datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=x e^{x(1-x)}, x \in R$, then $f(x)$ is

Options:
A
decreasing on $[-1 / 2,1]$

B
decreasing on $R$

C
increasing on $[-1 / 2,1]$

D
increasing on $R$

The correct answer is:
increasing on $[-1 / 2,1]$

Solution:
$$
\begin{aligned}
& f(x)=x e^{x(1-x)}, x \in R \\
& f^{\prime}(x)=e^{x(1-x)} \cdot\left[1+x-2 x^2\right]
\end{aligned}
$$
$$
\begin{aligned}
& =-e^{x(1-x} \cdot\left[2 x^2-x-1\right] \\
& =-2 e^{x(1-x} \cdot\left[\left(x+\frac{1}{2}\right)(x-1]\right) \\
& f^{\prime}(x)=-2 e^{x(1-x)} A \\
& \text { where } A=\left(x+\frac{1}{2}\right)(x-1)
\end{aligned}
$$
Now, exponential function is always $+$ ve and $f^{\prime}(x)$ will be opposite to the sign of $A$
which is -ve in $\left[-\frac{1}{2}, 1\right]$
Hence, $f^{\prime}(x)$ is $+$ ve in $\left[-\frac{1}{2}, 1\right]$
$\therefore f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (12 May Online)
If $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, for $\mathrm{x} \in(-1,2)$, then $\mathrm{f}$ is discontinuous at (where $[\mathrm{x}]$ represents floor function)

Options:
A $\mathrm{x}=-1,0,1,2$
B $\mathrm{x}=-1,0,1$
C $\mathrm{x}=0,1$
D $\mathrm{x}=2$

The correct answer is:
$\mathrm{x}=0,1$

Solution:
We have $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$
Let $[\mathrm{x}]=\mathrm{K}$, an integer.
$$
\therefore \lim _{x \rightarrow K^{+}} f(x)=K \text { and } \lim _{x \rightarrow K^{-}} f(x)=K-1
$$
Thus given function is not continuous at all integral values in its domain.
$\because \mathrm{f}$ is discontinuous at $\mathrm{x}=0,1$.

Topic: Mathematics, Functions, MHT CET, MHT CET 2021 (20 Sep Shift 1)
If $f(x)=\left\{\begin{array}{l}x, \text { if } x \text { is rational } \\ -x, \text { if } x \text { is irrational then } \lim _{x \rightarrow 0} f(x) \text { is }\end{array}\right.$

Options:
A Equal to 0
B Equal to 1
C Equal to -1
D Indeterminate

The correct answer is:
Equal to 0

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0-h)=0 \\ & \text { and } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}-(0+h)=0 \\ & \therefore \lim _{x \rightarrow 0} f(x)=0,\left(\because \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)\right) .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
If $f(x)=\left\{\begin{array}{cl}\frac{\log (1+2 a x)-\log (1-b x)}{x}, & x \neq 0 \\ k & , x=0\end{array}\right.$
continuous at $x=0$, then value of $k$ is

Options:
A $b+a$
B $b-2 a$
C $2 a-b$
D $2 a+b$

The correct answer is:
$2 a+b$

Solution:
Given, $f(x)=\left\{\begin{array}{cl}\frac{\log (1+2 a x)-\log (1-b x)}{x}, & x \neq 0 \\ k & , x=0\end{array}\right.$
is continuous at $x=0$. $\therefore f(0)=\lim _{x \rightarrow 0} \frac{\log (1+2 a x)-\log (1-b x)}{x}$
$\left(\frac{0}{0}\right.$ form $)$
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{\frac{1}{2 a x+1}(2 a)-\frac{1}{1-b x}(-b)}{+1}$
(by 'L' Hospital's rule) $\Rightarrow k=\frac{2 a}{0+1}+\frac{b}{1-0}=2 a+b$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=x^n$, then the value of $f(1)-\frac{f^{\prime}(1)}{1}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime}(1)}{3 !}+\ldots \ldots \ldots \ldots . \frac{(-1)^n f^{\prime \prime}(1)}{n !}$ is

Options:
A
1

B
$2^n$

C
$2^{\mathrm{n}}-1$

D
0

The correct answer is:
0

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \Rightarrow \mathrm{f}(1)=1$
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{nx} x^{\mathrm{n}-1} \Rightarrow \mathrm{f}^{\prime}(1)=\mathrm{n}$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1) \mathrm{x}^{\mathrm{n}-2} \Rightarrow \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1)$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \mathrm{f}^{\mathrm{n}}(\mathrm{x})=\mathrm{n} ! \Rightarrow \mathrm{f}^{\mathrm{n}}(1)=\mathrm{n} !$
$=1-\frac{\mathrm{n}}{1 !}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !}+\ldots \ldots \ldots+(-1)^{\mathrm{n}} \frac{\mathrm{n} !}{\mathrm{n} !}$
$={ }^{\mathrm{n}} \mathrm{C}_0-{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2-{ }^{\mathrm{n}} \mathrm{C}_3+\ldots \ldots \ldots+(-1)^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=0$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2003
If fx=xp+1cos1x:xâ 00:x=0, then at x=0 the function fx is

Options:
A Continuous if p>-1 and differentiable if p>0
B Continuous if p>0 and differentiable if p>-1
C Continuous and differentiable if p>-1
D None of these

The correct answer is:
Continuous if p>-1 and differentiable if p>0

Solution:
Continuity at x=0,
L.H.L.=limh→0+f0-h=limh→0+-hp+1cos-1⁡1h
=0 if p>-1
R.H.L.=limh→0+f0+h=limh→0+hp+1cos-1⁡1h
=0 if p>-1
and f0=0.
∴fx is continuous at x=0 if p>-1
Differentiability at x=0:
L.H.D.=limh→0+f0−h−f0−h
=limh→0+-hp+1cos-1⁡1h-0-h
=limh→0+-hpcos-1⁡1h=0 if p>0,
R.H.D.=limh→0+hp+1cos−11h−0h
=limh→0+hpcos-1⁡1h=0 if p>0
∴fx is differentiable at x=0 if p>0

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=xpcos⁡1x,x≠00,x=0 is differentiable at x=0, then

Options:
A p<0
B 0<p<1
C p=1
D p>1

The correct answer is:
p>1

Solution:
Given, fx=xpcos⁡1x, x≠0 0, x=0
Since, f(x) is differentiable at x=0, therefore it is continuous at x=0
∴ limx→0 fx=f0=0
⇒ limx→0 xpcos⁡1x=0 ⇒ p>0
As fx is differentiable at x=0
∴ limx→0 fx-f(0)x-0 exists finitely
⇒ limx→0 xpcos⁡1x-0x exists finitely
⇒limx→0 xp-1cos⁡1x exists finitely
⇒ p-1>0 ⇒ p>1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{r}\frac{\sin x}{x}+\cos x, \text { when } x \neq 0 \\ 2, \text { when } x=0\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 0^+} f(x) \neq 2$
B $\lim _{x \rightarrow 0^-} f(x)=0$
C $f(x)$ is continuous at $x=0$
D None of these

The correct answer is:
$f(x)$ is continuous at $x=0$

Solution:
$f(0^+)=f(0^-)=2$ and $f(0)=2$
Hence $f(x)$ is continuous at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=|x|+|\sin x|$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then its left hand derivative at $x=0$ is

Options:
A 0
B -1
C -2
D -3

The correct answer is:
-2

Solution:
$\begin{aligned} & f(x)=|x|+|\sin x| \\ & \text { LHD }=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h} \\ & =\lim _{h \rightarrow 0} \frac{|0-h|+|\sin (0-h)|-(0+0)}{0-h} \\ & =\lim _{h \rightarrow 0} \frac{h+\sin h}{-h}=-\lim _{h \rightarrow 0}\left(1+\frac{\sin h}{h}\right) \\ & =-(1+1)=-2\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $\mathrm{f}(\mathrm{x})=\mathrm{x}^{\tan \mathrm{x}}+(\tan \mathrm{x})^{\mathrm{x}}$, then $\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=$

Options:
A $1+\frac{\pi}{2} \log \left(\frac{\mathrm{e} \pi}{4}\right)$
B $\frac{\pi}{2}\left(\log \frac{\pi}{4}+1\right)$
C $1$
D $0$

The correct answer is:
$1+\frac{\pi}{2} \log \left(\frac{\mathrm{e} \pi}{4}\right)$

Solution:
$$
f(x)=x^{\tan x}+(\tan x)^x
$$
$$
\begin{gathered}
f_1=x^{\tan x} \Rightarrow \log f_1=\tan x \cdot \log x \\
\Rightarrow \frac{d}{d x} \log f_1=\frac{d}{d x}(\tan x \cdot \log x)
\end{gathered}
$$
$\Rightarrow \frac{d f_1}{d x}=f_1\left(\frac{\tan x}{x}+\left(\sec ^2 x\right) \log x\right)$ ...(1)
$\begin{aligned} & f_2=(\tan x)^x \Rightarrow \log f_2=x \cdot \log (\tan x) \\ & \frac{d}{d x} \log f_2=\frac{d}{d x} x \cdot \log (\tan x) \\ & \Rightarrow \frac{1}{f_2} \cdot \frac{d f_2}{d x}=\frac{x \cdot 1\left(\sec ^2 x\right)}{\tan x}+\log (\tan x)\end{aligned}$
$\Rightarrow \frac{d f_2}{d x}=f_2\left(\frac{x \sec ^2 x}{\tan x}+\log \tan x\right)$ ...(2)
$\begin{aligned} & \Rightarrow f^{\prime}(x)=\frac{d f_1}{d x}+\frac{d f_2}{d x} \\ & \Rightarrow f^{\prime}(x)=x^{\tan x}\left(\frac{\tan x}{x}+\left(\sec ^2 x\right) \log x\right) \\ & +(\tan x)^x\left(\frac{x \sec ^2 x}{\tan x}+\log \tan x\right) \\ & \Rightarrow f^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi}{4}\left(\frac{1}{\frac{\pi}{4}}+(\sqrt{2})^2 \cdot \log \frac{\pi}{4}\right)+(1)^{\pi / 4}\left(\frac{(\sqrt{2})^2 \cdot \frac{\pi}{4}}{1 \cdot}+\log 1\right) \\ & =\frac{\pi}{4}\left(\frac{4}{\pi}+2 \log \frac{\pi}{4}\right)+\frac{\pi}{2}=1+\frac{\pi}{2} \log \frac{\pi}{4}+\frac{\pi}{2} \\ & =1+\frac{\pi}{2}\left(\log \frac{\pi}{4}+1\right)=1+\frac{\pi}{2}\left(\log \frac{\pi}{4}+\log e\right) \\ & =1+\frac{\pi}{2} \log \left(\frac{e \pi}{4}\right) . \\ & \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=x \tan ^{-1} x$, then $f(1)=$

Options:
A $1+\frac{\pi}{4}$
B $\frac{1}{2}+\frac{\pi}{4}$
C $\frac{1}{2}-\frac{\pi}{4}$
D $2$

The correct answer is:
$\frac{1}{2}+\frac{\pi}{4}$

Solution:
$f(x)=x \tan ^{-1} x$
Differentiating w.r.t $x$, we get
$f^{\prime}(x)=x \frac{1}{1+x^2}+\tan ^{-1} x$
Now put $x=1$, then $f^{\prime}(1)=\frac{1}{2}+\tan ^{-1}(1)=\frac{\pi}{4}+\frac{1}{2}$.

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=|x|$, then $f(x)$ is

Options:
A Continuous for all x
B Differentiable at $x=0$
C Neither continuous nor differentiable at $x=0$
D None of these

The correct answer is:
Continuous for all x

Solution:
It is obvious that $|x|$ is continuous for all $x$.
Now,
$\begin{aligned} & \quad R f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{|0+h|-0}{h}=1 \\& \text { Now, } \\
& L f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{|0-h|-0}{-h}=-1\end{aligned}$
Hence $f(x)=|x|$ is not differentiable at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{l}x, \text { when } 0 \lt x \lt 1 / 2 \\ 1, \text { when } x=1 / 2 \\ 1-x, \text { when } 1 / 2 \lt x \lt 1\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 1 / 2}^{+} f(x)=2$
B $\lim _{x \rightarrow 1 / 2}^{-} f(x)=2$
C $f(x)$ is continuous at $x=\frac{1}{2}$
D $f(x)$ is discontinuous at $x=\frac{1}{2}$

The correct answer is:
$f(x)$ is discontinuous at $x=\frac{1}{2}$

Solution:
Since $\lim _{x \rightarrow 1 / 2} f(x) \neq f\left(\frac{1}{2}\right)$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\int_x^{x+1} e^{-t^2} d t$, then the interval in which $f(x)$ is decreasing is

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

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

Solution:
We have,
$$
\begin{aligned}
& f(x)=\int_x^{x+1} e^{-t^2} d t \Rightarrow f^{\prime}(x)=e^{-(x+1)^2}-e^{-x^2} \\
& f^{\prime}(x)=\frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}}
\end{aligned}
$$
Since, $f(x)$ is decreasing function.
$$
\begin{array}{ll}
\therefore & f^{\prime}(x) < 0 \\
\therefore \quad & \frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}} < 0 \Rightarrow e^{x^2} < e^{(x+1)^2} \\
& x^2 < (x+1)^2 \Rightarrow x^2+2 x+1-x^2>0 \\
& 2 x+1>0 \Rightarrow x>-\frac{1}{2} \\
\therefore & x \in\left(\frac{-1}{2}, \infty\right)
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main
If fx=xx-1, then the points of discontinuity of the function f15x, where fn=fof…….of (n times), are

Options:
A x=2,1
B x=0,1
C x=1,2,0
D continuous everywhere except x=1

The correct answer is:
continuous everywhere except x=1

Solution:
Clearly, x=1 is a point of discontinuity of the function fx=xx-1 .
If x≠1 , then
fofx=ffx=fxx-1=x , which is continuous ∀x∈R .
If x≠1 , then fofofx=fx=xx-1
Hence, f15x=fofof………15x=fx which is continuous everywhere except x=1
So, the only point of discontinuity is x=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x-x4, x∈R, where x denotes the greatest integer function, then:

Options:
A limx→4+fx exists but limx→4-fx does not exist
B f is continuous at x=4
C limx→4-fx exists but limx→4+fx does not exist
D Both limx→4-fx and limx→4+fx exist but are not equal

The correct answer is:
f is continuous at x=4

Solution:
R.H.L.=limx→4+x-x4=4-1=3

L.H.L.=limx→4-x-x4=3-0=3

fx=4-1=3

L.H.L.=f4=R.H.L.

Hence, given function is continuous at x=4.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If $f(x)=|x|/x$ for x≠0 and 1 for X=0, then the function is

Options:
A continuous but not differentiable at $x=0$
B differentiable but not continuous at $x=0$
C neither continuous nor differentiable at $x=0$
D continuous and differentiable at $x=0$

The correct answer is:
neither continuous nor differentiable at $x=0$

Solution:
$f(x)=\left\{\begin{array}{cc}
1 & x>0 \\
-1 & x < 0 \\
1 & x=0
\end{array}\right.$
So function discontinuous and non differentiable

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{cc}{[x]+[-x],} & x \neq 2 \\ K, & x=2\end{array}\right.$, then $f(x)$ is continuous at $x=2$, provided $K$ is equal to

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

The correct answer is:
$-1 \quad$

Solution:
$f(x)=\left\{\begin{array}{cc}{[x]+[-x],} & x \neq 2 \\ K, & x=2\end{array}\right.$
Since, $f(x)$ is continuous at $x=2$
$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}[x]+[-x]$
$=1+(-2)=-1$
$\quad \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}[x]+[-x]=2-3=-1$
$\Rightarrow \quad K=-1$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=xx+xx+sin⁡πx, where [·] and {·} represent the greatest integer function and the fractional part function respectively, then f'72 is equal to

Options:
A 3lnÂ 3+Ï4
B 3lnÂ 3+3Ï4
C 3lnÂ 3+Ï+34
D 3ln 3+34

The correct answer is:
3ln 3+34

Solution:
fx=xx+xx+sin⁡πx
fx=3x-3+x-33+sin⁡πx,x∈3, 4
f'x=3x-3ln⁡3+3x-32+πcos⁡πx
∴ f'72=3ln3+34

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f(x+y)=f(x) f(y) and Σx=1∞fx=2, x,y∈N, where N is the set of all natural numbers, then the value of f(4)f(2) is

Options:
A 23
B 19
C 13
D 49

The correct answer is:
49

Solution:
f(x)=ax satisfies the functional relation, so
Σx=1∞fx=2
⇒a+a2+a3……∞ terms =2
⇒a1-a=2

⇒a=23
⇒fx=23x
Now, f(4)f(2)=234232=49

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (06 Sep Shift 1)
If $f(x+y)=f(x) f(y)$ for all $x$ and $y$ and $f(5)=2, f^{\prime}(0)=3$, then $f^{\prime}(5)$ is

Options:
A 5
B 6
C 0
D None of these

The correct answer is:
6

Solution:
We have, $f(x+y)=f(x) f(y)$
Putting $x=y=0$
$$
\begin{array}{ll}
\Rightarrow & f(0+0)=f(0) f(0) \\
\Rightarrow & f(0)=[f(0)]^{2} \\
\Rightarrow & f(0)=1 \\
\text { Now, } f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\
= & \lim _{h \rightarrow 0} \frac{f(5)}{f(h)-f(5)}=\lim _{h \rightarrow 0} f(5)\left[\frac{f(h)-1}{h}\right] \\
= & f(5) \lim _{h \rightarrow 0}\left[\frac{f(h)-f(0)}{h}\right]\quad \text{f(0)=1} \\
= & 2 \times f^{\prime}(0)=2 \times 3=6
\end{array}
$$

Topic: Mathematics, Differentiation, JEE Main
If $f(x+y)=f(x) \cdot f(y) \forall x \cdot y$ and $f(5)=2, f^{\prime}(0)=3$ then $f^{\prime}(5)$ is

Options:
A
0

B
1

C
6

D
2

The correct answer is:
6

Solution:
$$
f(x+y)=f(x) \times f(y)
$$
Differentiate with respect to $x$, treating $y$ as constant
$$
f^{\prime}(x+y)=f(x) f(y)
$$
Putting $x=0$ and $y=x$, we get $f^{\prime}(x)=f^{\prime}(0) f(x) ; \Rightarrow f^{\prime}(5)=3 f(5)=3 \times 2=6$

Topic: Mathematics, Functions, JEE Main, JEE Main 2002
If f x + y, x - y = xy, then the arithmetic mean of fx, y and fy, x is

Options:
A x
B y
C 0
D x2-y22

The correct answer is:
0

Solution:
Let, x + y = p, x - y = q
Then, f p, q = p + q 2 Â· p - q 2 = p 2 - q 2 4
â´ f x, y = x 2 - y 2 4 and f y, x = y 2 - x 2 4
A.M.=fx, y+fy, x2
=x2-y24+y2-x242=0

Topic: Mathematics, Functions, JEE Main
If $f(y)=1-(y-1)+(y-1)^2-(y-1)^3$ $+\ldots-(y-1)^{17}$ then the coefficient of $y^2$ in it is

Options:
A
${ }^{17} \mathrm{C}_2$

B
${ }^{17} \mathrm{C}_3$

C
${ }^{18} \mathrm{C}_2$

D
${ }^{18} \mathrm{C}_3$

The correct answer is:
${ }^{18} \mathrm{C}_3$

Solution:
Given function is
$$
\begin{array}{r}
f(y)=1-(y-1)+(y-1)^2-(y-1)^3 \\
+\ldots \ldots \ldots-(y-1)^{17}
\end{array}
$$
In the expansion of $(y-1)^n$
$$
T_{r+1}={ }^n \mathrm{C}_r y^{n-r}(-1)^r
$$
coeff of $y^2$ in $(y-1)^2={ }^2 \mathrm{C}_0$
coeff of $y^2$ in $(y-1)^3={ }^3 \mathrm{C}_1$
coeff of $y^2$ in $(y-1)^4={ }^4 \mathrm{C}_2$
So, coeff of termwise is
$$
\begin{aligned}
& { }^2 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15} \\
& =1+{ }^3 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots . .+{ }^{17} \mathrm{C}_{15} \\
& =\left({ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1\right)+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots . .+{ }^{17} \mathrm{C}_{15} \\
& ={ }^4 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15}
\end{aligned}
$$
$$
\begin{aligned}
& ={ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15} \\
& ={ }^{18} \mathrm{C}_{15}={ }^{18} \mathrm{C}_3
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2012 (07 May Online)
If $\mathrm{f}(\mathrm{y})=\mathrm{e}^{\mathrm{y}}, \mathrm{g}(\mathrm{y})=\mathrm{y} ; \mathrm{y}>0$ and $F(t)=\int_0^{\mathrm{t}} \mathrm{f}(\mathrm{t}-\mathrm{y}) \mathrm{g}(\mathrm{y})$, then

Options:
A
$\mathrm{F}(\mathrm{t})=\mathrm{te}^{-t}$

B
$\mathrm{F}(\mathrm{t})=1-\mathrm{te}^{-t}(1+\mathrm{t})$

C
$\mathrm{F}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}-(1+\mathrm{t})$

D
$\mathrm{F}(\mathrm{t})=\mathrm{te}^{\mathrm{t}}$

The correct answer is:
$\mathrm{F}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}-(1+\mathrm{t})$

Solution:
$F(t)=\int_0^t f(t-y) g(y) d y$
$=\int_0^t e^{t-y} y d y=e^t \int_0^t e^{-y} y d y$
$=e^t\left[-y e^{-y}-e^{-y}\right]_0^t=-e^t\left[y e^{-y}+e^{-y}\right]_0^t$
$=-e^t\left[t e^{-t}+e^{-t}-0-1\right]=e^t\left[\frac{t+1-e^t}{e^t}\right]=e^t-(1+t)$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
If $f: Z \rightarrow N$ is defined by
$f(n)=\left\{\begin{array}{cll}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0, \text { then } f \text { is } \\ -2 n-1, & \text { if } & n < 0\end{array}\right.$

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

The correct answer is:
onto but not one-one

Solution:
We have,
$f(n)= \begin{cases}2 n, & \text { if } n>0 \\ 1, & \text { if } n=0 \\ -2 n-1, & \text { if } n < 0\end{cases}$
When, $n>0$
$f(n)=2,4,6,8 \ldots .$
When, $n < 0$
$f(n)=1,3,5,7 \ldots . .$
$\therefore$ Range of $f(n)$ is $\{1,2,3,4, \ldots .$.$\} .$
$\therefore \quad$ Codomain $=$ Range
Hence, $f(n)$ is onto.
Here, $f(0)-f(-1)=1$
$\therefore f$ is not one one.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
If first excitation potential of a hydrogen-like atom is V electron volt, then the ionization energy of this atom will be

Options:
A V electron volt

B 3V4 electron volt

C 4V3 electron volt

D Cannot be calculated by given information

The correct answer is:
4V3 electron volt

Solution:
First excitation energy =Rhc112-122=Rhc34
∴ 34Rhc=V eV
∴ Rhc=4V3 eV

Topic: Physics, Atomic Physics, JEE Main
If five G.M.'s are inserted between 486 and $2 / 3$ then fourth G.M. will be

Options:
A $4$
B $6$
C $12$
D $-6$

The correct answer is:
$6$

Solution:
Let $G_1, G_2, G_3, G_4, G_5$ be the G.M.'s are inserted between 486 and $2 / 3$. So total terms are 7 .
$T_n=a r^{n-1} \Rightarrow 2 / 3=486(r)^6 \Rightarrow r=1 / 3$
Hence $4^{\text {th }}$ G.M. will be, $T_5=a r^4=486(1 / 3)^4=6$

Topic: Mathematics, Sequences and Series, JEE Main
If for a>0, the feet of perpendiculars from the points Aa,-2a,3 and B0,4,5 on the plane lx+my+nz=0 are points C0,-a,-1 and D respectively, then the length of line segment CD is equal to :

Options:
A 31
B 41
C 55
D 66

The correct answer is:
66

Solution:
C lies on plane ⇒-ma-n=0⇒mn=-1a ….(1)

Also, CA→ will be parallel to the normal to the plane

So, CA→ ‖ li^+mj^+nk^

⇒a-0l=-am=4n⇒mn=-a4 ...(2)

From (1) and (2)

-1a=-a4⇒a2=4⇒a=2 (since a>0)

From (2), mn=-12

Let m=-t⇒n=2t

2l=-2-t⇒l=t

So plane : tx-y+2z=0

BD=66=6

C≡0,-2,-1

CD=BC2-BD2

=02+62+62-62

=66

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If for a continuous function $\mathrm{f}(\mathrm{x})$, $\int_{-\pi}^t(f(x)+x d x)=\pi^2-t^2$, for all $t \geq-\pi$, then $\mathrm{f}\left(-\frac{\pi}{3}\right)$ is equal to:

Options:
A
$\pi$

B
$\frac{\pi}{2}$

C
$\frac{\pi}{3}$

D
$\frac{\pi}{6}$

The correct answer is:
$\pi$

Solution:
Let $\int_{-\pi}^t(f(x)+x) d x=\pi^2-t^2$
$$
\begin{aligned}
&\Rightarrow \int_{-\pi}^t f(x) d x+\int_{-\pi}^t x d x=\pi^2-t^2 \\
&\Rightarrow \int_{-\pi}^t f(x) d x+\left(\frac{t^2}{2}-\frac{\pi^2}{2}\right)=\pi^2-t^2 \\
&\Rightarrow \int_{-\pi}^t f(x) d x=\frac{3}{2}\left(\pi^2-t^2\right)
\end{aligned}
$$
differentiating with respect to $t$
$\frac{d}{d t}\left[\int_{-\pi}^t f(x) d x\right]=\frac{3}{2} \frac{d}{d t}\left(\pi^2-t^2\right)$
$f(t) \cdot \frac{d t}{d t}-f(-\pi) \frac{d}{d t}(-\pi)=-3 t$
$f(t)=-3 t$
$f\left(-\frac{\pi}{3}\right)=-3\left(-\frac{\pi}{3}\right)=\pi$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (12 Apr Online)
If for a given substance, melting point is $T_B$ and freezing point is $T_A$ then correct variation of entropy by graph between entropy change and temperature is

Options:
A

B

C

D

The correct answer is:

Solution:
For a pure substance $T_A$ and $T_B$ are same. Above to this temperature, entropy will increase and below this temperature entropy remains constant.

Topic: Chemistry, Thermodynamics (C), JIPMER, JIPMER 2015
If for a plane, the intercepts on the coordinate axes are 8, 4, 4 then the length of the perpendicular from the origin on to the plane is

Options:
A $8 / 3$
B $3 / 8$
C 3
D $4 / 3$

The correct answer is:
$8 / 3$

Solution:
Equation of plane is, $\frac{x}{8}+\frac{y}{4}+\frac{z}{4}=1 \Rightarrow x+2 y+2 z=8$
Length of the perpendicular from the origin on the plane
$=\left|\frac{-8}{\sqrt{1^2+2^2+2^2}}\right|=\frac{8}{3}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If for a plane, x, y, z intercepts are 8,â4,â4 respectively, then the length of the perpendicular from the origin on to the plane is

Options:
A 83 units
B 38 units
C 3 units
D 45 units

The correct answer is:
83 units

Solution:
Equation of plane is x8+y4+z4=1
â x+2y+2z=8
Length of perpendicular from origin to x+2y+2z-8=0
=-81+4+4=83

Topic: Mathematics, Vector Algebra, JEE Main
If, for a positive integer n, the quadratic equation,
xx+1+x+1x+2+...+x+n-1¯x+n=10n
has two consecutive integral solutions, then n is equal to:

Options:
A 12
B 9
C 10
D 11

The correct answer is:
11

Solution:
On simplifying we get the quadratic equations as
x2+x2+...+x2⏟n times+1+3+5+...+2n-1x+1.2+2.3+...+n-1n=10n
nx2+n2x+nn2-13=10n
x2+nx+n2-313=0

Let, α, β are the roots of the above equation

∴ α+β=-n, αβ=n2-313
Now, the difference of roots α-β =1

⇒ α-β2=1

⇒α+β2-4αβ=1
⇒n2-43n2-31=1
⇒n2=121
⇒n=11

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2017 (02 Apr)
If for a sample size of 10, ∑i=110xi-52=350 and ∑i=110xi-6=20, then the variance is

Options:
A 23
B 24
C 25
D 26

The correct answer is:
26

Solution:
âi=110xi-6=20âÎ£xi=80
âi=110xi2-10xi+25=350
ââi=110xi2=350+800-250=900
Then, variance =90010-82
=90-64=26

Topic: Mathematics, Statistics, JEE Main
If for a variable line $\frac{x}{a}+\frac{y}{b}=1$, the condition $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is

Options:
A $x^2+y^2=c^2 / 2$
B $x^2+y^2=2 c^2$
C $x^2+y^2=c^2$
D $x^2-y^2=c^2$

The correct answer is:
$x^2+y^2=c^2$

Solution:
$\begin{array}{ll}
\Rightarrow b y-a x=0 \Rightarrow \frac{x}{b}-\frac{y}{a}=0 & {\left[\boxtimes m \text { of given line }=\frac{-b}{a}, \therefore m \text { of perpendicular }=\frac{a}{b}\right]}
\end{array}$
Now, the locus of foot of perpendicular is the intersection point of line $\frac{x}{a}+\frac{y}{b}=1$
and
$\frac{x}{b}-\frac{y}{a}=0$
To find locus, squaring and adding (i) and (ii)
$\begin{gathered}
\left(\frac{x}{a}+\frac{y}{b}\right)^2+\left(\frac{x}{b}-\frac{y}{a}\right)^2=1 \\
\Rightarrow \quad x^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+y^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=1
\end{gathered}$
$\begin{aligned} & x^2\left(\frac{1}{c^2}\right)+y^2\left(\frac{1}{c^2}\right)=1,\left[0 \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\right] \\ \Rightarrow & x^2+y^2=c^2\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main