datavorous/entrance-exam-dataset huggingface dataset:
If f:R→-1,1 be a function defined as fx=sinx2−8x2+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, f(x)=f-x
∴f is not one-one
Let, y=x2-8x2+2
⇒x2=8+2y1-y
For x to be real, 8+2y1-y≥0 and 1-y≠0

⇒y+4y-1≤0 and y≠1
⇒-4≤y<1
∵ y is neither equal to π2 nor -3π2
∴ Range of fx∉-1,1
∴fx is into function

Topic: Mathematics, Functions, JEE Main
If f:RâA defined as fx=tan-14x2+x+1 is surjective, then A is equal to

Options:
A -π2,π2
B 0,π2
C π3,π2
D 0,π3

The correct answer is:
π3,π2

Solution:
∵x2+x+1=x+122+34
∴4x2+x+1≥3
⇒4x2+x+1≥3
∴ Range of fx is π3,π2

Topic: Mathematics, Functions, JEE Main
If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$, then $f$ is (where $C$ denotes the set of all complex numbers)

Options:
A one-one
B onto
C one-one and onto
D neither one-one nor onto

The correct answer is:
neither one-one nor onto

Solution:
Given that, $f(x)=e^{2 i x}$ and $f: R \rightarrow C$. Function $f(x)$ is not one-one, because after some values of $x(i e, \pi)$ it will give the same values.
Also, $f(x)$ is not onto, because it has minimum and maximum values $-1-i$ and $1+i$ respectively.
Hence, option (d) is correct.

Topic: Mathematics, Functions, JEE Main
If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\left\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\right\}$ is equal to

Options:
A $\{0,1\}$
B $\{1,2\}$
C $\{-3,-2\}$
D $\{2,3\}$

The correct answer is:
$\{-3,-2\}$

Solution:
Given that, $f(x)=|x|$ and $g(x)=[x-3]$
For $\quad-\frac{8}{5} < x < \frac{8}{5}, 0 \leq f(x) < \frac{8}{5}$
Now, for $\quad 0 \leq f(x) < 1$,
$$
\begin{aligned}
g(f(x)) & =[f(x)-3] \\
& =-3[\because-3 \leq f(x)-3 < -2]
\end{aligned}
$$
Again, for $1 \leq f(x) < 1.6$
$$
g(f(x))=-2
$$
Hence, required set is $\{-3,-2\}$.

Topic: Mathematics, Functions, JEE Main
If f:R→R be a function such that fx=x3+x2+4x+sin⁡x. Then, the function fx is

Options:
A one-one and onto
B one-one and into
C many-one and onto
D many-one and into

The correct answer is:
one-one and onto

Solution:
fx=x3+x2+4x+sin⁡x
⇒f'x=3x2+2x+4+cos⁡x
f'x=3x+132+119--cos⁡x>0 as 3 x+132+119min=113
and -cos⁡x has the maximum value 1.
⇒fx is strictly increasing and hence it is one-one
Also, limx→∞fx⇒∞ and limx→-∞fx⇒-∞.

Thus, the range of fx is R, hence it is onto.

Topic: Mathematics, Functions, JEE Main
If f:RâR be defined as fx=e2x-e-2x2, then

Options:
A f is many-one
B f is into
C f-1x=12logx-x2+1
D f-1x=12logx+x2+1

The correct answer is:
f-1x=12logx+x2+1

Solution:
Let us check for invertibility of fx
A one-one: we have, fx=e2x-e-2x2
⇒ f ′ x = e 4x +1 e 2x , which is strictly increasing as e4x>0 for all x.
Thus, f is one-one
B Onto: Let y=fx
⇒dydx=e2x+e-2x, where y is strictly monotonic
Hence, the range of fx=f-∞,f∞
⇒ range of fx=-∞,∞
So, the range of fx= co-domain
Hence, fx is one-one and onto
C To find f-1:y=e4x-12e2x
⇒e4x-2e2xy-1=0
⇒e2x=2y±4y2+42
⇒2x=logy±y2+1
⇒f-1y=logy±y2+12
Since, ef-1x is always positive, so neglecting negative sign.
Hence, f-1x=logx+x2+12

Topic: Mathematics, Functions, JEE Main
If $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined by
$f(x)=x+2|x+1|+2|x-1|$, then the element in the co-domain, which has unique pre image in the domain is

Options:
A 3
B 1
C 2
D 5

The correct answer is:
3

Solution:
We have,
$\begin{aligned}
& f(x)=x+2|x+1|+2|x-1| \\
& f(x)=\left\{\begin{array}{cc}
x+2(-x-1)+2(-x+1), & x \leq-1 \\
x+2(x+1)+2(-x+1), & -1 < x \leq 1 \\
x+2(x+1)+2(x-1), & x \geq 1
\end{array}\right. \\
& f(x)=\left\{\begin{array}{cc}
-3 x, & x \leq-1 \\
x+4, & -1 < x < 1 \\
5 x, & x \geq 1
\end{array}\right.
\end{aligned}$
Graph of $f(x)$ is
Clearly, 3 has unique image in the domain.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
If $f: R \rightarrow R$ be mapping defined by $f(x)=x^{3}+5$, then $f^{-1}(x)$ is equal to

Options:
A $(x+5)^{1 / 3}$
B $(x-5)^{1 / 3}$
C $(5-x)^{1 / 3}$
D $5-x$

The correct answer is:
$(x-5)^{1 / 3}$

Solution:
Let
$$
y=f(x)=x^{3}+5
$$
$$
\begin{array}{lr}
\Rightarrow \quad x=(y-5)^{1 / 3} \\
\therefore \quad f^{-1}(x)=(x-5)^{1 / 3} \\
\text { Given, } & f(x)=\frac{a x+b}{c x+d}
\end{array}
$$

Topic: Mathematics, Functions, JEE Main
If $f: R \rightarrow R$ defined by
$f(x)=\left\{\begin{array}{cc}\frac{1+3 x^2-\cos 2 x}{x^2}, & \text { for } x \neq 0 \\k & , \text { for } x=0\end{array}\right.$
is continuous at $x=0$, then $k$ is equal to

Options:
A 1
B 5
C 6
D 0

The correct answer is:
5

Solution:
$f(x)=\left\{\begin{array}{rr}\frac{1+3 x^2-\cos 2 x}{x^2}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.$
RHL
$f(0+h)=\lim _{h \rightarrow 0} \frac{1+3(0+h)^2-\cos 2(0+h)}{(0+h)^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\left(1-2 \sin ^2 h\right)}{h^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-1+2 \sin ^2 h}{h^2}$
$=\lim _{h \rightarrow 0}\left\{3+2\left(\frac{\sin ^2 h}{h^2}\right)\right\}$
$=3+2 \cdot \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2$
$=3+2 \cdot(1)^2 \quad\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}$
$=3+2=5$
LHL
$f(0-h)=\lim _{h \rightarrow 0} \frac{1+3(0-h)^2-\cos 2(0-h)}{(0-h)^2}$
$=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$
$=5$
Since, the function is continuous at $x=0$, then LHL $=$ RHL $=f(0) \because f(0)=k \Rightarrow k=5$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f:R→R is a differentiable function and f2=6, then limx→2∫6fx2tdtx-2 is:

Options:
A 0
B 2f'2
C 24f'2
D 12f'2

The correct answer is:
12f'2

Solution:
The given limit can be written as

l=limx→2∫6fx2tdtx-2

Applying L' Hospital’s rule i.e. if l=limx→agxhx and limx→agx→0 & limx→ahx→0, then l=limx→ag'xh'x

l=limx→2ddx∫6fx2tdt1

Now, applying Newton's Leibnitz rule i.e. ddx∫abgxdx=gb·b'-ga·a', we get

⇒l=limx→22fx·f'x-0

⇒l=2f2⋅f'2

Put the given value, to get

⇒l=2×6×f'2=12f'2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If $f: R \rightarrow R$ is a differentiable function such that $f(x+y)=f(x) . f(y)$ for all $x, y \in R$ and if $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$, then $f(4)=$

Options:
A 72
B 5
C 11
D 8

The correct answer is:
8

Solution:
$\begin{aligned} & \text { Let } x=4, y=0 \Rightarrow f(4+0)=f(4) \cdot f(0) \\ & \Rightarrow \quad \begin{aligned} f(4) & =f(4) f(0) \Rightarrow f(0)=1 \\ f^{\prime}(4) & =\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(4) f(h)-f(4)}{h}=\lim _{h \rightarrow 0}\left(\frac{f(h)-1}{h}\right) f(4)\end{aligned}\end{aligned}$
$\begin{aligned} & f^{\prime}(4)=f(4) \lim _{h \rightarrow 0} \frac{(f(h)-f(0)}{h} \\ & f^{\prime}(4)=f(4) \cdot f^{\prime}(0) \\ & f^{\prime}(4)=24, f^{\prime}(0)=3 \Rightarrow \quad f(4)=\frac{f^{\prime}(4)}{f^{\prime}(0)}=\frac{24}{3}=8 .\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f:R→R is a function defined as fx3=x5, ∀x∈R-0 and fx is differentiable ∀x∈R, then the value of 14f'27 is equal to (here f' represents the derivative of f)

Options:

The correct answer is:
3.75

Solution:
fx3=x5
On differentiating with respect to x
f'x3â3x2=5âx4
f â² x 3 = 5 3 x 2
Putting x=3, we get,
f â² 27 = 5 3 9 =15

Topic: Mathematics, Differentiation, JEE Main
If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ is a function defined by $f(x)=[\mathrm{x}] \cos \left(\frac{2 x-1}{2}\right) \pi$, where $[\mathrm{x}]$ denotes the greatest integer function, then $f$ is

Options:
A
continuous for every real $x$

B
discontinuous only at $x=0$

C
discontinuous only at non-zero integral values of $x$

D
continuous only at $x=0$

The correct answer is:
continuous for every real $x$

Solution:
$f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi=[x] \cos \left(x-\frac{1}{2}\right) \pi$
$=[x] \sin \pi x$ is continuous for every real $x$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (Offline)
If f:R→R is a function defined by f(x)=xsin2x-12π, where x denotes the greatest integer function, then f is

Options:
A continuous for every real x
B discontinuous only at x=0
C discontinuous only at integral values of x
D continuous only at x=0

The correct answer is:
discontinuous only at integral values of x

Solution:
Doubtful points are x=n, ∀n∈Ι
L.H.L=limx→n-xsin2x-12π
=n-1sin2n-12π=±n-1
R.H.L=limx→n+xsin2x-12π
=nsin2n-12π=±n
fn=±n
Clearly, LHL,RHL, fn all cannot be equal for any n∈Ι
Hence, discontinuous at all integeral values of x

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f: RâR is a function satisfying the equation f3x+1+f3x+10=10, âxâR, then the period of fx is

Options:

The correct answer is:
18

Solution:
∵f3x+1+f3x+10=10 … 1
Replacing x→x+3
f3x+3+1+f3x+3+10=10
f3x+10+f3x+19=10 … 2
equation 2-1
f3x+19-f3x+1=0
f3x+1=f3x+19
Now, replace x→x3-13
f3x3-13+1=f3x3-13+19
fx=fx+18 ∀x∈R
Hence, period of fx is 18

Topic: Mathematics, Functions, JEE Main
If f:RâR is a function such that f5x+f5x+1+f5x+2=0, âxâR,
then the period of fx is

Options:

The correct answer is:
3

Solution:
∵ f5x+f5x+1+f5x+2=0 … 1
Replacing x by x+15 in 1, we get,
f5x+15+f5x+15+1+f5x+15+2=0
⇒f5x+1+f5x+2+f5x+3=0 … 2
On subtracting 2 from 1, we get,
f5x+3=f5x
Now, x→x5⇒fx+3=fx
∴period=3

Topic: Mathematics, Functions, JEE Main
If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
Let the even function be
$f(x)=\cos x$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=-\sin x$
Again differentiating w.r.t. $x$, we get
$f^{\prime \prime}(x)=-\cos x$
at $x=\pi$
$f^{\prime \prime}(\pi)=-\cos \pi=1$
$\therefore$ Our assumption is true.
at
$x=-\pi$
$f^{\prime \prime}(-\pi)=-\cos (-\pi)=1$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x^2-2 x-3$ then $f$ is

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

The correct answer is:
neither one-one nor onto

Solution:
$f(x)=x^2-2 x-3$
for one one
Let $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$
$\begin{aligned} & x_1^2=2 x_1-3=x_2^2-2 x_2-3 \\ & \Rightarrow x_1^2-x_2^2-2\left(x_1-x_2\right)=0 \\ & \Rightarrow\left(x_1-x_2\right)\left[x_1+x_2-2\right]=0 \\ & \Rightarrow x_1=x_2 \text { or } x_1+x_2=2\end{aligned}$
So, $f(x)$ is not one-one
for onto
Let $y=x^2-2 x-3=0$
$\begin{aligned} & x^2-2 x-3-y=0 \\ & x=\frac{2 \pm \sqrt{4-4(3+y)}}{2}=1 \pm \sqrt{4+y} \\ & \therefore 4+y \geq 0=y \geq-4 \\ & \therefore \text { range }=-4 \uparrow \text { codomain }(R)\end{aligned}$
So, $f(x)$ is not onto

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=2 x+3$, then $\mathrm{f}^{-1}(\mathrm{x})$

Options:
A is given by $\frac{x-3}{2}$
B is given by $\frac{1}{2 x+3}$
C does not exist because ' $P$ ' is not injective
D does not exist because ' $\mathrm{P}$ is not surjective

The correct answer is:
is given by $\frac{x-3}{2}$

Solution:
Let $\mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$
$$
\begin{array}{lrl}
\text { Let } & \mathrm{f}\left(\mathrm{x}_{1}\right) & =\mathrm{f}\left(\mathrm{x}_{2}\right) \\
\Rightarrow & 2 \mathrm{x}_{1}+3 & =2 \mathrm{x}_{2}+3 \\
\Rightarrow & \mathrm{x}_{1} & =\mathrm{x}_{2}
\end{array}
$$
Hence, $\mathrm{f}$ is injective.
Let $\mathrm{y} \in \mathrm{codomain} \mathrm{R}$
$\begin{array}{ll}\text { Let } & y=f(x) \\ \Rightarrow & y=2 x+3 \\ \Rightarrow & x=\frac{y-3}{2} \in \text { domain } R\end{array}$
Hence, $f$ is onto (surjective). $\therefore \mathrm{f}^{-1}$ exists.
$$
\therefore \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-3}{2}, \mathrm{x} \in \mathrm{R}
$$

Topic: Mathematics, Functions, KCET, KCET 2012
If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=3 x-4$, then $f^{-1}: \mathbb{R} \rightarrow \mathbb{R}$ is

Options:
A $4-3 x$
B $\frac{x+4}{3}$
C $\frac{1}{3 x-4}$
D $\frac{3}{x+4}$

The correct answer is:
$\frac{x+4}{3}$

Solution:
$\begin{aligned} & \text { Given } f(x)=3 x-4 \text {. Now let } y=f^{-1}(x) \Rightarrow f(y)=x \\ & \Rightarrow 3 y-4=x \Rightarrow 3 y=x+4 \\ & \Rightarrow y=\frac{x+4}{3} \Rightarrow f^{-1}(x)=\frac{x+4}{3}\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined by
$$
f(x)=\left\{\begin{array}{cl}
x-1, & \text { for } x \leq 1 \\
2-x^2, & \text { for } 1 < x \leq 3 \\
x-10, & \text { for } 3 < x < 5 \\
2 x, & \text { for } x \geq 5
\end{array}\right.
$$
then the set of points of discontinuity of $f$ is

Options:
A $\mathbf{R}-\{1,5\}$
B $\{1,3,5\}$
C $\{1,5\}$
D $\mathbf{R}-\{1,3,5\}$

The correct answer is:
$\{1,5\}$

Solution:
Given function
$$
f(x)= \begin{cases}x-1, & \text { for } x \leq 1 \\ 2-x^2, & \text { for } 1 < x \leq 3 \\ x-10, & \text { for } 3 < x < 5 \\ 2 x, & \text { for } x \geq 5\end{cases}
$$
So, $f(x)$ will be continuous in the intervals $(-\infty, 1),(1,3),(3,5)$ and $(5, \infty)$
Now, let us check the continuity at $x=1,3$ and 5
Here,
(i) At $x=1$,
$$
\lim _{x \rightarrow 1^{-}} f(x)=1-1=0 \text { and } \lim _{x \rightarrow 1^{+}} f(x)=2-1=1
$$
therefore $\mathrm{F}$ is not continuous at $x=1$
(ii)
$$
\begin{aligned}
& \text { At } x=3, \\
& \lim _{x \rightarrow 3^{-}} f(x)=2-9=-7 \text { and } \\
& f(3)=-7 \text { and } \lim _{x \rightarrow 3^{+}} f(x)=3-10=-7
\end{aligned}
$$
therefore $f$ is continuous at $x=3$
(iii) At $x=5, \lim _{x \rightarrow 5^{-}} f(x)=5-10=-5$ and,
$$
\lim _{x \rightarrow 5^{+}} f(x)=2 \times 5=10
$$
therefore $f$ is not continuous at $x=5$ Hence, points of discontinuity of $f$ are $\{1,5\}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f: R \rightarrow R$ is defined by
$f(x)=\left\{\begin{array}{ccc}
\frac{x+2}{x^2+3 x+2} & \text { if } & x \in R-\{-1,-2\} \\
-1 & \text { if } & x=-2 \\
0 & \text { if } & x=-1
\end{array}\right.$
then $f$ is continuous on the set

Options:
A $R$
B $R-\{-2\}$
C $R-\{-1\}$
D $R-\{-1,-2\}$

The correct answer is:
$R-\{-1\}$

Solution:
Given that
$f(x)=\left\{\begin{array}{ccc}
\frac{x+2}{x^2+3 x+2}, & \text { if } & x \in R-\{-1,-2\} \\
-1, & \text { if } & x=-2 \\
0, & \text { if } & x=-1
\end{array}\right.$
Now, we have to check the continuity
$\begin{aligned}
& \text { at } x= \\
& \text { at } x=-2,-1 \\
& \begin{aligned}
\text { LHL } & =\lim _{h \rightarrow 0} \frac{(-2-h)+2}{(-2-h)^2+3(-2-h)+2} \\
& =\lim _{h \rightarrow 0} \frac{-h}{4+h^2+4 h-6-3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{-h}{h^2+h}=\lim _{h \rightarrow 0} \frac{-1}{h+1}=-1
\end{aligned}
\end{aligned}$
$\begin{aligned}
\text { RHL } & =\lim _{h \rightarrow 0} \frac{(-2+h)+2}{(-2+h)^2+3(-2+h)+2} \\
& =\lim _{h \rightarrow 0} \frac{h}{4+h^2-4 h-6+3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{h}{h^2-h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h-1}=-1 \\
\Rightarrow \text { LHL } & =\text { RHL }=f(-2)
\end{aligned}$
$\therefore$ It is continuous at $x=-2$
Now, check for $x=-1$
$\begin{aligned}
\text { LHL } & =\lim _{h \rightarrow 0} \frac{(-1-h)+2}{(-1-h)^2+3(-1-h)+2} \\
& =\lim _{h \rightarrow 0} \frac{1-h}{1^2+h^2+2 h-3-3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{1-h}{h^2-h}=\lim _{h \rightarrow 0} \frac{-1}{2 h-1}=1 \\
\text { RHL } & =\lim _{h \rightarrow 0} \frac{(-1+h)+2}{(-1+h)^2+3(-1+h)+2} \\
& =\lim _{h \rightarrow 0} \frac{1+h}{1+h^2-2 h-3+3 h+2} \\
& =\lim _{h \rightarrow 0} \frac{1+h}{h^2+h}=1
\end{aligned}$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL} \neq f(-1)$
$\therefore$ It is not continuous at $x=-1$
The required function is continuous in $R-\{-1\}$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f:R→R is given by f(x)=x+1, then the value of limn→∞1nf0+f5n+f10n+…..+f5(n-1)n is:

Options:
A 32
B 52
C 12
D 72

The correct answer is:
72

Solution:
Let

I=limn→∞1nf0+f5n+f10n+…..+f5(n-1)n

⇒I=∑r=0n-1f5rn1n

⇒I=∫01f5xdx

⇒I=∫015x+1dx

⇒I=5x22+x01

⇒I=52+1=72

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If $f: R \rightarrow R$ satisfies $f(x+y)=f(x)+f(y)$, for all $x, y \in R$ and $f(1)=7$, then $\sum_{r=1}^n f(r)$ is

Options:
A
$\frac{7 n(n+1)}{2}$

B
$\frac{7 \mathrm{n}}{2}$

C
$\frac{7(\mathrm{n}+1)}{2}$

D
$7 n+(n+1)$

The correct answer is:
$\frac{7 n(n+1)}{2}$

Solution:
$\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) . \quad$ Let $\mathrm{f}(\alpha)=\mathrm{m} \alpha$
$\mathrm{f}(1)=7 ; \quad \therefore \mathrm{m}=7, \mathrm{f}(\mathrm{x})=7 \mathrm{x}$
$\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{r})=7 \sum_1^{\mathrm{n}} \mathrm{r}=\frac{7 \mathrm{n}(\mathrm{n}+1)}{2}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2003
If $f: R \rightarrow R$, then $f(x)=|x|$ is

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

The correct answer is:
None of these

Solution:
$\quad f(-1)=f(1)=1$; $\therefore$ function is many-one function. Obviously, $f$ is not onto so $f$ is neither one-one nor onto.

Topic: Mathematics, Functions, JEE Main
If \( f: R \rightarrow A \) defined as \( f(x)=\tan ^{-1}\left(\sqrt{4\left(x^{2}+x+1\right)}\right) \) is surjective, then \( A \) is equal to

Options:
A \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
B \( \left[0, \frac{\pi}{2}\right) \)
C \( \left[\frac{\pi}{3}, \frac{\pi}{2}\right) \)
D \( \left(0, \frac{\pi}{3}\right] \)

The correct answer is:
\( \left[\frac{\pi}{3}, \frac{\pi}{2}\right) \)

Solution:
∵ x2+x+1=x+122+34

∴ 4x2+x+1≥3

⇒4x2+x+1≥3

⇒tan-14x2+x+1≥tan-1(3)

⇒f(x)≥π3
∴ Range of fx is π3,π2

Topic: Mathematics, Functions, JEE Main
If \( f: R \rightarrow R \) is defined by \( f(x)=\left\{\begin{array}{cc}\frac{x+2}{x^{2}+3 x+2}, & \text { if } x \in R-\{-1,-2\} \\ -1, & \text { if } \quad x=-2 \quad \text { then } f \text { is continuous on the se } \\ 0, & \text { if } \quad x=-1\end{array}\right. \)

Options:
A \( R \)
B \( R-\{-2\} \)
C \( R-\{-1\} \)
D \( R-\{-1,-2\} \)

The correct answer is:
\( R-\{-1\} \)

Solution:
Since, fx is continuous for R, not sure about -1,-2.

Now, we have check continuity at these points.

At x=-2,

LHL=limn→0-2-n+2-2-n2+3-2-n+2

=limn→0-nn2+n=-1

RHL=limn→0-2+n+2-2+n2+3-2+n+2

=limn→0nn2-n=-1

LHL=RHL=f-2

It is continuous at x=-2

Now, check for x=-1

LHL=limn→0-1-n+2-1-n2+3-1-n+2

=limn→01-nn2-n=∞

RHL=limn→0-1+n+2-1+n2+3-1+n+2

=limn→01+nn2+n=∞

LHL=RHL≠f-1

It is not continuous at x=-1

The required function is continuous in R-{-1}

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f: R \rightarrow S$ defined by $f(x)=\sin x-\sqrt{3} \cos x+1$ is onto, then the interval of $S$ is

Options:
A $[-1,3]$
B $[1,1]$
C $[0,1]$
D $[0,-1]$

The correct answer is:
$[-1,3]$

Solution:
$\begin{aligned}
& -\sqrt{1+(-\sqrt{3})^2} \leq(\sin x-\sqrt{3} \cos x) \leq \sqrt{1+(-\sqrt{3})^2} \\
& -2 \leq(\sin x-\sqrt{3} \cos x) \leq 2 \\
& -2+1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 2+1
\end{aligned}$
$-1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 3$ i.e., range $=[-1,3]$
$\therefore$ For $f$ to be onto $S=[-1,3]$.

Topic: Mathematics, Functions, JEE Main
If $f: R \rightarrow S$, defined by $f(x)=\sin x-\sqrt{3} \cos x+1$, is onto, then the interval of $S$ is

Options:
A
$[0,3]$

B
$[-1,1]$

C
$[0,1]$

D
$[-1,3]$

The correct answer is:
$[-1,3]$

Solution:
$-2 \leq \sin x-\sqrt{3} \cos x \leq 2 \Rightarrow-1 \leq \sin x-\sqrt{3} \cos x+1 \leq 3$
$\Rightarrow$ range of $f(x)$ is $[-1,3]$.
Hence $S$ is $[-1,3]$

Topic: Mathematics, Functions, JEE Main, JEE Main 2004
If fθ=sinθ+∫-π2π2sinθ+tcosθ·ftdt, then ∫0π2fθdθ is

Options:

The correct answer is:
1

Solution:
Given fθ=sinθ+∫-π2π2sinθ+tcosθftdt

fθ=sinθ+sinθ∫-π2π2ftdt+cosθ∫-π2π2tftdt

Let A=∫-π2π2ftdt, B=∫-π2π2tftdt

So fθ=sinθ+Asinθ+Bcosθ

i.e. fθ=A+1sinθ+Bcosθ

A=∫-π2π2 A+1sint+Bcostdt

⇒A=A+1∫-π2π2 sintdt+B∫-π2π2 costdt

⇒A=2B …1

B=∫-π2π2tA+1sint+Bcostdt

⇒B=∫-π2π2tA+1sintdt

⇒B=A+12∫0π2tsintdt

⇒B=A+12

⇒2A+2-B=0 ⋯2

After solving

B=-23, A=-43

∫0π2fθdθ=∫0π2-13sinθ-23cosθdθ

=-13∫0π2sinθdθ-23∫0π2cosθdθ=1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If $f(t)=\int_0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0 < \mathrm{t} < \pi$, then the value of $\int_0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}$ equals_________

Options:

The correct answer is:
1

Solution:
$f(t)=\int_0^\pi \frac{2 x}{1-\cos ^2 t \sin ^2 x} d x$ .....(1)
$=2 \int_0^\pi \frac{(\pi-x) d x}{1-\cos ^2 \sin ^2 x}$ .....(2)
$\begin{aligned}
& 2 f(t)=2 \int_0^\pi \frac{\pi}{1-\cos ^2 \sin ^2 x} d x \\
& f(t)=\int_0^\pi \frac{\pi}{1-\cos ^2 t \sin ^2 x} d x
\end{aligned}$
divide \& by $\cos ^2 \mathrm{x}$
$\begin{aligned}
& f(t)=\pi \int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 x} \\
& f(t)=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 \tan ^2 x} \\
& \tan x=z \\
& \sec ^2 x d x=d z \\
& f(t)=2 \pi \int_0^{\infty} \frac{d z}{1+\sin ^2 t \cdot z^2} \\
& =\frac{\pi^2}{\sin t}
\end{aligned}$
Then $\int_0^{\pi / 2} \frac{\pi^2}{f(t)} d t$
$\begin{aligned}
& =\int_0^{\pi / 2} \sin t d t \\
& =1
\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (05 Apr Shift 2)
If $f(t)=\int_{-t}^t \frac{d x}{1+x^2}$, then $f^{\prime}(1)$ is

Options:
A Zero
B $2 / 3$
C -1
D 1

The correct answer is:
1

Solution:
Given $f(t)=\int_t^t \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_{-t}^t=2 \tan ^{-1} t$
Differentiating with respect to $t, \quad f^{\prime}(t)=\frac{2}{1+t^2}$
$\Rightarrow \quad f^{\prime}(1)=\frac{2}{2}=1$

Topic: Mathematics, Definite Integration, JEE Main
If ftanâ¡x=cosâ¡2x, xâ (2n+1)Ï2, nâI then incorrect statement is

Options:
A f x is an even function
B f x is an odd function
C Range of f x is −1,1
D Domain of f x is xâR

The correct answer is:
f x is an odd function

Solution:
∵1−tan2x1+tan2x=cos2x

Let, tanx=t
∴f t = 1− t 2 1+ t 2
∴f x = 1− x 2 1+ x 2

Topic: Mathematics, Functions, JEE Main
If ftan⁡x=sin⁡2x:x≠2n+1π2,n∈Ι, then which of the following is an incorrect statement?

Options:
A Domain of fx is R-2n+1π2,n∈Ι
B Range of fx is -1,1
C fx is odd function
D fx is many-one function

The correct answer is:
Domain of fx is R-2n+1π2,n∈Ι

Solution:
ftanâ¡x=2tanâ¡x1+tan2â¡xâfx=2x1+x2
Now, domain of fx is xâR
For the range of fxây=2x1+x2âyx2-2x+y=0
âµxâRâ´Dâ¥0
4-4y2â¥0âyâ-1,1 is the range
Also, f-x=2-x1+-x2=-fx
â´fx is an odd function
f'x=1+x22-2x2x1+x22=21-x21+x22
f'x can be positive & negative âxâR
â´fx is many-one

Topic: Mathematics, Functions, JEE Main
If fx=∫0x5+1-tdt,x>25x+1,x≤2, then

Options:
A fx is not continuous atx=2
B fx is everywhere differentiable
C fx is continuous but not differentiable at x=2
D fx is not differentiable at x=1

The correct answer is:
fx is continuous but not differentiable at x=2

Solution:
We have, fx=∫0x5+1-tdt,x>25x+1,x≤2

Therefore, fx=∫015+1-tdt+∫1x5+t-1dt

=6-12+4t+t22|1x

=112+4x+x22-4-12

=x22+4x+1

Now, f2+=2+8+1=11

and f2=f2-=5×2+1=11

Here, f2+=f2-

So, fx is continuous at x=2

Clearly, fx is differentiable at x=1

Now, LHD f'2-=5 and RHD f'2+=6

Therefore, fx is not differentiable at x=2

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If fx=∫0xtsinx-sintdt, then

Options:
A f'''x-f''x=cosx-2xsinx
B f'''x+f''x-f'x=cosx
C f'''x+f''x=sinx
D f'''x+f'x=cosx-2xsinx

The correct answer is:
f'''x+f'x=cosx-2xsinx

Solution:
Given fx=∫0xtsinx-sintdt

⇒fx=sinx∫0xtdt-∫0xtsintdt

On differentiating, we get

⇒f'x=sinxx+cosx∫0xtdt-xsinx

⇒f'x=cosx∫0xtdt

Differentiating the above equation again, we get

⇒f''x=cosxx-sinx∫0xtdt

⇒f'''x=x-sinx+cosx-sinxx-cosx∫0xtdt

⇒f'''x+f'x=cosx-2xsinx.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (16 Apr Online)