Mathematics Mock Test - 9

Given:

\(\tan x=-\frac{1}{\sqrt{3}}\)

Concept:

The principal solutions of a trigonometric equation are those solutions that lie between 0 and 2π.

Formula:

General solution of tan(x) = tan(α) is  given as;

x = nπ + α where α ∈ (-π/2 , π/2) and n ∈ Z.

Calculation:

∵ \(\tan x=-\frac{1}{\sqrt{3}}\)

⇒ tan(x) = tan(-π/6)

∴ α = -π/6

⇒ x = nπ + (-π/6) , n ∈ Z

Putting n = 1 and 2, we get - 

x = 5π/6 and 11π/6

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\matrix{ {0,\ x = 0} \cr {x-3,\ x>0} \cr } } \right.\)

• Since for x > 0, the value of function f(x) will verry as a line  (x -3) which is strictly increasing as the first derivative of (x -3) will be 1 which is positive for all the values of x > 0

∴ It is strictly increasing when x > 0.

Concept:

Multiplication of matrices:

• Let A be a matrix of m × n, B be a matrix of p × q then AB is a matrix of m × q, and for multiplication to exist n = p.
• Let A, B, and C be the matrices a, b, and c be scalars and the sizes of matrices are such that the operations can be performed then

Calculation:

Given:

Here, the order of  U → (1×3), X → (1×3), V → (3×1), and Y → (3×1) 

⇒ UV = [4]  and XY = [16]; ∴ UV + XY = [20].

Concept:

Degree: The degree of a differential equation is the power of the highest derivative.

Calculation:

We have, \(\rm \left(\frac {d^3y}{dx^3}\right)^{3/2} + \left(\frac {d^2y}{dx^2}\right)^{2} = 0\)

⇒ \(\rm \left(\frac {d^3y}{dx^3}\right)^{3/2} = - \left(\frac {d^2y}{dx^2}\right)^{2}\)

Squaring both the sides, we get 

\(\rm \left(\frac {d^3y}{dx^3}\right)^3 = \left(\frac {d^2y}{dx^2}\right)^{4}\)

Here highest derivative is \(\rm \left(\frac {d^3y}{dx^3}\right)^3\)

∴Degree = power of \(\rm \left(\frac {d^3y}{dx^3}\right)^3\)= 3

Hence, option (1) is correct. 

Calculation:

Let sin^-1 x = y. Then x = sin y

Since, \(- 1 \le x \le 0,\) 

therefore  \(\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0\) and so \(\frac{{ - \pi }}{2}\)\(\le y \le 0\)

We have,

cos y = \(\sqrt {1 - {{\sin }^2}y}\)

 \(\Rightarrow \cos y = \sqrt {1 - {x^2}} ,\) for \(0 \le y \le \pi\)       ....(i)

Now, \(- \frac{\pi }{2} \le y \le 0 \Rightarrow \frac{\pi }{2} \ge - y \ge 0\)      [From (i)]       

\(\Rightarrow \cos ( - y) = \sqrt {1 - {x^2}}\)

\(\Rightarrow - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}}\)

\( \Rightarrow y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} \)

Calculation:

Given:

f(x) = e^x and g(x) sin^-1 x and h(x) = f(g(x))

⇒ h(x) = f(sin^-1 x) = \( e^{sin^{-1}x}\)

∴ h(x) = \( e^{sin^{-1}x}\)

⇒ h'(x) = \( e^{sin^{-1}x}\frac{1}{{\sqrt {1 - {x^2}} }} \)

\(\Rightarrow \frac{{h'(x)}}{{h(x)}} = \frac{1}{{\sqrt {1 - {x^2}} }}.\)

Concept:

\(\int_a^b f(x)dx = f(a + b - x)dx\)

Calculation:

⇒ Let, I = \(\int \limits_0^\frac{π}{2} \frac{f(x)}{f(x) + f{(\frac{π}{2} - x)}}dx\)    ----- equation(1)

⇒ I = \(\int \limits_0^\frac{π}{2} \frac{f(\frac{π}{2}- x )}{f(\frac{π}{2} - x) + f{(\frac{π}{2} -(\frac{π}{2} - x))}}dx\)

⇒ I = \(\int \limits_0^\frac{π}{2} \frac{f(\frac{π}{2}- x )}{f(x) + f{(\frac{π}{2} - x)}}dx\)     ---- equation(2)

On adding equation (1) and (2)

⇒ 2I = \(\int \limits_0^\frac{π}{2} \frac{ f(x) + f(\frac{π}{2}- x )}{f(x) + f{(\frac{π}{2} - x)}}dx\)

⇒ 2I = \(\int \limits_0^\frac{π}{2} dx\)

⇒ 2I = \(\frac{π}{2}\)

⇒ I = \(\frac{π}{4}\)

∴ The value of \(\int_{\rm{0}}^{\frac{{\rm{\pi }}}{{\rm{2}}}} {\frac{{{\rm{f(x)}}}}{{{\rm{f(x) \,+\, f}}\left( {\frac{{\rm{\pi }}}{{\rm{2}}}\,{\rm{ - }}\,{\rm{x}}} \right)}}{\rm{dx}}} \) is \(\frac{π}{4}\)

Calculation:

\(A\, = \,\left[ {\begin{array}{*{20}{c}} 2&4&6\\ 0&2&6\\ 0&0&4 \end{array}} \right]\,\, = 2I\)

\(\therefore AB = 2IB = 2B = \left[ {\begin{array}{*{20}{c}} 2&4&6\\ 0&2&6\\ 0&0&4 \end{array}} \right]\)

\(\Rightarrow \left| {AB} \right|\, = \left[ {\begin{array}{*{20}{c}} 2&4&6\\ 0&2&6\\ 0&0&4 \end{array}} \right]\, = 2\,(8)\, = \,16\)

Alternate Method 

\(\begin{array}{l} \,\left| A \right|\, = \,2\, \times \,2\, \times 2\, = 8,\,\left| B \right|\, = \,1\, \times \,1\, \times 2\, = 2.\, \end{array}\)

\(\\ \therefore \left| {AB} \right| = \,\left| A \right|\left| B \right|\, = \,2\, \times \,8\, = \,16\)

Concept:

The statement of Lagrange's Mean Value Theorem is that -

• If a function is continuous in the closed interval [a , b],
• If a function is differentiable in the open interval (a , b)
• Then there exists some 'x = c' ϵ (a , b) such that

\( f'(c) =\frac{ [f(b) - f(a)]}{(b -a)}\)

• A function is non-differentiable at sharp turns.

Calculation:

Given: f(x) = |x| and Interval = [-1 , 1]

∘ f(x) = |x| is continuous in the closed interval [-1 , 1]

• From graph, we can see that f(x) = |x| is not differentiable at x = 0 due to sharp turn, 
• So, f(x) = |x| is not differentiable in the open interval (-1 , 1)
• Hence, condition for Lagrange's theorem fails.
• The value of c is non-existent.

CONCEPT:

• For global maxima and minima, we have to find the value of the function for the stationary points and endpoints.

CALCULATION:

Given:  \(\frac{{{d^2}y}}{{d{x^2}}}\) = 6x – 4

• Integrating both sides w.r.t. x

\(\Rightarrow \frac{{dy}}{{dx}}\) = 3x2 – 4x + c

At x = 1, \(\frac{{dy}}{{dx}}\) = 0

⇒ c = 1

⇒ \(\frac{{dy}}{{dx}}\) = 3x2 – 4x + 1 ......(1)

• Integrating both sides w.r.t. x,

⇒ y = x3 – 2x2 + x + c1

• At x = 1, y = 5

⇒ c1 = 5

⇒ y = x3 – 2x2 + x + 5

• From equation (1) we get the critical points x = \(\frac{1}{3}\), 1

f(1) = 5

• At the end points and the stationary points,

f(0) = 5, f(2) = 7, \(f\left( {\frac{1}{3}} \right) = \frac{{112}}{{27}}\)

• Hence the global maximum value = 7

Concept:

If, we are given a square matrix A then how to prove that 

⇒ adj(kA) = k^(n-1)adj(A)

Calculation:

Also, by multiplication, 

\(Let\,I = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\ then,\ \,kI = \left[ {\begin{array}{*{20}{c}} k&0&0\\ 0&k&0\\ 0&0&k \end{array}} \right]\)

\( ⇒ adj\,(kI) = \left[ {\begin{array}{*{20}{c}} {{k^2}}&0&0\\ 0&{{k^2}}&0\\ 0&0&{{k^2}} \end{array}} \right] \)

\(⇒ adj(kI)= {k^2}I\)

CONCEPT:

The angle between the lines with direction ratios \(\left\langle {{a_1},\;{b_1},\;{c_1}} \right\rangle \) and \(\left\langle {{a_2},\;{b_2},\;{c_2}} \right\rangle \) is given by: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

Calculation:

Given:

There are four cards from a set of 52 playing cards.

There are four units:- Diamond, Spade, Heart, Club 

13 cards of each unit.

Since they are different cases.

So, we add a number of ways.

Required number of ways = ¹³C₄ + 13C4 + 13C4 + 13C4

⇒ 4 × 13C4

⇒ \(4 \times \frac{13!}{4!(13 - 4)}\)

⇒ \(4 \times \frac{13!}{4! \times 9!}\)

⇒ \(4 \times \frac{13 \times 12 \times 11 \times 10 \times 9!}{ 4 \times 3 \times 2 \times 1 \times 9!}\)

⇒ \(4 \times \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\)

= 2860

∴ There are 2860 ways.

Given:

\(\overrightarrow a = 2\widehat i + 2\widehat j + 3\widehat k,\overrightarrow b = - \widehat i + 2\widehat j + ̂ k \rm \ and\overrightarrow c = 3\widehat i + \widehat j\)

 \(\overrightarrow a + γ \overrightarrow b \) is perpendicular to \(\vec{c}\)

Concept:

If two vectors are perpendicular then their dot product is zero.

Formula:

\(\vec{A}.\vec{B}\) = A_xB_x + A_yB_y + A_zB_z

Solution:

\(\overrightarrow a + γ \overrightarrow b \) = (2 - γ)î + (2 + 2γ)ĵ + (3 + γ)k̂ 

Now, (\(\overrightarrow a + γ \overrightarrow b \)).(\(\vec{c}\)) = 3(2 - γ) + (2 + 2γ)

∵ vectors are perpendicular 

⇒ (\(\overrightarrow a + γ \overrightarrow b \)).(\(\vec{c}\)) = 0

⇒ 3(2 - γ) + (2 + 2γ) = 0

or γ = 8

Given:

The plane which passes through the point (5, 2, -4)

Plane is perpendicular to the line with direction ratios 2, 3, -1

Concept:

Cartesian equation of the plane passing through (x₁ , y₁ , z₁) and perpendicular to line having drs a, b, c is given by :

a(x - x₁) + b(y - y₁) + c(z - z₁) = 0

Solution:

Equation of plane :

2(x - 5) + 3(y - 2) -(z - (-4)) = 0

⇒ 2x + 3y - z = 20

Given:

The plane : 4x - 2y + 3z - 6 = 0

Concept:

For a plane having equation ax + by + cz = d(a, b, c) are direction ratios of the normal to the plane.

Equation of a line passing through (x₁, y₁, z₁) and having direction ratios a, b, c in the cartesian form :

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}= \frac{z-z_{1}}{c}\)

Calculation:

Direction ratios of normal to the plane are 4, -2, and 3. 

Equation of a line passing through the origin (0, 0, 0) and having direction ratios 4, -2, 3 is:

\(\frac{x-0}{4}=\frac{y-0}{-2}= \frac{z-0}{3} = λ\)

⇒ x = 4λ , y = -2λ and z = 3λ

Satisfying equation of the plane with above coordinates :

⇒ 4(4λ) - 2(-2λ) + 3(3λ) - 6 = 0

⇒ λ = 6/29

∴ Foot of the perpendicular,

⇒ x = 24/29, y = -12/29 and z = 18/29

Concept:

Scaler triple product of the vectors:

The scalar triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

Coplanar vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplanar if the scalar triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

Calculation:

\(\left| {\begin{array}{*{20}{c}} a&a&c\\ 1&0&1\\ c&c&b \end{array}} \right| = 0 ⇒ \left| {\begin{array}{*{20}{c}} a&a&c\\ 1&0&1\\ c&c&b \end{array}} \right| = 0\)         [C₂ → C2 - C2]

⇒ a(-b) + c(c) = 0 

⇒ c² = ab.

Hence c is the geometric mean of a and b.

Concept:

One-One Function / Injective Function:

A function f: A → B is said to be a one-one function if different elements in A have different images or are associated with different elements in B.

i.e., if f(x1) = f(x2) then x1 = x2, ∀ x1, x2 ∈ R

Onto Function / Surjective Function:

A function f: A → B is said to be an onto function if each element in B has at least one pre-image in A.

i.e., If-Range of function f = Codomain of function f, then f is onto.

Bijective Function:

A function f: A → B is said to be a B a bijective function if it is both one-one and onto function.

Calculation:

Given: f(x) = x + √x2

The given function can also be written as,

f(x) = x + |x|

This function can be converted into a piecewise defined function.

For, x > 0

⇒ f(x) = x + (+ x) = 2x

For, x < 0

⇒ f(x) = x + (- x) = 0

Clearly, f(x) ≥ 0 for all x. So, it is not surjective or onto as,

Range of function f ≠  Codomain of function f.

Also, f(x) = x + |x|. Clearly, f(-1) = f(-2) = 0

This means for one tow values of x, we get the same value of y as 0  so, f is not one or we can say that it is many -one

Given:

Total number of aspirants = 1000

Total number of females = 430

Females are graduate = 10% of 430

Formula Used:

P(A|B) = P(A ∩ B)/P(B)

Calculation:

We have,

A : Aspirant is graduate

B : Aspirant is female

⇒ Females are graduate = 10% 0f 430

⇒ Females are graduate = \(\frac{10}{100} \times 430\)

⇒ Females are graduate = 43

⇒ P(E|F) = \(\frac{43}{430}\)

⇒ P(E|F) = 0.1

∴ The probability that an aspirant selected randomly is graduate given that the selected aspirant is a female is 0.1

Concept:

Since a, b, and c form a right-handed system 

∴ \(\vec c = \vec b × \vec a \)

Calculation:

Given:

The vectors \(\vec a, \vec c, \ and \ \vec b\) from a right-handed system.

\(\vec c= \hat j × (x\hat i + y\hat j + z\hat k)\)

\(\Rightarrow \vec c= x (\hat j × \hat i) + z (\hat j × \hat k) \)

\(\Rightarrow \vec c= - x\hat k + z\hat i - x\hat k\)

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 sin x negative in the interval. This is because x in the above equation is always positive but the value sin x 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 = sin x dx, so v = - cos x,

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

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

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

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

Hence, π - (-3π) = 4π

Therefore, k = 4

Concept:

If the equation of the line through (1, 2, -1) and (-1, 0, 1) then these points will satisfy the given line,

Calculation:

The general equation of the line,

x = lt+1, y = mt+2, z = nt -1

It is obvious that points (1, 2, -1) is satisfying the line,

For point (-1, 0, 1),

-1 = lt+1, 0 = mt+2, 1 = nt -1

Equating the values of t,

\(\Rightarrow \frac{{ - 2}}{l} = \frac{{ - 2}}{m} = \frac{2}{n}\); 

∴ (l, m, n) are (1, 1, -1)

Concept

Since g(x) is the inverse of function f(x), therefore

gof(x) = I(x) for all x.

Where I(x) is the identity function.

Calculation:

Given:

Now gof(x) = I(x), ∀ x

⇒ gof(x) = x, ∀ x 

⇒(gof)' (x) = 1, ∀ x

⇒ g'(f(x))f'(x) = 1, ∀ x

⇒ g'(f(x)) = \(\frac{1}{{f'(x)}}\) ∀ x 

⇒ g'(f(x))= \(\frac{1}{{f'(x)}}\).

⇒ g'(f(c))= \(\frac{1}{{f'(c)}}\).

CONCEPT:

• if the tangent to a given curve y at a point is perpendicular to the given line then the multiplication of slopes of that tangent and the given line will be equal to -1 as both are perpendicular to each other.
•  The slope of tangent on a given curve y can be calculated as,

⇒ y' at given point = dy/dx at a given point

CALCULATION:

Given:

Slop of line \( = - \frac{a}{b}\) 

\(y = \frac{4}{x} \)

\(\Rightarrow \frac{{dy}}{{dx}} = - \frac{4}{{{x^2}}} \)

\(\Rightarrow - \frac{a}{b} = - \frac{4}{{{x^2}}} \)

So, the ratio a/b is positive for all the values of x which is only possible if the a and b both are positive or both are negative.

\( \Rightarrow a < 0,b < 0.\)

Given:

Urn A consists 3 blue and 4 green balls 

Urn B consists 5 blue and 6 green balls

One ball drawn at random from one of the urns was blue

Concept:

Baye's Theorem :

Solution:

Let B be the event that the ball drawn is blue and E₁ be the event that the ball is drawn from urn 1 and E₂ be the event that the ball is drawn from urn 2.

∴ P(B) = P(B ∩ E1) + P(B ∩ E₂)

⇒ P(B) = \(\frac{1}{2}\times\frac{3}{7} + \frac{1}{2}\times\frac{5}{11} \)

= 34/77

∴ P(E₂/B) = \(\frac{P(E_1 \cap B)}{P(B)} = \frac{P(E_2) P(B/E_2))}{P(B)}\)

= \(\frac{\frac{1}{2}\times \frac{5}{11}}{\frac{34}{77}}\)

= 35/68

Concept:

In linear equations in two variables:- a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 

Here, a₁, a₂, b₁, b₂, c₁, c₂ are real numbers.

1. If \(\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}\) then the system of equations has a unique solution. Hence, the given system of equations is consistent.

In graphical representation, if the equations are consistent, then the lines intersect at only one point.

2. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), then the system of equations has infinitely many number of solutions. Hence, the given system of equations is consistent.

In graphical representation, if the equations are consistent, then the lines coincide with each other.

Calculation:

We have,

⇒ 7x - 5y = - 3   ----- equation (1)

⇒ 2x + 3y = 8    ----- equation (2)

On multiplying equation (1) by 3 and equation (2) by 5, we get

⇒ 21x - 15y = -9    ----- equation (3)

⇒ 10x + 15y = 40   ----- equation (4)

On adding equation (3) and (4), we get

⇒ 31x = 31

⇒ x = 1

Putting the value of x in equation (2)

⇒ 2 × 1 + 3y = 8

⇒ 2 + 3y = 8

⇒ 3y = 6

⇒ y = 2 

Putting the value of x and y in equation,

⇒ 4x - 6y + λ = 0

⇒ 4×1 - 6×2 + λ = 0

⇒ 4 - 12 + λ = 0

⇒ λ = 8

∴ The value of λ is 8

Given:

The vertices A,B,C of a triangle ABC are (1, 1, 3), (-1, 0, 0), (0, 1, 2) respectively.

Concept:

\(\vec{PQ}\) = p.v.(\(\vec{Q}\)) - p.v.(\(\vec{P}\))

Formula:

The angle between two vectors \(\vec{P}\) and \(\vec{Q}\) is given by : 

\(\theta = cos^{-1}(\frac{\vec{P}.\vec{Q}}{|\vec{P}||\vec{Q}|})\)

Solution:

\(\overrightarrow {BA} \) = (1, 1, 3) - (-1, 0, 0)

= 2î + ĵ + 3k̂ 

\(\overrightarrow {BC} \) = (0, 1, 2) - (-1, 0, 0) = î + ĵ + 2k̂ 

∴ \(\theta = cos^{-1}(\frac{\vec{BA}.\vec{BC}}{BA.BC})\)

\(\Rightarrow \theta = cos^{-1}(\frac{9}{\sqrt 14 \sqrt 6})\))

\(\Rightarrow \theta ={{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)

Concept:

A discontinuous function  f has a local maximum if \(f(a)\geq f (a+h),f(a)\geq f(a-h)\).

Calculation:

Given:

\(f(x)=\left\{\begin{matrix} 7x-x^3+\log(b^2-4b+4) &0\leq x<3 \\ x-9&x\geq3 \end{matrix}\right.\)

Since f attains maxima at x = 3.

⇒ \(f(3)\geq f(3-0) \ or\ f(3^-) \)

⇒ \(3-9\geq 21-27+\log(b^2-4b+4)\)

⇒\(-6\geq -6+\log(b^2-4b+4)\)

⇒ \(\log(b^2-4b+4)\leq 0\)

Therefore option 1 is correct.

Concept

Let the system of equations be,

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

\({\rm{\;}} ⇒ \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\)

⇒ If det (A) ≠ 0, system is consistent has unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given:

(1 - α)x + 3y + 5z = 0      ...(i)

5x + (1 - α)y + 3z = 0       ...(ii)

3x + 5y + (1 - α)z = 0      ...(iii)

Here, B = O

⇒ X = A-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\) = O

For any value of α.

For an infinite no. of solutions,

|A| = 0

\(\left| {\begin{array}{*{20}{c}} 1-α &3&5\\ 5&1-α &3\\ 3&5&1-α \end{array}} \right| = 0\)

After solving this,

(9-α)(α² + 6α) + 12) = 0

That means only one real value of α as the other factor is having imaginary roots because the discriminant of α2 + 6α) + 12 will be negative.

From question, the matrix given is:

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

Now,

\(\Rightarrow det\left( A \right) = \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&b&1\\ b&{{b^2} + 1}&b\\ 1&b&2 \end{array}} \right|\)

⇒ det (A) = 2[2(b² + 1) - b²] - b(2b - b) + 1(b² - b² - 1)

⇒ det (A) = 2[2b² + 2 - b²] - b² - 1

⇒ det (A) = 2b² + 4 - b² - 1

∴ det (A) = b² + 3

From question,

\(\Rightarrow \frac{{\det \left( A \right)}}{b} = \frac{{{b^2} + 3}}{b}\)

\(\therefore \frac{{det\left( A \right)}}{b} = b + \frac{3}{b}\)

From question, b > 0,

∴ Arithmetic Mean ≥ Geometric Mean

Now,

\(\Rightarrow \frac{{b + \frac{3}{b}}}{2} \ge {\left( {b \times \frac{3}{b}} \right)^{1/2}}\)

\(\Rightarrow b + \frac{3}{b} \ge 2\sqrt 3 \)

CONCEPT:

• Since the constant terms for the given equations are non-zero so the values Δ­1, Δ2, and Δ3 may be zero or may not be 0.
• For the unique solution, the value of the determinant of coefficients means Δ should be non-zero.

Δ  ≠  0

Where Δ is the determinant formed by the coefficient of the given homogeneous equation, 

CALCULATION

Given:

\(D= \,\left| {\begin{array}{*{20}{c}} 0&1&{ - 1}\\ { - 1}&0&2\\ 1&{ - 2}&0 \end{array}} \right|\, = 0\)

\({D_1}\, = \left| {\begin{array}{*{20}{c}} 0&1&{ - 1}\\ { - 2}&0&2\\ 3&{ - 2}&0 \end{array}} \right| = 14 \Rightarrow {D_1} \ne 0\)

\(D = 0\,and\,{D^1} \ne 0,\)

hence the system is inconsistent, so it has no solution.

Calculation:

Let \(Δ \) = be the value of the given determinant. Then

\(\begin{array}{l} Δ = \left| {\begin{array}{*{20}{c}} {x!}&{(x + 1)!}&{(x + 2)!}\\ {(x + 1)x!}&{(x + 2)(x + 1)!}&{(x + 3)(x + 2)!}\\ {(x + 2)(x + 1)x!}&{(x + 3)(x + 2)(x + 1)!}&{(x + 4)(x + 3)(x + 2)!} \end{array}} \right| \end{array}\)

\(\\ ⇒ Δ = x!(x + 1)!(x + 2)!\left| {\begin{array}{*{20}{c}} 1&1&1\\ {x + 1}&{x + 2}&{x + 3}\\ {(x + 1)(x + 2)}&{(x + 2)(x + 3)}&{(x + 3)(x + 4)!} \end{array}} \right|\)

\(\\ ⇒ Δ = x!(x + 1)!(x + 2)!\left| {\begin{array}{*{20}{c}} 1&0&0\\ {x + 1}&1&2\\ {(x + 1)(x + 2)}&{2(x + 2)}&{4x + 10} \end{array}} \right|\\\)

⇒ Δ = 2x!(x + 1)!(x + 2)!.

Shortcut TrickPut x = 1 and match the alternate.

Given:

\(a = \frac{2sinθ}{1 + cosθ + sinθ}\), then \(\frac{1+ \sin \theta - \cos \theta}{1+ \sin \theta}\)

Concept:

Rationalization:- It is a process that finds application in elementary algebra, where it is used to eliminate the irrational number in the denominator.

Formula:

cos²θ = 1 - sin²θ

Calculation:

\(\frac{1+ \sin \theta - \cos \theta}{1+ \sin \theta}\)

⇒ \(\frac{1 - cosθ + sinθ}{1 + sinθ} \times \frac{1+ cosθ + sinθ}{1 + cosθ + sinθ}\)

⇒ \(\frac{(1 + sinθ)^2 - cos^2θ}{(1 + sinθ)(1 + cosθ + sinθ)}\)     [(a² - b²) = (a - b)(a + b)]

⇒ \(\frac{(1 + sinθ)(1 + sinθ) - cos^2θ}{(1 + sinθ)(1 + cosθ + sinθ)}\)    [cos²θ = 1 - sin²θ]

⇒ \(\frac{(1 + sinθ)(1 + sinθ) - (1^2 - sin^2θ)}{(1 + sinθ)(1 + cosθ + sinθ)}\)  [(a² - b²⁾ = (a - b)(a + b)]

⇒ \(\frac{(1 + sinθ)(1 + sinθ) - [(1+ sinθ)(1 - sinθ)]}{(1 + sinθ)(1 + cosθ + sinθ)}\)

⇒ \(\frac{(1 + sinθ)[1 + sinθ - (1 - sinθ)]}{(1 + sinθ)(1 + cosθ + sinθ)}\)

⇒ \(\frac{1 + sinθ - 1 + sinθ}{1 + cosθ + sinθ}\)

⇒ \(\frac{2sinθ}{1 + cosθ + sinθ}\)

⇒ a       [given]

∴ \(\frac{1+ \sin \theta - \cos \theta}{1+ \sin \theta}\) is a.

Concept:

Let \(\vec{u}~and~\vec{v}\) be two vectors. Then the scalar projection of the vector \(\vec{u}~on~\vec{v}\) is given by: \(\frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|}\)

Calculation:

\(\overrightarrow {\left| a \right|} = 8,\overrightarrow {\left| b \right|} = 7,\overrightarrow {\left| c \right|} = 10,\)

Now, by using the cosine formula,

\(\cos \theta = \frac{{{{\left| {\vec b} \right|}^2} + {{\left| {\vec c} \right|}^2} - {{\left| {\vec a} \right|}^2}}}{{2\left| {\vec b} \right|\left| {\vec c} \right|}} = \frac{{17}}{{28}}\)

Projection of \({\vec c}\) on  \({\vec b}\),

\(\left| {\vec c} \right|\,\cos \theta = 10 \times \frac{{17}}{{28}} = \frac{{85}}{{14}}.\)

Calculation:

Given:

E(x) = 10, V(x) = 25 and E(Y) = 0, V(Y) = 1

E(y) = aE(x) - E(b) = a x10 - b = 0

⇒ a = 10a - b

⇒ b = 10a    ......(1)

∵ V(y) = V(ax - b)

⇒ 1 = a² V(x) + 0

⇒ a = ± 1/5

Using equation 1,

⇒ b = ± 2

\(\Rightarrow (a,b)= \left(\pm \frac{1}{5},\pm 2\right)\)

Concept:

Let \(\vec{u}~and~\vec{v}\) be two vectors. Then the vector component of the vector \(\vec{u}~on~\vec{v}\) is given by:

 \(\frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|}(\hat v) = \frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|^2}( \vec v)\)

Calculation:

Given:

\(\vec a = 4\hat i + 6 \hat j\) and \(\vec b = 3 \hat j + 4 \hat k\)

\(\vec a . \vec b=(4, 6, 0). (0, 3, 4)\)

\(\vec a . \vec b= 18\)

|\(\vec a\)| = √52 

\(|\vec b| = 5\) 

The component of vector a along b is \(\frac{{18}}{{25}}(3j + 4k)\)

Concept:

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing (0,0) else the opposite side of (0,0).

Calculation:

Given: \(Z=2x_1+5x_2\) subjected to,

x1+ 3x2 ≤ 40
x1+ x2 ≤ 10

x1 ≥ 0, x2 ≥ 0

• Draw the constraints,

• From Graph feasible region is AOC,
• Finding coordinates of all corner points of feasible region,

⇒ A(0.10), C(10.0) and O(0.0)

• Finding the value of the objective function Z at A, O, and C,

⇒ Z(C) = (10,0) = 2 × 10 + 0 = 20 

⇒ Z(A) = Z(0,10) = 50

⇒ Z(O) = Z(0,0) = 0

• So profit is maximum at A and the maximum value is 50.
• So, the correct answer is option 1

Concept:

Referring to the diagram,

From the law of addition of vectors,

\(\overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AD} \,\) and \(\overrightarrow {AC} + \overrightarrow {CD} = \overrightarrow {AD} \,\)

Calculation:

\(2\overrightarrow {AD} = \overrightarrow {AB} + \overrightarrow {BD} \,\, + \overrightarrow {AC} + \overrightarrow {CD} \)

\(\overrightarrow {AD} \, = \frac{{\overrightarrow {AB} + \overrightarrow {AC} }}{2}\,\,[\overrightarrow {BD} = - \overrightarrow {CD} ]\)

\(\overrightarrow {AD} = \frac{{(3 + 5)i + (0 - 2)j + (4 + 4)k}}{2} = 4i - j + 4k\)

\(\overrightarrow {|AD|} = \sqrt {16 + 16 + 1} = \sqrt {33.} \)

Calculation:

Given:

α, β are roots of the equation 2x² + 3x + 5 = 0

Therefore the Sum of roots (α, β) = \( - \frac{3}{2}\)

And product of roots (α,β) = \(\frac{5}{2}\) 

Now, \(\left| {\begin{array}{*{20}{c}} 0&{ β}&{ β}\\ α&0&α\\ β&a&0 \end{array}} \right|\)

= 0 | 0 - α² | - β | 0 - αβ | + β | α ² - 0 | = αβ(α + β)

= αβ²  + βα2 = αβ(α + β) 

⇒ \(\frac{5}{2}\left( {\frac{{ - 3}}{2}} \right) = \frac{{ - 15}}{4}\)

Concept:

If \(P(X=x)=f(x)\) is a probability mass function, then 

\(\sum_{x\in Range(X)}f(x)=1\)

and the Expected value of X is given by \(E(X)=\sum_{x_{i}}x_{i}f(x_{i})\)

Calculations:

Given:

Therefore the expected value of X is given by,

\(E(X)=\sum_{x_{i}}x_{i}f(x_{i})\)

\(\Rightarrow E(X)=\sum_{x=1}^{\infty}x\frac{1}{2^x}\)

\(\Rightarrow E(X)=(1.\frac{1}{2})+(2.\frac{1}{2^2})+(3.\frac{1}{2^3})+...\)----(1)

It is arithmetic geometric series. Let us multiply it by the common ratio \(\frac{1}{2}\)of geometric series. We get,

\(\Rightarrow \frac{1}{2}E(X)=(\frac{1}{2}.\frac{1}{2})+(\frac{2}{2}.\frac{1}{2^2})+(\frac{3}{2}.\frac{1}{2^3})+...\)---(2)

Subtracting (2) from (1), We get

\(\Rightarrow \frac{1}{2}E(X)=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\)

\(\Rightarrow \frac{1}{2}E(X)=\frac{\frac{1}{2}}{1-\frac{1}{2}}\)    sum of infinite of G.P. with first term a and common ratio r is given by \(S_{\infty}=\frac{a}{1-r}\)

\(\Rightarrow \frac{1}{2}E(X)=\frac{\frac{1}{2}}{\frac{1}{2}}\)

\(\Rightarrow \frac{1}{2}E(X)=1\)

\(\therefore E(X)=2\)

Therefore option 1 is correct.

Concept:

The angle made by the tangent to the curve y = f(x) at a point (a, b), with the x-axis, is given by,

m = tan θ = \(\rm \left[\frac{dy}{dx}\right]_{(a, b)}\).

Calculation:

Given:

\(\rm \left[\frac{dy}{dx}\right]_{(a, b)} = y \ (ordinate)\)

\(y = {x^3}\)

\( ⇒ \frac{{dy}}{{dx}} = 3{x^2}\)

Now, 3x² = y (Ordinate)

⇒ 3x³ = x³

⇒ x = 0, 3

Thus the two point are (0, 0) and (3, 27).

\(Area = S = 2\int\limits_0^1 {\left( {\sqrt x -x} \right)} \;dx = \frac{1}{3}\)

Explanation:

Factors to determine a plane:-

1. Three non-collinear points determine a plane.
2. A-line and a point not on the line determine a plane.
3. Two intersecting lines determine a plane.
4. Two parallel lines determine a plane.

∴ The equation of a plane lying in extraordinary form does not determine the plane.

Given:

A(2, 3, 2) and B(5, 1, 6) are two points.

Concept:

The cartesian equation of a line passing through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is given by,

\(\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2- z_1}\)

Calculation:

Here, x₁ = 2, y₁ = 3, z₁ = 2, x₂ = 5, y₂ = 1 and z₂ = 6

The cartesian equation of a required line is: 

⇒ \(\frac{x - 2}{5 - 2} = \frac{ y - 3}{1 -3} = \frac{z - 2}{6 - 2}\)

⇒ \(\frac{ x - 2}{3} = \frac{ y - 3}{-2} = \frac{ z - 2}{4}\)    ---- equation (1)

The line crosses xy - plane , i.e., z = 0

So, by substituting z = 0 in equation (1), we get

⇒ \(\frac{ x - 2}{3} = \frac{ y - 3}{-2} = \frac{ 0 - 2}{4}\)

⇒ \(\frac{ x - 2}{3} = \frac{ y - 3}{-2} = \frac{-1}{2}\)

⇒ \(\frac{ x - 2}{3} = \frac{-1}{2}\) and \( \frac{ y - 3}{-2} = \frac{-1}{2} \)

⇒ 2x - 4 =  - 3 and 2y - 6 = 2

⇒ 2x = 1 and 2y = 8

⇒ x = \(\frac{1}{2}\) and y = 4 

∴ The line AB crosses the XY-plane at (1/2, 4. 0)

y = 3 – x and x² + y² = 9

⇒x² + (3 – x)² = 9

⇒ 2x² – 6x = 0 ⇒ x = 0, 3

And then y = 3, 0

Point of intersection are (0, 3) and (3, 0).

If x varies from 0 to 3 while y varies from 3 – x to \(\sqrt {9 - {x^2}} \)

\(\begin{array}{l} \smallint dA = \mathop \smallint \limits_0^3 \mathop \smallint \limits_{3 - x}^{\sqrt {9 - {x^2}} } dxdy = \mathop \smallint \limits_0^3 \left[ y \right]_{3 - x}^{\sqrt {9 - {x^2}} }dx\\ = \mathop \smallint \limits_0^3 \left\{ {\sqrt {9 - {x^2}} - \left( {3 - x} \right)} \right\}dx\\ = \left[ {\frac{{x\sqrt {9 - {x^2}} }}{2} + \frac{9}{2}{{\sin }^{ - 1}}\frac{x}{3} - 3x + \frac{{{x^2}}}{2}} \right]_0^3\\ \Rightarrow A = \left[ {\frac{{3\sqrt {9 - 9} }}{2} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{3}{3}} \right) - 3 \times 3 + \frac{9}{2} - 0 - \frac{9}{2}{{\sin }^{ - 1}}\left( 0 \right) + 0 - 0} \right]\\ = \frac{9}{2}{\sin ^{ - 1}}\left( 1 \right) - 9 + \frac{9}{2} - \frac{9}{2}{\sin ^{ - 1}}\left( 0 \right)\\ = \frac{9}{2}.\frac{\pi }{2} - \frac{9}{2} = \frac{9}{2}\left( {\frac{\pi }{2} - 1} \right) \end{array}\)

Hence, (d) is the correct choice.

Concept:

For a function fx(x) to be a valid PDF, it must satisfy the following conditions:

1) fx(x) ≥ 0

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

Calculation:

Given: \({f_x}\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\left| x \right|,\;\left| x \right| < 1}\\ {0,\;otherwise} \end{array}} \right.\)

fx(x) ≥ 0 for all x.

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

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

\(=\left. {\frac{{ - {x^2}}}{2}} \right|_{ - 1}^0 + \left. {\frac{{{x^2}}}{2}} \right|_0^1\)

\(=\frac{{ - 1}}{2}\left( { - 1} \right) + \frac{1}{2}\left( {1 - 0} \right)\)

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

• Since this function also satisfies both the conditions.
• So, fx(x) is also a valid PDF.

Explanation:

If X is said to be a binomial random variable for n random trials with parameters n, p if,

The  probability mass function,

B (X : n, p) = p(x) = \(\rm{^n}C_xp^xq^{n-x}\)

Also, p + q = 1 

⇒ q = 1 - p

0 ≤ x ≤ n

⇒ Binomial distribution is applicable for independent Events only. (Trial are different but the experiment is the same)

⇒ Constant probability for all success and failure.

Also, the mean of the binomial random variable,

E(x) = np

Also, the variance of the binomial random variable,

V(x) = npq

From the above definitions,

⇒ E(x) > V(x)

Condition to use binomial distribution:

• The observation is independent
• The probability of success is constant
• The mean is greater than variance.

Concept: 

If a vector P makes an angle α with x-axis, β with y-axis, and γ with z-axis.

then cos²α + cos²β  + cos²γ = 1

and its components are:

|P| cos α , |P| cos β , |P| cos γ 

Where |P| is the magnitude of the given vector.

Calculation:

Since given vector a is the unit vector 

⇒ |a| = 1

The vector a is making angle π/3 with î, π/4 with ĵ and an acute angle θ with k̂

⇒ cos2π/3 + cos2π/4  + cos2θ = 1

⇒ (1/2)² + (1/√2)2  + cos2θ = 1

⇒ 1/4 + 1/2 + cos2θ = 1

⇒ cos2θ = 1 - 3/4

⇒ cos2θ = 1/4

⇒ cos θ = 1/2

⇒ The value of θ is π/3.

So, Statement I is correct.

And components of a are

|a| cos π/3 , |a| cos π/4 , |a| cos π/3

⇒ \(\frac{1}{2}, \frac{1}{\sqrt 2}, \frac{1}{2}\)

So, Statement II is incorrect.

∴ The correct answer is option (1).

Concept:

The maximum or minimum value of an objective function occurs at one of the corner points of the feasible region.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing (0,0) else the opposite side of (0,0).

Calculation:

Given: 

The objective function: z = 5x - 7y

The constraints:  x + y ≤ 7,

2x - 3y + 6 ≥ 0, and x ≥ 0, y ≥ 0.

Explanation:

The graph of the given constraints is shown below -

Here the corner points are (0 , 0) , (7 , 0) , (0 , 2) and (3 , 4).

Values of the objective function at corner points :

z = 5x - 7y

at (0 , 0), z = 0

at (7 , 0), z = 35

at (0 , 2), z = -14   →   MINIMUM

at (3 , 4), z = -13

Concept:

If \(\rm f(A) = b \;then \;f^{-1}(b) = a\)

Calculations:

Given A = {x ∈  R | x ≥ 0}. A function f : A → is defined by f(x) = x².

Here, \(\rm f(x) = {x^2}\)

⇒ y = x²

∴ x = √y 

⇒ \(\rm f^{-1}(x) = √{x}\)

hence, \(\rm f^{-1}\) exist.

Hence, The function has an inverse but f is not its own inverse