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The institute has not released the average package details as of yet. Check out the table for the highest packages offered between 2021-2023:

This is a multiple choice answer as classified in NCERT Exemplar

(i) Mole fraction

Explanation:  The mole fraction (x) is important for determining the relationship between a solution's concentration and its vapour pressure. Rault's law states that p₁ = x₁ po₁ in a binary solution of two volatile liquids, where p₁ is the vapour pressure of component 1, x1 is its mole fraction in solution, and po₁ is the vapour pressure of pure solvent.

The equation of the given curve is $9y^2=x^3$

Differentiate with respect to x, we have:

$\begin{array}{l}9\left(2y\right)\frac{dy}{dx}=3x^2\\ \Rightarrow \frac{dy}{dx}=\frac{x^2}{6y}\end{array}$

The slope of the normal to the given curve at point $\left(x_1,y_1\right)$ is

$\frac{-1}{{\frac{dy}{dx}]}_{\left(x_1,y_1\right)}}=-\frac{6y_1}{{x_1}^2}$

$\therefore$ The equation of the normal to the curve at $\left(x_1,y_1\right)$ is

$\begin{array}{l}y-y_1=-\frac{6y_1}{{x_1}^2}(x-x_1)\\ \Rightarrow {x_1}^{2}y-{x_1}^2{y}_1=-6x{y}_1+6x_1{y}_1\\ \Rightarrow 6x{y}_1+{x_1}^{2}y=6x_1{y}_1+{x_1}^2{y}_1\\ \Rightarrow \frac{6x{y}_1}{6x_1{y}_1+{x_1}^2{y}_1}+\frac{{x_1}^{2}y}{6x_1{y}_1+{x_1}^2{y}_1}=1\\ \Rightarrow \frac{x}{\frac{x_1(6+x_1)}{6}}+\frac{y}{\frac{y_1(6+x_1)}{x_1}}\end{array}$

It is given that the normal makes intercepts with the axes.

Therefore, we have:

$\begin{array}{l}\therefore \frac{x_1(6+x_1)}{6}=\frac{y_1(6+x_1)}{x_1}\\ \Rightarrow \frac{x_1}{6}=\frac{y_1}{x_1}\\ \Rightarrow {x_1}^2=6y_1..........(i)\end{array}$

Also, the point $\left(x_1,y_1\right)$ lies on the curve, so we have

$9{y_1}^2={x_1}^3..........(ii)$

From (i) and (ii), we have:

$9\left(\frac{{x_1}^2}{6}\right)={x_1}^3\Rightarrow \frac{{x_1}^4}{4}={x_1}^3\Rightarrow x_1=4$

From (ii), we have:

$\begin{array}{l}9{y_1}^2={(4)}^3=64\\ \Rightarrow {y_1}^2=\frac{64}{9}\\ \Rightarrow y_1=\pm \frac{8}{3}\end{array}$

Hence, the required points are $\left(4,\pm \frac{8}{3}\right)$ .

Therefore, option (A) is correct.

The median salary trends witnessed during IndSearch Institute of Management Studies and Research placements over the past three years are presented below:

IndSearch Institute of Management Studies and Research average package data is not released. However, according to the NIRF report 2025 released, the median package received during MBA placement 2024 was INR 7.5 LPA.

The equation of the given curve is $x^2=4y$

Differentiating with respect to x, we have:

$\begin{array}{l}2x=4.\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}=\frac{x}{2}\end{array}$

The slope of the normal to the given curve at point (h,k) is given by,

$\frac{-1}{{\frac{dy}{dx}]}_{(h,k)}}=-\frac{2}{h}$

$\therefore$ Equation of the normal at point (h,k) is given as:

$y-k=\frac{-2}{h}(x-h)$

Now, it is given that the normal passes through the point (1,2).

Therefore, we have:

$2-k=\frac{-2}{h}(1-h)or,k=2+\frac{2}{h}(1-h)..........(i)$

Since (h,k) lies on the curve $x^2=4y$ ,we have $h^2=4k$

$\Rightarrow k=\frac{h^2}{4}$

From equation (i), we have:

$\begin{array}{l}\frac{h^2}{4}=2+\frac{2}{h}(1-h)\\ \Rightarrow \frac{h^3}{4}=2h+2-2h=2\end{array}$

$\begin{array}{l}\Rightarrow h^3=8\\ \Rightarrow h=2\\ \therefore k=\frac{h^2}{4}\Rightarrow k=1\end{array}$

Hence, the equation of the normal is given as:

$\begin{array}{l}\Rightarrow y-1=\frac{-2}{2}(x-2)\\ \Rightarrow y-1=-(x-2)\\ \Rightarrow x+y=3\end{array}$

Therefore, option (A) is correct.

The equation of the given curve is $2y+x^2=3$ .

Differentiate with respect to x, we have:

$\begin{array}{l}\frac{2dy}{dx}+2x=0\\ \Rightarrow \frac{dy}{dx}=-x\\ \therefore {\frac{dy}{dx}]}_{(1,1)}=-1\end{array}$

The slope of the normal to the given curve at point (1,1) is

$\frac{-1}{{\frac{dy}{dx}]}_{(1,1)}}=-1$

Hence, the equation of the normal to the given curve at (1,1) is given as:

$\begin{array}{l}\Rightarrow y-1=1(x-1)\\ \Rightarrow y-1=x-1\\ \Rightarrow x-y=0\end{array}$

Therefore, option (B) is correct.

The S.P. Group of Colleges have been receiving good placement offers in recent years. Check out the tabulated data given below to know more about S.P. Group of Colleges placement in 2021-2023:

Note- the placement report for 2024 is yet to be released. 

Yes, Taxila Business School does offer internships to MBA students. Additionally, Taxila Business School offers two internships during the PGDM programme. First is the winter internship that happens during the months of January and February. Moreover, the school offers a summer internship that happens from July to August. 

The list of top recruiters that visited the campus during IndSearch Institute of Management Studies and Research placements in recent past is presented below:

The equation of the tangent to the given curve is $y=mx+1.$

Now, substituting $y=mx+1.$ in $y^2=4x,$ we get:

$\begin{array}{l}\Rightarrow {\left(mx+1\right)}^2=4x\\ \Rightarrow m^2{x}^2+1+2mx-4x=0\\ \Rightarrow m^2{x}^2+x(2m-4)+1=0..........(i)\end{array}$

Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:

Discriminant = 0

$\begin{array}{l}{(2m-4)}^2-4{(m)}^2(1)=0\\ \Rightarrow 4m^2+16-16m-4m^2=0\\ \Rightarrow 16-16m=0\\ \Rightarrow m=1\end{array}$

Hence, the required value of m is 1.

Therefore, option (A) is correct.

The S.P. Group of Colleges has had a strong placement record since its inception. Check out the tabulated data given below to know more about S.P. Group of Colleges placement in 2021-2023:

Note- the placement report for 2024 is yet to be released.

A total of 451 students got placed during the S.P. Group of Colleges placements in 2023. Refer to the table for total students placed details during placement season 2021-2023:

The number of students placed over the past three years during IndSearch Institute of Management Studies and Research placements are presented below:

Let x be the depth of the wheat inside the cylindrical tank is with radius = 10 cm

Then, Volume V of the cylindrical tank is

V = π (10)²x = 100πpx m³

As $\frac{d}{dt}=314\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$

$\Rightarrow 100\pi \frac{dx}{dt}=314$

$\frac{dx}{dt}=\frac{314}{100\pi }=\frac{314}{100\pi 3.14}=\frac{314}{314}=1.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ i e, rate of increasing of depth

option (A) is correct

The highest package offered at S.P. Group of Colleges placement in 2023 stood at INR 4.2 LPA. Refer to the table to know the highest package provided in placement 2021-2023

The median package received during IndSearch Institute of Management Studies and Research placements over the past three years is presented below:

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