41/41+x41+x41/4=x1+2x41/4

And fffx=fx1+2f4x1/4=x1+x41/41+2x41+x41/4=x1+3x41/4

⇒ffffx=x1+4x41/4=gx

Now finding,

18∫025x2gxdx

=18∫025x31+4x41/4dx

Let 1+4x4=t4

⇒16x3dx=4t3dt

So, the integral becomes

184∫13t3dtt

=92∫13t2dt

=92t3313

=32·26=39

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function given by
$f(x)=\left\{\begin{array}{ll}
\frac{1-\cos 2 x}{x^2}, & x < 0 \\
\alpha, & x=0, \\
\frac{\beta \sqrt{1-\cos x}}{x}, & x>0
\end{array}\right.$
where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :

Options:
A 3
B 12
C 48
D 6

The correct answer is:
12

Solution:
$\begin{aligned} & f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 x}{x^2}=2=\alpha \\ & f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{2 \frac{x}{2}}=\frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \beta=2 \sqrt{2} \\ & \alpha^2+\beta^2=4+8=12\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let f:R→R be a function such that f2=4 and f'2=1. Then, the value of limx→2x2f2-4fxx-2 is equal to:

Options:
A 4
B 8
C 16
D 12

The correct answer is:
12

Solution:
Given, f2=4

So, the given limit is in 00 form.

So, limx→2x2f2-4fxx-2

Apply L'Hospital Rule

limx→22xf2-4f'x1

=44-41=12

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let f:R→R be a function such that f2-x=f2+x and f4-x=f4+x, for all x∈R and ∫02fxdx=5. Then the value of ∫1050fxdx is

Options:
A 100
B 125
C 80
D 200

The correct answer is:
100

Solution:
f2-x=f2+x and f4-x=f(4+x)
Then curve is symmetric about x=2 and x=4
∴ The curve is periodic with period 4
Given ∫02fxdx=5
∫04fxdx=2∫02fxdx=10
∫1050fxdx=∫24+2124+2fxdx
=12-2∫04fxdx
=10×10=100

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2015 (11 Apr Online)
Let f:R→R be a function such that fx=x2+2x+1x2+1. Then

Options:
A fx is many-one in -∞,-1
B fx is many-one in 1,∞
C fx is one-one in [1,∞) but not in -∞,∞
D fx is one-one in -∞,∞

The correct answer is:
fx is one-one in [1,∞) but not in -∞,∞

Solution:
Given,

fx=x2+2x+1x2+1

f'(x)=x2+1(2x+2)-(x2+2x+1)2xx2+12

⇒f'(x)=2x3+2x2+2x+2-2x3-4x2-2xx2+12

⇒f'(x)=2(1-x2)x2+12=-2(x+1)(x-1)x2+12

Clearly f(x) is one-one in -∞,-1 and also in (1,∞) but f(x) is not one-one in -∞,∞

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (29 Jan Shift 1)