datavorous/entrance-exam-dataset huggingface dataset:
$\frac{d}{d x}\left[\log \sqrt{\frac{1-\cos x}{1+\cos x}}\right]=$

Options:
A $\sec x$
B $\operatorname{cosec} x$
C $\operatorname{cosec} \frac{x}{2}$
D $\sec \frac{x}{2}$

The correct answer is:
$\operatorname{cosec} x$

Solution:
$\frac{d}{d x}\left[\log \sqrt{\frac{1-\cos x}{1+\cos x}}\right]=\frac{d}{d x}\left[\log \left(\tan \frac{x}{2}\right)\right]=\operatorname{cosec} x$ (using half angle formula)

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left\{\log \left(\frac{e^x}{1+e^x}\right)\right\}=$

Options:
A $\frac{1}{1-e^x}$
B $-\frac{1}{1+e^x}$
C $-\frac{1}{1-e^x}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} \frac{d}{d x} \log \left(\frac{e^x}{1+e^x}\right)= & \frac{1+e^x}{e^x} \times \frac{d}{d x}\left(\frac{e^x}{1+e^x}\right) \\ & =\frac{1+e^x}{e^x} \times \frac{e^x}{\left(1+e^x\right)^2}=\frac{1}{1+e^x}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\left.\frac{d}{d x}\left(\log _e x\right)\left(\log _3 x\right)\right]=$

Options:
A $\frac{\log _\theta x}{x}$
B $\frac{\log _x x}{x}$
C $\frac{2 \log x}{x}$
D $\frac{2 \log _3 x}{x}$

The correct answer is:
$\frac{2 \log _3 x}{x}$

Solution:
$\begin{aligned} & \frac{d}{d x}\left(\log _e x \cdot \log _a x\right)=\frac{d}{d x}\left[\log x \cdot \frac{\log x}{\log a}\right] \\ & =\frac{2 \log x}{x \cdot \log a}=\frac{2 \log _a x}{x} . \\ & \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \log (\log x)=$

Options:
A $\frac{x}{\log x}$
B $\frac{\log x}{x}$
C $(x \log x)^{-1}$
D None of these

The correct answer is:
$(x \log x)^{-1}$

Solution:
$\frac{d}{d x} \log (\log x)=\frac{1}{x} \cdot \frac{1}{\log x}=(x \log x)^{-1}$.

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=$

Options:
A $\operatorname{cosec} x$
B $-\operatorname{cosec} x$
C $\sec x$
D $-\sec x$

The correct answer is:
$\sec x$

Solution:
$\begin{aligned} & \frac{d}{d x} \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)} \sec ^2\left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \frac{1}{2} \\ & =\frac{1}{2} \cdot \frac{1}{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}=\frac{1}{\sin \left(\frac{\pi}{2}+x\right)}=\frac{1}{\cos x}=\sec x .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \log (\sqrt{x-a}+\sqrt{x-b})=$

Options:
A $\frac{1}{2[\sqrt{(x-a)}+\sqrt{(x-b)}]}$
B $\frac{1}{2 \sqrt{(x-a)(x-b)}}$
C $\frac{1}{\sqrt{(x-a)(x-b)}}$
D None of these

The correct answer is:
$\frac{1}{2 \sqrt{(x-a)(x-b)}}$

Solution:
$\begin{aligned} & \frac{d}{d x} \log (\sqrt{x-a}+\sqrt{x-b}) \\ & \quad=\left(\frac{1}{\sqrt{x-a}+\sqrt{x-b}}\right) \frac{1}{2}\left[\frac{1}{\sqrt{x-a}}+\frac{1}{\sqrt{x-b}}\right]\end{aligned}$
$=\left[\frac{\sqrt{x-a}+\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}\right] \frac{1}{2 \sqrt{(x-a)(x-b)}}=\frac{1}{2 \sqrt{(x-a)(x-b)}}$.

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right)=$

Options:
A $\frac{2 \cos x}{(1-\sin x)^2}$
B $\frac{\cos x}{(1-\sin x)^2}$
C $\frac{2 \cos x}{1-\sin x}$
D None of these

The correct answer is:
$\frac{2 \cos x}{(1-\sin x)^2}$

Solution:
$\frac{d}{d x}\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right)=\frac{d}{d x}\left(\frac{1+\sin x}{1-\sin x}\right)$\(=\frac{\cos x(1-\sin x)-(-\cos x)(1+\sin x)}{(1-\sin x)^2}=\frac{2 \cos x}{(1-\sin x)^2}\)

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \sqrt{\sec ^2 x+\operatorname{cosec}^2 x}=$

Options:
A $4 \operatorname{cosec} 2 x \cdot \cot 2 x$
B $-4 \operatorname{cosec} 2 x \cdot \cot 2 x$
C $-4 \operatorname{cosec} x \cdot \cot 2 x$
D None of these

The correct answer is:
$-4 \operatorname{cosec} 2 x \cdot \cot 2 x$

Solution:
$\begin{aligned} & \frac{d}{d x}\left[\sqrt{\sec ^2 x+\operatorname{cosec}^2 x}\right]=\frac{d}{d x}\left[\sqrt{\left(\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\right)}\right] \\ & =\frac{d}{d x}[2 \operatorname{cosec} 2 x]=-4 \operatorname{cosec} 2 x \cot 2 x .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^3\right)$ is equal to

Options:
A $\frac{3}{\sqrt{4-x^2}}$
B $\frac{3}{\sqrt{1-x^2}}$
C $\frac{1}{\sqrt{4-x^2}}$
D $-\frac{1}{\sqrt{4-x^2}}$

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

Solution:
$\begin{aligned} & \frac{d}{d x}\left\{\sin ^{-1}\left(3 x-4 x^3\right)\right\} \\ & \text { Put } x=\sin \theta \\ & \Rightarrow \quad \theta=\sin ^{-1} x \\ & =\frac{d}{d x}\left\{\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right)\right\} \\ & =\frac{d}{d x}\left\{\sin ^{-1} \sin 3 \theta\right\} \\ & =\frac{d}{d x}\{3 \theta\}=3 \frac{d}{d x} \sin ^{-1} x=\frac{3}{\sqrt{1-x^2}} \\ & \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^3\right)=$

Options:
A $\frac{3}{\sqrt{1-x^2}}$
B $\frac{-3}{\sqrt{1-x^2}}$
C $\frac{1}{\sqrt{1-x^2}}$
D $\frac{-1}{\sqrt{1-x^2}}$

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

Solution:
Put $x=\sin \theta$, we get $\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^3\right)$
$=\frac{d}{d x} \sin ^{-1}(\sin 3 \theta)=\frac{3}{\sqrt{1-x^2}}$.

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \left(\tan ^{-1} \left(\frac{4 \sqrt{x}}{1-4 x}\right)\right)=$

Options:
A $\frac{1}{\sqrt{x}(1+4 x)}$
B $\frac{2}{\sqrt{x}(1+4 x)}$
C $\frac{4}{\sqrt{x}(1+4 x)}$
D None of these

The correct answer is:
$\frac{2}{\sqrt{x}(1+4 x)}$

Solution:
$\begin{aligned} & \frac{d}{d x} \left(\tan ^{-1} \left(\frac{4 \sqrt{x}}{1-4 x} \right)\right) \\ & =\frac{1}{1+\left(\frac{4 \sqrt{x}}{1-4 x}\right)^2} \cdot\left[\frac{(1-4 x) 4\left(\frac{1}{2 \sqrt{x}}\right)-4 \sqrt{x}(-4)}{(1-4 x)^2}\right] \\ & =\frac{2(1+4 x)}{\sqrt{x}(1+4 x)^2}=\frac{2}{\sqrt{x}(1+4 x)} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)$ is equal to

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

The correct answer is:
$-\frac{1}{4}$

Solution:
$\begin{aligned} & \text { Let } y=\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}=\tan ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{4}}{2 \sin ^2 \frac{x}{4}}} \\ & y=\tan ^{-1} \cot \frac{x}{4}=\tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{x}{4}\right)=\frac{\pi}{2}-\frac{x}{4} \\ & \therefore \frac{d y}{d x}=-\frac{1}{4} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\right]=$

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

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

Solution:
$\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\right]=\frac{d}{d x}\left[\tan ^{-1} \tan \frac{x}{2}\right]=\frac{1}{2}$.

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(\tan ^{-1} \frac{\cos x}{1+\sin x}\right)=$

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

The correct answer is:
$-\frac{1}{2}$

Solution:
$\begin{aligned} & =\frac{d}{d x}\left[\tan ^{-1}\left(\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\left.\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}\right)\right] \\ & =\frac{d}{d x}\left[\tan ^{-1}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\right]=\frac{d}{d x}\left[\tan ^{-1} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]=-\frac{1}{2}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \tan ^{-1}\left(\frac{a x-b}{b x+a}\right)=$

Options:
A $\frac{1}{1+x^2}-\frac{a^2}{a^2+b^2}$
B $\frac{-1}{1+x^2}-\frac{a^2}{a^2+b^2}$
C $\frac{1}{1+x^2}+\frac{a^2}{a^2+b^2}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} \frac{d}{d x} \tan ^{-1}\left(\frac{a x-b}{b x+a}\right) & =\frac{1}{1+\left(\frac{a x-b}{b x+a}\right)^2} \cdot \frac{d}{d x}\left(\frac{a x-b}{b x+a}\right) \\ & =\frac{a^2+b^2}{a^2+b^2+a^2 x^2+b^2 x^2}=\frac{1}{1+x^2} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(x^2 e^x \sin x\right)=$

Options:
A $x e^x(2 \sin x+x \sin x+x \cos x)$
B $x e^x(2 \sin x+x \sin x-\cos x)$
C $x e^x(2 \sin x+x \sin x+\cos x)$
D None of these

The correct answer is:
$x e^x(2 \sin x+x \sin x+x \cos x)$

Solution:
$\begin{aligned} \frac{d}{d x}\left(x^2 e^x \sin x\right)= & x^2 \frac{d}{d x}\left(e^x \sin x\right)+e^x \sin x \frac{d}{d x}\left(x^2\right) \\ & =x e^x(2 \sin x+x \sin x+x \cos x) .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(x^2 \sin \frac{1}{x}\right)=$

Options:
A $\cos \left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right)$
B $2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$
C $\cos \left(\frac{1}{x}\right)-2 x \sin \left(\frac{1}{x}\right)$
D None of these

The correct answer is:
$2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$

Solution:
$\begin{aligned} & \frac{d}{d x}\left(x^2 \sin \frac{1}{x}\right)=x^2 \cos \left(\frac{1}{x}\right) \frac{d}{d x}\left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right) \\ & =-\frac{1}{x^2} \cdot x^2 \cos \left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right)=2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(\frac{1}{x^4 \sec x}\right)=$

Options:
A $\frac{x \sin x+4 \cos x}{x^5}$
B $\frac{-(x \sin x+4 \cos x)}{x^5}$
C $\frac{4 \cos x-x \sin x}{x^5}$
D None of these

The correct answer is:
$\frac{-(x \sin x+4 \cos x)}{x^5}$

Solution:
$\begin{aligned} & \frac{d}{d x}\left(\frac{1}{x^4 \sec x}\right)=\frac{d}{d x}\left(\frac{\cos x}{x^4}\right)=\frac{x^4(-\sin x)-\cos x\left(4 x^3\right)}{\left(x^4\right)^2} \\ & =\frac{-x^3(x \sin x+4 \cos x)}{x^8}=\frac{-(x \sin x+4 \cos x)}{x^5}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]=$

Options:
A $x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]+(\sin x)^x$
$$
$x^{\sin x}\left[\frac{\sin x}{x}+\sin x \log x\right]+(\sin x)^x[\log \cos x+x \cot x]$
$$
B $x^{\sin x}[x \tan x+\cos x \log x]+(\sin x)^x$
$$
\left[\frac{\sin x}{x}+\log (\sin x)\right]
$$
C $x^{\sin x}\left[\frac{x}{\sin x}+\cos x \log x\right]+(\sin x)^x$
$$
[x \cot x+\log (\sin x)]
$$
D $x^{\sin x}\left[\frac{\sin x}{x}+\sin x \log x\right]+(\sin x)^x$
$$
\text { [ } x \cot x+\log (\cos x)]
$$

The correct answer is:
$x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]+(\sin x)^x$
$$
$x^{\sin x}\left[\frac{\sin x}{x}+\sin x \log x\right]+(\sin x)^x[\log \cos x+x \cot x]$
$$

Solution:
Given, $\frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]$
Let $U=x^{\sin x}, V=(\sin x)^x$
$$
\begin{aligned}
\log U & =\sin x \log x \\
\frac{1}{U} \cdot \frac{d U}{d x} & =\frac{\sin x}{x}+\cos x \log x \\
\frac{d U}{d x} & =x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]
\end{aligned}
$$
and
$$
V=(\sin x)^x
$$
$$
\begin{aligned}
& \log V=x \log \sin x \\
& \frac{1}{V} \cdot \frac{d V}{d x}=\log \sin x+x \cdot \frac{1}{\sin x} \cos x \\
& \frac{d V}{d x}=(\sin x)^x[\log \sin x+x \cot x] \\
& \Rightarrow \quad \frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]=\frac{d U}{d x}+\frac{d V}{d x} \\
&=x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right] \\
&+(\sin x)^x[\log \sin x+x \cot x] .
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x} \sqrt{x \sin x}=$

Options:
A $\frac{\sin x+x \cos x}{2 \sqrt{x \sin x}}$
B $\frac{\sin x+x \cos x}{\sqrt{x \sin x}}$
C $\frac{x \sin x+\cos x}{\sqrt{2 \sin x}}$
D $\frac{x \sin x+\cos x}{\sqrt{2 x \sin x}}$

The correct answer is:
$\frac{\sin x+x \cos x}{2 \sqrt{x \sin x}}$

Solution:
\(\begin{aligned}
& \frac{d}{d x}(\sqrt{x \sin x}) \\
& \frac{1}{2 \sqrt{x \sin x}} \frac{d}{d x}(x-\sin x) \\
& \Rightarrow \frac{1}{2 \sqrt{x \sin x}}(\sin x+x \cdot \cos x) \\
& \Rightarrow \frac{\sin x+x-\cos x}{2 \sqrt{x \sin x}}
\end{aligned}\)

Topic: Mathematics, Differentiation, JEE Main
$\frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=$

Options:
A $1-\frac{1}{x^2}$
B $1+\frac{1}{x^2}$
C $1-\frac{1}{2 x}$
D None of these

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

Solution:
$\frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=\frac{d}{d x}\left[x+\frac{1}{x}+1\right]=1-\frac{1}{x^2}$.

Topic: Mathematics, Differentiation, JEE Main
$\left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x$, find general solution

Options:
A
$y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)$

B
$y=\sec ^2 x+\tan x+c$

C
$y=\log |\sec x+\tan x|+\operatorname{cosec} x+c$

D
$y=\tan ^2 x+\sin x+c$

The correct answer is:

$y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)$

Solution:
Given that
$$
\begin{aligned}
& \left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x \\
& \tan x\left(\frac{d y}{d x}\right)-y \sec ^2 x=\sin x \\
& \frac{d y}{d x}+\left(\frac{-\sec ^2 x}{\tan x}\right) y=\frac{\sin x}{\tan x} \\
& \text { Here, } P=-\frac{\sec ^2 x}{\tan x} \text { and } Q=\cos x \\
& \text { IF }=e^{\int P d x}=e^{\int \frac{-\sec ^ 2 x}{\tan x} d x} \\
& \text { IF }=e^{-\log (\tan x)}=e^{\log \cot x}=\cot x \\
& \text { Now, } y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \\
& y \cdot \cot x=\int \cos x \cdot \cot x+C \\
& y \cdot \cot x=\int \frac{1-\sin ^2 x}{\sin x}+C \\
& y \cdot \cot x=\log |\operatorname{cosec} x-\cot x|+\cos x+C
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
∫dxex+e-x2=

Options:
A 12e2x+1+c
B -12e2x+1+c
C 13e2x+1+c
D 1e2x+1+c

The correct answer is:
-12e2x+1+c

Solution:
Let I=∫dxex+e-x2=∫dxex+1ex2

=∫e2x·dxe2x+12

=∫122e2xdxe2x+12

Let e2x+1=t⇒e2x·2dx=dt

So, I=12∫dtt2

=12-1t+c

=12-1e2x+1+c

∴∫dxex+e-x2=-12e2x+1+c

Topic: Mathematics, Indefinite Integration, JEE Main
$\frac{d^2 x}{d y^2}$ equals

Options:
A
$\left(\frac{d^2 y}{d x^2}\right)^{-1}$

B
$-\left(\frac{d^2 y}{d x^2}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}$

C
$\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}$

D
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

The correct answer is:
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

Solution:
Since,
$$
\begin{aligned}
\Rightarrow \quad \frac{d}{d y}\left(\frac{d x}{d y}\right) & =\frac{d}{d x}\left(\frac{d y}{d x}\right)^{-1} \frac{d x}{d y} \\
\frac{d^2 x}{d y^2} & =-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}\left(\frac{d x}{d y}\right) \\
& =-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
$\frac{d^2 x}{d y^2}$ equals

Options:
A
$-\left(\frac{d^2 y}{d x^2}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}$

B
$\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}$

C
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

D
$\left(\frac{d^2 y}{d x^2}\right)^{-1}$

The correct answer is:
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

Solution:
$\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right)=-\frac{1}{\left(\frac{d y}{d x}\right)^2} \frac{d}{d y}\left(\frac{d y}{d x}\right)^2$
$=-\left(\frac{d y}{d x}\right)^{-2} \frac{1}{\left(\frac{d y}{d x}\right)} \frac{d}{d x}\left(\frac{d y}{d x}\right)=-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2011
Dacron is continuous filament yarn used in curtains, dress fabrics and pressure fire hoses. The reaction for preparing dacron is by the combination of which of the following?

Options:
A Hexamethylene diamene and adipic acid
B Caprolactum
C Phenol and formaldehyde
D Ethylene glycol and terephthalic acid

The correct answer is:
Ethylene glycol and terephthalic acid

Solution:
The continuous filament yarn is used in curtains, dress fabrics, high-pressure fire hoses, men's shirts, and thread. The staple fiber is ideal for mixing with wool in men's and women's suits, as well as in dress fabrics, knitted wear, and washable woven sportswear.

Topic: Chemistry, Polymers, JEE Main
Dalda is prepared from oils by

Options:
A Oxidation
B Reduction
C Hydrolysis
D Distillation

The correct answer is:
Reduction

Solution:
Dalda is an ester of saturated fatty acids, whereas oils are esters of unsaturated fatty acids.

Hence oils (liquid glycerides) react with hydrogen in the presence of a metal catalyst (like nickel) it gives saturated glycerides (semi-solid glycerides) i.e., fats. Thus, vegetable ghee (dalda) is obtained by the hydrogenation (reduction) of oils.
Oils + H2→ Dalda

Topic: Chemistry, Hydrogen, JEE Main
Daniel cell is a electrochemical cell, which of the following statement is true about it?

Options:
A Cations move toward zinc electrode
B Current flows from zinc electrode to copper electrode
C Cations move toward copper electrode which is cathode
D Electrons flow from copper electrode to zinc electrode

The correct answer is:
Cations move toward copper electrode which is cathode

Solution:
In daniel cell copper rod acts as cathode so there cations move towards copper electrode and reduction take place on copper rod.

Topic: Chemistry, Electrochemistry, JEE Main
Darwinism explains all the following except

Options:
A within each species, there are variations
B organisms tend to produce more number of offspring that can survive
C offspring with better traits that overcome competition are best suited for the environment
D variations are inherited from parents to offspring through genes

The correct answer is:
variations are inherited from parents to offspring through genes

Solution:
Variation is one of the main postulates of Darwinism. Darwin recognised two types of variations-continous and dis-continous variation, but he could not explain the inheritance of variations.

Topic: Biology, Evolution, JIPMER, JIPMER 2014
Darwin's finches are a good example of

Options:
A industrial melanism
B connecting link
C adaptive radiation
D convergent evolution

The correct answer is:
adaptive radiation

Solution:
Darwin's finches are good example of adaptive radiation. It is an evolutionary process starting from a point in a geographical area, giving rise to new species depending upon habitat. Main Darwin finch was in South America, some flew to Galapagos islands and some variations got selected and gave rise to new species.

Topic: Biology, Evolution, NEET, NEET 2010 (Screening)
Darwin's Finches are an excellent example of

Options:
A Connecting links
B Adaptive radiation
C Seasonal migration
D Brood parasitism

The correct answer is:
Adaptive radiation

Solution:
Darwins Finches are examples of evolutionary divergence giving rise to new species, arising from one commmon ancestor, depending upon the habitat.

Topic: Biology, Evolution, NEET, NEET 2008 (Mains)
Darwin's finches are an excellent example of

Options:
A adaptive radiation
B seasonal migration
C brood parasitism
D connecting links

The correct answer is:
adaptive radiation

Solution:
Adaptive radiation represents evolution of new forms in several directions from the common ancestral type. In 1831 Darwin got an opportunity to travel by $\mathrm{H}$ M S Beagle for a voyage of world exploration. Beagle sailed to the Galapagos Islands, here Darwin found a living laboratory of evolution. The common birds of Galapagos Islands, the finches were markedly different from the finches of main land. The closely related species of finches had beaks of different shapes and sizes and adapted for feeding on completely different diet showing adaptive radiation.
The transitional fossil forms, which show characteristic of two different groups of living animals are called connecting links, eg, Archaeapteryx, Seymouria etc.

Topic: Biology, Evolution, NEET, NEET 2008 (Screening)
Darwin's finches are an excellent example of

Options:
A adaptive radiation
B seasonal migration
C brood parasitism
D connecting links

The correct answer is:
adaptive radiation

Solution:
Adaptive radiation represents evolution of new forms in several directions from the common ancestral type. In 1831, Darwin got an opportunity to travel by HMS Beagle for a voyage of world exploration. Beagle sailed to the Galapagos Islands, here Darwin found a living laboratory of evolution. The common birds of Galapagos Islands, the finches were markedly different from the finches of main land. The closely related species of finches had beaks of different shapes and sizes and adapted for feeding on completely different diet showing adaptive radiation.
The transitional fossil forms, which show characteristic of two different groups of living animals are called connecting links, e.g., Archaeopteryx, Seymouria, etc.

Topic: Biology, Evolution, JIPMER, JIPMER 2012
Data given for the following reaction is as follows :

FeO(s)+C(graphite)⟶Fe(s)+CO(g)

Substance	ΔfH°kJmol-1	ΔS°Jmol-1 K-1

FeO(s)	-266.3	57.49

C(graphite)	0	5.74

Fe(s)	0	27.28

CO(g)	-110.5	197.6

The minimum temperature in K at which the reaction becomes spontaneous is_____.(Integer answer)

Options:

The correct answer is:
964

Solution:
Tmin=Δ°HΔ°S

ΔoHrxn=Δf°H(Fes)+Δf°H(COg)-Δf°H(FeOs)+Δf°H(Cgraphite)

=[0-110.5]-[-266.3+0]

=155.8 kJ/mol

Δ°Srxn=Δ°S(Fes)+Δ°S(COg)-Δ°S(FeOs)+Δ°SC(graphite )

=[27.28+197.6]-[57.49+5.74]

=161.65 J/mol.K

Tmin=155.8×103 J/mol161.65 J/mol-K=963.8 K

≈964 K (nearest integer)

Topic: Chemistry, Thermodynamics (C), JEE Main
Data given for the following reaction is as follows :

FeO(s)+C(graphite)⟶Fe(s)+CO(g)

Substance	ΔfH°kJmol-1	ΔS°Jmol-1 K-1

FeO(s)	-266.3	57.49

C(graphite)	0	5.74

Fe(s)	0	27.28

CO(g)	-110.5	197.6

The minimum temperature in K at which the reaction becomes spontaneous is_____.(Integer answer)

Options:

The correct answer is:
964

Solution:
Tmin=Δ°HΔ°S

ΔoHrxn=Δf°H(Fes)+Δf°H(COg)-Δf°H(FeOs)+Δf°H(Cgraphite)

=[0-110.5]-[-266.3+0]

=155.8 kJ/mol

Δ°Srxn=Δ°S(Fes)+Δ°S(COg)-Δ°S(FeOs)+Δ°SC(graphite )

=[27.28+197.6]-[57.49+5.74]

=161.65 J/mol.K

Tmin=155.8×103 J/mol161.65 J/mol-K=963.8 K

≈964 K (nearest integer)

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (27 Aug Shift 2)
David Tilman's experiment is related to?

Options:
A Role of species diversity in ecosystem stability.
B Role of dominant species in stability of ecosystem.
C Role of human being in maintaining ecosystem.
D More than one option is correct.

The correct answer is:
Role of species diversity in ecosystem stability.

Solution:
A community is said to be stable when its populations size of all species remains same over a period of time. In other words, the population is not showing much change or variation and is able to withstand any hardships or changes in the environment. A stable community is not affected by invasion of other organisms including alien species. It resists loss of biodiversity.
David Tilman is an American ecologist whose work with outdoor plots in study of ecosystem led to the observation that in an area with more variety of species showed less variation in total biomass. His experiments showed that increased diversity contributed to higher productivity. His study helped in better understanding the role of species diversity in ecosystem stability.

Topic: Biology, Biodiversity and its Conservation, NEET
Davison and Germer experiments show that the electron has a behavior

Options:
A like particle
B like wave
C similar to both particle and wave
D none of these

The correct answer is:
like wave

Solution:
- Davison Germer experiment was conducted in 1927 to study the wave nature of the electron.
- The electrons are emitted by an electron gun and are accelerated by applying a suitable potential difference between anode and cathode.
- The finely collimated beam is directed towards the Ni crystal.
- The scattered electrons are collected using a movable detector and a graph is plotted between applied potential and angle of reflection.
- Analysis of those graphs showed the wave nature of the electron.
So, the correct option is like wave.

Topic: Physics, Dual Nature of Matter, JEE Main