Engineering Mathematics Test 2

\(\mathop \smallint \limits_0^{2a} f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}}{2\mathop \smallint \limits_0^a f\left( x \right)dx,\;\;f\left( {2a - x} \right) = f\left( x \right)}\\{0,\;\;f\left( {2a - x} \right) = - f\left( x \right)}\end{array}} \right.\)

From the property of integration as mentioned above, the given integral can be reduced to

\( = 2\mathop \smallint \limits_0^\pi \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}{{\sin }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9{{\sec }^2}\theta + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9\left( {1 + {{\tan }^2}\theta } \right) + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9 + 10{{\tan }^2}\theta }}} \right)d\theta \)

Put tan θ = t

⇒ sec²θ dθ = dt

\( = 12\mathop \smallint \limits_0^\infty \left( {\frac{1}{{9 + 10{t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}}\mathop \smallint \limits_0^\infty \left( {\frac{1}{{\frac{9}{{10}} + {t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}} \times \frac{1}{{\frac{3}{{\sqrt {10} }}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{t}{{\frac{3}{{\sqrt {10} }}}}} \right)} \right]_0^\infty \)

\( = \frac{4}{{\sqrt {10} }}\left[ {\frac{\pi }{2} - 0} \right] = \frac{{2\pi }}{{\sqrt {10} }}\)

Concept:

The determinant of a matrix is the product of its eigenvalues.

The determinant of a matrix is same as its transpose.

The determinant of a matrix is reciprocal to its inverse.

Calculation:

Eigenvalues of matrix A are 4, 16 and 64.

Determinant of matrix A = 4 × 16 × 64 = 4096

Determinant of inverse of A = det (A-1) = \(\frac{1}{{{\rm{det}}\left( A \right)}} = \;\frac{1}{4086} \)

det (A-1) = det (A-1)T = \(\frac{1}{4086}\)

2000 ×(M-1)T = 0.488

Data:

No. of Students in class = 30 students

30 students have to sit on 40 seats

Calculation:

Total seats = (5 row) × (8 seats/row) = 40 seats

Total no. of ways = ⁴⁰C₃₀

For 6^th seat of 5^th row to be empty, all 30 students have to sit on remaining 39 seats.

No. of favourable ways = ³⁹C₃₀

Probability (sixth seat in the fifth row will be empty) =

\(= \frac{{{\rm{NO}}.{\rm{\;of\;favourable\;ways}}}}{{{\rm{Total\;no}}.{\rm{\;of\;ways}}}}\)

= ³⁹C₃₀ ÷ ⁴⁰C₃₀

= ¼ 

Concept:

Expected length E(X) = ∑ (X × P(X))

where X is random variable denoting length of word drawn

Explanation:

Each of the nine words has equal probability.

\(\frac{1}{8}\left( {2 + 3 + 5 + 4 + 8 + 6 + 4 + 5 } \right) = \frac{{37}}{8} = 4.625\)

Hence 4.62 is the expected length.

Given:  f(x) = x + ln x, a = 1, b = e

f’(x) = 1 + \(\frac{1}{{{\rm{x\;}}}}\)

Lagrange’s Mean Value Theorem: If a function f(x) is

• Continuous in closed interval [a, b] and
• Differentiable in open interval (a, b), then

There exist at least one c value of x lying in the open interval (a, b) such that f’(c) = \(\frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( a \right)}}{{{\rm{b}} - {\rm{a}}}}\)

f’(c) = \(\frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( a \right)}}{{{\rm{b}} - {\rm{a}}}}\; = \;\frac{{{\rm{f}}\left( {\rm{e}} \right) - {\rm{f}}\left( 1 \right)}}{{{\rm{e}} - 1}}\)

f’(c) = \(\frac{{{\rm{e\;}} + {\rm{\;}}ln{\rm{\;e}} - (1 + \ln 1){\rm{\;}}}}{{{\rm{e}} - 1}}\) 

= \(\frac{{{\rm{e\;}} + {\rm{\;}}1 - 1 + 0{\rm{\;}}}}{{{\rm{e}} - 1}}\) 

= \(\frac{{{\rm{e\;}}}}{{{\rm{e}} - 1}}\)

f’(c) = \(\frac{{{\rm{e\;}}}}{{{\rm{e}} - 1}}\)

1 + \(\frac{1}{{\rm{c}}}\) = \(\frac{{{\rm{e\;}}}}{{{\rm{e}} - 1}}\)

\(\frac{1}{{\rm{c}}}\) = \(\frac{{{\rm{e\;}}}}{{{\rm{e}} - 1}} - 1\)

\(\frac{1}{{\rm{c}}}\) = \(\frac{{{\rm{e}} - {\rm{e\;}} + 1{\rm{\;}}}}{{{\rm{e}} - 1}}\)

c = e – 1

Concept:

Expected length E(X) = ∑ (X × P(X))

where X is random variable denoting length of word drawn

Explanation:

Each of the nine words has equal probability.

\(\frac{1}{9}\left( {3 + 5 + 5 + 3 + 5 + 4 + 3 + 4 + 3} \right) = \frac{{35}}{9} = 3.9\) 

Hence 3.9 is the expected length.

Since given system of equations is homogeneous and has infinitely many solutions.

Therefore, given matrix is consistent and determinant of the coefficient matrix should be equal to 0

\(\left| {\begin{array}{*{20}{c}} 2&3&1\\ 4&6&6\\ 8&{12a}&2 \end{array}} \right| = 0\)

\(2 \times \left( {12\; - 72a} \right) - 3 \times \left( {8 - 48} \right) + 1\left( {48a - 48} \right) = 0\)

24 – 144a – 24 + 144 + 48a – 48 = 0

- 96a + 96 = 0

Since y is a random variable:

\(\mathop \sum \limits_{i = 0}^7 p\left( {{y_i}} \right) = 1\)

\(= 0 + m + m + 3m + 3m + {m^2} + 4{m^2} + 5{m^2} + m = 1\)

\(10{m^2} + 9m - \;1 = 0\)

\(\left( {m + 1} \right)\left( {10m - \;1} \right) = 0\)

\(\therefore m = - 1\;\;or\;m = \frac{1}{{10}}\)

since probability cannot be in negative

\(\therefore \;m \ne \; - 1\)

\(p\left( {y \ge 5} \right) = p\left( 5 \right) + p\left( 6 \right) + p\left( 7 \right) \)

\(\therefore p\left( {y \ge 5} \right)\; = \;{m^2} + 4{m^2} + 5{m^2} + m\;\;\)

\(put\;m = \frac{1}{{10}}\)

\(\therefore p\left( {y \ge 5} \right) = \frac{1}{5}\)

P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.7

\(P\left( {A{\rm{|}}B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{1}{3}\)

\(P\left( {B{\rm{|}}A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = \frac{1}{5}\)

\(P\left( {A{\rm{|}}A \cup B} \right) = \frac{{P\left( A \right)}}{{{\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right)}} = \frac{5}{7}\)

P(A|A ∩ B) = P(A ∩ B)/P(A ∩ B) = 1

\(P\left( {A \cap B{\rm{|}}A \cup B} \right) = \frac{{P\left( {A \cap B} \right)}}{{{\rm{P}}({\rm{A}} \cup {\rm{B}}}} = \frac{1}{7}\)

Concept:

In case of Poisson distribution, mean and variance are same.

Calculation:

Given, mean = 5

E[(X + 2)²] = E [X² + 4X + 4] = E[X²] + E[4X] + E[4]

Variance = E[X²] - (E[X])²

As, mean = variance = 5

Mean = E[X]

5 = E[X²] -  (5)²

5 = E[X²] - 25

E[X²] = 30

A = \(\left[ {\begin{array}{*{20}{c}} 5\\ 4 \end{array}\begin{array}{*{20}{c}} { - 1}\\ 1 \end{array}} \right]\)

Characteristic Equation: |A – λ I | = 0

So, \(\left| {\begin{array}{*{20}{c}} {5 - \;{\rm{\lambda }}}&{ - 1}\\ 4&{1 - \;{\rm{\lambda }}} \end{array}} \right|\) = 0

5 - 5λ – λ + λ2 + 4 = 0

λ2 - 6 λ + 9 = 0

λ = 3, 3

Product of all the eigen values is = 3 × 3 = 9

An algebraic multiplicity of eigenvalue 3 is 2.

Also, it has only one independent eigenvector that exists.

Therefore options 2 and 3 are not TRUE (FALSE).

Given:

μ₁ = μ₂ =1, σ₁ = 1, σ₂ = 2

Calculation:  X₁, X₂ are two independent random variables

Y = X₁ - X₂

μ(Y) = μ(X₁ - X₂)

= μ(X₁) - μ(X₂)

= μ₁ - μ₂ = 1 – 1 = 0

Hence, Mean = 0

Since X₁, X₂ are independent variables, Var(Y) is

Var(Y) = Var(X₁) – Var(X₂)

= σ₁² + σ₂² = 1 + 4 = 5

Hence, Variance = 5

\(let\;L\; = \;\mathop {{\rm{lim}}}\limits_{x \to 0\;} \frac{{{{({p^x} - 1)}^3}}}{{({q^x} - 1).sin2x.\;{\rm{log}}\left( {1\; + \;x} \right)}}\)

Since x → 0 ∴ x³ → 0

Divide number and denominator by x³

\(L\; = \;\mathop {{\rm{lim}}}\limits_{x \to 0\;} \frac{{\frac{{{{\left( {{p^x} - 1} \right)}^3}}}{{{x^3}}}}}{{\;\frac{{({q^x} - 1).sin2x.\;{\rm{log}}\left( {1\; + \;x} \right)}}{{{x^3}}}\;}}\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{a^x} - 1}}{x}\; = \;lo{g_e}a\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{sinx}}{x}\; = \;1\)

\(\therefore \;\mathop {{\rm{lim}}}\limits_{2x \to 0} \frac{{2sin2x}}{{2x}}\; = \;2\;\left( {\because as\;x \to 0\;then\;2x \to 0\;} \right)\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1\; + \;logx}}{x} = 1\)

\(L = \frac{{{{(\log p)}^3}\;}}{{2logq}}\)

Confusion points:

(log(a))² ≠ 2log(a)

log(a²) = 2log(a)

f(x) = 3x3 – 40.5x2 + 180x + 7

f’(x) = 9x2 – 81x + 180 = 0

put f’(x) = 0 to find the point at which maximum and minimum value exists

9x2 – 81x + 180 = 0

x2 – 9x + 20 = 0

(x – 5)(x – 4 ) = 0

∴ x = 5 or x =4

Interval [4, 5]

f'’(x) = 2x – 9

put x = 4

f’’(x) = -1 < 0 (maximum value might exist)

put x = 5

f’’(x) = 1 > 0 (minimum value might exist)

Also check border values:

The minimum value is 269.5

Characteristic equation of given matrix

\(\left| {\begin{array}{*{20}{c}}{1 - \lambda }&0&0\\1&{ - \;\lambda }&1\\0&1&{ - \;\lambda }\end{array}} \right| = 0\)

(1 – λ) × (λ² – 1) = 0

λ² – 1 – λ³ + λ = 0

λ³ = λ² + λ – 1

Every matrix satisfies its own characteristic equation.

A³ = A² + A – I

Where I is identity matrix

Multiplying by A on both sides

A⁴ = A³ + A² – A = A² + A – I + A² – A

∴ A⁴ = 2A² – I

Multiplying by A² on both sides

A⁶ = 2A⁴ – A²

A⁶ = 2(2A² – I) – A² = 4A² – 2I – A²

∴ A⁶ = 3A² – 2I

Multiplying by A² on both sides

A⁸ = 3A⁴ – 2A² = 3(2A² – I) – 2A²

A⁸ = 4A² – 3I

By mathematical induction

A²ⁿ = nA² – (n – 1)I

∴ A¹⁰² = 51A² – 50I

\({A^2} = \;\left[ {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right] \times \;\left[ {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right]\;\)

\({A^2} = \;\left[ {\begin{array}{*{20}{c}}1&0&0\\1&1&0\\1&0&1\end{array}} \right]\)

A¹⁰² = 51A² – 50I

\({A^{102}} = 51 \times \left[ {\begin{array}{*{20}{c}}1&0&0\\1&1&0\\1&0&1\end{array}} \right] - 50\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\)

\({A^{102}} = \;\left[ {\begin{array}{*{20}{c}}1&0&0\\{51}&1&0\\{51}&0&1\end{array}} \right]\;\)

Option 1 is correct.

Calculation:

Given matrix → \(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&0&1&{ - 1}&0\\ 0&1&{ - 1}&0&0\\ { - 1}&0&0&0&1\\ 0&0&0&1&{ - 1} \end{array}} \right]\) 

Apply now transformation, R₄ → R₄ + R₁, we get,

\(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&0&1&{ - 1}&0\\ 0&1&{ - 1}&0&0\\ 0&{ - 1}&0&0&1\\ 0&0&0&1&{ - 1} \end{array}} \right]\)

R₂ ↔ R₃

\(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&1&{ - 1}&0&0\\ 0&0&1&{ - 1}&0\\ 0&{ - 1}&0&0&1\\ 0&0&0&1&{ - 1} \end{array}} \right]\)

R₄ → R₄ + R₂

A = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&1&{ - 1}&0&0\\ 0&0&1&{ - 1}&0\\ 0&0&{ - 1}&0&1\\ 0&0&0&1&{ - 1} \end{array}} \right]\)

R₄ → R₄ + R₃

\(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&1&{ - 1}&0&0\\ 0&0&1&{ - 1}&0\\ 0&0&0&{ - 1}&1\\ 0&0&0&1&{ - 1} \end{array}} \right]\)

R₅ → R₅ + R₄

\(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0&0&0\\ 0&1&{ - 1}&0&0\\ 0&0&1&{ - 1}&0\\ 0&0&0&{ - 1}&1\\ 0&0&0&0&0 \end{array}} \right]\) → This matrix is in triangular form with last row = 0,

So, Rank = 4

Let D is the event that the coin is the one with two heads.

Let H is the event that we get a head when we toss the coin.

The required probability is P(D|H)

By using Bayes theorem, \(P\left( {D{\rm{|}}H} \right) = \frac{{P\left( {H{\rm{|}}D} \right)P\left( D \right)}}{{P\left( H \right)}}\)

P(H|D) = 1

P(D) = 1/10 = 0.1

P(H) = P(H ∩ D) + P(H ∩ D^C)

= P(H|D) P(D) + P(H|D^C) P(D^C)

= (1) (0.1) + (0.5) (0.9) = 11/20

\(P\left( {D{\rm{|}}H} \right) = \frac{{\frac{1}{{10}}}}{{\frac{{11}}{{20}}}} = \frac{2}{{11}}\)

\(f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = {x^2}\)

\(f\left( {\frac{1}{x}} \right) + 2f\left( x \right) = \frac{1}{{{x^2}}}\)

By solving the above two equations,

\(2f\left( {\frac{1}{x}} \right) + 4\;f\left( x \right) - f\left( x \right) - 2f\left( {\frac{1}{x}} \right) = \frac{2}{{{x^2}}} - {x^2}\)

\(\Rightarrow 3f\left( x \right) = \frac{2}{{{x^2}}} - {x^2}\)

\(\Rightarrow {x^2}f\left( x \right) = \frac{1}{3}\left( {1 - {x^4}} \right)\)

\(\mathop \smallint \limits_1^2 {x^2}f\left( x \right)dx = \mathop \smallint \limits_1^2 \frac{1}{3}\left( {2 - {x^4}} \right)dx\)

\( = \frac{1}{3}\left[ {2x - \frac{{{x^5}}}{5}} \right]_1^2\)

\(= \frac{1}{3}\left[ {\left( {2 - 1} \right) - \frac{1}{5}\left( {32 - 1} \right)} \right] = \frac{{ - 7}}{5}\)

Concept:

A square matrix ‘A’ is said to be orthogonal if it follows the following condition.

AAT = ATA = I

Calculation:

\(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\)

\({M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right]\)

\(M{M^T} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ a&b \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&a\\ {\frac{1}{{\sqrt 2 }}}&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right]\)

For orthogonal matrix, MM^T = I

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}\\ {\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }}}&{{a^2} + {b^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

By comparing both the sides, we get

\(\frac{a}{{\sqrt 2 }} + \frac{b}{{\sqrt 2 }} = 0 \Rightarrow a + b = 0\)

a² + b² = 1

\(\Rightarrow b = \pm \sqrt {1 - {a^2}}\)

Therefore, \(b = \sqrt {1 - {a^2}}\) is not always true.

⇒ (a + b)² – 2ab = 1

\(\Rightarrow ab = - \frac{1}{2}\)

By using the relation, a + b = 0 i.e. b = -a, we get

\( \Rightarrow a\left( { - a} \right) = - \frac{1}{2} \Rightarrow a = \pm \frac{1}{{\sqrt 2 }}\)

For \(a = \frac{1}{{\sqrt 2 }},b = - \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)

\({M^2} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{1}{{\sqrt 2 }}}&{\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {I_2}\)

For \(a = \frac{{ - 1}}{{\sqrt 2 }},b = \frac{1}{{\sqrt 2 }}\), \(M = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\({M^2} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&0 \end{array}} \right] \ne {I_2}\)

Therefore, M² = I₂ is not always true.

Given:

\({x_1} + 2{x_2} = {b_1}\)     …1)

\(2{x_1} + 4{x_2} = {b_2}\)     …2)

\(3{x_1} + 7{x_2} = {b_3}\)     …3)

\(3{x_1} + 9{x_2} = {b_4}\)     …4)

From equations (1) and (2), it is clear that:

\({b_2} = 2{b_1}\)

So options (2) and (4) are incorrect.

Evaluating option (3) now, we are given:

3b₁ – 6b₃ + b ₄ = 0

Using Equation 1, 3 and 4, we can write:

\(3({x_1} + 2{x_2}) - 6\left( {3{x_1} + 7{x_2}} \right) + \;3{x_1} + 9{x_2} \ne 0\) 

Hence option 3 is also incorrect.

Analysing option 1, we are given:

6b₁ – 3b₃ + b₄ = 0

Put the value from equation 1, 3 and 4

\(6({x_1} + 2{x_2}) - 3\left( {3{x_1} + 7{x_2}} \right) + \;3{x_1} + 9{x_2} = 0\)

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \;f{\left( x \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right)\; - \;1} \right)}}\;if\;it\;is\;of\;the\;form\;{1^\infty }\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;{\frac{{\left( {{a^x} + {b^x} + {c^x} + {d^x}\;} \right)}}{4}^{\frac{1}{x}}}\) 

It is of the form 1^∞

\(= {e^{\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}} \right)\left( {\frac{{{a^x} + {b^x} + {c^x} + {d^x}}}{4} - \;1} \right)}}\)

\(= {e^{\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{4}\frac{{({a^x}\; - \;1) + ({b^x} - 1) + \left( {{c^x} - 1} \right)\; + ({d^x} - 1)\;}}{x}} \right)}}\)

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = lo{g_e}a\)

\(= {e^{\left( {\frac{{loga + logb + logc + logd}}{4}\;} \right)}}\)

\(= {e^{\frac{{{{\log }_{\rm{e}}}\left( {abcd} \right)}}{4}\;}}\)

\(= {e^{{{\log }_{\rm{e}}}{{\left( {abcd} \right)}^{\frac{1}{4}}}\;}}\)

\(= {\left( {abcd} \right)^{\frac{1}{4}}}\)

\(= \sqrt[4]{{abcd}}\)

From diagonalization,

\({{\rm{P}}^{ - 1}}{\rm{AP}} = \left( {\begin{array}{*{20}{c}} 3&0&0\\ 0&3&0\\ 0&0&{ - 3} \end{array}} \right)\)

Comparing with

\({{\rm{P}}^{ - 1}}{\rm{AP}} = \left( {\begin{array}{*{20}{c}} {{\rm{\lambda }}1}&0&0\\ 0&{{\rm{\lambda }}2}&0\\ 0&0&{{\rm{\lambda }}3} \end{array}} \right)\)

where λ1, λ2, and λ3 is an eigenvalue of A

∴ the eigenvalue of matrix A is 3, 3, -3

Using the property of eigenvalue:

The determinant of matrix A = product of eigenvalues

\(\left| {\begin{array}{*{20}{c}} 3&0&0\\ 0&2&{ - 5}\\ 0&{\rm{\alpha }}&{ - 2} \end{array}} \right| = 3 \times 3 \times - 3\)

∴ 3(-4 + 5α) = -27

Concept –

Following steps to solve the equation \(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = k\pi ,\)

1. To remove the modulus
2. To keep sin x positive in the interval 0 to π to 2π and to keep the sinx negative in the interval

• This is because x in the above equation is always positive but the value sinx changes in the two mentioned intervals.

Explanation –

Solving the equation as per the steps,

\(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = \mathop \smallint \nolimits_0^\pi xsinxdx + \left( { - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx} \right)\)

\(= \mathop \smallint \nolimits_0^\pi xsinxdx - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\)

Keeping u = x, du = dx, dv = sinxdx, so v = - cosx,

\(= \mathop \smallint \nolimits_0^\pi udv = uv - \mathop \smallint \nolimits_0^\pi vdu\)

\( = \mathop \smallint \nolimits_0^\pi xsinxdx = \left[ { - xcosx} \right]_0^\pi + \mathop \smallint \nolimits_0^\pi cosxdx\;\)

\( = \pi + \left[ {sinx} \right]_0^\pi \)

Now, repeating the same with \(- \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\), we get -3

Hence, π - (-3π) = 4π

Therefore, k = 4

Probability density function

\(\mathop \smallint \limits_{ - \infty \;}^\infty f\left( x \right)dx = 1\)

\(\mathop \smallint \limits_{ - \infty \;}^0 f\left( x \right)dx + \mathop \smallint \limits_0^3 f\left( x \right)dx + \mathop \smallint \limits_3^\infty f\left( x \right)dx = 1\)

\(0 + \mathop \smallint \limits_0^3 \left( {\alpha x + \beta {x^2} - 1} \right)dx + 0 = 1\)

\(\left( {\frac{{a{x^2}}}{2} + \frac{{\beta {x^3}}}{3} - x} \right)_0^3 = 1\)

\(9\alpha + 18\beta = \;8\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\)

\(E\left( X \right) = \;\mathop \smallint \limits_{0\;}^3 xf\left( x \right)dx = \frac{5}{4}\)

\(\mathop \smallint \limits_{0\;}^3 x\left( {\alpha x + \beta {x^2} - 1} \right)dx = \frac{5}{4}\)

\(\left( {\frac{{a{x^3}}}{3} + \frac{{\beta {x^4}}}{4} - \frac{{{x^2}}}{2}} \right)_0^3 = \frac{5}{4}\)

\(36\alpha + 81\beta = \;23\;\;\;\;\;\;\;\;\;\left( 2 \right)\)

Solving (1) and (2) we get

\(\therefore \alpha = \; \frac{{26}}{9}\;and\;\beta = {-1}\)

Concept:

If the matrix is symmetric then the product of transpose of one eigenvector to the other eigenvector should be zero. 

Calculation:

The eigenvectors of a symmetric matrix A corresponding to different eigenvalues are orthogonal to each other.

The given matrix,

\(A = \left[ {\begin{array}{*{20}{c}} {\frac{3}{2}}&0&{\frac{1}{2}}\\ 0&{ - 1}&0\\ {\frac{1}{2}}&0&{\frac{3}{2}} \end{array}} \right]\)

A^T = A

Hence A is a symmetric matrix.

\({\left[ {\begin{array}{*{20}{c}} { 1}\\ 0 \\1 \end{array}} \right]^T} = \;\left[ {\begin{array}{*{20}{c}} { 1}&0&1 \end{array}} \right]\)

Option 3:

\(\left[ {\begin{array}{*{20}{c}} { 1}&0 &1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {1}\\ 0 \\-1\end{array}} \right] = \left[ {1 + 0 -1} \right] = \left[ 0 \right]\;\)

Therefore \(\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right]\)can be another Eigenvector of A.