datavorous/entrance-exam-dataset huggingface dataset:
Let f: R→R be a function such that fx≤x2, for all x∈R. Then, at x=0, f is

Options:
A differentiable but not continuous
B neither continuous nor differentiable
C continuous as well as differentiable
D continuous but not differentiable

The correct answer is:
continuous as well as differentiable

Solution:
We know that if y≤k2, ⇒-k≤y≤k, k∈R.

Given fx≤x2, ∀ x∈R

⇒-x2≤fx≤x2 ∀ x∈R

Now, limx→0-x2=limx→0x2=0,

Hence, by using sandwich theorem, limx→0fx=0

Further f0≤0 ⇒f0=0

Since, limx→0fx=f0=0

Hence, fx is continuous at x=0

Now, if x≠0, then -x≤fxx≤x

Thus, by using sandwich theorem, we have limx→0-x≤limx→0fxx≤limx→0x

⇒0≤limx→0fxx≤0

⇒limx→0fxx=0 ...i

We know that a function fx is differentiable at a point x=a, if limh→0fa-h-a-h=limh→0fa+h-ah

Now, the left-hand derivative at x=0 is

limh→0f0-h-f0-h=limh→0f-h-h=0, ( by using equation i)

And, the right-hand derivative at x=0 is

limh→0f0+h-f0h=limh→0fhh=0, ( by using equation i)
Hence, fx is differentiable at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2014 (19 Apr Online)
Let, f:R→R be a function such that fx=x3+x2f'1+xf''2+f'''3, ∀x∈R. Then f2 equals

Options:
A 30
B 8
C -4
D -2

The correct answer is:
-2

Solution:
Let, fx=x3+ax2+bx+c

⇒f'x=3x2+2ax+b

⇒f''x=6x+2a

⇒f'''x=6

According to the question,

a=f'1=3+2a+b⇒a+b=-3 ...i

b=f''2=12+2a⇒2a-b=-12 ...ii

c=f'''3⇒c=6

Solving equations i & ii, we get, a= -5 & b=2

⇒fx=x3-5x2+2x+6

∴ f2=8-20+4+6=-2.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let f:RâR be a function such that fx+y3=fx+fy3, f0=0 and f â² 0 =5, then

Options:
A fx is a quadratic function
B fx is continuous but not differentiable
C fx is differentiable in R
D fx is bounded in R

The correct answer is:
fx is differentiable in R

Solution:
We have, fx+y3=fx+fy3,f0=0 and f ′ 0 =5
f ′ x = lim h→0 f x+h −f x h
=limh→0⁡f3x+3h3-f3x+03h
=limh→0f3x+f3h3-f3x+f03h
= lim h→0 f 3h −f 0 3h = f ′ 0 =5
∴fx=5x+c∴f0=0⇒c=0
Hence, fx=5x

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in \mathbb{R}$

If $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$, then which one of the following is incorrect?

Options:
A $\mathrm{f}(\mathrm{x})$ is continuous, $\forall \mathrm{x} \in \mathrm{R}$
B $\mathrm{f}^{\prime}(\mathrm{x})$ is constant, $\forall \mathrm{x} \in \mathrm{R}$
C $\mathrm{f}(\mathrm{x})$ is differentiable, $\forall \mathrm{x} \in \mathrm{R}$
D $\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is differentiable only in a finite interval containing zera.

Solution:
Let $f(x+y)=f(x)+f(y), \forall x, y \in R$

Put $x=0=y$

$\Rightarrow f(0)=f(0)+f(0)$

$\Rightarrow f(0)=0$

Now, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$f^{\prime \prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)}{h}$

Now, $f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h}$

$$

\begin{array}{l}

\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}=\mathrm{f}^{\prime}(0) \\

\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{f}(0)+\mathrm{C}

\end{array}

$$

But $f(0)=0$

$$

\therefore \mathrm{C}=0

$$

Hence, $f(x)=x f(0), \forall x \in R$

Clearly, $\mathrm{f}(\mathrm{x})$ is everywhere continuous and differentiable and $\mathrm{f}^{\prime}(\mathrm{x})$ is constant.

$\forall \mathrm{x} \in \mathrm{R}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f: R \rightarrow R$ be a function such that $\lim _{x \rightarrow \infty} f(x)=M>0$. Then which of the following is false?

Options:
A $\lim _{x \rightarrow \infty} x \sin (1 / x) f(x)=M$
B $\lim _{x \rightarrow \infty} \sin (f(x))=\sin M$
C $\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x)=M$
D $\lim _{x \rightarrow \infty} \frac{\sin x}{x} \cdot f(x)=0$

The correct answer is:
$\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x)=M$

Solution:
$\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x)=\lim _{x \rightarrow \infty} \frac{\sin \left(e^{-x}\right)}{e^{-x}} \frac{x}{e^{x}} f(x)$
$=1 \times(0) \times M=0$

Topic: Mathematics, Limits, JEE Main
Let f:R→R be a function which satisfies fx+y=fx+fy, ∀x, y∈R . If f1=2 and gn=∑k=1n-1fk, n∈N then the value of n, for which gn=20, is

Options:
A 5
B 20
C 4
D 9

The correct answer is:
5

Solution:
∵ fx+y=fx+fy

∴ fx=kx, ∀x∈R

Now f1=2 ⇒ k=2

∴ fx=2x, ∀x∈R

gn=f1+f2+……..+fn-1

=21+2+3+……+n-1

=2·n-12n=20

⇒n2-n-20=0

⇒n=5.

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let f:R→R be a periodic function such that fT+x=1+1-3fx+3fx2-fx313 where T is a fixed positive number, then period of fx is

Options:
A T
B 2T
C 3T
D None of these

The correct answer is:
2T

Solution:
Given : fT+x=1+1-fx313
=1+1-fx
⇒fT+x+fx=2 ... (1)
⇒f2T+x+fT+x=2 ... (2)
Subtracting (1) from (2)
f2T+x-fx=0
f2T+x-fx=0
Also, T is positive therefore period of fx=2T

Topic: Mathematics, Functions, JEE Main
Let $f: R \rightarrow R$ be a positive increasing function with $\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1$. Then $\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=$

Options:
A
$\frac{2}{3}$

B
$\frac{3}{2}$

C
3

D
1

The correct answer is:
1

Solution:
$f(x)$ is a positive increasing function
$$
\begin{aligned}
& \Rightarrow 0 < f(x) < f(2 x) < f(3 x) \\
& \Rightarrow 0 < 1 < \frac{f(2 x)}{f(x)} < \frac{f(3 x)}{f(x)} \\
& \Rightarrow \lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}
\end{aligned}
$$
By sandwich theorem.
$$
\Rightarrow \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=1
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2010
Let f:R→R be continuous function satisfying fx+fx+k=n, for all x∈R where k>0 and n is a positive integer. If I1=∫04nkfxdx and I2=∫-k3kfxdx, then

Options:
A I1+2I2=4nk
B I1+2I2=2nk
C I1+nI2=4n2 K
D I1+nI2=6n2k

The correct answer is:
I1+nI2=4n2 K

Solution:
Given,

fx+fx+k=n ....(i)

Replacing x→x+k in above equation we get,

fx+k+fx+2k=n .....(ii)

Subtraction equation (i) from equation (ii) we get,

⇒fx=fx+2k

fx is periodic with period 2k

Now, I1=∫04nkfxdx=2n∫02kfxdx

I2=∫-k3kfxdx=2∫02kfxdx

Now integrating both side of fx+fx+k=n with limit ∫0k we get,

⇒∫0kfxdx+∫0kfx+kdx=nk

⇒∫0kfxdx+∫k2kfxdx=nk

⇒∫02kfxdx=nk

⇒I1=2n2k,I2=2nk

⇒I1+nI2=4n2k

Topic: Mathematics, Definite Integration, JEE Main
Let f:R→R be continuous function satisfying fx+fx+k=n, for all x∈R where k>0 and n is a positive integer. If I1=∫04nkfxdx and I2=∫-k3kfxdx, then

Options:
A I1+2I2=4nk
B I1+2I2=2nk
C I1+nI2=4n2 K
D I1+nI2=6n2k

The correct answer is:
I1+nI2=4n2 K

Solution:
Given,

fx+fx+k=n ....(i)

Replacing x→x+k in above equation we get,

fx+k+fx+2k=n .....(ii)

Subtraction equation (i) from equation (ii) we get,

⇒fx=fx+2k

fx is periodic with period 2k

Now, I1=∫04nkfxdx=2n∫02kfxdx

I2=∫-k3kfxdx=2∫02kfxdx

Now integrating both side of fx+fx+k=n with limit ∫0k we get,

⇒∫0kfxdx+∫0kfx+kdx=nk

⇒∫0kfxdx+∫k2kfxdx=nk

⇒∫02kfxdx=nk

⇒I1=2n2k,I2=2nk

⇒I1+nI2=4n2k

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let f:R→R be defined as fx=2sin-πx2, if x<-1ax2+x+b, if -1≤x≤1sinπx, if x>1

If fx is continuous on R, then a+b equals :

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

The correct answer is:
-1

Solution:
fx is continuous on R

⇒f1-=f1=f1+

a+1+b=limx→1+sinπx

a+1+b=0⇒a+b=-1

⇒ Also

f-1-=f-1=f-1+

limx→-12sin-πx2=a-1+b

a-1+b=2

Either a-1+b=2 or a-1+b=-2

a+b=3 ...2 or a+b=-1 ...3

from 1 and 2 ⇒a+b=3=-1 (reject)

from 1 and 3 ⇒a+b=-1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let f:R→R be defined as fx=2 x-1 and g:R-1→R. be defined as gx=x-12x-1. Then the composition function fgx is:

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

The correct answer is:
one-one but not onto

Solution:
fgx=2gx-1=22x-12x-1-1.

=xx-1=1+1x-1

Range of fgx=R-1

Range of fgx is not onto

& fgx is one-one

So fgx is one-one but not onto.

one-one, into

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Let $f: R \rightarrow R$ be defined as $f(x)=3 x$. Choose the correct answer.
(a) f is one-one onto
(b) fis many-one onto
(c) f is one-one but not onto
(d) fis neither one-one nor onto

Options:

None

Solution:
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $\mathrm{f}(\mathrm{x})=3 \mathrm{x}$
(a) $f\left(x_1\right)=3 x_1, f\left(x_2\right)=3 x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 3 x_1=3 x_2 \Rightarrow x_1=x_2 \Rightarrow f$ is one-one
(b) for every member $y$ belonging to co-domain has pre-image $x$ in domain of $\mathrm{f}$.
$\because \quad y=3 x \Rightarrow x=\frac{y}{3}, \therefore$ fis onto
$\therefore \mathrm{f}$ is one-one and onto. Option (a) is correct.

Topic: Mathematics, Relations and Functions (Class 12)
Let f:R→R be defined as fx=-43x3+2x2+3x, x>0 3xex , x≤0. Then f is increasing function in the interval

Options:
A -12, 2
B 0, 2
C -1, 32
D -3, -1

The correct answer is:
-1, 32

Solution:
We know that, ddxxn=nxn-1 and ddxex=ex

Hence, ddx-43x3+2x2+3x=-43×3x2+2×2x+3=-4x2+4x+3

And, using product rule, we getddx3xex=3xddxex+3exddxx=3xex+3ex=3exx+1

⇒f'x=-4x2+4x+3,x>03ex(1+x),x≤0

For x>0, f'(x)=-4x2+4x+3=-2x+12x-3

The sign scheme of f'x is

We know that, if f'x>0 then for those value of x, fx is increasing.

⇒f(x) is increasing in -12, 32.

For x≤0, f'(x)=3ex(1+x)

As ex>0, ∀ x∈R

⇒f'(x)>0 ∀ x∈(-1, 0).

So, in complete domain, f(x) is increasing in -1, 0∪-12, 32=-1, 32.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let f:R→R be defined as

fx=-55x,if x<-52x3-3x2-120x,if-5≤x≤42x3-3x2-36x-336,if x>4

Let A={x∈R:f is increasing}. Then A is equal to:

Options:
A -5,∞
B -5, -4∪4, ∞
C -∞,-5∪-4,∞
D -∞,-5∪4,∞

The correct answer is:
-5, -4∪4, ∞

Solution:
fx=-55x,if x<-52x3-3x2-120x,if-5≤x≤42x3-3x2-36x-336,if x>4

f'x=-55,if x<-56x2-6x-120,if-5≤x≤46x2-6x-36,if x>4

f'x=-55;x<-56x-5x+4;-5<x<46x-3x+2;x>4

fx is increasing

⇒f'x>0

So, x∈-5,-4∪4,∞

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let f:R→R be defined as fx=e-xsinx. If F:0, 1→R is a differentiable function such that Fx=∫0xftdt, then the value of ∫01F'(x)+f(x)exdx lies in the interval

Options:
A 327360,329360
B 330360,331360
C 331360,334360
D 335360,336360

The correct answer is:
330360,331360

Solution:
Given f(x)=e-xsinx

Now, Fx=∫0xftdt

Using Newton Leibnitz rule i.e. ddx∫uxvxftdt=fvx·v'x-fux·u'x,

⇒F'(x)=f(x)

Now, I=∫01F'(x)+f(x)exdx

⇒I=∫01(f(x)+f(x))·exdx

⇒I=2∫01f(x)·exdx

⇒I=2∫01e-xsinx·exdx

⇒I=2∫01sinxdx

⇒I=2-cosx01

⇒I=2(1-cos1)

Using the expansion of cosx=1-x22!+x44!-x66!+...

⇒I=21-1-12+14!+16!+...

⇒I=1-24!+26!-29!...

⇒1-24!<I<1-24!+26!

⇒1112<I<331360

⇒I∈1112,331360

⇒I∈330360,331360.

Topic: Mathematics, Definite Integration, JEE Main
Let f:R→R be defined as fx=e-xsinx. If F:0, 1→R is a differentiable function such that Fx=∫0xftdt, then the value of ∫01F'(x)+f(x)exdx lies in the interval

Options:
A 327360,329360
B 330360,331360
C 331360,334360
D 335360,336360

The correct answer is:
330360,331360

Solution:
Given f(x)=e-xsinx

Now, Fx=∫0xftdt

Using Newton Leibnitz rule i.e. ddx∫uxvxftdt=fvx·v'x-fux·u'x,

⇒F'(x)=f(x)

Now, I=∫01F'(x)+f(x)exdx

⇒I=∫01(f(x)+f(x))·exdx

⇒I=2∫01f(x)·exdx

⇒I=2∫01e-xsinx·exdx

⇒I=2∫01sinxdx

⇒I=2-cosx01

⇒I=2(1-cos1)

Using the expansion of cosx=1-x22!+x44!-x66!+...

⇒I=21-1-12+14!+16!+...

⇒I=1-24!+26!-29!...

⇒1-24!<I<1-24!+26!

⇒1112<I<331360

⇒I∈1112,331360

⇒I∈330360,331360.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\sin (|\mathrm{x}|)$
Which one of the following is correct?

Options:
A fis not differentiable only at 0
B fis differentiable at 0 only
C fis differentiable everywhere
D fis non-differentiable at many points

The correct answer is:
fis not differentiable only at 0

Solution:
Given function is $: \mathrm{f}(\mathrm{x})=\sin |\mathrm{x}|$
$=\left\{\begin{array}{ll}\sin (\mathrm{x}), & \mathrm{x} \geq 0 \\ \sin (-\mathrm{x}), & \mathrm{x} < 0\end{array}\right.$
$=\left\{\begin{array}{cl}\sin \mathrm{x}, & \mathrm{x} \geq 0 \\ -\sin \mathrm{x}, & \mathrm{x} < 0\end{array}\right.$
$\mathrm{LHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{0-\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (-h)-0}{-h}=-1$
$\mathrm{RHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{0+\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (\mathrm{h}-0)}{\mathrm{h}}=1$
$\mathrm{LHD} \neq \mathrm{RHD}$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

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

fx=λx2-5x+6μ5x-x2-6x<2etan(x-2)x-[x]x>2μx=2

where x is the greatest integer less than or equal to x. If f is continuous at x=2, then λ+μ is equal to :

Options:
A e(-e+1)
B e(e-2)
C 1
D 2e-1

The correct answer is:
e(-e+1)

Solution:
If fx is continuous at x=2, then limx→2+fx=limx→2-fx=limx→2fx

Here,

limx→2+fx=limx→2+etan(x-2)x-2=e1=e

limx→2-fx=limx→2-λx2-5x+6μ5x-x2-6=limx→2--λ(x-2)(x-3)μ(x-2)(x-3)=-λμ

∵x2-5x+6=x-2x-3 if x < 2 or x > 3-x-2x-3 if 2 < x < 3

For continuity μ=e=-λμ⇒μ=e, λ=-e2

λ+μ=e(-e+1)

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let f:R→R be defined as fx=x3(1-cos2x)2loge1+2xe-2x1-xe-x2, x≠0α, x=0

If f is continuous at x=0, then α is equal to:

Options:
A 1
B 3
C 0
D 2

The correct answer is:
1

Solution:
A function fx is continuous at a point x=a, if limx→afx=fa.

Thus, for continuity of the given function, we have

limx→0fx=α

⇒ limx→0x3(1-cos2x)2loge1+2xe-2x1-xe-x2=α

Using, cos2x=1-2sin2x, logemn=logem-logen & logemn=nlogem,

we get

limx→0x32sin2x2loge1+2xe-2x-loge1-xe-x2=α

⇒ limx→0x34sin4xloge1+2xe-2x-2loge1-xe-x=α

⇒ limx→0x44xsin4x2xe-2x·loge1+2xe-2x2xe-2x+2xe-x·loge1-xe-x-xe-x=α

Now, using the standard limits limx→0sinxx=1 & limx→0loge1+xx=1, we get

⇒ limx→014x2xe-2x+2xe-x=α⇒ limx→02x4xe-2x+e-x=α

⇒ limx→012e-2x+e-x=α

⇒ 122=α

⇒ α=1.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let $f: R \rightarrow R$ be defined as $f(x)=x^4$. Choose the correct answer.
(a) $f$ is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) fis neither one-one nor onto

Options:

None

Solution:
$\mathrm{f}(-1)=(-1)^4=1, \mathrm{f}(1)=1^4=1$
$\therefore-1,1$ have the same image $1 \Rightarrow f$ is not one-one
Further $-2$ in the codomain of $f$ has no pre-image in its domain.
$\therefore \mathrm{f}$ is not onto i.e. f is neither one-one nor onto
Option $(d)$ is correct.

Topic: Mathematics, Relations and Functions (Class 12)
Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{1+x^{2}}, x \in R .$ Then the range of $f$ is

Options:
A $\left[-\frac{1}{2}, \frac{1}{2}\right]$
B $R-[-1,1]$
C $R-\left[-\frac{1}{2}, \frac{1}{2}\right]$
D $(-1,1)-\{0\}$

The correct answer is:
$\left[-\frac{1}{2}, \frac{1}{2}\right]$

Solution:
$f(x)=\frac{x}{1+x^{2}}, x \in R$

Let, $y=\frac{x}{1+x^{2}}$

$\Rightarrow y x^{2}-x+y=0 \Rightarrow x=\frac{1 \pm \sqrt{1-4 y^{2}}}{2}$

$\Rightarrow 1-4 y^{2} \geq 0$

$\Rightarrow 1 \geq 4 y^{2}$

$\Rightarrow|y| \leq \frac{1}{2}$

$\Rightarrow \quad-\frac{1}{2} \leq y \leq \frac{1}{2}$

$\Rightarrow$ The range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let $f: R \rightarrow R$ be defined by $f(x)=\left\{\begin{array}{ll}k-2 x, & \text { if } x \leq-1 \\ 2 x+3, & \text { if } x>-1\end{array}\right.$. If $f$ has a local minimum at $x=-1$, then a possible value of $\mathrm{k}$ is

Options:
A
0

B
$-\frac{1}{2}$

C
$-1$

D
1

The correct answer is:
$-1$

Solution:
$f(x)=k-2 x \quad$ if $x \leq-1$
$=2 x+3 \quad$ if $x>-1$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2010
Let f: R→R be defined by fx=x-1x+1, then f is

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

The correct answer is:
neither one-one nor onto

Solution:
f: R→R, fx=x-1x+1
For x=0, fx=-1

And fx=1-1x1+1x

If x→±∞, fx→1

Thus, the graph of the function fx is given by

If a line parallel to the x-axis intersects the graph of a function at more than one point, then the function is many-one.

And, it is clear from the graph that, the given function is many-one.

Also, the range of the function is [-1, 1) which is not same as the co-domain which is R and if the co-domain of a function and range are not same then the function is not onto.

Hence, the given function is neither one-one nor onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2014 (19 Apr Online)
Let $f: R \rightarrow R$ be defined by $f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}$ for all $x \in R$. Then,

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

The correct answer is:
$f$ is neither one-one nor onto

Solution:
We have,
$$
\begin{aligned}
f(x) &=x^{2}-\frac{x^{2}}{1+x^{2}} \\
f(-x) &=(-x)^{2}-\frac{(-x)^{2}}{1+(-x)^{2}} \\
&=x^{2}-\frac{x^{2}}{1+x^{2}}=f(x) \\
\therefore f(-x)=f(x) &
\end{aligned}
$$
So, $f(x)$ is many one Again, $f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}$
$$
=\frac{x^{2}+x^{4}-x^{2}}{1+x^{2}}=\frac{x^{4}}{1+x^{2}}
$$
$\therefore$ Range of $f(x)$ will be $[0, \infty)$ But, co-domain of $f(x)=R$
$\therefore$ Range $\neq$ co-domain. So, $f(x)$ is onto.

Topic: Mathematics, Functions, WBJEE, WBJEE 2019
Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{ccc}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
where, $[x]$ denotes the integral part of $x$. If $f$ continuous at $x=0$, then $\beta-\alpha$ is equal to

Options:
A $-1$
B $1$
C $0$
D $2$

The correct answer is:
$1$

Solution:
Given.
$f(x)=\left\{\begin{array}{lll}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
Since, $f$ is continuous at $x=0$.
$\therefore \mathrm{IHI}=f(0)=$ RHI.
Now, I.HI. $=\lim f(x)$
$\begin{aligned} & =\lim _{x \rightarrow 0}\left[\beta+\left(\frac{\sin x-x}{x^3}\right)\right] \\ & =\lim _{h \rightarrow 0}\left[\beta+\left(\frac{\sin h+h}{-h^3}\right)\right] \\ & =\beta+0\end{aligned}$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}}\left[\alpha+\frac{\sin [x]}{x}\right]$
$\begin{aligned} & =\lim _{h \rightarrow 0}\left[\alpha+\frac{\sin [h]}{h}\right] \\ & =\alpha+1\end{aligned}$
and $f(0)=2$
$\therefore$ From Eq. (i), we get
$\begin{array}{ll}\quad \beta+0=2=\alpha+1 \\ \Rightarrow \quad & \beta-\alpha=1\end{array}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:R→R be defined fx=ae2x+bex+cx. If f(0)=-1,f'loge2=21 and ∫0log4fx-cxdx=392, then the value of |a+b+c| equals:

Options:
A 16
B 10
C 12
D 8

The correct answer is:
8

Solution:
Given: fx=ae2x+bex+cx, f0=-1, f'log2=21, ∫0log4fx-cx=392

⇒f0=a+b=-1 ...i

⇒f'x=2ae2x+bex+c

⇒f'log2=2a4+b2+c=21

⇒8a+2b+c=21 ...ii

Also, ∫0log4ae2x+bex+cx-cx=392

⇒∫0log4ae2x+bex=392

⇒ae2x2+bex0log4=392

⇒ae2log42+belog4-a2-b=392

⇒15a2+3b=392

⇒15a+6b=39

Using equation i,

⇒15a+6-1-a=39

⇒15a-6-6a=39

⇒9a=45

⇒a=5

⇒b=-1-5=-6

⇒c=21-8a-2b

⇒c=21-40+12=-7

⇒c=-7

⇒a+b+c=8

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let f:R→R be differentiable at c∈R and fc=0. If gx=fx, then at x=c, g is:

Options:
A not differentiable
B not differentiable if f'c=0
C differentiable if f'c=0
D differentiable if f'c≠0

The correct answer is:
differentiable if f'c=0

Solution:
Given gx=fx and fc=0

g'c+=limx→c+fx-fcx-c=limx→c+fx-f(c)x-c, as x>c

⇒g'c+=f'c

gc-=limx→c-fx-fcx-c=-limx→c-fx-f(c)x-c, as x<c

⇒g'c-=-f'c

For gx to be differentiable at x=c, f'c must be 0. Else, it is non-differentiable.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Let f:R→R be such that for all x∈R21+x+21-x, fx and 3x+3-x are in A.P., then the minimum value of fx is

Options:
A 2
B 3
C 0
D 4

The correct answer is:
3

Solution:
fx=21-x+21+x+3x+3-x2

⇒fx=23-x+3x+3x+3-x2

⇒fx=3-x+3x+3x+3-x2

Using AM≥GM
fx≥3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let $f: R \rightarrow R$ be twice continuously differentiable. Let $f(0)=f(1)=f^{\prime}(0)=0 .$ Then,

Options:
A $f^{\prime \prime}(x) \neq 0$ for all $x$
B $f^{\prime \prime}(c)=0$ for some $c \in R$
C $f^{\prime \prime}(x) \neq 0$ if $x \neq 0$
D $f^{\prime}(x)>0$ for all $x$

The correct answers are:
$f^{\prime \prime}(c)=0$ for some $c \in R$

Solution:
Let function $f(x)=x^{2}(x-1)$
$\Rightarrow \quad f^{\prime}(x)=3 x^{2}-2 x$
and $\quad f^{\prime \prime}(x)=6 x-2$
$\begin{array}{ll}\text { Now, } & f(0)=f(1)=f^{\prime}(0)=0\end{array}$
Then, according to question, at $x=\frac{1}{3}, \quad f^{\prime \prime}(x)=0$
(i.e. $c=1 / 3$ for some $C \in R$ )

Topic: Mathematics, Differentiation, JEE Main
Let f:R→R,fx=x4-8x3+22x2-24x+c. If sum of all local extremum values of fx is 1, then c is equal to

Options:
A 8
B 9
C 10
D 11

The correct answer is:
9

Solution:
f â² x =4 x 3 â6 x 2 +11xâ6
=4x-1x-2x-3
Hence, x=1,2,3 are points of extrema
So, f1+f2+f3=1
c-9+c-8+c-9=1âc=9

Topic: Mathematics, Application of Derivatives, JEE Main
Let f: R →R is a function defined as fx= x-x:x is oddx-x+1 :x is even, where [.] denotes the greatest integer function, then ∫-24fx dx is equal to

Options:

The correct answer is:
3

Solution:
We know that,
x–[x]=x⇒x–[x+1]=x–1
Hence, ∫-24fx dx = 6.121.1=3

Topic: Mathematics, Definite Integration, JEE Main
Let f : R→R satisfy the equation f(x+y)=f(x)·f(y) for all x,y∈R and f(x)≠0 for any x∈R. If the function f is differentiable at x=0 and f'(0)=3, then limh→01hfh-1 is equal to ___ .

Options:

The correct answer is:
3

Solution:
If f(x+y)=f(x)·f(y) & f'(0)=3 then

f(x)=ax⇒f'x=ax·lna

⇒f'0=lna=3⇒a=e3

⇒fx=e3x=e3x

limx→0f(x)-1x=limx→0e3x-13x×3=1×3=3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let \( f: R \rightarrow R \) be a positive increasing function
with \( \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1 \). Then \( \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}= \)

Options:
A \( 1 \)
B \( \frac{2}{3} \)
C \( \frac{3}{2} \)
D \( 3 \)

The correct answer is:
\( 1 \)

Solution:
As f is a positive increasing function, we have
fx<f2x<f3x
Dividing by fx leads to 1 < f⁡ 2 x f⁡ x < f⁡ 3 x f⁡ x
As lim x → ∞ f⁡ 3 x f⁡ x = 1 , we have by squeeze theorem
or sandwich theorem, lim x → ∞ f⁡ 2 x f⁡ x = 1

Topic: Mathematics, Limits, JEE Main
Let fx=∫0xetftdt+ex be a differentiable function for all x∈R. Then fx equals :

Options:
A eex-1
B eex-1
C 2eex-1
D 2eex-1-1

The correct answer is:
2eex-1-1

Solution:
Given fx=∫0xetftdt+ex⇒f0=1

Differentiating with respect to x we get,

f'x=exfx+ex

f'x=exfx+1

f'xfx+1=ex

Integrate both the sides we get,

∫0xf'xfx+1dx=∫0xexdx

lnfx+10x=ex0x

lnfx+1-lnf0+1=ex-1

lnfx+12=ex-1 {as f0=1}

fx=2eex-1-1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let fx=∫0xgtdt, where g is a non-zero even function. If fx+5=gx, then ∫0xf(t)dt equals

Options:
A ∫5x+5g(t)dt
B ∫x+55g(t)dt
C 5∫x+55g(t)dt
D 2∫5x+5g(t)dt

The correct answer is:
∫x+55g(t)dt

Solution:
Given

: gx is an even function, then
fx=∫0xg(t)dt will be an odd function,
⇒f-x=-fx
Let I=∫0xftdt put t=z+5
⇒I=∫-5x-5fz+5dz
⇒I=∫-5x-5g(z)dz ∵fx+5=gx
⇒I=∫-5x-5f'(z)dz                  (∵f'(x)=g(x))
⇒I=f(z)|-5x-5=f(x-5)-f-5
⇒I=f(5)-f5+x ( f is an odd function and g is given as an even function)
⇒I=∫5+x5f'(t)dt=∫5+x5g(t)dt

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (08 Apr Shift 2)