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To get admission in Imarticus Learning you need to fill the eligibility criteria and then apply online for the desired programme. Those who get selected will get a callback from the institute. Selected students are then required to pay the admission fees to confirm a seat.

Yes, Imarticus Learning admissions are open and students can register online at the website of the institute. To apply, you need to provide the required information including course of choice, location, etc. In case of any queries, you can connect with admission desk of the institute.

u = (2z+i)/ (z-ki)
= (2x² + (2y+1) (y-k)/ (x²+ (y-k)²) + I (x (2y+1) - 2x (y-k)/ (x²+ (y-k)²)
Since Re (u) + Im (u) = 1
⇒ 2x² + (2y+1) (y-k) + x (2y+1) - 2x (y-k) = x² + (y-k)²
P (0, y? )
Q (0, y? )
⇒ y² + y - k - k² = 0 {y? + y? = -1, y? = -k-k²}
∴ PQ = 5
⇒ |y? - y? | = 5 ⇒ k² + k - 6 = 0
⇒ k = -3, 2
So, k = 2 (k > 0)

Let TV (r) denotes truth value of a statement r.
Now, if TV (p) = TV (q) = T
⇒ TV (S? ) = F
Also, if TV (p) = T and TV (q) = F
⇒ TV (S? ) = T

As per the NTA, 22,76,069 candidates registered for NEET 2025. Out of these, 22,09,318 appeared for the exam. The number of NEET applicants exam reduced in comparison to the previous year.

1 + (1 - 2²⋅1) + (1 - 4²⋅3) + . + (1 - 20²⋅19)
= α - 220β
= 11 - (2²⋅1 + 4²⋅3 + . + 20²⋅19)
= 11 - 2² ⋅ Σ? (r=1) r² (2r-1) = 11 - 4 (110²/2) - 35 x 11)
= 11 - 220 (103)
⇒ α = 11, β = 103

Compared to last year, the number of NEET 2025 applicants reduced. According to NTA, 22,09,318 candidates appeared for the exam. In the year 2024, 23,33,297 candidates had appeared. 

(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0

Adichunchanagiri University BTech/BE programme can be compared with other popular institutes on parameters such as infrastructure, placements, faculty student ratio, and industry exposure. The university offers modern infrastructure, experienced faculty, and exposure to industry practices, which helps students gain practical knowledge along with academics. Placements are improving year-on-year, with recruiters from multiple sectors. However, a detailed comparison may vary depending on the specific college being evaluated.

To take admission to Le Cordon Bleu NZ Bachelor's courses, students must submit an IELTS score of 6.0, with no band lower than 5.5 & at least 6.0 in writing Or equivalent. The other test scores accepted for this course are: 

• TOEFL iBT: 60

• PTE: 50 with no band score lower than 42

• OET: Minimum of Grade C or 200 in all sub-tests 

For the other courses at Le Corden Bleu NZ like Certificates, Diploma, Grand Diploma, an IELTS score of 5.0 Basic level is required. An IELTS score of 5.5 is required for Intermediate and Superior level Certificates. 

f (x) = a? ⋅ (b? × c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22

For admission to Le Cordon Bleu Bachelor of Culinary Arts & Business course, the applicants must be 17 years old. Also, they must submit an IELTS score of 6.0. An academic equivalent of to New Zealand University entrance is required. This course prepares graduates to enter and manage world-class culinary businesses in a globally competitive environment. 

 x? = 10
⇒ x? = (63 + a + b)/8 = 10
⇒ a + b = 17
Since, variance is independent of origin.
So, we subtract 10 from each observation.
So, σ² = 13.5 = (79 + (a-10)² + (b-10)²)/8
⇒ a² + b² - 20 (a+b) = -171
⇒ a² + b² = 169
From (1) and (2) ; a = 12 and b = 5

 x²/a² + y²/b² = 1 (a > b); 2b²/a = 10 ⇒ b² = 5a
Now, φ (t) = -5/12 + t - t² = 8/12 - (t - 1/2)²
φ (t)max = 8/12 = 2/3 = e ⇒ e² = 1 - b²/a² = 4/9
⇒ a² = 81 (From (i) and (ii)
So, a² + b² = 81 + 45 = 126

f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28