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SOU Devibai Narayandas Chhabada Rural Education Society

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x + y i \\ | z_{1} | = \sqrt[]{x^{2} + y^{2}} = 1 [G i v e n t h a t | z_{1} | = 1] \\ \Rightarrow x^{2} + y^{2} = 1 \dots (i) \\ N o w z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + y i - 1}{x + y i + 1} \\ = \frac{(x - 1) + y i}{(x + 1) + y i} = \frac{(x - 1) + y i}{(x + 1) + y i} \times \frac{x + 1 - y i}{x + 1 - y i} \\ = \frac{(x - 1) (x + 1) - y (x - 1) i + y (x + 1) i - y^{2} i^{2}}{{(x + 1)}^{2} - y^{2} i^{2}} \\ = \frac{x^{2} - 1 + y i (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} \\ = \frac{(x^{2} + y^{2} - 1) + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ = \frac{(1 - 1)}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ = 0 + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ H e n c e, t h e r e a l p a r t o f z_{2} i s 0 . \end{array}$

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9 months ago

Is direct admission available at SOU Devibai Narayandas Chhabada Rural Education Society?

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saurya snehal

Contributor-Level 10

Yes, SOU Devibai Narayandas Chhabada Rural Education Society does offer direct admission to a few courses. Candidates applying for the Medical programmes are required to appear for the NEET exams. Admission to some of the programmes are done based on entrance. Candidates applying to courses with merit-based admissions can get direct admission to the programmes. Candidates are required to follow the prospectus of UG or PG admissions on the official website.

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9 months ago

What factors should I consider when comparing BA colleges like PK University?

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

When comparing BA colleges, students should evaluate aspects such as available specialisations, faculty experience, practical training opportunities, campus infrastructure, placement support, and industry collaborations. These factors significantly impact the overall learning experience and future career prospects.

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9 months ago

If $c o s θ = \frac{x}{x + 1}$ , then $θ$ is:

(a) An acute angle

(b) A right angle

(c) An obtuse angle

(d) Not possible

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = x + \frac{1}{x} \Rightarrow c o s θ = \frac{x^{2} + 1}{x} \\ \Rightarrow x^{2} + 1 = x c o s θ \Rightarrow x^{2} - x c o s θ + 1 = 0 \\ F o r r e a l v a l u e o f x, b^{2} - 4 a c \geq 0 \\ \Rightarrow {(- c o s θ)}^{2} - 4 \times 1 \times 1 \geq 0 \\ \Rightarrow c o s^{2} θ - 4 \geq 0 \Rightarrow c o s^{2} θ \geq 4 \\ \Rightarrow c o s θ \geq \pm 2 [- 1 \leq c o s θ \leq 1] \\ S o, t h e v a l u e o f θ i s n o t p o s s i b l e . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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9 months ago

$z_{1}$ and $z_{2}$ are two complex numbers such that $? z_{1} ? = ? z_{2} ? = 1$ and $arg (z_{1}) + arg (z_{2}) = π$ , then show that $z_{1} = - \frac{}{z_{2}}$ .

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Government & Defence Entrance Exam TNPSC Group 4 Admit Card

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) a r e p o l a r f o r m o f t w o c o m p l e x \\ n u m b e r s z_{1} a n d z_{2} . \\ G i v e n t h a t : | z_{1} | = | z_{2} | \Rightarrow r_{1} = r_{2} \\ a n d a r g (z_{1}) + a r g (z_{2}) = π \\ \Rightarrow θ_{1} + θ_{2} = π \\ \Rightarrow θ_{1} = π - θ_{2} \\ N o w z_{1} = r_{1} [c o s (π - θ_{2}) + i s i n (π - θ_{2})] \\ \Rightarrow z_{1} = r_{1} [- c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow z_{1} = - r_{1} [c o s θ_{2} - i s i n θ_{2}] \dots (i) \\ \Rightarrow z_{2} = r_{2} [c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow {\bar{z}}_{2} = r_{1} [c o s θ_{2} - i s i n θ_{2}] [? r_{1} = r_{2}] \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t, \\ z_{1} = - {\bar{z}}_{2} . H e n c e p r o v e d . \end{array}$

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9 months ago

When was TNPSC Group 4 Admit Card 2025 released?

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Raj Shukla

Contributor-Level 6

The Tamil Nadu Public Service Commission conducted the TNPSC Group 4 exam 2025 on July 13, 2025. The Commission released the TNPSC Group 4 admit card 2025 on July 3, 2025. The TNPSC Group 4 2025 exam was conducted in all the districts of Tamil Nadu State. The candidates were required to download their TNPSC Group 4 admit card from the portal - tnpsc.gov.in.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $t a n θ = \frac{a}{b}$ , then $b c o s 2 θ + a s i n 2 θ$ is equal to:

(a) a

(b) b

(c) $\frac{b}{a}$

(d) None

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = \frac{a}{b} \\ b c o s 2 θ + a s i n 2 θ = b [\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}] + a [\frac{2 t a n θ}{1 + t a n^{2} θ}] \\ = b [\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}] + a [\frac{\frac{2 a}{b}}{1 + \frac{a^{2}}{b^{2}}}] \\ = b [\frac{b^{2} - a^{2}}{b^{2} + a^{2}}] + [\frac{\frac{2 a^{2}}{b}}{\frac{b^{2} + a^{2}}{b^{2}}}] \\ = \frac{b^{3} - a^{2} b}{b^{2} + a^{2}} + \frac{2 a^{2} b}{b^{2} + a^{2}} = \frac{b^{3} - a^{2} b + 2 a^{2} b}{b^{2} + a^{2}} \\ = \frac{b^{3} + a^{2} b}{b^{2} + a^{2}} = \frac{b (b^{2} + a^{2})}{b^{2} + a^{2}} = b \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

If $\frac{z - 1}{z + 1}$ is a purely imaginary number ( $z \neq - 1$ ), then find the value of $? z ?$ .

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Medicine Entrance Exam Medicine

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \frac{z - 1}{z + 1} i s p u r e l y i m a g i n a r y n u m b e r . \\ L e t z = x + y i \\ ∴ \frac{x + y i - 1}{x + y i + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} \\ \Rightarrow \frac{(x - 1) (x + 1) - i y (x - 1) + (x + 1) i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} \\ \Rightarrow \frac{x^{2} - 1 + i y (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ \Rightarrow \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ Since, t h e n u m b e r i s p u r e l y i m a g i n a r y, t h e n r e a l p a r t = 0 \\ ∴ \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} = 0 \\ \Rightarrow x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 1 ∴ | z | = 1 \end{array}$

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9 months ago

What are the top UP NEET accepting MBBS Colleges in UP?

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Tasbiya Khan

Contributor-Level 10

Listed belore are the top UP NEET accepting MBBS colleges in Uttar Pradesh along with their tuition fees:

Top Colleges	Tuition Fee
KGMU - King George's Medical University	INR 2.14 lakh
Integral University - IUL	INR 81 lakh
Teerthanker Mahaveer University	INR 72.9 lakh
Era's Lucknow Medical College and Hospital	INR 74.7 lakh
SVSU - Swami Vivekanand Subharti University	INR 53.33 lakh

Source: Official site and may vary.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $t a n α = \frac{1}{7}$ , $t a n β = \frac{1}{3}$ , then $c o s 2 α$ is equal to:

(a) $s i n 2 β$

(b) $s i n 4 β$

(c) $s i n 3 β$

(d) $c o s 2 β$

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n α = \frac{1}{7} a n d t a n β = \frac{1}{3} \\ c o s 2 α = \frac{1 - t a n^{2} α}{1 + t a n^{2} α} = \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} = \frac{48}{50} = \frac{24}{25} \\ t a n 2 β = \frac{2 t a n β}{1 - t a n^{2} β} = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4} \\ ∴ t a n 2 β = \frac{3}{4} \\ s i n 4 β = \frac{2 t a n 2 β}{1 + t a n^{2} 2 β} = \frac{2 \times \frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} = \frac{\frac{3}{2}}{1 + \frac{9}{16}} = \frac{3}{2} \times \frac{16}{25} = \frac{24}{25} \\ c o s 2 α = s i n 4 β = \frac{24}{25} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

The value of $c o s^{2} 4 8^{?} - s i n^{2} 1 2^{?}$ is:

(a) $\frac{\sqrt[]{5} + 1}{4}$

(b) $- \frac{\sqrt[]{5} + 1}{4}$

(c) $\frac{1}{2}$

(d) $- \frac{1}{2}$

Hint : Use of $c o s^{2} A - s i n^{2} B = c o s (A + B) C o s (A - B)$

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} \\ c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} = c o s (4 8^{0} + 1 2^{0}) . c o s (4 8^{0} - 1 2^{0}) [? c o s^{2} A - s i n^{2} B = c o s (A + B) . c o s (A - B)] \\ = c o s 6 0^{0} . c o s 3 6^{0} = \frac{1}{2} \times \frac{\sqrt[]{5} + 1}{4} = \frac{\sqrt[]{5} + 1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Show that $? \frac{z - 2}{z + 3} ? = 2$ represents a circle. Find its centre and radius.

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Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{z - 2}{z - 3} | = 2 \\ L e t z = x + i y \\ ∴ | \frac{x + i y - 2}{x + i y - 3} | = 2 \\ \Rightarrow | \frac{(x - 2) + i y}{(x - 3) + i y} | = 2 \\ \Rightarrow \frac{| (x - 2) + i y |}{| (x - 3) + i y |} = 2 [? | \frac{a}{b} | = \frac{| a |}{| b |}] \\ \Rightarrow | (x - 2) + i y | = 2 | (x - 3) + i y | \\ \Rightarrow \sqrt[]{{(x - 2)}^{2} + y^{2}} = 2 \sqrt[]{{(x - 3)}^{2} + y^{2}} [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x - 2)}^{2} + y^{2} = 4 [{(x - 3)}^{2} + y^{2}] \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} = 4 [x^{2} + 9 - 6 x + y^{2}] \\ \Rightarrow x^{2} + y^{2} - 4 x + 4 = 4 x^{2} + 4 y^{2} - 24 x + 36 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 20 x + 32 = 0 \\ \Rightarrow x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ H e r e, g = - \frac{10}{3}, f = 0 \\ r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{\frac{100}{9} - 0 - \frac{32}{3}} = \sqrt[]{\frac{4}{9}} = \frac{2}{3} \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f a c i r c l e i s \\ x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ C e n t r e = (- g, - f) = (\frac{10}{3}, 0) a n d r = \frac{2}{3} . \end{array}$

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9 months ago

What documents are required for BA admission at PK University?

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

Applicants need to upload essential documents such as their Class 12 mark sheet, passport-sized photograph, valid ID proof, and any other documents as specified in the application instructions. Ensure all files are in the required format before submission.

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9 months ago

If $A$ lies in the second quadrant and $3 t a n A + 4 = 0$ , then the value of $2 c o t A - 5 c o s A + s i n A$ is equal to:

(a) $- \frac{53}{10}$

(b) $- \frac{23}{10}$

(c) $- \frac{37}{10}$

(d) None

0 Follower 2 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : 3 t a n A + 4 = 0, A l i e s i n second q u a d r a n t \\ ∴ t a n A = \frac{- 4}{3} \\ c o s A = \frac{- 3}{5} [A l i e s i n second q u a d r a n t] \\ a n d s i n A = \frac{4}{5} a n d c o t A = \frac{- 3}{4} \\ ∴ 2 c o t A - 5 c o s A + s i n A = 2 (\frac{- 3}{4}) - 5 (\frac{- 3}{5}) + \frac{4}{5} \\ = \frac{- 3}{2} + 3 + \frac{4}{5} = \frac{- 15 + 30 + 8}{10} = \frac{23}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

What is the final step to confirm BA admission at PK University?

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Hanyang University Study Abroad

Contributor-Level 10

After receiving an admission offer from PK University, candidates must confirm their seat by paying the BA course tuition fee of INR 30,000 through the official payment portal. Retaining the payment receipt is crucial for verification and future reference.

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9 months ago

What are the programs opted by Hanyang University alumni?

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Rachit Sharma

Contributor-Level 10

Hanyang University is renowned for offering the best degree programs for its international students. Hanyang offers various programs at undergraduate and postgraduate levels for its international students. Hanyang University offers over 170 programs for international students. The university offers various specialisation in fields such as Energy Engineering, Mechanical Engineering, Electronics Engineering and Data Science. Some of the popular courses at Hanyang are Engineering, Humanities, Computer Science, Computing, Medicine, Pharmacy, and more. Some top programs opted by Hanyang University alumni is listed below:

Business Administrat

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9 months ago

What is the selection process for BA admission at PK University?

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