Mathematics Mock Test - 7

Concept:

Property of definite integral: 

1) If f(x) is an odd function then,

 \(\int\limits_{ - a}^a {f(x)dx} = 0\)

2) If f(x) is an even function then,

 \(\int\limits_{ - a}^a {f(x)dx} =2\int\limits_0^a {f(x)dx}\)

Concept:

• A differential equation is called linear if there are no multiplications among dependent variables and their derivatives.
• Integrating factors of \(\frac{dx}{dy} + Px =Q\) is \(e^{∫Pdy}\)

Calculation:

• Consider the differential equation of type,

 \(\frac{dx}{dy} + P(y)x = Q(y)\)

• Then, the integrating factor (I.F.) = \(e^{∫P(y).dy}\)
• Hence multiplying the I.F. in the equation

⇒ \(e^{∫P(y)dy}.\frac{dx}{dy} + e^{∫P(y)dy}.P(y)x = e^{∫P(y)dy}.Q(y)\)

⇒ \(e^{∫P(y)dy}.dx+ P(y)xe^{∫P(y)dy}.dy = e^{∫P(y)dy}.Q(y)dy\)

⇒ \(d(xe^{∫P(y)dy}) = e^{∫P(y)dy}.Q(y)dy\)

• On integrating both sides

⇒ \(∫d(xe^{∫P(y)dy}) = ∫e^{∫P(y)dy}Q(y).dy\)

⇒ \(xe^{∫P(y)dy} = ∫e^{∫P(y)dy}Q(y).dy + c\)

⇒ \(x.e^{∫Pdy} = ∫Qe^{∫Pdy}dy + c\)

∴ The solution of the  differential equation \(\frac{dx}{dy} + Px =Q\) is \(x.e^{∫Pdy} = ∫Qe^{∫Pdy}dy + c\)

Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

For mutually exclusive events A and B, P(A ∩ B) = 0

Calculation:

We know, for mutually exclusive events A and B,  P(A ∩ B) = 0

∴ P(A ∪ B) = P(A) + P(B) - 0

=  P(A) + P(B)

⇒ 0.6 = 0.4 + P(B)

∴ P(B) = 0.2

Hence, the correct option is 1. 

Concept:

A matrix can only be added to another matrix if the two matrices have the same dimensions. Their sum is obtained by summing each element of one matrix to the corresponding element of the other matrix.

Determinant:- It is the scalar value or number calculated using a square matrix. It is denoted by |A|, det A

If det A = \(\begin{bmatrix} a_{11} \ a_{12} \\\ a_{21}\ a_{22} \end{bmatrix}\) 

det A = a₁₁ × a₂₂ - a₁₂ × a₂₁

Formula:

sin²x + cos²x = 1

Calculation:

\(A = \begin{bmatrix} \cos^2 x & \sin^2 x \\\ \sin^2 x & \cos^2 x \end{bmatrix}\) and \(B = \begin{bmatrix} \sin^2 x & \cos^2 x \\\ \cos^2 x & \sin^2 x \end{bmatrix}\)

⇒ \( \begin{bmatrix} \cos^2 x & \sin^2 x \\\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\\ \cos^2 x & \sin^2 x \end{bmatrix}\)

⇒  \( \begin{bmatrix} \cos^2 x + sin^2 x& \sin^2 x +cos^2 x \\\ \sin^2 x + cos^2x & \cos^2x + sin^2 x \end{bmatrix}\)

⇒   \( \begin{bmatrix} 1 & 1 \\\ 1 & 1 \end{bmatrix} \)

⇒ det (A + B) =  1 × 1 - 1 × 1

⇒ det (A + B) = 0

∴ The value of det (A + B) = 0.

Concept:

tan-1 x + cot-1 x = \(\rm \frac {π}{2}\)

Calculation:

To Find: Value of cos (2tan-1 x + 2cot-1 x)

cos (2tan-1 x + 2cot-1 x) = cos 2(tan-1 x + cot-1 x)

As we know, tan-1 x + cot-1 x = \(\rm \frac {π}{2}\)

cos (2tan-1 x + 2cot-1 x) = cos [2 × \(\rm \frac {π}{2}\) ]

= cos π 

= -1

Explanation:

• Statements A and B are true by definition of a critical point. Statement D is also true.
• Statement C is false: f(x) = 1 / x has no critical points. The point x = 0 is not considered.as a critical point since it is not included in the domain of \(f\left( x \right)\) 

i.e. \(f'\left( x \right) = - 1/{x^2}\) is neither 0 nor undefined, for all x in the domain of f(x) as the domain of f(x) = 1/x is R - {0}.

CONCEPT:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) is a square matrix of order 3, then determinant of A is given by |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• If A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 3&4&9\\ {11}&6&7\\ 8&9&5 \end{array}} \right]\) and |2A| = k

⇒ |A| = 3 × (30 - 63) - 4 × (55 - 56) + 9 × (99 - 48)

⇒ |A| = - 99 + 4 + 459 = 364

As we know that, if A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

⇒ |2A| = 2³ ⋅ 364 = 2912

Hence, the correct option is 3.

Given:

\(I = \mathop \smallint \nolimits_1^2 ydx\)

Since the integration is done w.r.t 'x' variable, 'y' in terms of x can be written as:

y = mx + c

The slope of the given line will be:

\(m=\frac{1-0}{0-(-1)}=1\)

y = x + c

when x = 0, y = 1, i.e.

c = 1

∴ y = x + 1

\(I = \mathop \smallint \nolimits_1^2 ydx= \mathop \smallint \nolimits_1^2 \left( {x + 1} \right)dx\)

\(= \left[ {\frac{{{x^2}}}{2} + x} \right]_1^2 = 4 - \frac{3}{2}\)

= 2.5

CONCEPT:

The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b\)

​CALCULATION:

Let \(\vec a = \hat i - \hat j \ and \ \vec b = \vec i + \vec j\)

Here, we have to find the component of \(\vec{a}\) along \(\vec{b}\) .

⇒ \(\vec a \cdot \vec b = 1 - 1 = 0 \ and \ |\vec b| = \sqrt 2\)

⇒ The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b = \vec O\)

Hence, option 2 is correct.

Concept:

f(x) = |x| ⇒ f(x) = x if x > 0,  and f(x) =  -x, x < 0

A function f(x) is continuous at x = a, if \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

A function f(x) is differentiable at x = a, if  LHD = RHD

\(\begin{array}{l} \rm L H D=\lim _{x \rightarrow a^{-}}f'(x)=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{x \rightarrow a^{+}}f'(x)=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

Calculation:

Here, f(x) =  e-|x| 

 \(\text {LHL =}\rm\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}e^{-|x|}=e^0=1\\ \text {RHL =}\lim _{x \rightarrow 0^{+}} f(x)=lim _{x \rightarrow 0^+} e^{-|x|}=e^0=1\\ lim _{x \rightarrow 0}f(x)=lim _{x \rightarrow 0}e^{-|x|}=e^0=1\)

So, the function is continuous at x = 0

f(x) =  e-|x| 

f'(x) = ex  for x < 0 and f'(x) =  -e-x  for x > 0

 \(\rm LHD=\rm \lim _{x \rightarrow 0^{-}}f'(x)=\lim _{x \rightarrow 0^{-}}e^x=e^0=1\\ \rm RHD=\rm \lim _{x \rightarrow 0^{+}}f'(x)=\lim _{x \rightarrow 0^{+}}-e^{-x}=-e^0=-1\)

Here, LHD ≠ RHD so f(x) is not differentiable at x = 0

Hence, option (1) is correct.

Concept:

If f(-x) = f(x) then the function is said to be even function

If f(-x) = - f(x) then the function is said to be odd function.

For an odd function:

\(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 0

For an even function

\(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 2 × \(\mathop \smallint \nolimits_{ 0}^a x\;{f(x)}dx\)

Calculation:

Given:

Function f(x) = x e|x|

Integral is -1 to 1.

Here,

f(-x) = -x e|-x| = -x e|x| = - f(x)

∴ The given function is an odd function.

Now, as the function is odd,

\(\mathop \smallint \nolimits_{ - 1}^1 x\;{e^{\left| x \right|}}dx\) = 0

Concept:

• If the parametric equations x = x(t) and y = y(t), then the equation of the tangent line at t = a is found using the following steps:
  • Find the point at which the tangent is drawn, (x_o, y_o) by substituting  = a in the given parametric equations. i.e., (x_o, y_o) = [x(a), y(a)
  • Find the derivative of the function using,

\(\frac{dy}{dx} =\frac{\frac{dy}{dt} }{\frac{dx}{dt} }\)

• Find the slope of the tangent (m) by substituting either t = a in the above derivative
• Find the equation of the tangent line using,

y - y_o = m(x - x_o)

Formula:

\(\frac{d}{dx}(x^n) = nx^{n -1} \)

Calculation:

x = a sin³t, y = a cos³t

Differentiate 'x' with respect to 't'

⇒ \(\frac{dx}{dt} = a\frac{d}{dt}(sint)^3\)

⇒ \(\frac{dx}{dt} = a.3sin^2t\frac{d}{dt}(sint)\)

⇒ \(\frac{dx}{dt} = 3asin^2t.cost\)   ----- equation (1)       [∵ \(\frac{d}{dt}(sint) = cost\) ]

⇒ \(\frac{dx}{dt} = 3asin^2t.cost\)

Differentiate 'y' with respect to 't'

⇒ \(\frac{dy}{dt} = b.\frac{d}{dt}(cost)^3\)

⇒ \(\frac{dy}{dt} = b.3cos^2t\frac{d}{dt}(cost)\)

⇒ \(\frac{dy}{dt} = -3bcos^2tsint\)  ----- equation(2)       [∵ \(\frac{d}{dt}(cost) = - sint\) ]

Now, divide equation (2) by equation (1)

⇒ \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

⇒ \(\frac{dy}{dx} = \frac{- 3bcos^2t.sint}{3asin^2t.cost}\)

⇒ \(\frac{dy}{dx} = - \frac{b}{a}. \frac{cost}{sint}\) is the slope of the tangent

The slope at t = \(\frac{\pi}{2}\)

⇒ \(\frac{dy}{dx}|_{t = \frac{\pi}{2}} = - \frac{b}{a}. \frac{cos\frac{\pi}{2}}{sin\frac{\pi}{2}}\)

⇒ \(\frac{dy}{dx} = - \frac{b}{a}.\frac{0}{1}\)    [∵ \(cos\frac{\pi}{2} = 0\) and \(sin\frac{\pi}{2} = 1\)]

⇒ \(\frac{dy}{dx} = 0\)

So, the tangent to the given curve will be parallel to the x-axis.

Additional InformationFor equation of tangent:

At t = \(\frac{\pi}{2}\)

⇒ x = asin³t

⇒ \(x = a(sin\frac{\pi}{2})^3\)

⇒ x = a

⇒ y = bcos³t

⇒ \(y = b(cos\frac{\pi}{2})^3\)

⇒ y = 0

⇒ (x₁, y₁) = (a, 0)

Equation of tangent passing through (a, 0) and slope is 0

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 0(x - a)

⇒ y = 0

∴ The required equation of tangent is y = 0

Concept:

Mean value theorem for integrals:

Let f be continuous on [a, b]. Then there is a point x_o in (a, b) such that

\(f\left( {{x_o}} \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

Calculation:

\(f\left( \xi \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

\(\mathop \smallint \limits_a^b f\left( x \right)dx = f\left( \xi \right)\left( {b - a} \right)\)

Calculation:

f(x) = x + 1

g(x) = (x² - 7)

t(x) = x + 1 + x² - 7 = x² + x - 6

Let y = x² + x - 6

⇒ y = x² + x + 1/4 - 1/4 - 6

⇒ y = (x + 1/2)² - 25/4

⇒ y has minimum value when (x + 1/2) = 0 at x = -1/2

∴ Minimum value of t(x) = -25/4

Alternate Method

f(x) = x + 1

g(x) = (x² - 7)

t(x) = x + 1 + x² - 7 = x² + x - 6

⇒ y has minimum value at = (-1)/(2 × 1) = -1/2

Formula used: 

Selection of r things out of total n things:

\(^nC_r = \frac{n!}{r!\times (n-r)!}\)

Probability of occurrence of the event:

P(E) =  \(\frac{n(E)}{n(S)}\)

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

Calculation:

Given digits are 1, 2, 3,  4, and 5.

As discussed above we know, that two digits can be chosen out of 5 in ⁵C₂ ways.

⇒ n(S) = ⁵C₂ = \(\frac{5!}{2!\times 3!}\)

 ⇒ n(S) = \(\frac{5 \times 4 \times 3!}{2 \times 1\times 3!} = 10\)

⇒ n(S) = 10

Since the last digit in the product appears as 0 which is only possible when any of 2 or 4 is multiplied by 5 so possible required result,

(E) = {(2, 5), (5, 2), (4, 5), (5, 4)}

⇒ n(E) = 4

Hence, required probability

P(E) = n(E)/n(S) = 4/10

⇒ P(E) =  2/5

∴ Required probability is \(\frac{2}{5}\).

Formula:

\(tan^{-1}x + tan^{-1}y = tan^{-1}\frac{x + y}{1 - xy}\)

\(\frac{π}{2} - tan^{-1}z = cot^{-1}z\)

\(cot^{-1}z = tan^{-1}\frac{1}{z}\)

Calculation:

tan-1 x + tan-1 y + tan-1 z = \(\frac{\pi}{2}\)

⇒ \(tan^{-1}\frac{x + y}{1 - xy} + tan^{-1}z = \frac{π}{2}\)

⇒ \(tan^{-1}\frac{x + y}{1 - xy} = \frac{π}{2} - tan^{-1}z\)

⇒ \(tan^{-1}\frac{x + y}{1 - xy} = cot^{-1}z\)

⇒ \(\frac{x + y}{1 - xy} = tan(cot^{-1}z)\)

⇒ \(\frac{x + y}{1 - xy} = tan(tan^{-1}\frac{1}{z})\)

⇒ \(\frac{x + y}{1 - xy} = \frac{1}{z}\)

⇒ xz + yz = 1 - xy

⇒ xz + yz + xy = 1

⇒ xz + yz + xy - 1 = 0

∴ The required solution is xz + yz + xy - 1 = 0

Concept:

Let the variable is x which is varying with time t.

\(\frac{dx}{dt} \propto x\)

Calculation:

Given:

\(\frac{dx}{dt} \propto x\)

\(⇒ \frac{dx}{dt} =kx\)

\(⇒ \int\frac{dx}{x} =\int kdt\)

\(⇒ ln x =kt + p\)

Where p is a constant.

Now, take antilog,

⇒ x = e^{(kt +p)}

⇒ x = e(kt) e^p

Let c = ep

⇒ x = c e(kt) 

So, the variation will be exponential.

Concept:

For a given probability density function f(x) to be valid, the integration of it over the complete range must equal 1.

Mathematically, it must satisfy:

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

Calculation:

\(P\left( {0.5 < x<5} \right) = \mathop \smallint \limits_{0.5}^5 f\left( x \right)dx\)

\(= \mathop \smallint \limits_{0.5}^1 0.6dx + \mathop \smallint \limits_1^4 0.2dx + \mathop \smallint \limits_4^5 0dx\)

= 0.6 × 0.5 + 0.2 × 3 = 0.90

Concept:

Binomial Distribution: A Binomial Distribution is an experiment in which only ONE OUT OF TWO outcomes are possible. For example Head or Tail, Yes or No, 1 or 0, etc.

If ‘n’ and ‘p’ are the parameters, then ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event.

The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula:

P(X = k) = nCk pk (1 - p)n - k

The mean and variance of Binomial distribute,

Mean (μ) = np, Variance (σ²) = npq, S. D.(σ) \( = \sqrt {npq}\)

Calculation:

Mean = np = 3, S. D. \( = \sqrt {npq} = \frac{3}{2}\) 

\( \Rightarrow q = \frac{{npq}}{{np}} = \frac{9}{{4 \times 3}} = \frac{3}{4} \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}\)

Hence binomial distribution is (q + p)ⁿ =\({\left( {\frac{3}{4} + \frac{1}{4}} \right)^{12}}.\)

\(\eqalign{ & y = mx\_\_\_\_\_\_\_\_\_\_\left( 1 \right) \cr & {{dy} \over {dx}} = m \cr}\)

Substitute m in eqn. (1)

\(\eqalign{ & y = {{dy} \over {dx}}x..........(2) \cr & To{\rm{ }}\ get{\rm{ }}\ orthogonal{\rm{ }}\ curve,{\rm{ }}replace\ {{dy} \over {dx}}\ by - {{dx} \over {dy}}\ in{\rm{ }}\ equation{\rm{ }}\left( 2 \right) \cr & y = - {{dx} \over {dy}}x \cr & y{\rm{ }}dy = - xdx \cr & \mathop \smallint y{\rm{ }}dy = - \mathop \smallint x{\rm{ }}dx \cr & {{{y^2}} \over 2} = {{ - {x^2}} \over 2} + c \cr & {x^2} + {y^2} = {r^2} \cr} \)

Given:

Let E₁ and E₂ be two independent variables.

Further, E^'₁ and E^'₂ denote the complements of E₁ and E₂ respectively.

Concept:

Multiplication rule of probability for independent events -

• If the outcome of one event does not affect the outcome of another, then those events are referred to as independent events. 
• The multiplication rule of probability for dependent events can be extended to the independent events.
• We have P(A ∩ B) = P(A)P(B|A), so if the events A and B are independent then P(B|A) = P(B), and thus P(A ∩ B) = P(A).P(B).

Calculation:

Let P(E₁) = m 

⇒ P(E₁^') = 1 - m

⇒ \(P(E'_1\ {\cap}\ E_2) = P(E'_1).P(E_2) = \frac{2}{15}\) 

⇒ \(P(E'_1).P(E_2) = \frac{2}{15}\)

⇒ \((1 - m)P(E_2) = \frac{2}{15}\)

⇒ \(P(E_2) = \frac{2}{15(1 - m)}\)

⇒ \(\rm P(E_1\cap E_2^{'})=\frac{1}{6} \)

⇒ \(\rm P(E_1\cap E_2^{'}) = P(E_1).(E'_2)=\frac{1}{6}\)

⇒ \( P(E_1).(E'_2)=\frac{1}{6}\)

⇒ \(m[1 - \frac{2}{15(1 - m)}] = \frac{1}{6}\)

⇒ 6m[15(1 - m) - 2] = 15(1 - m)

⇒ 2m(13 - 15m) = 5 - 5m

⇒ 30m² - 31m + 5 = 0

⇒ \(m = \frac{5}{6} \ or\ \frac{1}{5}\)

⇒ \(P(E_1) = \frac{1}{5}\ or \ \frac{5}{6}\)

∴ The value of P(E₁) is 1/5

Concept:

Let * be a binary operation on a non-empty set S. If there exists an element e in S such that a * e = e * a = a ∀ a ∈ S. Then the element e is said to be an identity element of S with respect to *.

Calculation:

Given: * is a binary operation on Q where Q is set of rational numbers such that a * b = ab/2 ∀ a, b ∈ Q.

Let e be the identity element of Q with respect to *.

As we know that if e is an identity element of a non-empty set S with respect to a binary operation * then a * e = e * a = a ∀ a ∈ S.

Let a ∈ Q and because e is the identity element of Q with respect to given operation *.

.e a * e = a =  e * a ∀ a ∈ Q.

According to the definition of *, we have

⇒ a * e = ae/2 = a

⇒ e = 2 ∈ Q

Hence, 2 is the identity element of Q with respect to given operation *

The given line is

6x - 8 = 3y + 2 = 2z + 4

\(6\left( {x - \frac{8}{6}} \right) = 3\left( {y + \frac{2}{3}} \right) = 2\left( {z + 2} \right) \Rightarrow 6\left( {x - \frac{4}{3}} \right) = 3\left( {y + \frac{2}{3}} \right) = 2\left( {z + 2} \right) \Rightarrow \frac{{x - \frac{4}{3}}}{{\frac{1}{6}}} = \frac{{y\; + \;\frac{2}{3}}}{{\frac{1}{3}}} = \frac{{z\; + \;2}}{{\frac{1}{2}}}\)

Direction Ratios of the line are 1/6, 1/3, 1/2

\(\sqrt {\frac{1}{{{6^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^2}}}} = \sqrt {\frac{{1 + 4 + 9}}{{36}}} = \sqrt {\frac{{14}}{{36}}}\)

Direction Cosines of the line are

\(\frac{{\frac{1}{6}}}{{\sqrt {\frac{{14}}{{36}}} }},\;\frac{{\frac{1}{3}}}{{\sqrt {\frac{{14}}{{36}}} }},\;\frac{{\frac{1}{2}}}{{\sqrt {\frac{{14}}{{36}}} }} = \frac{{\frac{1}{6}}}{{\frac{{\sqrt {14} }}{6}}},\;\frac{{\frac{1}{3}}}{{\frac{{\sqrt {14} }}{6}}},\;\frac{{\frac{1}{2}}}{{\frac{{\sqrt {14} }}{6}}} = \frac{1}{6} \times \frac{6}{{\sqrt {14} }},\;\frac{1}{3} \times \frac{6}{{\sqrt {14} }},\;\frac{1}{2} \times \frac{6}{{\sqrt {14} }} = \frac{1}{{\sqrt {14} }},\;\frac{2}{{\sqrt {14} }},\;\frac{3}{{\sqrt {14} }}\)

Concept:

Domain of a Relation: 

Let R be a relation from set A to set B. Then, the set of all first components of the ordered pair belonging to relation R forms the domain of the relation R

i.e Domain (R) = {a : (a, b) ∈ R}.

Calculation:

Given: A = {1, 2, 3}, B = {1, - 1} and R is a relation from A to B such that {(x, y) : x ∈ A, y ∈ B and y = ix where i is the imaginary number}

As we know that, i² = - 1

When x = 0 ∈ A then y = i⁰ = 1 ϵ B ⇒ (0, 1) ϵ R

When x = 1 ∈ A then y = i¹ = i ∉ B ⇒ (1, i) ∉ R

When x = 2 ∈ A then y = i² = - 1 ∈ B ⇒ (2, - 1) ∈ R

When x = 3 then y = i³ = - i ∉ B ⇒ (1, - i) ∉ R

So, the given relation R can be re-written in roaster form as R = {(0, 1), (2, - 1)}

As we know that, Domain (R) = {a : (a, b) ∈ R}.

⇒ Domain (R) = {0, 2}

Hence, the correct option is 4.

Concept:

The general equation of a straight line is Y = m X + C , where m is the slope/gradient and ( 0 , C ) the coordinates of the Y-intercept and C is called the Y-intercept as show in the below image.

• The thick continuous line represents that the line is included then we use ≤ or ≥  to write the inequalities
• The dotted line represents that the line is excluded then we use < or > to write the inequalities.

Calculations:

The given image has three straight lines which together enclose the shaded region as shown in the image.

Let us consider one by one to find their inequalities

First, let us consider the green thick line

the y-intercept of this line is C = 3

and it has a negative slope m = \(\frac{-3}{1}\)  = - 3

And since the shaded region from the line is towards the origin and the line is thick we can write the equation of straight line as y ≤  - 3x + 3, as shown in the below image

Next, let us consider the violet thick line

the y-intercept of this line is C = - 4

and it has a no i.e., slope m = 0  

And since the shaded region from the line is towards the origin and the line is thick we can write the equation of straight line as y ≥ - 4, as shown in the below image.

Finally, 

consider the pink dotted line

the y-intercept of this line is C = + 3

and it has a positive slope m = \(\frac{3}{3}\) = 1

And since the shaded region from the line is towards the origin and the line is dotted i.e it is excluded, so we can write the equation of straight line as y < x + 3 as shown in the below image.

∴ The shaded region in the graph is shown by the inequalities y ≤ -3x + 3;  y ≥ - 4;  y < x + 3.

Hence, the correct answer is option 1).

Concept:

Linear Programming Problem (LPP):

• Identify the ‘n’ number of decision variables that govern the behavior of the objective function (which needs to be optimized).
• Identify the set of constraints on the decision variables and express them in the form of linear equations /inequations. This will set up our region in the n-dimensional space within which the objective function needs to be optimized.
• Don’t forget to impose the condition of non-negativity on the decision variables i.e. all of them must be positive since the problem might represent a physical scenario, and such variables can’t be negative.
• Express the objective function in the form of a linear equation in the decision variables.
• Optimize the objective function either graphically (corner method) or mathematically.

Calculation:

Objective function is profit Z = 3x + 4y.

Constraints:

2x + y ≤ 4 ...... (1)

x + 2y ≥ 12...... (2)

x, y ≥ 0... ... (3)

Graph:

Solving the lines in equation (1) and (2) simultaneously:

Multiplying equation (2) by 2 we get, 2x + 4y ≥ 24 ..... (4)

Subtracting equation 4 from equation 1

-3y = -20

⇒ y = \(\frac{20}{3}\)

Substituting this in equation (1), we get:

x = \(\frac{-4}{3}\)

∴ The lines intersect at x = \(\frac{-4}{3}\), y = \(\frac{20}{3}\).

All the points are shown in the graph below:

As you can see that there is no common feasible reason, so the given equation will have no feasible solution.

Concept:

Geometric Progression:

• The ratio between any two consecutive terms of a geometric progression is a fixed constant, called the common ratio (r) of the progression.

Logarithms:

• log a - log b = \(\rm \log {a\over b}\).

Determinants:

• A linear combination of the rows/columns does not affect the value of the determinant.
• If two rows/columns of a given matrix are interchanged, then the value of the determinant gets multiplied by -1.
• If a row/column of a given matrix is multiplied by a scalar k, then the value of the determinant is also multiplied by k.

Calculation:

Let the common ratio of the given G.P. be r.

Let D = \(\begin{vmatrix} \rm \ln a_1 & \rm \ln a_2 & \rm \ln a_3 \\ \rm \ln a_4 & \rm \ln a_5 & \rm \ln a_6 \\ \rm \ln a_7 & \rm \ln a_8 & \rm \ln a_9 \end{vmatrix}\).

Using C₁ → C₁ - C₂ and C2 → C2 - C3, we get:

⇒ D = \(\begin{vmatrix} \rm \ln a_1 -\ln a_2& \rm \ln a_2 -\ln a_3& \rm \ln a_3 \\ \rm \ln a_4 -\ln a_5& \rm \ln a_5 -\ln a_6& \rm \ln a_6 \\ \rm \ln a_7-\ln a_8 & \rm \ln a_8-\ln a_9 & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln {a_1 \over a_2}& \rm \ln {a_2 \over a_3}& \rm \ln a_3 \\ \rm \ln {a_4 \over a_5}& \rm \ln {a_5 \over a_6}& \rm \ln a_6 \\ \rm \ln {a_7 \over a_8} & \rm \ln {a_8 \over a_9} & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln r& \rm \ln r& \rm \ln a_3 \\ \rm \ln r& \rm \ln r& \rm \ln a_6 \\ \rm \ln r & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Using C₁ → C₁ - C₂, we get:

⇒ D = \(\begin{vmatrix} 0& \rm \ln r& \rm \ln a_3 \\0& \rm \ln r& \rm \ln a_6 \\ 0 & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Expanding along C₁, we get:

⇒ D = 0.

Concept:

• Here we have to use the multiplication and addition properties of the matrixes.
• Also, AI = IA      ...(1)

∵  (A - B)2 = (A - B) (A - B) 

⇒ (A - B)2 = (A2 - AB - BA + B2)      ....(2)

Calculation: 

Given:

A is a square matrix such that,

A² = A.

Now, (I - A)³ + A = (I - A)² (I - A) + A

By 1 and 2,

= (I² - 2AI + A²) (I - A) + A       [∵ A² = A]

= (I - 2A + A) (I - A) + A

= (I - A) (I - A) + A = (I² - 2AI + A²) + A

= (I - 2A + A) + A = I - A + A = 1       [∵ A² = A]

Concept:

Some properties of inverse

• A-1A = I where I is the identity matrix
• A-1 I = A^-1 where I is the identity matrix

Calculation:

∴ A satisfies the equation x2 - 4x + 3 = 0

⇒ A2 - 4A + 3I = 0

Multiply the above equation by A^-1, we get 

⇒ A^-1A2 - 4A^-1A + 3A^-1I = 0

⇒ A - 4I + 3A^-1 = 0

⇒ 3A-1 = 4I - A

⇒ A-1 = (4I - A) / 3

Hence, option 3 is correct.

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C

Calculation:

I = \(\rm\int{dx\over x^5 +x^3}\)

I = \(\rm\int{dx\over x^3(x^2 +1)}\) 

By partial fraction

I = \(\rm \int {x\over x^2+1}-{1\over x}+{1\over x^3}\) dx

I = \(\rm \int {x\over x^2+1}dx-\int {dx\over x}+\int{dx\over x^3}\)

Let x² + 1 = t ⇒ 2x dx = dt

I = \(\rm \int {dt\over2t}-\ln x+\int x^{-3}dx + C\)

I = \(\rm {1\over2}\ln t-\ln x+{x^{-2}\over-2} + C\)

Putting the value of t

I = \(\rm {1\over2}\ln (x^2+1)-\ln x-{1\over 2x^2} + C\)

Concept:

The tangent to a circle equation x² + y² = a² at (x₁, y₁) is xx₁ + yy₁ = a²

Area of triangle  \(\rm = \dfrac{1}{2} \times \text{Base} \times {Height}\)

Calculation:

The tangents to the circle x² + y² = 25 at (3, 4) is 3x + 4y = 25

I

Tangent cut on coordinate axes at A(\(\frac{25}{3},0\)) and B(\(0, \frac{25}{4}\))

AO = Base = \(\frac{25}{3}\)

OB = Height = \(\frac{25}{4}\)

Now,

Area of triangle \(\rm = \dfrac{1}{2} \times \text{Base} \times {Height}\)

\(\rm = \dfrac{1}{2} \times \dfrac{25}{3} \times \dfrac{25}{4} \\= \dfrac{625}{24}\)

Calculation:

From diagram, the two points are in feasible region A and B,

At A (0 , 1.33) and B (2 , 0)

Z_max = 3x + 7y

Therefore, Z_A = 3 × 0 + 7 × 1.33 = 9.33  and

Z_B = 3 × 2 + 7 × 0 = 6

∴ Zmax = Z_A = 9.33

Therefore only one having maximum value. So, it is having exactly one optimal solution. 

Calculation:

Following figure can be drawn for the solution of above question

The area of the region contained by the lines |x| - y ≤ 1, y ≥ 0 and y ≤ 1 is the shaded region

Required area = Area of ABCD

⇒ 1/2(2 + 4) × 1

⇒ 1/2 × 6

⇒ 3

∴ Answer is 3

Concept:

Let A = (x1 ,y1) and B = (x2 ,y2) be any two-point. let P (x,y) be any point on AB and 

By section formula we have, 

\(\rm x = \frac{m_2x_1+m_1x_2}{m_1+m_2}\) and \(\rm y = \frac{m_2y_1+m_1y_2}{m_1+m_2}\)

Calculations:

Given, the line y = 0 divides the line joining the points (3, -5) and (-4, 7).

consider, the line y = 0 divides the line joining the points (3, -5) and (-4, 7) at point P(x, y) in the ratio \(\lambda : 1\).

Now find the value of \(\rm\lambda\)

A = (3 ,5) = (x₁ ,y₁) and B = (- 4, 7) = (x2 ,y₂) 

By section formula we have, 

\(\rm x = \frac{m_2x_1+m_1x_2}{m_1+m_2}\) and \(\rm y = \frac{m_2y_1+m_1y_2}{m_1+m_2}\)

Consider, \(\rm y = \frac{m_2y_1+m_1y_2}{m_1+m_2}\)

⇒\(\rm 0 = \frac{7\lambda-5}{\lambda+1}\)

⇒ \(7\lambda - 5 = 0\)

⇒ \(\rm \lambda = \frac{7}{5}\)

Hence, the line y = 0 divides the line joining the points (3, -5) and (-4, 7) at point P(x, y) in the ratio \(\rm\frac {7}{5}: 1\) i.e. 7 : 5

Concept:

The adjoint of a square matrix is the transpose of its cofactor matrix.

The determinant of a square matrix is the same as the determinant of its transpose.

Determinant of the adjoint of matrix = (determinant of the matrix)\(\rm ^{n-1}\), where n = order of the matrix

⇒ |Adj A| =  |A|^n-1

Calculation:

Given that determinant of the matrix

 \(Δ=\rm \begin{vmatrix}a_1&b_1&c_1\\\ a_2 &b_2&c_2\\\ a_3&b_3&c_3\end{vmatrix}\) 

and order = n = 3

We know that determinant formed by replacing each of its elements by its cofactor = determinant of the transpose of adjoint

⇒ \(\rm \begin{vmatrix}A_1&B_1&C_1\\\ A_2 &B_2&C_2\\ A_3&B_3&C_3\end{vmatrix}\) = (Δ)^3-1

⇒ \(\rm \begin{vmatrix}A_1&B_1&C_1\\\ A_2 &B_2&C_2\\ A_3&B_3&C_3\end{vmatrix}\) = Δ²

∴ The value of the determinant is Δ2.

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

A differential equation is said to be linear when

• Dependent variable and its derivative should have power ‘1’.
• Dependent variable and its derivatives can have the product with the independent variable.
• Dependent variable and its derivatives can’t have the product.

Calculation:

Given differential equation is y''' - sin(y') + y = 0

∵ In the given equation the polynomial equation cannot be formed in y', so order and degree cannot be defined

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:

x + 4y - z = 0,

3x - 4y - z = 0,

x-3y + z = 0 

The given system of equations is homogeneous.

So, column matrix B will be a null matrix.

⇒  (adj A). B = 0 for all the entries of adj A.

Now, check for |A|,

\(|A| = \left| {\begin{array}{*{20}{c}} 3&{ - 2}&1\\ \lambda &{ - 14}&{15}\\ 1&2&{ - 3} \end{array}} \right| \)

⇒ |A| = 1(- 4 - 3) -4(3 + 1) -1(-9 + 4)

= - 7 - 16 + 5 ≠ 0

Concept:

Vector triple product:- The value of the vector triple product can be found by the cross product of a vector with the cross product of the other two vectors. It gives a vector as a result. When we simplify the vector triple product it gives us an identity name as BAC - CAB identity.

Given:

Let \(\vec a\), \(\vec b\) and \(\vec c\) be three unit vectors

\(\vec a \times (\vec b \times \vec c)=\frac{\sqrt3}{2}(\vec b+ \vec c)\)

Formula:

\(\vec a \times (\vec b \times \vec c)= (\vec a . \vec c)\vec b - (\vec a . \vec b)\vec c \)

Calculation:

Let the angle between \(\vec a\) and \(\vec c\) is θ

\(\vec a \times (\vec b \times \vec c)=\frac{\sqrt3}{2}(\vec b+ \vec c)\)

\( (\vec a . \vec c)\vec b - (\vec a . \vec b)\vec c = \frac{\sqrt3}{2}(\vec b + \vec c)\)

⇒ \((\vec a . \vec c) = \frac{\sqrt3}{2} \) or \((\vec a . \vec b ) =- \frac{\sqrt3}{2}\)

⇒ Angle between \(\vec a\) and \(\vec c\)

⇒ \((\vec a . \vec c) = \frac{\sqrt3}{2} \)

⇒ \(|a|.|c|.cosθ =\frac{\sqrt3}{2}\)

⇒ \(1.1.cosθ = \frac{\sqrt3}{2}\)

⇒ \(cosθ = \frac{\sqrt3}{2}\)

⇒ \(θ = \frac{π}{6}\)

∴ The angle between \(\vec a \) and \(\vec c\) is \(\frac{π}{6}\)

Concept:

The cartesian equation of line with direction ratio's a, b, c and passing through point A (x₁, y₁, z₁) is given by: \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

Calculation:

The given line is \(\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}\)

Let \(\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3} = λ\)

⇒ x = λ, y = 2λ + 1 and z = 3λ + 2

So, the general form of the point lying on the given line is: (λ, 2λ + 1, 3λ + 2)

Let N be the foot of the perpendicular drawn from point P (1, 6, 3) to the given line.

∵ Point N lies on the given line so it is of the form: (λ, 2λ + 1, 3λ + 2)

The direction ratios of PN is : (λ - 1), (2λ - 5), (3λ - 1)

The direction ratios of the given line is: 1, 2, 3

∵ PN perpendicular to the given line

⇒ 1 ⋅ (λ - 1) + 2 ⋅ (2λ - 5) + 3 ⋅ (3λ - 1) = 0

⇒ λ = 1

 So, the foot of perpendicular on the given line is N (1, 3, 5)

So, the distance between P (1, 6, 3) and N (1, 3, 5) is: \(\sqrt {{{\left( {1 - 1} \right)}^2} + {{\left( {3 - 6} \right)}^2} + {{\left( {5 - 3} \right)}^2}} = \sqrt {13} \;units\)

Concept:

The equation of a plane parallel to plane ax + by + cz + d = 0 is given by ax + by + cz + k = 0;

The distance of point p = (x₁, y₁, z₁) from a plane ax + by + cz + d = 0 is given by

\(d = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\)

Calculation:

Given equation of a plane parallel to x + 2y + 2z = 5

⇒ equation of plane = x + 2y + 2z + k = 0;

Given unit distance from origin i.e. (0, 0, 0);

\( \Rightarrow 1 = \frac{{\left| k \right|}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}\)

⇒ k = 3 or -3;

The final equation of plane will be

 x + 2y + 2z + 3 = 0 or x + 2y + 2z -3 = 0

CONCEPT:

The distance between two parallel lines such as  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\) is given by \(d = \left| {\frac{{\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right)}}{{\left| {\vec b} \right|}}} \right|\)

CALCULATION:

Given: Equation of lines L1 and L2 are \(\vec r = \;\hat i + \;2\hat j + \lambda \times \left( {2\hat i + \;3\hat j + \;6\hat k} \right)\;and\;\vec r = 3\hat i + \;3\hat j - 5\hat k + \mu \times \left( {2\hat i + 3\hat j + 6\hat k} \right)\)

As we can see that, the given equation of lines are parallel.

So, by comparing the given equation of lines with  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\)

Here, we have \(\vec b = \;2\hat i + 3\hat j + 6\hat k,\;\overrightarrow {{a_1}} = \;\hat i + 2\hat j\;and\;\overrightarrow {{a_2}} = \;3\hat i + 3\hat j - \;5\hat k\)

As we know that, distance between two parallel lines such as  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\) is given by \(d = \left| {\frac{{\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right)}}{{\left| {\vec b} \right|}}} \right|\)

Given:

(x₁, y₁) = (a, 0)

(x₂, y₂) = (0, b)

(x₃, y₃) = (1, 1)

Formula Used:

Area of triangle = (1/2)[x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

Concept:

If points are collinear, then the area of triangle is zero.

Calculation: 

Area of triangle = (1/2)[x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)] = 0

⇒ Area of triangle = (1/2)[a (b - 1) + 0 (1 - 0) + 1 (0 - b)] = 0

⇒ Area of triangle = (1/2)[ab - a - b] = 0

⇒ Area of triangle = ab - a - b = 0

⇒ ab = a + b

\(⇒ \frac{a + b}{ab}\) = 1

Hence, \( \frac{a + b}{ab}=1\)

Concept:

Symmetric Matrix:

• Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself
• AT = A or A’ = A

Where,

AT or A’ denotes the transpose of matrix

• A square matrix A is said to be symmetric if aij = aji for all i and j

Where aij and aji is an elements present in the matrix.

Skew-Symmetric Matrix or Anti-symmetric:

• Square matrix A is said to be skew-symmetric if aij = −aji for all i and j.
• Square matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A ⇔ AT = −A
• Note that all the main diagonal elements in the skew-symmetric matrix are zero.

Calculation:

Since A is a skew-symmetric matrix, therefore

\({A^T} = - A \)

\(\Rightarrow {({A^T})^n} = \,{(A)^n} = {({A^n})^T} = \left\{ {\frac{{{A^n},\,if\,n\,\,is\,even}}{{ - {A^n},\,if\,n\,is\,odd}}} \right..\)

Concept:

The probability mass function,  \(P(X=x)=f(x)\) , of a discrete random variable  X  is a function that satisfies the following properties:

\(P(X=x)=f(x)\geq 0\) for \(x\in Range (X)\)

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

Calculation:

Given:

Probability distribution of random variable X.,

⇒ Sum of probabilities = 1

Here, p(X = - 5) + p(X = - 4) + ... + p(X = 5) =1

i. e. P + 2p + 3p + ... +12p = 1

⇒ 72p = 1 

⇒ P = 1 / 72

CONCEPT:

\(\rm \frac{{d\left( {\sin x} \right)}}{{dx}} = \; \cos x\)

\(\rm \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)

\(\rm \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

CALCULATION:

Given: y = sin (log x)

First let's find out y₁

As we know that, \(\rm \frac{{d\left( {\sin x} \right)}}{{dx}} = \; \cos x\) and \(\rm \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

⇒ \(\rm y_1 = \frac{dy}{dx} = \frac{cos\ (log x)}{x}\)

Now, again by differentiating the above equation with respect to x we get,

As we know that, \(\rm \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)

⇒ \(\rm y_2 =\frac{d^2y}{dx^2} = \frac{-sin \ (log \ x) \ - \ cos \ (log \ x)}{x^2}\)

Now, x²y₂ + xy₁ = -y

Hence, correct option is 2.

Concept:

If the equation of the line is an \(\rm \vec r = \;\vec a + \lambda \;\vec b\) and equation of the plane is \(\rm \overrightarrow{r}. \overrightarrow{n}=d\)  then the angle α  between line and direction parallel to the plane is, 

\(\rm \sin α = \left | \frac{\overrightarrow{b}.\overrightarrow{n}}{\left | \overrightarrow{b} \right |\left | \overrightarrow{n} \right |} \right |\)   

Calculation: 

Equation of line with Z-axis is,

\(\rm \overrightarrow{b}= \hat{k}\)

Equation of plane is  x -3y + 5z -8 = 0 

or in  vector form,  \(\rm \overrightarrow{n}= \hat{i}-3\hat{j}+5\hat{k}\)   

∴   \(\rm \sin α = \left | \frac{\overrightarrow{b}.\overrightarrow{n}}{\left | \overrightarrow{b} \right |\left | \overrightarrow{n} \right |} \right |\)

⇒ \(\rm \sin α = \frac{\left ( \hat{i}-3\hat{j}+5\hat{k} \right ). \left ( 0\hat{i}+ 0\hat{j}+\hat{k} \right )}{\sqrt{1^{2}+(-3)^{2}+5^{2}}. \sqrt{1^{2}}}\) = \(\frac{5}{\sqrt{35}}\)

⇒ sin α =  \(\frac{5}{\sqrt{35}}\)

⇒ α = sin-1 (\(\frac{5}{\sqrt{35}}\) )  

⇒ α = \(\frac{5}{\sqrt{35}}\) 

The correct option is 3.

Concept:

A function f : A → B is called an onto function if every element of B is an image of at least one element of A i.e.

Calculation:

⇒ If n(P) = m and n(Q) = 2, then

⇒ (2^m - 2) onto functions are possible from P to Q, Hence m = 4

⇒ Thus 2⁴ - 2 = 14

∴ The required result will be 14.

Concept:

A function attains a minimum value at a given point if the double derivative of that function at that point is positive.

A function attains a maximum value at a given point if the double derivative of that function at that point is negative.

Calculations:

Given function is \(y = x^2 + \dfrac{250}{x}\)

Differentiating both sides w.r.t. x, we get,

\(dy \over dx \) = 2x - \(250 \over x^2\)

Putting \(dy \over dx \) = 0 , we get ,

2x - \(250 \over x^2\) = 0

Solving the above equation, we get, 

x = 5

Now, \(d^2y \over d^2x\) = 2 + \(500 \over x^3\)

Putting x = 5 in the above equation, we ge ,

\(d^2y \over d^2x\) = 6, which is positive. 

So, the function attains a minimum value at x = 5

concept:

• If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then the determinant of A is given by,

|A| = (a­11 × a22) – (a12 – a21)

Calculation:

Given:

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

⇒ det (A) = 2 log x - log x

⇒ 2 log x² - log x

⇒ 2 log \(\left( {\frac{{{x^2}}}{x}} \right ) \)⇒ 2 log x

⇒ log x² = 2 

⇒ x = e²

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = nCk pk (1 - p)(n - k)

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

Properties:

• Mean of the distribution (μX) = n × p
• The variance (σ2x) = n × p × (1 - p)
• Standard deviation (σx) = \(\sqrt{np(1-p)}\)

Calculation:

Given:

mean of BD = np = 15

And variance of BD = npq = 10

⇒ np(1 – p) = 10 (∵ p + q = 1)

\(\Rightarrow 1 - {\rm{p}}\frac{{10}}{{15}} = \frac{2}{3} \Rightarrow {\rm{P}} = 1 - \frac{2}{3}\)

\(\Rightarrow {\rm{p}} = \frac{1}{3}\)