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Humanities & Social Sciences B.A

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : a r g (z - 1) = a r g (z + 3 i) \\ \Rightarrow a r g (x + y i - 1) = a r g (x + y i + 3 i) \\ \Rightarrow a r g [(x - 1) + y i] = a r g [x + (y + 3) i] \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} = t a n^{- 1} \frac{y + 3}{x} \\ \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \\ \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \\ \Rightarrow 3 x - y = 3 \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 (x - 1) = y \Rightarrow \frac{(x - 1)}{y} = \frac{1}{3} \Rightarrow x - 1 : y = 1 : 3 \\ H e n c e, x - 1 : y = 1 : 3 . \end{array}$

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9 months ago

How can I apply for the BA programme at PK University?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

The value of $s i n \frac{2 π}{18} + s i n \frac{5 π}{18} + s i n \frac{7 π}{18} + s i n \frac{4 π}{9}$ is given by:

(a) 1

(b) -1

(c) $c o s \frac{3 π}{9} + s i n \frac{π}{18}$

(d) $c o s \frac{2 π}{9} + s i n \frac{π}{18}$

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Is CUET final answer key out?

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s s i n \frac{π}{18} + s i n \frac{π}{9} + s i n \frac{2 π}{9} + s i n \frac{5 π}{18} \\ = (s i n \frac{5 π}{18} + s i n \frac{π}{18}) + (s i n \frac{2 π}{9} + s i n \frac{π}{9}) \\ = 2 s i n (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) . c o s (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 s i n (\frac{\frac{2 π}{9} + \frac{π}{9}}{2}) . c o s (\frac{\frac{2 π}{9} - \frac{π}{9}}{2}) \\ = 2 s i n \frac{π}{6} . c o s \frac{π}{9} + 2 s i n \frac{π}{6} . c o s \frac{π}{18} \\ = 2 \times \frac{1}{2} c o s \frac{π}{9} + 2 \times \frac{1}{2} c o s \frac{π}{18} = c o s \frac{π}{9} + c o s \frac{π}{18} \\ = s i n (\frac{π}{2} - \frac{π}{9}) + s i n (\frac{π}{2} - \frac{π}{18}) = s i n \frac{7 π}{18} + s i n \frac{8 π}{18} \\ = s i n \frac{7 π}{18} + s i n \frac{4 π}{9} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Science CUET Answer Key

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Nishtha Arora

Contributor-Level 7

Yes, National Testing Agency has releasd the CUET 2025 final answer key on July 01. Candidates can now download the revised answer key PDF at cuet.nta.nic.in. As per the CUET 2025 final answer key:

27 questions have been dropped
75 questions have multiple correct options

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 \times 2 \times 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = \end{array}$

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9 months ago

Number of solutions of the equation $t a n x + s e c x = 2 c o s x$ lying in the interval $[0, 2 π]$ is:

(a) 0

(b) 1

(c) 2

(d) 3

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Hanyang University Study Abroad

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n x + s e c x = 2 c o s x \\ \Rightarrow \frac{s i n x}{c o s x} + \frac{1}{c o s x} = 2 c o s x \\ \Rightarrow 1 + s i n x = 2 c o s^{2} x \Rightarrow 2 c o s^{2} x - s i n x - 1 = 0 \\ \Rightarrow 2 (1 - si n^{2} x) - s i n x - 1 = 0 \Rightarrow 2 - 2 si n^{2} x - s i n x - 1 = 0 \\ \Rightarrow - 2 si n^{2} x - s i n x + 1 = 0 \Rightarrow 2 si n^{2} x + s i n x - 1 = 0 \\ Since t h e e q u a t i o n i s q u a d r a t i c e q u a t i o n i n s i n x . S o i t w i l l h a v e 2 s o l u t i o n s . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

How many programs does the Hanyang University have?

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Kartik Bisht

Contributor-Level 10

The University is known for offering the best degree programs for its students. Hanyang offers more than 170 programs at undergraduate and postgraduate levels for its international students. There are 60 departments and 15 colleges on the Seoul campus. It consists of numerous research centres that are centred around various academic disciplines. Some of the popular courses at Hanyang are Engineering, Humanities, Computer Science, Computing, Education, and more.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $s i n θ = - \frac{2}{5}$ and $θ$ lies in the third quadrant, then the value of $c o s θ$ is:

(a) $- \frac{1}{5}$

(b) $- \frac{\sqrt[]{21}}{5}$

(c) $\frac{1}{5}$

(d) $\frac{\sqrt[]{21}}{5}$

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Has CUET 2025 counselling registration commenced?

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ = - \frac{4}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = \sqrt[]{1 - s i n^{2} θ} = \sqrt[]{1 - {(- \frac{4}{5})}^{2}} = \sqrt[]{1 - \frac{16}{25}} = \sqrt[]{\frac{9}{25}} = \frac{+ 3}{- 5} \\ ∴ c o s θ = - \frac{3}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = 2 c o s^{2} \frac{θ}{2} - 1 [? π < θ < \frac{3 π}{2}, ∴ \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ \Rightarrow \frac{- 3}{5} = 2 c o s^{2} \frac{θ}{2} - 1 \\ \Rightarrow 2 c o s^{2} \frac{θ}{2} = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow c o s^{2} \frac{θ}{2} = \frac{2}{5 \times 2} = \frac{1}{5} \\ \Rightarrow c o s \frac{θ}{2} = \pm \frac{1}{\sqrt[]{5}} \Rightarrow c o s \frac{θ}{2} = - \frac{1}{\sqrt[]{5}} [? \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

Science CUET

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Dhanya Dhyani

Contributor-Level 7

Participating universities have commenced the CUET counselling 2025 registration. Check below to know the list of universities who have commenced the CUET admission 2025:

Delhi University started the counselling registration on June 17 at CSAS UG portal
Allahabad University scheduled CUET counselling registration 2025 to commence from June 30. Due to technical issues, it has been delayed.
Banaras Hindu University has activated the CAP UG 2025 portal and is expected to commence CUET counselling registration soon.

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9 months ago

Madras Medical College, Chennai Pharmacy

How can a GPAT qualified student apply for M.Pharma. in Pharmaceutics at Madras Medical College, Chennai?

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9 months ago

Political Science Humanities & Social Sciences

What are the top CUET PG accepting MA in Public Policy colleges in India?

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Are there any government MA Public Policy colleges in India?

Contributor-Level 10

Listed below are the top CUET PG accepting MA Public Policy colleges in India along with their tuition fees:

Top Colleges	Tuition Fee
Galgotias University	INR 92,000
TISS Mumbai	INR 68,000
Central University of Kerala	INR 5,200
Central University of Rajasthan	INR 10,000
Tata Institute of Social Sciences, Hyderabad	INR 94,600

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

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9 months ago

Public Policy M.A

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

Yes, there are many government MA Public Policy colleges in India. Some of them are mentioned below along with their tuition fees:

Public/Govt. Colleges	Tuition Fee
TISS - Tata Institute of Social Sciences, Mumbai Admission	INR 68,000
University of Mumbai Admission	INR 68,810 – INR 80,000
IIT Bombay Admission	INR 2.69 lakh
Central University of Kerala Admission	INR 5,200
Guru Nanak Dev University Admission	INR 61,550

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

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9 months ago

If $α + β = \frac{π}{4}$ , then the value of $(1 + t a n α) (1 + t a n β)$ is:

(a) 1

(b) 2

(c) -2

(d) Not defined

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Political Science Humanities & Social Sciences

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : α + β = \frac{π}{4} \\ W e k n o w t h a t \\ t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β} \\ \Rightarrow t a n α + t a n β = 1 - t a n α . t a n β \\ \Rightarrow t a n α + t a n β + t a n α . t a n β = 1 \\ \Rightarrow 1 + t a n α + t a n β + t a n α . t a n β = 1 + 1 \\ \Rightarrow 1 (1 + t a n α) + t a n β (1 + t a n α) = 2 \\ \Rightarrow (1 + t a n α) (1 + t a n β) = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Is it worth joining the top MA in Public Policy colleges in India?

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Hanyang University Study Abroad

Contributor-Level 10

Yes, joining the top MA in Public Policy colleges in India can be worthwhile for you. For that, you need to check its ROI. Listed below are some colleges along with their tuition fees and average placement package:

Top Colleges	Tuition Fee	Average Placement Package
St. Xavier's College, Mumbai MA	INR 1.5 lakh	INR 6.23 LPA
Christ University Bangalore MA	INR 4.8 lakh	INR 6.14 LPA
TISS Mumbai MA	INR 68,000	INR 6.5 LPA
University of Mumbai MA	INR 68,810 – INR 80,000	INR 18 LPA
University of Madras MA	INR 3,000	IN 4.2 LPA

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

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9 months ago

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Mohit Thapa

Contributor-Level 10

Hanyang University is among the top choices of international students for pursuing higher education in the South Korea. Hanyang is known for offering the best academic programs all over the world. Hanyang University offers more than 170 degree programs at UG and PG levels for its international students. Some of the famous programs at Hanyang University opted by international students are Pharmacy, Music, Computer Science, Medicine, Humanities, Engineering, Computing, and more. Furthermore, the top majors at the Hanyang are as follows:

Medicine
Humanities
Social Sciences
Performing Art

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9 months ago

Political Science Humanities & Social Sciences

Are there any affordable MA in Public Policy colleges in India?

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Contributor-Level 10

Yes, Joining the top MA Public Policy colleges in India can be affordable as most colleges have tuition fees of less than INR 1 lakh. Some of them are mentioned below along with their tuition fees:

Low-cost Colleges	Tuition fee
Galgotias University	INR 92,000
Central University of Kerala	INR 5,200
TISS Mumbai	INR 68,000
University of Mumbai	INR 68,810 – INR 80,000
University of Madras	INR 3,000

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

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9 months ago

If $s i n θ + c o s θ = 1$ , then the value of $s i n 2 θ$ is:

(a) 1

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

(c) 0

(d) -1

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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:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s θ = 1 \\ \Rightarrow {(s i n θ + c o s θ)}^{2} = {(1)}^{2} \\ \Rightarrow s i n^{2} θ + c o s^{2} θ + 2 s i n θ c o s θ = 1 \\ \Rightarrow 1 + s i n 2 θ = 1 \\ \Rightarrow s i n 2 θ = 1 - 1 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

The value of $s i n 5 0^{\circ} - s i n 7 0^{\circ} + s i n 1 0^{\circ}$ is equal to:

(a) 1

(b) 0

(c) $- \frac{1}{4}$

(d) 1

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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:

$\begin{array}{l} G i v e n t h a t : sin 5 0^{0} - sin 7 0^{0} + sin 1 0^{0} \\ (sin 5 0^{0} - sin 7 0^{0}) + sin 1 0^{0} = 2 c o s \frac{5 0^{0} + 7 0^{0}}{2} . s i n \frac{5 0^{0} - 7 0^{0}}{2} + sin 1 0^{0} \\ = 2 c o s 6 0^{0} . (- sin 1 0^{0}) + sin 1 0^{0} \\ = - 2 \times \frac{1}{2} sin 1 0^{0} + sin 1 0^{0} \\ = - sin 1 0^{0} + sin 1 0^{0} = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Solve the equation $z = \overset{?}{z} + 1 + 2 i$ .

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Maths NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : | z | = z + 1 + 2 i \\ L e t z = x + y i \\ | z | = (z + 1) + 2 i \\ S q u a r i n g b o t h s i d e s \\ {| z |}^{2} = {(z + 1)}^{2} + 4 i^{2} + 4 (z + 1) i \\ \Rightarrow {| z |}^{2} = {| z |}^{2} + 1 + 2 z - 4 + 4 (z + 1) i \\ \Rightarrow 0 = - 3 + 2 z + 4 (z + 1) i \\ \Rightarrow 3 - 2 z - 4 (z + 1) i = 0 \\ \Rightarrow 3 - 2 (x + y i) - 4 (x + y i + 1) i = 0 \\ \Rightarrow 3 - 2 x - 2 y i - 4 x i - 4 y i^{2} - 4 i = 0 \\ \Rightarrow 3 - 2 x + 4 y - 2 y i - 4 x i - 4 i = 0 \\ \Rightarrow (3 - 2 x + 4 y) - i (2 y + 4 x + 4) = 0 \\ \Rightarrow 3 - 2 x + 4 y = 0 \Rightarrow 2 x - 4 y = 3 \dots (i) \\ \Rightarrow 2 y + 4 x + 4 = 0 \Rightarrow 2 x + y = - 2 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i), w e g e t y = - 1 a n d x = - \frac{1}{2} \\ H e n c e, t h e v a l u e o f z = x + y i = (- \frac{1}{2} - i) . \end{array}$

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9 months ago

Show that the complex number $z$ , satisfying the condition $arg (\frac{1 - z}{1 + z}) = \frac{π}{4}$ , lies on a circle.

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