datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2$Question:
Which of the following is true?

Options:
A
$f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$

B
$f(x)$ is increasing on $(-1,1)$ and has a local maximum at $x=1$

C
$f(x)$ is increasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$

D
$f(x)$ is decreasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$

The correct answer is:
$f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$

Solution:
When $x \in(-1,1)$
$$
x^2 < 1 \Rightarrow x^2-1 < 0
$$
$\therefore \quad f^{\prime}(x) < 0 \Rightarrow f(x)$ is decreasing.
Also, at $x=1, f^{\prime \prime}(1)=\frac{4 a}{(a+2)^2>0}$
$[\because 0 < a < 2]$
$\therefore f(x)$ has a local minimum at $x=1$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2$Question:
Which of the following is true ?

Options:
A
$(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$

B
$(2-a)^2 f^{\prime \prime}(1)-(2+a)^2 f^{\prime \prime}(-1)=0$

C
$f^{\prime}(1) f^{\prime}(-1)=(2-a)^2$

D
$f^{\prime}(1) f^{\prime}(-1)=-(2+a)^2$

The correct answer is:
$(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$

Solution:
\because f(x)= & \frac{\left(x^2+a x+1\right)-2 a x}{x^2+a x+1}=1-\frac{2 a x}{x^2+a x+1} \\
\therefore \quad f^{\prime}(x) & =-\left[\frac{\left(x^2+a x+1\right) \cdot 2 a-2 a x(2 x+a)}{\left(x^2+a x+1\right)^2}\right] \\
& =-\left[\frac{-2 a x^2+2 a}{\left(x^2+a x+1\right)^2}\right]=2 a\left[\frac{\left(x^2-1\right)}{\left(x^2+a x+1\right)^2}\right]
\end{aligned}
$$
$$
\begin{aligned}
& \text { and } \begin{aligned}
f^{\prime \prime}(x) & =2 a\left[\frac{\left(x^2+a x+1\right)^2(2 x)-2\left(x^2-1\right)\left(x^2+a x+1\right)(2 x+a)}{\left(x^2+a x+1\right)^4}\right] \\
& =2 a\left[\frac{2 x\left(x^2+a x+1\right)-2\left(x^2-1\right)(2 x+a)}{\left(x^2+a x+1\right)^3}\right]
\end{aligned} \\
& \text { Now, } f^{\prime \prime}(1)=\frac{4 a(a+2)}{(a+2)^3}=\frac{4 a}{(a+2)^2}
\end{aligned}
$$
and $f(-1)=\frac{4 a(a-2)}{(2-a)^3}=-\frac{4 a}{(a-2)^2}$
$$
\therefore(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=4 a-4 a=0
$$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question:
$\int_{-1}^1 g^{\prime}(x) d x$ is equal to

Options:
A
$2 g(-1)$

B
0

C
$-2 g(1)$

D
$2 g(1)$

The correct answer is:
$2 g(1)$

Solution:
Let $I=\int_{-1}^1 g^{\prime}(x) d x=[g(x)]_{-1}^1=g(1)-g(-1)$
Since, $\quad y^3-3 y+x=0$ and $\quad y=g(x)$
$$
\therefore(g(x))^3-3 g(x)+x=0
$$
[by Eq. (i)]
At $\quad x=1$,
$$
\begin{aligned}
(g(1))^3-3 g(1)+1 & =0 \\
\text { At } \quad x & =-1, \\
(g(-1))^3-3 g(-1)-1 & =0
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{array}{rlrl}
& & (g(1))^3+(g(-1))^3-3(g(1)+g(-1))=0 \\
\Rightarrow & & {[g(1)+g(-1)]\left[(g(1))^2+(g(-1))^2-g(1) g(-1)-3\right]=0} \\
\Rightarrow & g(1)+g(-1)=0 \\
\Rightarrow & g(1) & =-g(-1) \\
& \therefore & I & =g(1)-g(-1)=g(1)-(-g(1))=2 g(1)
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question:
If $f(-10 \sqrt{2})=2 \sqrt{2}$, then $f^{\prime \prime}(-10 \sqrt{2})$ is equal to

Options:
A
$\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

B
$-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

C
$\frac{4 \sqrt{2}}{7^3 \cdot 3}$

D
$-\frac{4 \sqrt{2}}{7^3 \cdot 3}$

The correct answer is:
$-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

Solution:
Given, $y^3-3 y+x=0 \Rightarrow 3 y^2 \frac{d y}{d x}-3 \frac{d y}{d x}+1=0$
$$
\Rightarrow \quad 3 y^2\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2-3 \frac{d^2 y}{d x^2}=0
$$
On substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (i), we get
$$
3(2 \sqrt{2})^2 \cdot \frac{d y}{d x}-3 \cdot \frac{d y}{d x}+1=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{21}
$$
Again on substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (ii), we get
$$
\begin{aligned}
& 3(2 \sqrt{2})^2 \frac{d^2 y}{d x^2}+6(2 \sqrt{2}) \cdot\left(\frac{-1}{21}\right)^2-3 \cdot \frac{d^2 y}{d x^2}=0 \\
& \Rightarrow \quad 21 \cdot \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^2} \Rightarrow \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^3}=-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question:
The area of the region bounded by the curve $y=f(x)$, the $X$-axis and the lines $x=a$ and $x=b$, where $-\infty < a < b < -2$, is

Options:
A
$\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)$

B
$-\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x+b f(b)-a f(a)$

C
$\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x-b f(b)+a f(a)$

D
$-\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x-b f(b)-a f(a)$

The correct answer is:
$\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)$

Solution:
Required area $=\int_a^b y d x=\int_a^b f(x) d x$
$$
=[f(x) \cdot x]_a^b-\int_a^b f^{\prime}(x) \cdot x d x=b f(b)-a f(a)-\int_a^b f^{\prime}(x) \cdot x d x
$$
$$
=b f(b)-a f(a)+\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x
$$
As, $\quad f^{\prime}(x)=\frac{d y}{d x}=-\frac{1}{3\left(y^2-1\right)}=-\frac{1}{3\left[(f(x))^2-1\right]}$

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$Question:
The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L_1$ and $L_2$, is

Options:
A
$2 / \sqrt{75}$

B
$7 / \sqrt{75}$

C
$13 / \sqrt{75}$

D
$23 / \sqrt{75}$

The correct answer is:
$13 / \sqrt{75}$

Solution:
The equation of the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the given lines $L_1$ and $L_2$ may be written as
$$
\begin{array}{rlrl}
\Rightarrow \quad(x+1)+7(y+2)-5(z+1) & =0 \\
& x+7 y-5 z+10 & =0
\end{array}
$$
The distance of the point $(1,1,1)$ from the plane
$$
=\left|\frac{1+7-5+10}{\sqrt{1+49+25}}\right|=\frac{13}{\sqrt{75}}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$Question:
The shortest distance between $L_1$ and $L_2$ is

Options:
A
0

B
$17 / \sqrt{3}$

C
$41 / 5 \sqrt{3}$

D
$17 / 5 \sqrt{3}$

The correct answer is:
$17 / 5 \sqrt{3}$

Solution:
The shortest distance between $L_1$ and $L_2$ is
$$
\begin{array}{r}
\left|\frac{\{(2-(-1)) \hat{\mathbf{i}}+(2-2) \hat{\mathbf{j}}+(3-(-1)) \hat{\mathbf{k}}\} \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right| \\
=\left|\frac{(3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{5 \sqrt{3}}\right|=\frac{17}{5 \sqrt{3}}
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
Consider the lines: $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$Question:
The unit vector perpendicular to both $L_1$ and $L_2$ is

Options:
A
$\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}}{\sqrt{99}}$

B
$\frac{-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$

C
$\frac{-\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$

D
$\frac{7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{99}}$

The correct answer is:
$\frac{-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{5 \sqrt{3}}$

Solution:
The equations of given lines in vector form may be written as
$$
L_1: \mathbf{r}(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})
$$
and $L_2: \mathbf{r}=(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$
$\therefore$ The vector perpendicular to both $L_1$ and $L_2$ is
$$
\left|\begin{array}{rrr}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{array}\right|=-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}
$$
$$
\begin{aligned}
\therefore \text { Required unit vector } & =\frac{(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{\sqrt{(-1)^2+(-7)^2+(5)^2}} \\
& =\frac{1}{5 \sqrt{3}}(-\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
Consider the polynomial $f(x)=1+2 x+3 x^2+4 x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$.Question:
The area bounded by the curve $y=f(x)$ and the lines $x=0, y=0$ and $x=t$, lies in the interval

Options:
A
$\left(\frac{3}{4}, 3\right)$

B
$\left(\frac{21}{64}, \frac{11}{16}\right)$

C
$(9,10)$

D
$\left(0, \frac{21}{64}\right)$

The correct answer is:
$\left(\frac{3}{4}, 3\right)$

Solution:
$\int_0^{1 / 2} f(x) d x < \int_0^t f(x) d x < \int_0^{3 / 4} f(x) d x$ Now, $\int f(x) d x$
$$
\begin{gathered}
=\int\left(1+2 x+3 x^2+4 x^3\right) d x \\
=x+x^2+x^3+x^4 \\
\Rightarrow \quad \int_0^{1 / 2} f(x) d x=\frac{15}{16}>\frac{3}{4} \\
\int_0^{3 / 4} f(x) d x=\frac{530}{256} < 3
\end{gathered}
$$

Topic: Mathematics, Area Under Curves, JEE Main
Paragraph:
Consider the polynomial $f(x)=1+2 x+3 x^2+4 x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$.Question:
The area bounded by the curve $y=f(x)$ and the lines $x=0, y=0$ and $x=t$, lies in the interval

Options:
A
$\left(\frac{3}{4}, 3\right)$

B
$\left(\frac{21}{64}, \frac{11}{16}\right)$

C
$(9,10)$

D
$\left(0, \frac{21}{64}\right)$

The correct answer is:
$\left(\frac{3}{4}, 3\right)$

Solution:
$\int_0^{1 / 2} f(x) d x < \int_0^t f(x) d x < \int_0^{3 / 4} f(x) d x$ Now, $\int f(x) d x$
$$
\begin{gathered}
=\int\left(1+2 x+3 x^2+4 x^3\right) d x \\
=x+x^2+x^3+x^4 \\
\Rightarrow \quad \int_0^{1 / 2} f(x) d x=\frac{15}{16}>\frac{3}{4} \\
\int_0^{3 / 4} f(x) d x=\frac{530}{256} < 3
\end{gathered}
$$

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Consider the polynomial $f(x)=1+2 x+3 x^2+4 x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$.Question:
The function $f^{\prime}(x)$ is

Options:
A
increasing in $\left(-t,-\frac{1}{4}\right)$ and decreasing in $\left(-\frac{1}{4}, t\right)$

B
decreasing in $\left(-t,-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, t\right)$

C
increasing in $(-t, t)$

D
decreasing in $(-t, t)$

The correct answer is:
decreasing in $\left(-t,-\frac{1}{4}\right)$ and increasing in $\left(-\frac{1}{4}, t\right)$

Solution:
Figure is self explanatory

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Consider the polynomial $f(x)=1+2 x+3 x^2+4 x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$.Question:
The real number $s$ lies in the interval

Options:
A
$\left(-\frac{1}{4}, 0\right)$

B
$\left(-11,-\frac{3}{4}\right)$

C
$\left(-\frac{3}{4},-\frac{1}{2}\right)$

D
$\left(0, \frac{1}{4}\right)$

The correct answer is:
$\left(-\frac{3}{4},-\frac{1}{2}\right)$

Solution:
Given, $f(x)=4 x^3+3 x^2+2 x+1$
$$
\begin{aligned}
& f^{\prime}(x)=2\left(6 x^2+3 x+1\right) \\
& D=9-24 < 0
\end{aligned}
$$
Hence, $f(x)=0$ has only one real root.
$$
\begin{aligned}
f\left(-\frac{1}{2}\right) & =1-1+\frac{3}{4}-\frac{4}{8}>0 \\
f\left(-\frac{3}{4}\right) & =1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64} \\
& =\frac{64-96+108-108}{64} < 0
\end{aligned}
$$
$f(x)$ changes its sign in $\left(-\frac{3}{4}, \frac{-1}{2}\right)$, hence $f(x)=0$ has a root in $\left(\frac{-3}{4}, \frac{-1}{2}\right)$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\right)$, atacamite $\left(\mathrm{Cu}_2 \mathrm{Cl}(\mathrm{OH})_3\right)$, cuprite $\left(\mathrm{Cu}_2 \mathrm{O}\right)$, copper glance $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$ and malachite $\left(\mathrm{Cu}_2(\mathrm{OH})_2 \mathrm{CO}_3\right)$. However, 80\% of the world copper production comes from the ore chalcopyrite $\left(\mathrm{CuFeS}_2\right)$. The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.Question:
Partial roasting of chalcopyrite produces

Options:
A
$\mathrm{Cu}_2 \mathrm{~S}$ and $\mathrm{FeO}$

B
$\mathrm{Cu}_2 \mathrm{O}$ and $\mathrm{FeO}$

C
$\mathrm{CuS}$ and $\mathrm{Fe}_2 \mathrm{O}_3$

D
$\mathrm{Cu}_2 \mathrm{O}$ and $\mathrm{Fe}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{Cu}_2 \mathrm{~S}$ and $\mathrm{FeO}$

Solution:
$2 \mathrm{CuFeS}_2+4 \mathrm{O}_2 \longrightarrow \mathrm{Cu}_2 \mathrm{~S}$
$$
+2 \mathrm{FeO}+3 \mathrm{SO}_2
$$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\right)$, atacamite $\left(\mathrm{Cu}_2 \mathrm{Cl}(\mathrm{OH})_3\right)$, cuprite $\left(\mathrm{Cu}_2 \mathrm{O}\right)$, copper glance $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$ and malachite $\left(\mathrm{Cu}_2(\mathrm{OH})_2 \mathrm{CO}_3\right)$. However, 80\% of the world copper production comes from the ore chalcopyrite $\left(\mathrm{CuFeS}_2\right)$. The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.Question:
In self-reduction, the reducing species is

Options:
A
$\mathrm{S}$

B
$\mathrm{O}^{2-}$

C
$\mathrm{S}^{2-}$

D
$\mathrm{SO}_2$

The correct answer is:
$\mathrm{S}^{2-}$

Solution:
$\mathrm{S}^{2-}$ acts as reducing species $2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S}^{2-} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2$
Metallurgy
Conceptual
III, II, II

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\right)$, atacamite $\left(\mathrm{Cu}_2 \mathrm{Cl}(\mathrm{OH})_3\right)$, cuprite $\left(\mathrm{Cu}_2 \mathrm{O}\right)$, copper glance $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$ and malachite $\left(\mathrm{Cu}_2(\mathrm{OH})_2 \mathrm{CO}_3\right)$. However, 80\% of the world copper production comes from the ore chalcopyrite $\left(\mathrm{CuFeS}_2\right)$. The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.Question:
Iron is removed from chalcopyrite as

Options:
A
$\mathrm{FeO}$

B
$\mathrm{FeS}$

C
$\mathrm{Fe}_2 \mathrm{O}_3$

D
$\mathrm{FeSiO}_3$

The correct answer is:
$\mathrm{FeSiO}_3$

Solution:
Iron is removed in the form of slag of $\mathrm{FeSiO}_3$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature $T_C(0)$. An interesting property of superconductors is that their critical temperature becomes smaller than $T_C(0)$ if they are placed in a magnetic field i.e., the critical temperature $T_C(B)$ is a function of the magnetic field strength $B$. The dependence of $T_C(B)$ on $B$ is shown in the figure.
Question:
A superconductor has $T_C(0)=100 \mathrm{~K}$. When a magnetic field of $7.5$ Tesla is applied, its $T_C$ decreases to $75 \mathrm{~K}$. For this material one can definitely say that when (Note : $\mathrm{T}$ = Tesla)

Options:
A
$B=5 \mathrm{~T}, T_C(B)=80 \mathrm{~K}$

B
$B=5 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$
C
$B=10 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$
D
$B=10 \mathrm{~T}, T_C(B)=70 \mathrm{~K}$

The correct answer is:
$B=5 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$

Solution:
With increase in temperature, $T_C$ is decreasing.
$$
\begin{aligned}
T_C(0) & =100 \mathrm{~K} \\
T_C & =75 \mathrm{~K} \text { at } B=7.5 \mathrm{~T}
\end{aligned}
$$
Hence, at $B=5 \mathrm{~T}, \quad T_C$ should lie between $75 \mathrm{~K}$ and $100 \mathrm{~K}$. Hence, the correct option should be (b).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature $T_C(0)$. An interesting property of superconductors is that their critical temperature becomes smaller than $T_C(0)$ if they are placed in a magnetic field i.e., the critical temperature $T_C(B)$ is a function of the magnetic field strength $B$. The dependence of $T_C(B)$ on $B$ is shown in the figure.
Question:
A superconductor has $T_C(0)=100 \mathrm{~K}$. When a magnetic field of $7.5$ Tesla is applied, its $T_C$ decreases to $75 \mathrm{~K}$. For this material one can definitely say that when (Note : $\mathrm{T}$ = Tesla)

Options:
A
$B=5 \mathrm{~T}, T_C(B)=80 \mathrm{~K}$

B
$B=5 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$
C
$B=10 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$
D
$B=10 \mathrm{~T}, T_C(B)=70 \mathrm{~K}$

The correct answer is:
$B=5 \mathrm{~T}, 75 \mathrm{~K} < T_C(B) < 100 \mathrm{~K}$

Solution:
With increase in temperature, $T_C$ is decreasing.
$$
\begin{aligned}
T_C(0) & =100 \mathrm{~K} \\
T_C & =75 \mathrm{~K} \text { at } B=7.5 \mathrm{~T}
\end{aligned}
$$
Hence, at $B=5 \mathrm{~T}, \quad T_C$ should lie between $75 \mathrm{~K}$ and $100 \mathrm{~K}$. Hence, the correct option should be (b).

Topic: Physics, Current Electricity, JEE Main
Paragraph:
Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature $T_C(0)$. An interesting property of superconductors is that their critical temperature becomes smaller than $T_C(0)$ if they are placed in a magnetic field i.e., the critical temperature $T_C(B)$ is a function of the magnetic field strength $B$. The dependence of $T_C(B)$ on $B$ is shown in the figure.
Question:
In the graphs below, the resistance $R$ of a superconductor is shown as a function of its temperature $T$ for two different magnetic fields $B_1$ (solid line) and $B_2$ (dashed line). If $B_2$ is larger than $B_1$, which of the following graphs shows the correct variation of $R$ with $T$ in these fields?

Options:
A

B

C

D

The correct answer is:

Solution:
If $B_2>B_1$, critical temperature, (at which resistance of semiconductors abruptly becomes zero) in case-2 will be less than compared to case- 1 .
$\therefore$ The correct option is (a).

Topic: Physics, Current Electricity, JEE Main
Paragraph:
Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature $T_C(0)$. An interesting property of superconductors is that their critical temperature becomes smaller than $T_C(0)$ if they are placed in a magnetic field i.e., the critical temperature $T_C(B)$ is a function of the magnetic field strength $B$. The dependence of $T_C(B)$ on $B$ is shown in the figure.
Question:
In the graphs below, the resistance $R$ of a superconductor is shown as a function of its temperature $T$ for two different magnetic fields $B_1$ (solid line) and $B_2$ (dashed line). If $B_2$ is larger than $B_1$, which of the following graphs shows the correct variation of $R$ with $T$ in these fields?

Options:
A

B

C

D

The correct answer is:

Solution:
If $B_2>B_1$, critical temperature, (at which resistance of semiconductors abruptly becomes zero) in case-2 will be less than compared to case- 1 .
$\therefore$ The correct option is (a).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
The positive value of $k$ for which $k e^x-x=0$ has only one root is

Options:
A
$\frac{1}{e}$

B
1

C
$e$

D
$\log _e 2$

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

Solution:
Let $f(x)=k e^x-x$
$\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text { Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1>0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}$
For one root of given equation
$$
\begin{aligned}
1+\ln k & =0 \\
Hence,
k & =\frac{1}{e} .
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
The line $y=x$ meets $y=k e^x$ for $k \leq 0$ at

Options:
A
no point

B
one point

C
two points

D
more than two points

The correct answer is:
one point

Solution:
$$
\text { Let } y=x \text { intersect the curve } y=k e^x \text { at exactly one point when } k \leq 0 \text {. }
$$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
For $k>0$, the set of all values of $k$ for which $k e^x-x=0$ has two distinct roots, is

Options:
A
$\left(0, \frac{1}{e}\right)$

B
$\left(\frac{1}{e}, 1\right)$

C
$\left(\frac{1}{e}, \infty\right)$

D
$(0,1)$

The correct answer is:
$\left(0, \frac{1}{e}\right)$

Solution:
For two distinct roots $1+\ln k < 0(k>0)$
$\ln k < -1 \quad k < \frac{1}{e}$
Hence, $k \in\left(0, \frac{1}{e}\right)$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2007 (Paper 2)