Ask & Answer: India's Largest Education Community

Let A.B.C.D.be the square increased in a given fixed circle with radius x

Let ‘x’ and ‘y’ be the length and breadth of the rectangle

∴x, y> 0

In ABC, right angle at B,

x² + y² = (2x)²

$\Rightarrow$ x² + y² = 4x²

Students shortlisted through the merit list must visit MIT School of Vedic Sciences, MIT-ADT University in person for document verification. Upon successful verification, candidates are required to pay a portion of the tuition fee to secure their seat. Admission to the BA programme is officially confirmed only after both steps are successfully completed.

BSc specialisations at Shri Guru Ram Rai University are spread across different fields, such as Medical, Allied Health Sciences, Agriculture Science, and more. Listed below are the specialisations available at Shri Guru Ram Rai University for students to choose from:

• Nursing
• Agriculture
• Microbiology
• Yogic Science
• Biotechnology
• Optometry
• Food Nutrition & Dietetics
• PMD
• PCM

Many copanies were a part of the 2025 placement drive at Srusti Academy of Management and Technology. Refer to the table below:

Let ‘x’ cm be the length of side of the square to be cut off from the rectangular surface

Then, the volume v of the box is v = (45 - 2x) (24 - 2x) x

= 1080x - 138x² + 4x³

So,  $\frac{dv}{dx}=1080-276x+12x^2$

$\frac{d^2}{d{x}^2}=-276+24x$

At,  $\frac{dv}{dx}=0\to 1080-276x+12x^2=0$

$\Rightarrow$ x²- 23x + 90 = 0

$\Rightarrow$ x²- 5x - 18x + 90 = 0

$\Rightarrow$ x (x - 5) - 18 (x - 5) = 0

$\Rightarrow$ (x- 5) (x- 18) = 0

$\Rightarrow$ x = 5 and x = 18

At x = 18, breadth = 24 - 2 (18) = 24 - 36 = -12 which is not possible

At,  $x=5,\frac{d^{2}v}{d{x}^2}=-276+24(5)=-276+120=-150<0$

Hence, x = 5 is the point of maximum

So, ‘5’ cm length of square seeds to be cut from each corner of the secthgle

According to Shiksha Popularity Rankings, Catholic University of Leuven ranks at #1 in Belgium and #39 in the world. International students are drawn to this university for its welcoming high academic standing, vibrant international community, and advantageous.

MIT School of Vedic Sciences, MIT-ADT University selects candidates for its BA programme based primarily on Class 12 board exam performance. Once applications are reviewed, the university publishes up to three merit lists based on academic records. Shortlisted candidates from these lists are notified and invited to proceed with the next step in the admission process.

This is a  Assertion Type Questions as classified in NCERT Exemplar

Option (ii)

i.e., Both A and R are correct, but R is not the correct explanation of A is the answer since the increased amount of carbon dioxide in the air is the main cause for global warming. Excess of carbon dioxide gets released into the atmosphere and is removed by green plants which maintains an appropriate level of carbon dioxide into the atmosphere. 

Uttarakhand Open University BSc admission takes place through merit secured in Class 12 exams. Candidates who have a passing grade in their Class 12 are eligible for admission into the programme. To secure admission into the programme, candidates can apply through the institute website. Once candidates have applied for admission into the programme they will be intimated by the institute for admission into the BSc programme.

Side of the tin square piece = 18 cm

Let x cm be the thought of the square to be cut from each corner.

The volume v of the box after cutting

$\Rightarrow$ v = length × breadth × height

= (18 - 2x) (18 - 2x)  x x

= (18 - 2x)²x

= (324 + 4x²- 72x) x

= 4x³- 72x² + 324x

So,  $\frac{d}{dx}=12x^2-144x+324$

$\frac{d^{2}v}{d{x}^2}=24x-144$

As $\frac{d}{dx}=0\to 12x^2-144x+324=0$

$\Rightarrow$ x²- 12x + 27 = 0

$\Rightarrow$x²- 9x- 3x + 27 = 0

$\Rightarrow$ x (x - 9) - 3 (x - 9) = 0

$\Rightarrow$ (x - 9) (x - 3) = 0

$\Rightarrow$ x = 9 and x = 3

At x = 0, length of box = 18 - 2π9 = 18 - 18 = 0

Which is not possible

And at x = 3,  $\frac{d^{2}y}{d{x}^2}$ = 24 (3) - 144 = -72 < 0

∴x = 3 is a point of maximum

Hence, ‘3’ cm (square) is to be cut from each side of the square

So that volume of box is maximum

The total tuition fees for Shri Guru Ram Rai University BSc courses range from INR 38,000 to INR 5.1 lakh. This fee is provided in a range because the tuition fees differ from course to course. Moreover, in case a student fails to pay this fee at the time of admission, the university reserves the right to forfeit the admission.

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

Course wise placements for Srusti Academy of Management and Technology has not been released yet. However, the number of students placed in MBA were 141. Check out the table below to know more about the overall placement highlights at Srusti Academy of Management and Technology:

To apply for the BA at MIT School of Vedic Sciences, MIT-ADT University, students need to visit the official university website. The application process is entirely online and requires filling out personal and academic details accurately. After submitting the form, students must pay a registration fee to complete the process. Early registration is encouraged to avoid last-minute delays.

Let, x and y be the two positive number

Then, x + y = 16 y = 16 - x

Let p be the sun of the cubes then

p = x³ + y³ = x³ + (16 -x)³ = x³ + (16)³-x³- 48x (16 -x)

p = 16³ + 48x²- 76 8x

So,  $\frac{dp}{dx}=96x-768.$

$\Rightarrow \frac{d^{2}p}{d{x}^2}=96.$

At $\frac{dp}{dx}=0$

96x - 768 = 0

$\Rightarrow x=\frac{768}{96}=8.$

∴at x = 8,  $\frac{d^{2}p}{d{x}^2}=96\cdot >0$

So, x = 8 is a point of local minima

So, y = 16 - 8 = 8

Hence, x = 8, y = 8

In 2025, Srusti Academy of Management and Technology recorded a 80.9% placement rate.However, Srusti Academy of Management and Technology provides good placement to its students. The data tabulated below provides Srusti Academy of Management and Technology placement highlights for 2025:

We have, x + y = 35.

y = 35 - x

Let the product, P =x² y⁵

$\Rightarrow$ P = x² (35 -x)⁵

So,  $\frac{dp}{dx}$ = x² 5 (35 -x)⁴ (1) + (35 -x)⁵ 2x

= x (35 -x)⁴ [ - 5x + (35 -x) 2]

= x (35 -x)⁴ [ - 5x + 70 - 2x]

= x (35 -x)⁴ (70 - 7x)

= 7x (35 -x)⁴ (10 -x)

At $\frac{dp}{dx}=0$

$\Rightarrow$7x (35 -x)⁴ (10 -x) = 0

$\Rightarrow$ x = 0, 35, 10

As x is a (+) ve number we have only

x = 10, 35

And again (at x = 35) y = 35 = 0 but yis also a (+) ve number

we get, x = 10 (only)

whenx < 10,

$?\frac{dp}{dx}=(+ve)(+ve)(+ve)>0$

and when x > 10,

$\frac{dp}{dx}=(+ve)ve(+ve)(-ve)=(-ve)<0$

$\therefore \frac{dp}{dx}$ changes from (+ ve) to ( -ve) as x increases while passing through 10

Hence, x = 10 is a point of local maxima

So, y = 35 - 10 = 25

∴x = 10 and y = 25

DES Pune University campus is located in Pune, Maharashtra, and offers a safe environment to students. The college offers all the essential facilities for students, including academic blocks, a library, collaboration spaces, a café, a convenience store, etc. Take a look below at the facilities provided by the DES PU:

• Modern and smart classrooms
• Labs
• Free Wi-Fi
• Library
• Medical center
• A hostel for both boys and girls
• Seminar and lecture halls
• Sports ground

We have, x + y = 60.where x, y > 0

x = 60 - y.

Let the product P = xy³ = (60 -y) y³ = 60y³-y⁴

$\Rightarrow \frac{dp}{dy}=180y^2-4y^3$

= 4y² (45 -y)

At $\frac{dp}{dy}=0$

$\Rightarrow$ 4y² (45 -y) = 0

$\Rightarrow$ y = 0 and y = 45

As y > 0, y = 45

When, y > 45,  $\frac{dp}{dy}=(+ve)(-ve)$

= ve < 0

Ad y < 45,  $\frac{dp}{dy}=(+ve)(+ve)$

= (+ ve) > 0

∴p is maximum when y = 45 from + ve to- ve or y increases through 45.

So, x = 60 - y = 6Ø - 45 = 15.

Øx = 15 and y = 45.