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This is a short answer type question as classified in NCERT Exemplar

The vapour pressure of a liquid in comparison to air pressure determines its boiling point. At a constant atmospheric pressure, the lower the vapour pressure, the higher the boiling point of a liquid, and vice versa.

Because NaCl is a nonvolatile solute, it reduces the vapour pressure of water when added to it. The boiling point of water rises as a result. Methyl alcohol, on the other hand, is more volatile than water, therefore adding it to the solution raises the overall vapour pressure, lowering the boiling point of water.

Let ‘b’ and ‘x’ be the fixed base and equal side of isosceales triangle.

Then,  $\frac{dx}{dt}=-3$ cm/s (Ø decreasing).

Admission to the BA (Hons) Psychology and Philosophy & Yoga programmes at MIT School of Vedic Sciences, MIT-ADT University is purely merit-based. Students are shortlisted based on their Class 12 marks, with a minimum eligibility requirement of 50% or equivalent. No entrance exam is conducted for these courses, and selection is made through a merit list followed by document verification and partial fee payment.

Candidates willing to get admission must take part in the BNMIT admission process before the last date of application. Candidates must know their BNMIT eligibility criteria and apply accordingly. Follow the BNM Institute of Technology application process mentioned below:

Step 1: Visit the official website of BNM Institute of Technology.

Step 2: Select the required course.

Step 3: Pay the application fee.

Step 4: Fill out and submit the application form with the required information.

Because both components exist in the distillate and the liquid and vapour compositions are the same, this indicates that the liquids have formed an azeotropic combination that cannot be separated at this stage by fractional distillation.

We have, f(x) $\frac{logx}{x},x>0.$

f(x) = $\begin{array}{l}\underset{\_}{x\frac{d}{dx}logx-logx\frac{d}{dx}x}\\ x^2.\end{array}$

$=\frac{x\times \frac{1}{x}-logx}{x^2}=\frac{1-logx}{x^2}$

f(x) = $\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{d}{dx}{x}^2}{x^4}$

= $\frac{x^2\left(-\frac{1}{x}\right)-(1-logx)(2x)}{x^4}$

= $\frac{-1-2+2logx}{x^3}$ = $\frac{2logx-3}{x^3}$

At extreme points, f(x) = 0.

$\Rightarrow \frac{1-logx}{x^2}=0.$

$\Rightarrow logx=1=loge.$

$\Rightarrow x=e.$

At x = e, f"(e) = $\frac{2loge-3}{e^3}=\frac{2-3}{e^3}=-\frac{1}{e^3}<0$

x = e is a point of maximum.

There are several factors that Taxila Business School considers while releasing the cutoff for the PGDM course. Some of the factors that the school considers before releasing the cutoff are the number of applicants, scores candidates have achieved in the entrance examination, number of available seats, previous years' cutoff trends, etc. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) i.e., rich in dissolved oxygen is the answer since the total amount of oxygen consumed by microorganisms in decomposing organic matter present in a certain volume of water is known as Biochemical Oxygen Demand (BOD) of water. If the BOD value is less than 5 ppm, then the sample of water is considered to be pure and is rich in dissolved oxygen. 

Students who wish to take admission into the Shri Guru Ram Rai University courses must fill out an application form. After filling out the form, they are required to pay an application fee of INR 1,000. This fee is non-refundable under any circumstances. 

Note: The above-mentioned fee is as per official sources. However, it is subject to change.

(a) Consider $y=x^{\frac{1}{4}}.Let,x=\frac{16}{81}\&?x=\frac{1}{81}.$

$\begin{array}{l}Then,?y={\left(x+?x\right)}^{\frac{1}{4}}-x^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-{\left(\frac{16}{81}\right)}^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-\frac{2}{3}\\ \therefore {\left(\frac{17}{81}\right)}^{\frac{1}{4}}=\frac{2}{3}+?y\end{array}$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)?x=\frac{1}{4{\left(x\right)}^{\frac{3}{4}}}\left(?x\right)\left(as,y=x^{\frac{1}{4}}\right)\\ =\frac{1}{4{\left(\frac{16}{81}\right)}^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4\times 8}\times \frac{1}{81}=\frac{1}{32\times 3}=\frac{1}{96}=0.010\end{array}$

Hence, the approximate value of ${\left(\frac{17}{81}\right)}^{\frac{1}{4}}$ is $\frac{2}{3}+0.010=0.667+0.010$

=0.677

(b) ${(33)}^{\frac{-1}{5}}$

(b) Consider $y=x^{\frac{1}{5}}.Let,x=32\&?x=1$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)\left(?x\right)=\frac{-1}{5{\left(x\right)}^{\frac{6}{5}}}\left(?x\right)\left(as,y=x^{\frac{1}{5}}\right)\\ =-\frac{-1}{5{\left(x\right)}^6}\left(1\right)=-\frac{1}{320}=-0.003\end{array}$

Hence, the approximate value of ${\mathrm{33}}^{\frac{1}{5}}$
is $\frac{1}{2}+\left(-0.003=0.497\right)$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)

i.e., Its main components are produced by the action of sunlight on emissions of automobiles and factories is the answer since classical smog is a mixture of smoke, fog and sulphur dioxide and occurs in cold humid climates. While the gases released by automobiles and factories are not responsible for classical fog. 

We have,

$f^{\prime }(x)=[\mathrm{X(x\; -\; 1)}+$
${\mathrm{1]}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},0\le x\le 1$

$f(x)=\frac{1}{3}{\left[x^2-x+1\right]}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{d}{dx}\left(x^2-x+1\right)$

$=\frac{2x-1}{3{\left(x^2-x+1\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

At f(x) = 0.

2x – 1 = 0

$x=\frac{1}{2}\in [0,1]$

$f\left(\frac{1}{2}\right)={\left[\frac{1}{2}\left(\frac{1}{2}-1\right)+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\left[\frac{1}{4}-\frac{1}{2}+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

$={\left[\frac{1-2+4}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

= ${\left[\frac{3}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}?0.90$

$f(0)=[\mathrm{0(0\; -\; 1)}$
$\mathrm{+}{1]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

$f(1)=[1(1\; -\; 1)+\mathrm{1]}$
${}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

Option (B) is

Hence maximum value of f(x) = at x = 1 and x = 0.

Option (c) is correct.

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We have,

$f(x)=\frac{1-x+x^2}{1+x+x^2}.$

$\cancel{f}(x)=\frac{\left(1+x+x^2\right)\frac{d}{dx}\left(1-x+x^2\right)-\left(1-x+x^2\right)\frac{d}{dx}(1+x+2)}{{\left(1+x+x^2\right)}^2}$

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

$=\frac{-1+2x-x+2x^2-x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{{\left(1+x+x^2\right)}^2}$

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

At f(x) = 0.

$\Rightarrow \frac{-2(1+x)(1-x)}{{\left(1+x+x^2\right)}^2}=0$

x = 1 and x = -1.

At $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$ $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$

At x = -1, $f(-1)=\frac{1+1+1}{1-1+1}=3$

The maximum value of f(x) $=\frac{1}{3}x=1$

Hence, option (D) is correct.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

SO₂ is correct since photochemical fog is formed in the presence of sunlight in summers when NO and hydrocarbons are present in large amounts in the atmosphere. SO₂ is not responsible for photochemical fog. 

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The latest cutoff information for the MBA course at Taxila Business School has not been released yet. However, according to the cutoff of the previous three academic years, the course and school show consistency. For better understanding, check the below table: