datavorous/entrance-exam-dataset huggingface dataset:
Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) d t=3 x f(x)-x^3$ for all $x \geq 1$, then the value of $f(2)$ is

Options:

The correct answer is:
2.667

Solution:
Given, $f(1)=\frac{1}{3}$ and $6 \int_1^x f(t) d t$ $=3 x f(x)-x^3$, for all $x \geq 1$
Using (Newton-Leibnitz formula),
On Differentiating both sides,
$$
\begin{aligned}
& 6 f(x) \cdot 1-0-3 f(x)+3 x f^{\prime}(x)-3 x^2 \\
\Rightarrow & 3 x f^{\prime}(x)-3 f(x)=3 x^2 \\
\Rightarrow & f^{\prime}(x)-\frac{1}{x} f(x)=x
\end{aligned}
$$
$$
\Rightarrow \frac{x f^{\prime}(x)-f(x)}{x^2}=1 \Rightarrow \frac{d}{d x}\left\{\frac{f(x)}{x}\right\}=1
$$
On integrating both sides,
$$
\begin{array}{ll}
& \frac{f(x)}{x}=x+C \quad\left[\because f(1)=\frac{1}{3}\right] \\
\Rightarrow \quad & \frac{1}{3}=1+C \Rightarrow \quad C=-\frac{2}{3} \\
& \text { Now, } \quad f(x)=x^2-\frac{2}{3} x \\
\Rightarrow \quad & f(2)=4-\frac{4}{3}=\frac{8}{3}
\end{array}
$$
Note Here, $f(1)=2$ does not satisfy given function.
$$
\therefore \quad f(1)=\frac{1}{3}
$$
For that, $f(x)=x^2-\frac{2}{3} x$ and $\quad f(2)=4-\frac{4}{3}=\frac{8}{3}$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq 0$ and $f(x)=g(x) \sin x$.
Statement $1 \lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.
Statement $2 f^{\prime}(0)=g(0)$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
We have,
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0
\end{aligned}
$$
Since, $f(x)=g(x) \sin x$
$$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \\
& \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \\
& \Rightarrow \quad f^{\prime \prime}(0)=0
\end{aligned}
$$
Thus, $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$
$\Rightarrow$ Statement 1 is true.
Statement 2. $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$
$$
\Rightarrow \quad f^{\prime}(0)=g(0)
$$
Statement 2 is not a correct explanation of Statement 1.

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let $f$ be a function defined on $R$ (the set of all real numbers) such that $f^{\prime}(x)=2010(x-2009)$ $(x-2010)^2(x-2011)^3(x-2012)^4$, for all $x \in R$. If $g$ is a function defined on $R$ with values in the interval $(0, \infty)$ such that $f(x)=\ln (g(x))$, for all $x \in R$, then the number of points in $R$ at which $g$ has a local maximum is

Options:

The correct answer is:
1

Solution:
Let $g(x)=e^{f(x)}, \forall x \in R$
$$
\Rightarrow g^{\prime}(x)=e^{f(x)} \cdot f^{\prime}(x)
$$
$\Rightarrow f^{\prime}(x)$ changes its sign from positive to negative in the neighbourhood of $x=2009$
$\Rightarrow f(x)$ has local maxima at $x=2009$
So, the number of local maximum is one.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 2)
Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int_0^x \sqrt{1-\left\{f^{\prime}(t)\right\}^2} d t=\int_0^x f(t) d t, \quad 0 \leq x \leq 1$
$$
\text { and } f(0)=0 \text {, then }
$$

Options:
A
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
B
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

C
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$
D
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$

The correct answer is:
$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$

Solution:
Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$,
$$
0 \leq x \leq 1
$$
Applying Leibnitz theorem, we get
$$
\begin{array}{rlrl}
& & \sqrt{1-\left(f^{\prime}(x)\right)^2} & =f(x) \\
\Rightarrow & & 1-\left(f^{\prime}(x)\right)^2 & =f^2(x) \\
\Rightarrow & & \left(f^{\prime}(x)\right)^2 & =1-f^2(x) \\
\Rightarrow & & f^{\prime}(x) & =\pm \sqrt{1-f^2(x)} \\
\Rightarrow & & \quad \frac{d y}{d x} & =\pm \sqrt{1-y^2} \\
& \text { where } y=f(x) \Rightarrow & \frac{d y}{\sqrt{1-y^2}}=\pm d x
\end{array}
$$
On integrating both sides, we get
$$
\begin{aligned}
& \sin ^{-1}(y)=\pm x+C \\
& \because \quad f(0)=0 \Rightarrow C=0 \Rightarrow y=\pm \sin x \\
& y=\sin x=f(x) \text { given } f(x) \geq 0 \text { for } \\
& x \in[0,1]
\end{aligned}
$$
It is known that $\sin x < x, \forall x \in R^{+}$
$\therefore \quad \sin \left(\frac{1}{2}\right) < \frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right) < \frac{1}{2}$
and $\sin \left(\frac{1}{3}\right) < \frac{1}{3} \Rightarrow f\left(\frac{1}{3}\right) < \frac{1}{3}$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2009 (Paper 1)
Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to

Options:

The correct answer is:
9

Solution:
The equation of the tangent at $(x, y)$ to the given curve $y=f(x)$ is
$$
\begin{gathered}
Y-y=\frac{d y}{d x}(X-x) \\
Y \text {-intercept }=y-x \frac{d y}{d x}
\end{gathered}
$$
According to the question
$$
x^3=y-x \frac{d y}{d x}
$$
$$
\Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2
$$
which is linear in $x$.
$$
\text { IF }=e^{\int \frac{-1}{x} d x}=\frac{1}{x}
$$
$\therefore$ Required solution is
$$
\begin{aligned}
& y \cdot \frac{1}{x}=\int-x^2 \cdot \frac{1}{x} d x \\
& \Rightarrow \quad \frac{y}{x}=\frac{-x^2}{2}+c \\
& \Rightarrow \quad y=\frac{-x^3}{2}+c x \\
& \text { At } x=1, y=1 \text {, } \\
& 1=\frac{-1}{2}+c \\
& \Rightarrow \quad c=\frac{3}{2} \\
&
\end{aligned}
$$
Now, $\begin{aligned} f(-3) & =\frac{27}{2}+\frac{3}{2}(-3) \\ & =\frac{27-9}{2}=9\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $f$ be a real-valued function defined on the interval $(0, \infty)$, by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} d t$. Then which of the following statement(s) is (are) true ?

Options:
A
$f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$

B
$f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$

C
there exists $\alpha>1$ such that $\left|f^{\prime}(x)\right| < |f(x)|$ for all $x \in(\alpha, \infty)$
D
there exists $\beta>0$ such that $|f(x)|+\left|f^{\prime}(x)\right| \leq \beta \quad$ from all $x \in(0, \infty)$

The correct answers are:
$f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$
,
there exists $\alpha>1$ such that $\left|f^{\prime}(x)\right| < |f(x)|$ for all $x \in(\alpha, \infty)$

Solution:
Here, $f^{\prime}(x)=\frac{1}{x}+\sqrt{1+\sin x}, x>0$ but $f(x)$ is not differentiable in $(0, \infty)$ as $\sin x$ may be $-1$ and then $f^{\prime \prime}(x)=-\frac{1}{x^2}+\frac{\cos x}{2 \sqrt{1+\sin x}}$ will not exists.
$\Rightarrow f^{\prime}(x)$ is continuous for all $x \in(0, \infty)$ but $f^{\prime}(x)$ is not differentiable on $(0, \infty)$.
$\therefore$ Option (b) is true.
Also, $\quad f^{\prime}(x) \leq 3$, if $x>1$ and $\quad f(x)>3$, if $x>e^3$
$\therefore$ Let $\alpha=e^3$
$\Rightarrow$ Option (c) is true.
(d) is not possible as $f(x) \rightarrow \infty$ when $x \rightarrow \infty$.
Hence, (b, c) is the correct option.

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad$ for all $x \in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to

Options:
A
1

B
$\frac{1}{3}$

C
$\frac{1}{2}$

D
$\frac{1}{e}$

The correct answer is:
$\frac{1}{3}$

Solution:
We have, $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t x \in(-1,1)$
On differentiating w.r.t. $x$, we get
$$
\begin{array}{ll}
& e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\
\Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\
\because & f^{-1} \text { is the inverse of } f \\
\therefore & f^{-1}(f(x))=x \\
\Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\
\Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\
\Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\
\text { At } \quad & x=0, f(x)=2 \\
& f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3}
\end{array}
$$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2010 (Paper 2)
Let $f, g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f(x)=e^{x^2}+e^{-x^2}, \quad g(x)=x e^{x^2}+e^{-x^2}$ and $h(x)=x^2 e^{x^2}+e^{-x^2}$. If $a, b$ and $c$ denote respectively, the absolute maximum of $f, g$ and $h$ on $[0,1]$, then

Options:
A
$a=b$ and $c \neq b$

B
$a=c$ and $a \neq b$

C
$a \neq b$ and $c \neq b$

D
$a=b=c$

The correct answer is:
$a=b=c$

Solution:
Given function,
$$
\begin{aligned}
& f(x)=e^{x^2}+e^{-x^2}, \\
& g(x)=x e^{x^2}+e^{-x^2} \text { and } \\
& h(x)=x^2 e^{x^2}+e^{-x^2} \text { are } \quad \text { strictly }
\end{aligned}
$$
increasing on $[0,1]$. Hence, at $x=1$, the given function attains absolute maximum all equal to $e+\frac{1}{e}$.
$$
\Rightarrow \quad a=b=c
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $f: I R \rightarrow I R$ be defined as $f(x)=|x|+\left|x^{2}-1\right|$. The total number of points at which $f$ attains either a local maximum or a local minimum is

Options:

The correct answer is:
5

Solution:
$f(x)=|x|+\left|x^{2}-1\right|=\left\{\begin{array}{c}-x+x^{2}-1, x < -1 \\ -x-x^{2}+1,-1 \leq x \leq 0 \\ x-x^{2}+1,0 < x < 1 \\ x^{2}+x-1, \quad x \geq 1\end{array}\right.$

$\therefore \quad f^{\prime}(x)=\left[\begin{array}{ccc}

2 x-1 & , & x < -1 \\

-2 x-1 & , & -1 \leq x \leq 0 \\

-2 x+1 & , & 0 < x < 1 \\

2 x+1 & , & x>1

\end{array}\right.$

Critical points are $\frac{1}{2}, \frac{-1}{2},-1,0$ and 1 .

We observe at five points $f^{\prime}(x)$ changes its sign $\therefore$ There are 5 points at which either local maximum or local minimum.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $f: R \rightarrow R$ be a continuous function, which satisfies $f(x)=\int_0^x f(t) d t$. Then, the value of $f(\ln 5)$ is

Options:

The correct answer is:
0

Solution:
From given integral equation, $f(0)=0$. Also, differentiating the given integral equation w.r.t. $x$, we get
$$
\begin{gathered}
f^{\prime}(x)=f(x) \\
\text { If } \quad f(x) \neq 0 \Rightarrow \frac{f^{\prime}(x)}{f(x)}=1 \\
\Rightarrow \quad \log f(x)=x+C \Rightarrow f(x)=e^C e^x \\
\because f(0)=0 \Rightarrow e^C=0, \text { a contradiction } \\
\therefore f(x)=0, \forall x \in R \Rightarrow f(\ln 5)=0
\end{gathered}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2009 (Paper 2)
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in R$. If $f(x)$ is differentiable at $x=0$, then

Options:
A
$f(x)$ is differentiable only in a finite interval containing zero

B
$f(x)$ is continuous, $\forall x \in R$

C
$f^{\prime}(x)$ is constant, $\forall x \in R$

D
$f(x)$ is differentiable except at finitely many points

The correct answers are:
$f(x)$ is continuous, $\forall x \in R$
,
$f^{\prime}(x)$ is constant, $\forall x \in R$

Solution:
$f(x+y)=f(x)+f(y)$, as $f(x)$ is differentiable at $x=0$.
$$
\begin{aligned}
\Rightarrow & f^{\prime}(0)=k \\
\text { Now, } f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(h)}{h} \quad\left[\frac{0}{0} \text { from }\right]
\end{aligned}
$$
Given, $f(x+y)=f(x)+f(y), \forall x, y$
$$
\begin{aligned}
& \therefore & f(0) & =f(0)+f(0), \\
& \text { when } & x & =y=0 \Rightarrow f(0)=0
\end{aligned}
$$
Using L'Hospital's rule,
$$
\lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1}=f^{\prime}(0)=k
$$
$\Rightarrow f^{\prime}(x)=k$, on integrating both sides, $f(x)=k x+C$, as $f(0)=0 \Rightarrow C=0$
So, $f(x)=k x$
$\therefore f(x)$ is continuous for all $x \in R$ and $f^{\prime}(x)=k$, i.e. constant for all $x \in R$.
Hence, both (b) and (c) are correct.
Solutions (Q. Nos. 12-13)
$$
\left.\left(\begin{array}{l}
3 \mathrm{~W} \\
2 \mathrm{R}
\end{array}\right)(\underbrace{1 \mathrm{~W}}_{u_1})\right\} \text { Initial }
$$
Head appears.
$\left(\begin{array}{l}2 \mathrm{~W} \\ 2 \mathrm{R}\end{array}\right) \underset{u_1}{\mathrm{wW}}(\underbrace{2 \mathrm{~W}}_{u_2})$
or $\left.\left(\begin{array}{l}3 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right) \underset{u_1}{\stackrel{1 \mathrm{R}}{u_1}}\left(\begin{array}{c}1 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right)\right)_{u_2}^{u_2}$

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $f(\theta)=\sin \left[\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right]$, where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then, the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

Options:

The correct answer is:
1

Solution:
$f(\theta)=\sin \left(\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right),-\frac{\pi}{4} < \theta < \frac{\pi}{4}$
Let $\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}=\phi$
$$
\Rightarrow \quad \tan \phi=\frac{\sin \theta}{\sqrt{\cos 2 \theta}}
$$
$$
\begin{aligned}
\therefore \quad \sin \phi & =\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}} \\
& =\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}=\frac{\sin \theta}{\cos \theta}=\tan \theta
\end{aligned}
$$
$$
\begin{array}{ll}
\therefore & f(\theta)=\sin \phi=\tan \theta \\
\Rightarrow & \frac{d f(\theta)}{d(\tan \theta)}=1
\end{array}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$.

Question: Consider the statements:

$P$ : There exists some $x \in \mathrm{R}$ such that $f(x)+2 x$

$=2\left(1+x^{2}\right)$

$Q$ : There exists some $x \in \mathrm{R}$ such that $2 f(x)+1$ $=2 x(1+x)$

Then

Options:
A both $P$ and $Q$ are true
B $P$ is true and $Q$ is false
C $P$ is false and $Q$ is true
D both $P$ and $Q$ are false

The correct answer is:
$P$ is false and $Q$ is true

Solution:
For the statement $P, f(x)+2 x=2\left(1+x^{2}\right)$

$\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2\left(1+x^{2}\right)$

$\Rightarrow(1-x)^{2} \sin ^{2} x=x^{2}-2 x+1+1$

$\Rightarrow(1-x)^{2} \sin ^{2} x=(1-x)^{2}+1$

$\Rightarrow(1-x)^{2} \cos ^{2} x=-1$, which is not possible for any real value of $x$.

Hence $P$ is not true.

Let $H(x)=2 f(x)+1-2 x(1+x)$

$H(0)=2 f(0)+1-0=1$

and $H(1)=2 f(1)+1-4=-3$

Hence, $H(x)$ has a solution in $(0,1)$

Therefore, $Q$ is true.

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$.

Question: Which of the following is true?

Options:
A $g$ is increasing on $(1, \infty)$
B $g$ is decreasing on $(1, \infty)$
C $g$ is increasing on $(1,2)$ and decreasing on $(2, \infty)$
D $g$ is decreasing on $(1,2)$ and increasing on $(2, \infty)$

The correct answer is:
$g$ is decreasing on $(1, \infty)$

Solution:
$g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$, $\therefore \quad g^{\prime}(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right] f(x)$

Here $f(x)>0, \forall x \in(1, \infty)$

Let $h(x)=\frac{2(x-1)}{x+1}-\ln x$

$\therefore \quad h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{(x+1)^{2} x} < 0, x \in(1, \infty)$

$\Rightarrow h(x)$ is decreasing function.

Hence, for $x>1, h(x) < h(1) \Rightarrow h(x) < 0 \forall x>1$

$\Rightarrow g^{\prime}(x) < 0 \forall x \in(1, \infty)$

Therefore, $g(x)$ is decreasing on $(1, \infty)$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}}$ for $n \geq 2$ and $g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text { occurs } n \text { times }}(x)$. Then, $\int x^{n-2} g(x) d x$ equals

Options:
A
$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

B
$\frac{1}{n-1}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

C
$\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$

D
$\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$

The correct answer is:
$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

Solution:
Here, $\begin{array}{rlrl} & \text { and } & f f(x) & =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ & \therefore \quad f f f(x) & =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \text { Let } & g(x) & =\underbrace{(f \circ f o \ldots o f)}_{n \text { times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I & =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ & & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array}$
$\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$.

Topic: Mathematics, Indefinite Integration, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $f(x)=2+\cos x$ for all real $x$.
Statement I For each real $t$, there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$.
Statement II $f(t)=f(t+2 \pi)$ for each real $t$.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is True, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is True, Statement II is true; Statement II is not a correct explanation for Statement I

Solution:
$f(x)=2+\cos x \forall x \in R$
Statement I There exists a point $c \in[t, t+\pi]$, where $f^{\prime}(c)=0$ Hence, Statement 1 is true.
Statement II $f(t)=f(t+2 \pi)$ is true.
But Statement II is not a correct explanation for Statement I.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$, such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then,

Options:
A
$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$

B
$f^{\prime}\left(\frac{1}{2}\right)=0$

C
$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$

D
$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$

The correct answers are:
$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$
,
$f^{\prime}\left(\frac{1}{2}\right)=0$
,
$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$
,
$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$

Solution:
Given that, $f(x)=f(1-x)$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=-f^{\prime}(1-x)
$$
Let us put $\quad x=\frac{1}{2}$
$$
\Rightarrow \quad 2 f^{\prime}\left(\frac{1}{2}\right)=0 \Rightarrow f^{\prime}\left(\frac{1}{2}\right)=0
$$
Since, $f^{\prime}\left(\frac{1}{2}\right)=0$ and $f^{\prime}\left(\frac{1}{4}\right)=0$
$\Rightarrow f^{\prime \prime}(x)=0$ at two points in $[0,1]$.
Now, $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
As, $f\left(x+\frac{1}{2}\right) \sin x$ is an odd function which is clear from the following explanation.
Let $g(x)=f\left(x+\frac{1}{2}\right) \sin x$,
$$
g(-x)=f\left(\frac{1}{2}-x\right) \sin (-x)=-\sin x f\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin x f\left(\frac{1}{2}+x\right)=-g(x)
$$
Moreover, $\int_{1 / 2}^1 f(1-t) e^{\sin (\pi t)} d t=\int_0^{1 / 2} f(u) \cdot e^{\sin \pi u} d u$ where, $1-t=u$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
Statement I The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$.
Statement II $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
$$
\begin{aligned}
& F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2} d x \\
& \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c
\end{aligned}
$$
Since, $F(x+\pi) \neq F(x)$.
Hence, Statement I is false.
But Statement II is true as $\sin ^2 x$ is periodic with period $\pi$.

Topic: Mathematics, Indefinite Integration, JEE Advanced, JEE Advanced 2007 (Paper 1)
Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$, and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then, $f(x)$ is

Options:
A
$\frac{1}{3 x}+\frac{2 x^2}{3}$

B
$-\frac{1}{3 x}+\frac{4 x^2}{3}$

C
$-\frac{1}{x}+\frac{2}{x^2}$

D
$\frac{1}{x}$

The correct answer is:
$\frac{1}{3 x}+\frac{2 x^2}{3}$

Solution:
$$
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
$$
$$
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2007 (Paper 1)
Let $f(x)=\frac{x^2-6 x+5}{x^2-5 x+6}$.

Options:
A
A-p; B-q; C-q; D-p

B
A-r; B-s; C-q; D-r

C
A-s; B-p; C-q; D-p

D
A-r; B-q; C-p; D-r

The correct answer is:
A-p; B-q; C-q; D-p

Solution:
$$
f(x)=\frac{(x-1)(x-5)}{(x-2)(x-3)}
$$
The graph of $f(x)$ is shown
(A) If $-1 < x < 1 \Rightarrow 0 < f(x) < 1$
(B) If $1 < x < 2 \Rightarrow f(x) < 0$
(C) If $3 < x < 5 \Rightarrow f(x) < 0$
(D) If $x>5 \Rightarrow 0 < f(x) < 1$

Topic: Mathematics, Application of Derivatives, JEE Main
Let $f(x)=\frac{x^2-6 x+5}{x^2-5 x+6}$.

Options:
A
A-p; B-q; C-q; D-p

B
A-r; B-s; C-q; D-r

C
A-s; B-p; C-q; D-p

D
A-r; B-q; C-p; D-r

The correct answer is:
A-p; B-q; C-q; D-p

Solution:
$$
f(x)=\frac{(x-1)(x-5)}{(x-2)(x-3)}
$$
The graph of $f(x)$ is shown
(A) If $-1 < x < 1 \Rightarrow 0 < f(x) < 1$
(B) If $1 < x < 2 \Rightarrow f(x) < 0$
(C) If $3 < x < 5 \Rightarrow f(x) < 0$
(D) If $x>5 \Rightarrow 0 < f(x) < 1$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2007 (Paper 2)