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It is known that the values of sinx repeat after an interval of 2π or 360^∘
 
∴sin765^∘
 =sin(2×360^∘ +45^∘)
=sin45^∘   

\(= \frac{1}{\sqrt 2}\)

\(\underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}\) = f'(c)

\(\frac{\underset{x\rightarrow c}{lim}\,f(x)-f(c)}{\underset{x\rightarrow c}{lim}\,x-c}\) = f'(c)

\(\underset{x\rightarrow c}{lim}\,f(x)-f(c)\) = f'(c) . \(\underset{x\rightarrow c}{lim}\,(x-c)\)

\(\underset{x\rightarrow c}{lim}\,(f(x)-f(c))\) = f'(c) . (c - c)

\(\underset{x\rightarrow c}{lim}\,f(x)\) - \(\underset{x\rightarrow c}{lim}\,f(c)\) = f'(c).(0)

\(\underset{x\rightarrow c}{lim}\,f(x)-f(c)\) = 0

\(\underset{x\rightarrow c}{lim}\,f(x)\) = f(c)

Correct answer is: (d) 1

The total area under the normal distribution curve above the base line is 1

Concept:

Consider a complex number, z = a + ib

Conjugate of complex number = z̅ = a - ib

Modulus of complex number  = |z| = \(\rm \sqrt{(a^2 + b^2) }\)

Calculation: 

Let, z = a + ib,

zz̅ = (a + ib)(a - ib)

= \(\rm a^2-(ib)^2\)

= \(\rm a^2-i^2(b)^2\)

=\(\rm a^2-(ib)^2\)

=\(\rm a^2+b^2\cdots (\because i^2=-1)\)

And, |z|² = \(\rm (\sqrt{a^2+b^2})^2\)

= \(\rm {a^2+b^2}\)

∴ zz̅  = |z|²

Now, 

 \(\rm z^{-1}=\frac1 z=\frac{1}{a+ib}\)

=\(\rm \frac{1}{a+ib}\times \frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}\)

= \(\rm \frac{\bar z}{|z|^2}\) ≠ \(\rm \frac {z}{|z|^2}\)

Hence, option (1) is correct.

Concept:

Iota power:

• i² = -1
• i³ = -i
• i⁴ = 1

Law of exponents:

(a^m × aⁿ)  = a^m+n

\(\rm (a^m)^n=a^{mn}\)

Calculation:

Here, \(\rm [({i)^{25}+(\frac {1}{i})^{27}}]^2\)

= \(\rm [(i)^{25}+(\frac {1}{i})^{27}]^2\)                          ....(∵ √-1 = i)

= \(\rm [(i)^{24}i+(\frac {1}{i})^{24}(\frac 1 i)^3]^2\)                 .....(∵ (am × an)  = am+n)

= \(\rm [((i)^{6})^4i+((\frac {1}{i})^{6})^4(\frac 1 i)^3]^2\)           ....(∵ \(\rm (a^m)^n=a^{mn}\))

= \(\rm [i+(\frac 1 i)^3]^2\)                                 ....(∵ i4 = 1)

 \(=\rm [i-\frac 1 i]^2\\ =(\frac{i^2-1}{i})^2\)                                   ......(∵ i3 = -i)

\( \rm =(\frac{-1-1}{i})^2\\ =\frac{(-2)^2}{i^2}\)

= -4                                                ...(∵ i2 = -1)

Hence, option (4) is correct.

Concept:

For any two real numbers a and b, the result \(\rm \sqrt{a} × \sqrt{b} = \sqrt{ab}\)is true only when at least one of the given numbers is either zero or positive.

i = \(\rm \sqrt{-1}\),

i² = -1, 

i3 = -i, 

i⁴ = 1

Calculation:

1. We know, \(\rm \sqrt{a} × \sqrt{b} = \sqrt{ab}\) only when a, b ≥ 0

\(\rm \sqrt{-a} × \sqrt{-b} =\rm \sqrt{-1}\sqrt{a} × \sqrt{-1}\sqrt{b} \)

= \(\rm \sqrt{ab}\) × (i × i)                      ....(∵ i = \(\rm \sqrt{-1}\)) 

= -\(\rm \sqrt{ab}\)                                 ....(∵ i² = -1) 

2.  i4m+3

i4m+3 = i^4mi³

= i³                                                         ....(∵ i^4m = 1)

= -i                                           ....(∵ i³ = -i)

So, both the statements are not correct.

Hence, option (4) is correct.

Concept:

Let, Z1 = a1 + jb1 and Z2 = a2 + jb2

Z1 and Z2 are said to equal when,

a1 = a2 and b1 = b2

Calculation:

Let, Z = x+ iy

Given that,

|Z| - Z = 1 + 3i

⇒ |x + iy| - (x + iy) = 1 + 3i

\(⇒\ \sqrt{x^2\ +\ y^2}\ -\ x\ -\ iy\ =\ 1\ +\ 3i\)

Comparing real and imaginary parts of both sides, we will get

\(⇒\ \sqrt{x^2\ +\ y^2}\ -\ x\ =\ 1\)       -----(1)

- i y = 3i ⇒ y = - 3

Therefore, from equation (1)

\(\ \sqrt{x^2\ +\ (-3)^2}\ -\ x\ =\ 1\)

\(⇒ \ \sqrt{x^2\ +\ 9}\ \ =\ 1 + x\)

Taking square both side

x² + 9 = (1 + x)²

but we know that, (a + b)² = a² + 2ab + b²

⇒ x2 + 9 = 1 + x2 + 2x     

⇒ x = 4

Therefore, solution of given equation will be 

Z = x + iy = 4 - 3i

Hence, option 3 is correct.

Concept:

Formula used:

• i² = -1
• i³ = i² × i = -i
• i⁴ = i² × i² = 1

Calculation:

To find: i¹⁰

⇒ i¹⁰ = i⁴ × i⁴ × i²

⇒ i¹⁰ = 1 × 1 × (-1)

⇒ i¹⁰ = -1

CONCEPT:

Let z = a + ib be a complex number. Then, the modulus of z, denoted by |z|, is defined to be the non-negative real number \(\sqrt {{a^2} + {b^2}} .\)

CALCULATION:

Given expression is \(\frac{{4 + \sqrt 2 i}}{{3 - \sqrt 2 i}}\)

\( \Rightarrow \frac{{4 + \sqrt 2 i}}{{3 - \sqrt 2 i}} = \frac{{4 + \sqrt 2 i}}{{3 - \sqrt 2 i}} \times \frac{{3 + \sqrt 2 i}}{{3 + \sqrt 2 i}} = \frac{{12 + 4\sqrt 2 i + 3\sqrt 2 i - 2}}{{{3^2} - {{\left( {\sqrt 2 i} \right)}^2}}}\)

\( \Rightarrow \frac{{10 + 7\sqrt 2 i}}{11} = \frac{{10}}{11} + \frac{{7\sqrt 2 }}{11}i\)

Concept:

Properties of |Z|: If Z = x + iy is a complex number then the following properties are applicable for |Z|.

1. \(|Z|\ =\ |\bar{Z}|\)

2. \(|z|^2 \ =\ Z̅{Z}\)

3. \(|\overline{z_1\ +\ z_2}|\ =\ |\bar{Z_1}\ +\ \bar{Z_2}|\)

Calculation:

Given that,

|Z| = 1 

⇒ |Z|² = 1

⇒ Z Z̅  = 1                  (∵ \(|z|^2 \ =\ Z̅{Z}\))

\(⇒ Z = \frac{1}{\bar{Z}}\)        ----(1)

Therefore, the value of \(2(Z\ +\ \bar{Z})\ -\ 2(\frac{1}{Z}\ +\ \frac{1}{\bar{Z}})\)

=  \(2(Z\ -\ \frac{1}{\bar{Z}})\ -\ 2(\ \bar{Z}\ -\ \frac{1}{Z})\)

But, from equation (1) \( Z = \frac{1}{\bar{Z}}\)

⇒ \(2(Z\ -\ \frac{1}{\bar{Z}})\ -\ 2(\ \bar{Z}\ -\ \frac{1}{Z})\ =\ 0\)

Hence, option 2 is correct.

Calculation:

z = \(\rm {15-30i}\over{3+4i}\)

z = \(\rm {15-30i\over3+4i}\times {3-4i\over3-4i}\)

z = \(\rm {45-150i+120i^2\over3^2-(4i)^2}\)

z = \(\rm {-150i-75\over9+16}\)

z = \(\rm {-75(2i+1)\over25}\)

z = -3 - 6i

Concept:

A complex number (Z):  Complex number is the combination of a real number and an imaginary number. It is given by

Z = x + iy, where 'x' and 'y' are the real and imaginary part of Z and i = √-1 

Conjugate of a complex number: When the i of a complex number is replaced with - i, we get the conjugate of that complex number.

\(\bar{Z}\ =\ x\ -\ iy\)

Re(Z) = x

\Img(Z) = y

|Z| = \(\sqrt{x^2\ +\ y^2}\)

Formula used:

1. \( |\frac{Z_1}{Z_2}| = \frac{|Z_1|}{|Z_2|}\)

2. \(Z\bar{Z}\ = |Z|^2\)

3. (a - b)² = a2 + b2 - 2ab

Calculation:

Given that,

\(\sqrt{2ab}(\frac{Z}{\bar{Z}})\ =\ a + ib\)     -----(1)

Z = (b + ia)     ----(2)

Therefore, a conjugate of Z

Z̅ = b - ia      ----(3)

Hence, from equation (1)

\(\sqrt{2ab}(\frac{b\ +\ ia}{b\ -\ ia})\ =\ a + ib\)

Taking modulus of both sides,

\(\sqrt{2ab}|(\frac{b\ +\ ia}{b\ -\ ia})|\ =\ |a + ib|\)

\(\sqrt{2ab}\frac{|b\ +\ ia|}{|b\ -\ ia|}|\ =\ |a + ib|\)    

\(\sqrt{2ab}\frac{\sqrt{b^2\ +\ a^2}}{\sqrt{b^2\ +\ a^2}​​​​}\ = \sqrt{a^2\ +\ b^2}\)   

Taking square of both side

a² + b² - 2ab = 0

⇒ (a - b)² = 0

⇒ a = b

Hence, option 4 is correct.

Concept:

The argument of a complex number z = x + iy

arg(z) = tan-1\(\rm \left(y\over x\right)\)

The angle is according to the sign of the y and x

• Both positive then angle ∈ [0°, 90°]
• Negative x and positive y then angle ∈ [90°, 180°]

Calculation:

Let the complex number be z = x + iy

arg(z) = tan-1\(\rm \left(y\over x\right)\)

∵ y > 0 and x = -y

arg(z) = tan-1\(\rm \left(y\over (-y)\right)\)

arg(z) = tan^-1(-1)

arg(z) = 135° (∵ Negative x and positive y then angle ∈ [90°, 180°])

CONCEPT:

We know that for any integer k, \({i^{4k}} = 1,{\rm{\;}}{i^{4k + 1}} = i,{\rm{\;}}{i^{4k + 2}} = - 1,\;{i^{4k + 3}} = - i\)

Also (a + b)2 = a² + b² + 2ab

CALCULATION:

Given expression is (2 – i)⁴

∴ (2 – i)⁴ = [(2 – i)²]²

⇒ [2² + i² – 4i]² = [4 – 1 – 4i]² = [3 – 4i]²

⇒ [3² + (4i)² + 2 × 3 × (-4i)] = [9 – 16 – 24i]

correct option:

(B) b/a

Correct answer is (a) monkeys, owls and leopards

'माता का अँचल’ पाठ में वर्णित भोलानाथ के पिता की दिनचर्या के बारे में यह पता चलता है कि वे सुबह जल्दी उठते और नहा-धोकर पूजा-पाठ पर बैठ जाते थे। वे अकसर बालक भोलानाथ (अपने पुत्र) को भी अपने साथ बिठा लिया करते थे। वे प्रतिदिन रामायण का पाठ करते थे। पूजा के समय वे भोलानाथ को भभूत से तिलक लगा देते थे। पूजा-पाठ के उपरांत वे रामनामी बही पर एक हज़ार बार राम-राम लिखते थे और अपनी पाठ करने की पोथी में रख लेते थे। इसके उपरांत वे पाँच सौ बार कागज के टुकड़े पर राम-राम लिखते और उन्हीं कागजों पर आटे की छोटी-छोटी गोलियाँ रखकर लपेटते। उन गोलियों को लेकर वे गंगा जी के तट पर जाते और अपने हाथों से मछलियों को खिला देते थे। इस समय भी भोलानाथ उनके साथ ही हुआ करता था। आज के आम आदमी की सुबह इस सुबह की दिनचर्या से इसलिए भिन्न है क्योंकि आज इस भौतिकवादी युग में धन-दौलत के पीछे लगी भागम-भाग के कारण आम आदमी के पास इतना समय ही नहीं है।

• District Planning Committee (DPC) is the committee created as per article 243ZD of the Constitution of India at the district level for planning at the district and below.
• The Committee in each district should consolidate the plans prepared by the Panchayats and the Municipalities in the district and prepare a draft development plan for the district.
• The Legislative of a State may, by law, make provision with respect to—
• (a) the composition of the District Planning Committees;
• (b) the manner in which the seats in such Committees shall be filled: Provided that not less than four-fifths of the total number of members of such Committee shall be elected by, and from amongst, the elected members of the Panchayat at the district level and of the Municipalities in the district in proportion to the ratio between the population of the rural areas and of the urban areas in the district;
• (c) the functions relating to district planning which may be assigned to such Committees;
• (d) the manner in which the Chairpersons of such Committees be chosen.
• Elected and Ex-Officer members are there in District Planning Committee are 20 and 5 respectively. 

The correct answer is Muppavarapu Venkaiah Naidu.

• P.J. Kurien is former the Deputy Chairman of the Rajya Sabha (21 August 2012 – 1 July 2018).
• Muppavarapu Venkaiah Naidu is the current chairperson of Rajya Sabha (since 11 August 2017).
• Harivansh Narayan Singh is the current Deputy Chairman of Rajya Sabha, since 14 September 2020.

The correct answer is 8 November.

• On 8 November 2016, the Government of India announced the demonetization of all ₹500 and ₹1,000 banknotes of the Mahatma Gandhi Series.
• It also announced the issuance of new ₹500 and ₹2,000 banknotes in exchange for the demonetized banknotes.
• Prime Minister Narendra Modi claimed that the action would curtail the shadow economy and reduce the use of illicit and counterfeit cash to fund illegal activity and terrorism.
• According to a 2018 report from the Reserve Bank of India, approximately 99.3% of the demonetized banknotes, or ₹15.30 lakh crore (15.3 trillion) of the ₹15.41 lakh crore that had been demonetized.

• Rs 1,000 and higher denomination notes were first demonetized in January 1946 and again in 1978. 
• The highest denomination note ever printed by the Reserve Bank of India was the Rs 10,000 note in 1938 and again in 1954.
• But these notes were demonetized in January 1946.
• Demonetisation has been implemented thrice -1946, 1978, and 2016.

The National Highway - 1 connects Delhi to Amritsar.

The correct answer is Maritime Transport.

• India has a vast coastline of approximately 7,517 km, including islands.
• Oceanic routes play an important role in the transport sector of India’s economy.
• Approximately 95 percent of India’s foreign trade by volume and 70 percent by value moves through ocean routes.
• Apart from international trade, these are also used for the purpose of transportation between the islands and the rest of the country.

Correct option: c) Uttar Pradesh

(b) Five types.

Law of equi-marginal Utility is the First law of Gossen.

(A) Scarcity of factors

(a) Fiscal deficit

(B) Law of Diminishing marginal utility

(b) Washington DC

(D) All of these

(C) Both A and B

(b) World Bank

The inertia force is an imaginary force, which when acts upon a rigid body, brings it in an equilibrium position. 

Inertia force = -Accelerating force = -m.a 

(c) Ratio of prices of exports and imports

If components of a machine accelerate, inertia forces are produced due to their masses. If the magnitude of these forces are small compared to the externally applied loads, they can be neglected while analysis the mechanism. Such an analysis is known as static force analysis.

When the inertia effect due to the mass of the component is also considered, it is called dynamic force analysis. 

(B) Dynamic Concept

[b] Horizontal analysis

Correct option is: (A) Capital Reserve

(a) SDR – Special Drawing Rights

Balance of Trade means Import & Export of goods.

Premium on redemption of debentures A/C is Liability

(d) Import and export of visible items only