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No, Avodha does not give any job guarantee or employment assurance. Nonetheless, once a student completes a course, they are a part of Avodha's Placement Assistance process. Here, the Avodha team assists students in applying for placements and conducting practice interviews. Though there is complete assistance, employment is subject to the student's ability, work ethic, and performance.

The University of Calcutta popular programmes are BTech and MBA. Kolkata University tries to generate knowledge through research and interaction with industries and academia. The university tries to inspire the various students of the department to contribute to the sustainable development of society and nation through education, professionalism, research, and public and social services. The students are offered good placement assistance. Hence, students feel happy choosing this university. 

Yes, students can request for a course extension. Avodha mentions on its website that it understands that some learners may need more time to finish their course. Even if the original 3-month access period ends, students can ask for an extension. Extensions are only given to those who request it, and Avodha is happy to support learners who are committed to completing their course.

University of Calcutta BTech course admissions are based on the WBJEE, JEE Main, and various other entrance exam scores. Candidates willing to get admission must visit the official website and apply online. 

Calcutta University campus is spread across 2.7 acres. Some of the campus facilities include:

• A fully Wi-Fi campus
• Well-equipped Conference hall and Seminar hall.
• Library and full-time internet connectivity
• Hostels facility for boys & girls

40.

So, the only solution of the given equation is 0.

Hence, there is no non – zero integral solution of the given equation.

39. (x + iy)³ = u + iv

=>x³ + (iy)³ + 3.x.iy (x + iy) = u + iv   [since, (a + b)³ = a³ + b³ + 3ab (a + b)]

=>x³ – iy³ + 3x²yi + 3xy²i² = u + iv

=>x³ – iy³ + 3x²yi – 3xy² = u + iv                [since, i² = -1]

=> (x³ – 3xy²) + i (3x²y – y³) = u + iv

Equating real and imaginary part we get,

u = x³ – 3xy² and v = 3x²y – y³

Now,  $\frac{u}{x}$ + $\frac{v}{y}$

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

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

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4 (x² – y²)

Hence proved.

38. $\frac{1+i}{1-i}$ – $\frac{1-i}{1+i}$

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

= $\frac{1^2+i^2+2.1.i –\left(1^2+i^2- 2.1.i\right)}{1^2- i^2}$ [Since, (a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a² – b² = (a + b) (a – b)]

= $\frac{1-1+2i-1+1+2i}{1+1}$ [Since, i² = –1]

= $\frac{4i}{2}$

= 2i

37. Let z = (x – iy) (3 + 5i)

= 3x + 5xi – 3yi – 5yi²

= (3x + 5y) + (5x – 3y)i

Given,  $\overline{z}$ = –6 – 24i

=> (3x + 5y) – (5x – 3y)i = –6 – 24i

Equating real and imaginary part,

3x + 5y = –6 - (1)

5x – 3y = 24 - (2)

Multiplying (1) by 3 and (2) by 5 and adding them, we get

9x + 15y + 25x – 15y = –18 + 120

=> 34x = 102

=>x = 102/34 = 3

Putting x = 3 in (1) we get,

3 × 3 + 5y = –6

=> 9 + 5y = –6

=> 5y = –6 – 9

=> 5y = –15

=>y = –15/5 = –3

Hence, the values of x and y are 3 and –3 respectively.

36.  Z₁ = 2 – i, z₂ = –2 + i

Z₁z₂ = (2 – i)(–2 + i)

= –4 + 2i + 2i – i2

= –4 + 4i + 1[since, i² = –1]

= –3 + 4i

$\overline{z_1}$ = 2 + i

i. $\frac{z_1{z}_2}{\overline{z_1}}$ = $\frac{- 3+4i}{2+i}$

= $\frac{-3+4i}{2+i}$ × $\frac{2-i}{2-i}$ [multiply denominator and numerator by (2 – i)]

= $\frac{-6+3i+8i-4i^2}{2^2- i^2}$

= $\frac{-6+11i+4}{4-\left(-1\right)}$ [since, i2 = –1]

= $\frac{-6+4+11i}{4+1 }$

= $\frac{-2+11i}{5}$

= $\frac{-2}{5}$ + $\frac{11}{5}i$

So, Re( $\frac{z_1{z}_2}{\overline{z_1}}$ ) = $\frac{-2}{5}$

ii. $\frac{1}{z_1\overline{z_1}}$ = $\frac{1}{\left(2-i\right)\left(2+i\right)}$

= $\frac{1}{2^2- i^2}$

= $\frac{1}{4+1}$ [since, i2 = –1]

= $\frac{1}{5}$ + 0i

Therefore, Im $\left(\frac{1}{z_1\overline{z_1}}\right)$ = 0

Jhulelal Institute of Technology hasn't released an exact last date for 2025 admissions yet. It's best to check their website regularly for updates. Applying early improves your chances and reduces last-minute stress. Don't wait till the deadline 

You need to achieve at least 45% in Class 12 to be eligible for admission into Jhulelal Institute of Technology. An MBA or M.Tech candidate must have scored at least 50% in their graduation. Reaching the requirement is just the beginning; entrance exams are also important.

Yes, there are many government BA colleges in Mumbai (All). Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from official website and may vary.