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The pattern of the original number series is

⇒ 18 + 2¹ = 20

⇒ 20 + 2² = 24

⇒ 24 + 2³ = 32

⇒ 32 + 2⁴ = 48

⇒ 48 + 2⁵ = 80

According the pattern of the above series the number series starting from the given number would be as:

⇒ A = 19 + 2¹ = 21

⇒ B = 21 + 2² = 25

⇒ C = 25 + 2³ = 33

⇒ D = 33 + 2⁴ = 49

⇒ E = 49 + 2⁵ = 81

∴ The required number in place of E would be 81.

The pattern of the original number series is

⇒ 30 × 1 + 1 = 31

⇒ 31 × 2 + 2 = 64

⇒ 64 × 3 + 3 = 195

⇒ 195 × 4 + 4 = 784

⇒ 784 × 5 + 5 = 3925

According the pattern of the above series the number series starting from the given number would be as:

⇒ A = 19 × 1 + 1 = 20

⇒ B = 20 × 2 + 2 = 42

⇒ C = 42 × 3 + 3 = 129

⇒ D = 129 × 4 + 4 = 520

⇒ E = 520 × 5 + 5 = 2605

∴ The required number in place of E would be 2605.

Option : (B) z = 11x + 19y

Given surfaces are

z = x² + y² – 3

\(\Rightarrow\) x² + y² – z = 3

And x² + y² + z² = 9

let \(\phi_1\) = x² + y² – z

And \(\phi_2\) = x² + y² + z²

Given point (2, –1, 2).

\(\vec \nabla \phi_1 = \vec \nabla(\mathrm x^2+y^2-z)\)

\(=\hat i\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2-z)\) \(+\hat j\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2-z)\) \(+\hat k \frac{\partial }{\partial z}(\mathrm x^2+y^2-z)\)

\(=2\mathrm x \hat i + 2y\hat j-\hat k\)

\((\vec \nabla\phi_1)_{(2, -1, 2) }= 4 \hat i -2\hat j-\hat k\)

\(\vec \nabla \phi_2=\vec \nabla(\mathrm x^2+y^2+z^2)\)

\(=\hat i\frac{\partial}{\partial \mathrm x}(\mathrm x^2+y^2+z^2)\) \(+\hat j \frac{\partial}{\partial y}(\mathrm x^2+y^2+z^2)\) \(+\hat k\frac{\partial}{\partial z}(\mathrm x^2+y^2+z^2)\)

\(=2\mathrm x \hat i+2y\hat j + 2z\hat k\)

\((\vec \nabla \phi_2)_{(2, -1, 2)}=4\hat i-2\hat j+4\hat k\)

let angle between surfaces \(\phi_1\) and \(\phi_2\) is θ.

Then \(\vec \nabla\phi_1.\vec \nabla \phi_2\) \(=|\vec \nabla \phi_1||\vec \nabla \phi_2|\cos \theta\)

\(\Rightarrow \cos\theta = \frac{\vec \Delta \phi_1.\vec \nabla\phi_2}{|\vec \nabla \phi_1||\vec \nabla.\phi_2|}\)

\(=\frac{(4\hat i-2\hat j-\hat k)(4\hat i-2\hat j+4\hat k)}{|4\hat i-2\hat j-\hat k||4\hat i-2\hat j+4\hat k|}\)

\(=\frac{16+4-4}{\sqrt{16+4+1}\sqrt{16+4+16}}\)    \(\Big(\because \hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1\) & \(\hat i.\hat j=\hat i.\hat k=\hat j.\hat k=0\Big)\)

\(=\frac{16}{\sqrt{21}\sqrt{36}}=\frac{16}{6\sqrt{21}}=\frac{8}{3\sqrt{21}}\)

\(\Rightarrow \theta =\cos^{-1}\left(\frac{8}{3\sqrt{21}}\right)\)

Hence, angle between given surfaces is \(\cos^{-1}\frac{8}{3\sqrt{21}}\).

2x + y = 7          ... (1) 

And 4x – 3y + 1 = 0 

⇒ 4x – 3y = –1     ... (2) 

Now, multiplying equation (1) by 2, we get 

4x + 2y = 14 ... (3) 

Now subtracting equation (2) from (3), we get 

4x + 2y – (4x – 3y) = 14 – (–1) 

⇒ 5y = 15 ⇒ y = 3. 

Now, putting y = 3 in equation (1), we get 

2x + 3 = 7 

⇒ 2x = 7 – 3 = 4 

⇒ x = 4/2 = 2. 

Hence, the solution of given system of equations is x = 2 and y = 3.

Calculation:

\(\frac{2\times 4 \times 8 \times 16}{(log_24)^2 (log_48)^2(log _816)^4}\)

⇒ \(\frac{{2{\rm{\;}} \times 4 \times 8 \times 16}}{{{{({\rm{lo}}{{\rm{g}}_2}4{\rm{\;}} \times {\rm{\;lo}}{{\rm{g}}_4}8{\rm{\;}} \times {\rm{lo}}{{\rm{g}}_8}16)}^2}{\rm{\;}} \times \left( {{\rm{lo}}{{\rm{g}}_4}8 \times {\rm{lo}}{{\rm{g}}_8}16} \right) \times {\rm{lo}}{{\rm{g}}_8}16}}\)

We know \(lo{g_{b\;}}a = \frac{{loga}}{{logb}}\)

⇒ \(\frac{{\left( {2 \times 4 \times 8 \times 16} \right)}}{{{{(\frac{{{\bf{log}}4}}{{{\bf{log}}2}} \times \frac{{{\bf{log}}8}}{{{\bf{log}}4}} \times \frac{{{\bf{log}}16}}{{{\bf{log}}8}})}^{2\;}} \times \left( {\frac{{{\bf{log}}8}}{{{\bf{log}}4}} \times \frac{{{\bf{log}}16}}{{{\bf{log}}8}}} \right) \times {\rm{lo}}{{\rm{g}}_8}16}}\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{(lo{g_2}16)}^2} \times \;{\rm{lo}}{{\rm{g}}_4}16 \times {\rm{lo}}{{\rm{g}}_8}16}}\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{(lo{g_2}{2^4})}^2} \times \;{\rm{lo}}{{\rm{g}}_4}{4^2} \times {\rm{lo}}{{\rm{g}}_8}\left( {8 \times 2} \right)}}\)

We know \(lo{g_{a\;}}a = 1\;\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{\left( 4 \right)}^2} \times 2 \times {\rm{lo}}{{\rm{g}}_8}\left( {8 \times 2} \right)}}\)

We know  \(\log \left( {m\; \times n} \right)\; = logm + logn\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{\left( 4 \right)}^2} \times 2 \times \left[ {{\rm{lo}}{{\rm{g}}_8}8 + {\rm{lo}}{{\rm{g}}_8}2} \right]}}\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{\left( 4 \right)}^2} \times 2 \times \left[ {1 + \frac{{{\rm{log}}2}}{{{\rm{log}}8}}} \right]}}\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{\left( 4 \right)}^2} \times 2 \times \left[ {1 + \frac{{{\rm{log}}2}}{{{\rm{log}}{2^3}}}} \right]}}\)

⇒ \(\frac{{2\; \times 4 \times 8 \times 16}}{{{{\left( 4 \right)}^2} \times 2 \times \left[ {1 + \frac{1}{3}} \right]}}\)

⇒ 24

∴ \(\frac{2\times 4 \times 8 \times 16}{(log_24)^2 (log_48)^2(log _816)^4}\) is 24.

\(\phi = \mathrm x^2+2\mathrm xy \) at (1, –1, 3)

directional vector \(\vec p = \hat i+2\hat j+2\hat k\)

We know that direction derivative \(= \nabla\phi.\hat p = \nabla \phi\frac{\vec p}{|\vec p|}\)

Now, \(\nabla \phi = \nabla(\mathrm x^2+2\mathrm xy)\)

\(=\hat i \frac{\partial}{\partial \mathrm x}(\mathrm x^2+2\mathrm xy)\) \(+\hat j\frac{\partial}{\partial \mathrm x}(\mathrm x^2+2\mathrm xy)\) \(+\hat k\frac{\partial}{\partial z}(\mathrm x^2+2\mathrm xy)\)

\(=(2\mathrm x+2y)\hat i + 2\mathrm x\hat j\)

\((\nabla \phi)_{(1, -1, 3)}=(2-2)\hat i + 2\hat j=2\hat j\)

And \(\hat p = \frac{\vec p}{| \vec p|}\) \(=\frac{\hat i+2\hat j+2\hat k}{\sqrt{1^2+2^2+2^2}}\) \(=\frac{1}{3}(\hat i+2\hat j+2\hat k)\)

Now, directional derivative of \(\phi\) at (1, –1, 3) in the direction of vector p

\(=(\nabla \phi)_{(1, -1, 3)}\hat p \) \(=2\hat j.\left(\frac{1}{3}(\hat i+2\hat j+2\hat k)\right)\)

\(=\frac{4}{3}\)    \(\Big(\hat i.\hat j=\hat j.\hat k = 0\) but \(\hat j.\hat j =1\Big)\)

 Intersection point of y = 0 with first line is the point B(-p, 0)          

Intersection point of y = 0 with second line is the point A(-q, 0)          

Intersection point of the two lines is the point C(pq, (p + 1)(q + 1))         

 Altitude from C to AB is x = pq          

Altitude from B to AC is y = -q/1+q (x + p)          

On solving these two we get x = pq and y = -pq         

 => locus of orthocenter is x + y = 0.

Correct option is C) reveal

• Satyashodhak Samaj started by Jyotiba Phule opposed social evils and domination of priests.
• Started educational institutions for backward classes.

• To bring about changes in society. 
• To maintain unity.

Correct option is C) alimony

Correct option is C) adequate

The second language classroom is a confluence of varied languages. Teachers should give their students comfortable environments to develop requisite skills. In Krashen’s words, children require ‘comprehensible input’ in the second language. ‘Comprehensible input’ refers to using language which children are capable of understanding, and at the same time holds challenges for them.

• An important part of making this language comprehensible is providing it in natural, communicative situations that are meaningful to children, and this will help children in meeting the challenge. For example, if children in your class know some words in English, then ‘comprehensible input’ might mean using these words in sentences that are meaningful for them.
• ‘Comprehensible input’ and a ‘natural and communication friendly environment’ play an important role in acquiring a second language.
• In the second language, the teacher should provide a comfortable environment so that students can express themselves and learn the requisite skills such as listening, speaking, reading, and writing holistically.
• Adequate self-explanatory notes, summaries, and simplified versions of texts and Worksheets with a variety of tasks while covering the syllabus focused on writing and reading skills only in a fragmented manner. As the second language classroom is a confluence of varied languages, these do not help students to relate and express themselves.
• In the second language classroom, learning from authentic material and resources is a must.

Hence, we can conclude that The second language classroom is a confluence of varied languages. Teachers should give their students Comfortable environments to develop requisite skills.

The teacher’s cues for activities are given in the first language, in a second language class. This bilingual technique exploits the communicative potential of a given structure.

• The students’ L1 by incorporating it into the lesson. It does this by using bilingual technique all or part of the L1 within L2.
• C.J. Dodson (1967) was the proponent of the bilingual method. It is used for teaching a foreign language, and it is complementary to the audiovisual method. From the beginning, the sandwich approach is used to convey the meaning bilingually. The mother tongue is used for bilingual pattern drills.
• The bilingual technique exploits the communicative potential of a given structure as Learners may rely more on their L1, and it can turn out to be a translation method. The student, unable to use the skills in target language, hence lose the motivation and communicative potential of a given structure.
• Sandwich approach: this approach involves the use of a positive comment, followed by a negative, then closed off with another positive comment for constructive feedback.
• The structural technique is a technique wherein the learner masters the pattern of a sentence. It is based on the assumptions that language can be best learned through a scientific selection and grading of the structures or patterns of sentences and vocabulary.
• Communicative approach:. The communicative approach is based on the idea that learning a language successfully comes through having to communicate real meaning. This approach focus is on meaningful interaction rather than on making complete sentences and cues in the first language. It focuses on fluency but not accuracy.

Hence, we can conclude that the teacher’s cues for activities are given in the first language, in a second language class. This Bilingual technique exploits the communicative potential of a given structure.