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Concept:

 A sequence {a_n} is said to be 

• increasing if m < n implies a_m ≤ a_n
• strictly increasing if m < n implies am < an 
• decreasing if m < n implies  a_m ≥ a_n
• strictly decreasing if m < n implies am > an 
• monotone if it is either increasing or decreasing

Calculation:

 Let us look at the first few terms of this sequence. These are 

3, 9/2, 9/2, 27/8, 81/80, 243/560, ....

You can see that the first 5 terms of this sequence are in increasing order However, the 6^th term is less than the 5^th term. Hence, the sequence is neither increasing nor decreasing. That is the sequence is not monotone.

Concept:

A sequence {a_n} is said to be bounded below if there exists a number l such that l ≤ a_n ∀ n ≤ N. Such a number l is called a lower bound of {an}. 

• A sequence {an} s said to be bounded above if there exists a number u such that a_n ≤ u,  ∀ n ≤ N such a number u is called an upper bound of {an}. 
• When a sequence {a_n}, n∈N a is both bounded below and bounded above, it is called a bounded sequence.

Calculation:

First, let us check the boundedness of the sequence from below. Note that ∀n∈N

\((1+\frac{1}{n})>1\) ⇒ \((1+\frac{1}{n})^\frac{1}{n}>1\)

Hence the sequence is bounded below. Now we check the boundedness of the sequence from above. You can see that for all n ∈ N

\(\frac{1}{n}≤1\)

⇒ \(1+\frac{1}{n}≤2≤2^n \) 

⇒ \((1+\frac{1}{n})^\frac{1}{n}≤2\)

Thus the sequence is bounded above also. Hence the sequence is bounded.

For equilibrium:

\(\dfrac{P}{\sin p} =\dfrac{Q}{\sin q} = \dfrac{R}{\sin r}\)

Where,

p = Angle between Q and R = 120° 

r = Angle between P and Q = 150° 

q = Angle between P & R = 360° - (120° + 150°)

= 90° 

hence,

P ∶ Q ∶ R = sin p ∶ sin q ∶ sin r

= sin 120° ∶ sin 90° ∶ sin 150° 

\(=\dfrac{\sqrt{3}}{2} : \dfrac{1}{2} : \dfrac{1}{1}\)

\(\therefore \sqrt{3} : 1 : 2\)

Concept:

Cauchy’s Root test-

A series of positive terms ∑unis

(i) convergent if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} < 1\)

(ii) Divergent if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} > 1\)

(iii) Test fail if \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} = 1\)

Observation

A necessary condition for convergence of an infinite series \(\sum {{u_n}} \) is that \(\mathop {\lim }\limits_{n \to \infty } {u_n}< 1\)

None of the options are correct. Hence, option 4 is the correct answer.

Concept:

Derived Set- The set of all limit points of a set S is called the derived set of S and is denoted by S'.

Closed Set- A set is said to be closed if each of its limit points is a member of the set.

Dense Set- A subset A of the set of reals R is said to be dense (or dense in R or everywhere dense) if every point of R is a point of A or a limit point of A or equivalently if the closure of A is R. 

We have \({\rm{\;f\;}}\left( {\rm{x}} \right) = {{\rm{x}}^2} - {\rm{x}} - 2\)

\(\begin{array}{l} {{\rm{f}}^{'}}\left( {\rm{x}} \right) = 2{\rm{x}} - 1 = 0 \to {\rm{x}} = \frac{1}{2}\\ {{\rm{f}}^{''}}\left( {\rm{x}} \right) = 2 \end{array}\)

Since\({\rm{f'}}\left( {\rm{x}} \right) = 2 > 0\), thus \({\rm{x}} = \frac{1}{2}\) is minimum point. The maximum value in closed interval [-4, 4] will be at x = -4 or x = 4

Now maximum value \( = {\rm{max}}\left[ {{\rm{f\;}}\left( { - {\rm{\;}}4} \right),{\rm{\;f\;}}\left( 4 \right)} \right] = {\rm{max}}\left( {18,10} \right) = 18\)

Concept:

Continuity at an endpoint-

A function f defined on a closed interval [a, b] is said to be continuous at the endpoint 'a' if it is continuous from the right at a, i.e,

⇒ \(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaqa % aaaaaaaaWdbiGacYgacaGGPbGaaiyBaaWcpaqaa8qacaWG4bGaeyOK % H4QaamyyaiabgUcaRiaaicdaa8aabeaak8qacaWGMbWaaeWaa8aaba % WdbiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWGMbWaaeWaa8aabaWd % biaadggaaiaawIcacaGLPaaaaaa!46E3! \mathop {\lim }\limits_{x \to a + 0} f\left( x \right) = f\left( a \right)\)

Also, the function is continuous at the endpoint b of [a, b] if,

⇒ \(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaqa % aaaaaaaaWdbiGacYgacaGGPbGaaiyBaaWcpaqaa8qacaWG4bGaeyOK % H4QaamOyaiabgkHiTiaaicdaa8aabeaak8qacaWGMbWaaeWaa8aaba % WdbiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWGMbWaaeWaa8aabaWd % biaadkgaaiaawIcacaGLPaaaaaa!46F0! \mathop {\lim }\limits_{x \to b - 0} f\left( x \right) = f\left( b \right)\)

Concept:

Bolzano-Weierstrass Theorem: Every bounded sequence has a convergent subsequence.

Bolzano-Weierstrass Theorem (for a set): Every infinite bounded set has a limit point.

Concept:

Cauchy Sequence-

A sequence (s_n) of real numbers is called a Cauchy sequence or a fundamental sequence. if, for each ε > 0, there exist a natural number m such that 

⇒ |sn - sk| < ε,  for ,all n > k > m. 

•  Every Cauchy sequence is bounded.  
• Every Cauchy sequence is convergent. 

Concept:

Bounded Sequence-

A sequence S: N ---> R is said to be bounded if its range is bounded in R. In other words, a sequence ( s_n ) is bounded if there exists a number K > 0 such that 

|Sn| < K, for n = 1, 2, 3, ....

Bolzano Weierstrass Theorem:

The theorem says that there is a limit point for a bounded infinite set S.

Let Ø ≠ S ⊆ R and S be bounded and infinite. Then there exists a limit point of S (in R ). 

Another Statement- Every sequence has a monotone subsequence.

Concept:

Open Interval

• Let a, b ∈ R and a < b. Then the set of real numbers { y : a < y < b} is called an open interval and is denoted by (a, b). All the points between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval.
• Any open interval is an open set.
• Both R and the empty set are open.
• The union of open sets is an open set.

Closed Interval

• The interval which contains the endpoints also is called a closed interval and is denoted by [ a, b ]. Thus [ a, b ] = {x : a ≤ x ≤ b}

We can also have intervals closed at one end and open at the other, i.e.,

• [ a, b ) = {x : a ≤ x < b} is an open interval from a to b, including a but excluding b.
• ( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding a.

Formula used:

a^m × aⁿ = a^{(m + n)}

Calculation:

\(\frac{10^{22}+10^{20}}{10^{20}}\)

⇒ \(\frac{10^{20} × 10^2+10^{20}}{10^{20}}\)

⇒ \(\frac{10^{20} ( 10^2+1)}{10^{20}}\)

⇒ 100 + 1 = 101

∴ The value of \(\frac{10^{22}+10^{20}}{10^{20}}\) is 101.

Concept:

The direction cosines of a vector are the cosines of the angles between the vector and the three +ve coordinate axes i.e. X, Y, Z axes.

 l2 + m2 + n2 = 1

where l = the cosines of the angles between the vector and  X-axis.

m = the cosines of the angles between the vector and  Y-axis.

n = the cosines of the angles between the vector and  Z-axis.

Calculation:

Let the line makes an angle γ with the positive direction of Z-axis. Thus it makes angle 30°, 60° and γ with the three axes.

∴ the d.c's of line are cos 30°, cos 60°, cos γ i. e. \(\dfrac{\sqrt{3}}{2}\), \(\dfrac{1}{2}\), cos γ

we know that, l² + m² + n² = 1

∴ \({\left( {\sqrt 3 /2} \right)^2} + {\left( {1/2} \right)^2} + {\left( {\cos γ } \right)^2} = 1\)

or cos² γ = 1 - 1

⇒ cos² γ = 0

⇒ cos γ = 0

or, γ = 90°

Thus the line makes an angle of 90° with the Z-axis.

Calculation:

Area of the triangle formed by given points

= \(\frac{1}{2}\) |1(1-11) + x(11-5) + 4(5-1)|

= \(\frac{1}{2}\)|6x + 6|

= 3 |x + 1|

Concept:

If two lines with slopes m₁ and m₂ make an angle θ with each other, then:

⇒ tan θ = \(\rm \left|\frac{m_1-m_2}{1+m_1m_2}\right|\).

Calculation:

The given lines are \(\sqrt{3}x +y=1\) and \(x+\sqrt{3}y=1\)

Let m₁ and m₂ be the slopes of these lines, then

\(m_1=-\frac{\sqrt{3}}{1}\)= \(-\sqrt{3}\)

and \(m_2= -\frac{1}{\sqrt{3}}\)

If θ is the acute angle between the lines the,

⇒ tan θ = |\(\frac{m_1-m_2}{1+m_1m_2}\)| = |\(\frac{-\sqrt{3}-(-\frac{1}{\sqrt{3}})}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})}\)| = |\(\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+1}\)| = |\(\frac{-3+1}{2\sqrt{3}}\)| = |\(\frac{-2}{2\sqrt{3}}\)| = \(\frac{1}{\sqrt{3}}\)

⇒ tan θ = 30^o

Concept:

If p is any real number such that p > 1, then the sequence \(\sqrt{3}, \sqrt{p\sqrt{p}},\sqrt{p\sqrt{p}\sqrt{p}},...\)converges to p.

Calculation:

Given sequence \(\sqrt{3}, \sqrt{3\sqrt{3}},\sqrt{3\sqrt{3}\sqrt{3}},...\)

from the above statement, we can say that the given sequence converges to 3.

Calculation:

We have, 

⇒ \(a_n = \frac{1.2.3...n}{n.n.n...n}\)

⇒ \(a_n=(\frac{1}{n})(\frac{2}{n})(\frac{3}{n})...(\frac{n}{n})<\frac{1}{n}\)

Thus, 0 < a_n < 1/n

Taking limit as n-->∞, we have

⇒ 0 ≤ \(\mathop {\lim }\limits_{n \to \infty } a_n\) ≤ \(\mathop {\lim }\limits_{n \to \infty } (\frac{1}{n})\) = 0

Therefore, \(\mathop {\lim }\limits_{n \to \infty } a_n\) = 0

Hence the sequence  converges to zero.

Concept:

The general equation of a line is y = mx + c   -----(A)

Where m is the slope and c is any constant.

• The slope of parallel lines is equal.
• Slope of the perpendicular line have their product = -1

Calculation:

The given lines are, p(p2 + 1)x - y + q = 0    ------(i)

and (p2 + 1)2x + (p2 + 1)y + 2q = 0               -------(ii)

Slope of the lines (i) is, \(\frac{-p(p^2+1)}{-1}\) i.e, p(p² + 1)

Similarly slope of the line (ii) is, \(-\frac{(p^2+1)^2}{p^2+1}\) i.e, -(p² +1)

Since the given lines are perpendicular to the same line, therefore, these lines are parallel which gives the lines have equal slope.

⇒ p(p2 + 1) = -(p2 +1)

⇒ (p² + 1) (p + 1) = 0

but p² + 1 ≠ 0

⇒ p + 1 = 0

⇒ p = -1

Concept:

Cauchy's General Principle of Convergence- A necessary and sufficient condition for the convergence of a sequence {S_n} is that, for each € > 0 there exists a positive integer m such that, 

⇒ | Sn+p - Sn | < €, ∀ n ≥ m