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10 months ago

Tritya Air Hostess Academy Diploma

What is the total tuition fee for Diploma at Tritya Air Hostess Academy?

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Indrani Choudhury

Contributor-Level 10

The fee range for the Diploma programme at Tritya Air Hostess Academy is INR 85,000. This information is sourced from official website/ sanctioning body and is subject to change. The fee cover various academic expenses like tuition fees, examination fees, library fees, and other relevant charges. It is important to note that the fee structure is subject to change, and candidates are advised to refer to the official website of the university for the latest and most accurate fee structure.

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10 months ago

Study Abroad Exam

Are IELTS exam dates the same in all cities?

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Raushan Kumar

Contributor-Level 10

Yes, mostly the exam dates are the same across major Indian cities like Delhi, Mumbai, Bengaluru, Chennai, and others. But sometimes smaller cities may have fewer test dates, so always check before booking.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} H e r e, A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ | A | = 0 | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | - 1 | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | \\ = 0 - 1 (0 - 1) + 1 (1 - 0) \\ = 1 + 1 = 2 \neq 0 (n o n - singular m a t r i x) \\ N o w, c o - f a c t o r s, \\ a_{11} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{12} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{13} = + | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1 \\ a_{21} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{22} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{23} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1 \\ a_{31} = + | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1, a_{32} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1, a_{33} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1 \\ A d j (A) = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'} = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} A d j (A) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ H e n c e, A^{2} = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ N o w, w e h a v e t o p r o v e t h a t A^{- 1} = \frac{A^{2} - 3 I}{2} \\ R . H . S . = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}{2} \\ = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]}{2} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ = A^{- 1} = L . H . S . \\ H e n c e, p r o v e d . \end{array}$

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10 months ago

Sri Venkateswara University AP EAMCET

I want to know the cut-off rank for SV University, Tirupati.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} | \begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A + c o s A \\ - c o s^{2} B - c o s B \end{array} & \begin{array}{l} c o s^{2} B + c o s B \\ - c o s^{2} C - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} c o s^{2} A - c o s^{2} B \\ + c o s A - c o s B \end{array} & \begin{array}{l} c o s^{2} B - c o s^{2} C \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} 0 & 0 & 1 \\ c o s A - c o s B & c o s B - c o s C & 1 + c o s C \\ \begin{array}{l} (c o s A + c o s B) \times \\ (c o s A - c o s B) \\ + (c o s A - c o s B) \end{array} & \begin{array}{l} (c o s B + c o s C) \times \\ (c o s B - c o s C) \\ + c o s B - c o s C \end{array} & c o s^{2} C + c o s C \end{matrix} | = 0 \\ T a k i n g (c o s A - c o s B) a n d (c o s B - c o s C) c o m m o n f r o m C_{1} a n d C_{2} r e s p e c t i v e l y . \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & 1 + c o s C \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 & c o s^{2} C + c o s C \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [1 | \begin{matrix} 1 & 1 \\ c o s A + c o s B + 1 & c o s B + c o s C + 1 \end{matrix} |] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [(c o s B + c o s C + 1) - (c o s A + c o s B + 1)] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) [c o s B + c o s C + 1 - c o s A - c o s B - 1] = 0 \\ \Rightarrow (c o s A - c o s B) (c o s B - c o s C) (c o s C - c o s A) = 0 \\ \Rightarrow c o s A - c o s B = 0 o r c o s B - c o s C = 0 o r c o s C - c o s A = 0 \\ \Rightarrow c o s A = c o s B o r c o s B = c o s C o r c o s C = c o s A \\ \Rightarrow ∠ A = ∠ C o r ∠ B = ∠ C \Rightarrow ∠ A = ∠ B \\ H e n c e, Δ A B C i s a n i s o s c e l e s t r i a n g l e . \end{array}$

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10 months ago

UoH - University of Hyderabad Telangana

What all facilities are offered to the student of University of Hyderabad?

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Rashmi Shukla

Contributor-Level 10

The University of Hyderabad campus is spread across 2,300 acres. The university has a large area covered under forest and three seasonal lakes. Below mentioned are some of the major highlights of facilities offered by Hyderabad University:

Library: The library has a collection of over 4 Lacs books which include textbooks, journals, monographs etc. It also consists of more than 25,000 books, e-journals, e-books, reference books, serials, theses & dissertations, and back volumes of journals. The university provides wi-fi and other advanced facilities to its students.
Hostel: There are a total of 21 hostels for students, out of wh

...more

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10 months ago

Jain Deemed-to-be University, Bangalore Computer Science and Engineering

Can I get B.Tech CSE in Jain University, Bangalore, through management quota as I got 39% in PCM subjects in my 12th?

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Rupesh Katariya

Contributor-Level 10

With 39% in PCM, it might be tough to get B.Tech CSE at Jain University even through management quota, as the minimum required is 45%. You can still check directly with the college—they sometimes allow exceptions or may suggest other options.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f t h e g i v e n points l i e o n a s t r a i g h t l i n e, t h e n t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e \\ points p a i r w i s e i s z e r o . \\ S 0, | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 7 & - 7 & 0 \\ - 2 & 3 & 0 \\ a & a & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ \Rightarrow 1 . | \begin{matrix} 7 & - 7 \\ - 2 & 3 \end{matrix} | = 21 - 14 = 7 u n i t s \\ A s 7 \neq 0 . H e n c e, t h e t h r e e points d o n o t l i e o n a s t r a i g h t l i n e f o r a n y v a l u e o f a . \end{array}$

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10 months ago

Tritya Air Hostess Academy Diploma

Is Diploma available at Tritya Air Hostess Academy?

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Indrani Choudhury

Contributor-Level 10

Yes, Diploma course is available at Tritya Air Hostess Academy. The institute offers Diploma courses at the UG level. Candidate must meet the eligibility criteria to enrol for Diploma course admission set by the college. Aspirants must pass Class 12. Tritya Air Hostess Academy's admission process is merit-based.

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10 months ago

Study Abroad Exam

Can I book the next IELTS exam right now?

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Raushan Kumar

Contributor-Level 10

Yes, since the exam is just a few days away, you should go online to the IDP IELTS India portal, check for seat availability at your preferred centre, and book your slot now to avoid missing out.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} I f a_{1}, a_{2}, a_{3}, \dots a_{r} b e t h e t e r m s o f G . P ., t h e n \\ a_{n} = A R^{n - 1} (w h e r e A i s t h e f i r s t t e r m a n d R i s t h e c o m m o n r a t i o o f t h e G . P .) \\ ∴ a_{r + 1} = A R^{r + 1 - 1} = A R^{r}; a_{r + 5} = A R^{r + 5 - 1} = A R^{r + 4} \\ a_{r + 9} = A R^{r + 9 - 1} = A R^{r + 8}; a_{r + 7} = A R^{r + 7 - 1} = A R^{r + 6} \\ a_{r + 11} = A R^{r + 11 - 1} = A R^{r + 10}; a_{r + 15} = A R^{r + 15 - 1} = A R^{r + 14} \\ a_{r + 17} = A R^{r + 17 - 1} = A R^{r + 16}; a_{r + 21} = A R^{r + 21 - 1} = A R^{r + 20} \\ ∴ T h e determinant b e c o m e s \\ | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} | \\ T a k i n g A R^{r}, A R^{r + 6} a n d A R^{r + 10} c o m m o n f r o m R_{1,} R_{2,} a n d R_{3} r e s p e c t i v e l y . \\ A R^{r} . A R^{r + 6} . A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} | \\ = A R^{r} . A R^{r + 6} . A R^{r + 10} | 0 | [? R_{1} a n d R_{2} a r e i d e n t i c a l r o w s] \\ = 0 \\ H e n c e, t h e g i v e n determinant i s i n d e p e n d e n t o f r . \end{array}$

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10 months ago

Business Management Studies Business Analytics

What is the eligibility to apply for MBA in Business Analytics in India?

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Mohit Mishra

Contributor-Level 7

The Eligibility criteria for Business Analytics MBA is similar to other MBA programs as it is one the specialisation programme provided under MBA Course:

A bachelor's degree in any stream with at least 50% marks.
A valid score in entrance exams that are conducted for MBA such as CAT, XAT, MAT, CMAT, GMAT, NMAT, etc.
Some colleges may ask for work experience or conduct interviews/GD rounds.

You don't necessarily need a background in data or IT but having maths or stats in your UG course helps in understanding analytical concepts better.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] = 0 \\ | A | = | \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow | \begin{matrix} 12 + x & 12 + x & 12 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ T a k i n g (12 + x) c o m m o n f r o m R_{1} \\ \Rightarrow (12 + x) | \begin{matrix} 1 & 1 & 1 \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow (12 + x) | \begin{matrix} 0 & 0 & 1 \\ 2 x & - 2 x & 4 + x \\ 0 & 2 x & 4 - x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (12 + x) [1 . | \begin{matrix} 2 x & - 2 x \\ 0 & 2 x \end{matrix} |] = 0 \\ \Rightarrow (12 + x) (4 x^{2} - 0) = 0 \Rightarrow 12 + x = 0 o r 4 x^{2} = 0 \\ \Rightarrow x = - 12 o r x = 0 \end{array}$

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10 months ago

R.V. College of Engineering Colleges in Bangalore

Is it a good choice to pursue BTech from RV College of Engineering?

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Nishtha Pruthi

Contributor-Level 10

Yes, studying BTech at RV College of Engineering is a good option, particularly if candidates willing to pursue a career in computer science, IT, or electronics because of its high academic rating, and great placement record. The college has a rich campus life, though charges can be steep under the management quota, and opportunities might differ from branch to branch. Overall, it is a good bet if candidates are seeking for quality education and industry interaction beyond the IIT/NIT system.

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10 months ago

Dr. Ambedkar College of Law India

Can I get admission to LLB at Dr. Ambedkar College of Law through Telangana Law CET?

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10 months ago

Exam IELTS

Where can I check exact city wise dates and book my IELTS exam?

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Raushan Kumar

Contributor-Level 10

You can check a full city wise calendar on the IDP IELTS India registration page. There you will find exact IELTS dates by city (like Delhi, Hyderabad, Kolkata, etc.), test type (Academic/GT), and centre.

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10 months ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0 \\ | A | = | \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0 \\ T a k i n g 7 c o m m o n f r o m C_{1} \\ \Rightarrow = 7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} 1 & s i n 3 θ \\ - 7 & - 2 \end{matrix} | + 2 | \begin{matrix} 1 & s i n 3 θ \\ 3 & c o s 2 θ \end{matrix} | = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 (c o s 2 θ - 3 s i n 3 θ) = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 c o s 2 θ - 6 s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 c o s 2 θ + s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 4 s i n^{3} θ - 4 s i n^{2} θ + 3 s i n θ = 0 \\ \Rightarrow - s i n θ (4 s i n^{2} θ + 4 s i n θ - 3) = 0 \\ - s i n θ = 0 o r 4 s i n^{2} θ + 4 s i n θ - 3 = 0 \\ ∴ θ = n π o r 4 s i n^{2} θ + 6 s i n θ - 2 s i n θ - 3 = 0 w h e n n \in I \\ \Rightarrow 2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3) = 0 \\ \Rightarrow (2 s i n θ + 3) (2 s i n θ - 1) = 0 \\ \Rightarrow &thinsp \end{array}$

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10 months ago

Medicine & Health Sciences Pharmacy

Which are the pharmacy colleges that offer the best placements?

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10 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Ncert Solutions Class 12th

In many experimental set-ups, the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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