Mathematics Mock Test - 11

Given:

\(2f(x) + f(\frac{1}{x}) = logx\)

Concept:

In such questions, we generally try to reduce the expression to a single function f(x) by elimination method.

Formula:

log(ex) = x

log(1/ex) = log(e-x) = -x

Calculation:

\(2f(x) + f(\frac{1}{x}) = logx\) - (i)

Replacing x by 1/x in (i),

⇒ \(2f(\frac{1}{x}) + f({x}) = log(\frac{1}{x})\) - (ii)

Now, 2 × (i) - (ii) :

⇒ 3f(x) = 2log(x) - log(1/x)

⇒ f(x) = \(\frac{1}{3}\)(2log(x) - log(1/x)) - (iii)

Replacing x by ex in (iii) -

⇒ f(ex) =  \(\frac{1}{3}\)(2log(ex) - log(1/ex)) 

⇒ f(ex) =  \(\frac{1}{3}\)(2x + x)

⇒ f(ex) =  x

Concept:

• cot^-1 (-x) = π - cot^-1x

• cot (π/3) = 1/√3 

Calculation:

Let y = \(\cot^{-1}\left(-\frac{1}{√{3}}\right)\)

\(⇒ y=π -\cot ^{-1}\frac{1}{√{3}}\)

⇒ y = π - π/3

⇒ y = 2π/3

Since range of cot^-1 is (0, π)

Hence, the principal value is \(\frac{2π}{3}\).

Concept:

If two columns (or rows) of a determinant are equal, then its value is zero.

Solution:

Δ = \(\begin {vmatrix} 0 & b-a & c - a\\ a-b & 0& c-b\\a-c & b-c& 0\end {vmatrix}\)

Applying C1 → C1 - C3 and C2 → C2 - C3

Δ = \(\begin {vmatrix} a-c & b-c & c - a\\ a-c & b-c& c-b\\a-c & b-c& 0\end {vmatrix}\)

Δ = (a - c)(b - c)\(\begin {vmatrix} 1 & 1 & c - a\\ 1 & 1& c-b\\1 & 1& 0\end {vmatrix}\)

∵ Two columns C1 and C2 are equal hence Δ = 0

Given:

• f(x) = \(\frac{log(1-x+x^2) + log(1+x+x^2)}{secx-cosx}\), x ≠ 0
• f(x) is continuous at x = 0

Concept:

If a function f(x) is continuous at x = a then -

1. LHL = RHL at (x = a), Limit exists at x = a
2. f(a) = Limit at (x = a) = A finite number

Solution:

Limit at x = 0 -

⇒ \(lim_{x \rightarrow 0} f(x)\) = \(lim_{x \rightarrow 0} \)\(\frac{log(1-x+x^2) + log(1+x+x^2)}{secx-cosx}\)

⇒ \(lim_{x \rightarrow 0} f(x)\) = \(lim_{x \rightarrow 0} \)\(\frac{cosx[log(1+x^2+x^4)] }{sin^2x}\)

⇒ \(lim_{x \rightarrow 0} f(x)\) = \(lim_{x \rightarrow 0} \)\(\frac{cosx[\frac{(x^2+x^4)log(1+x^2+x^4)}{x^2+x^4}] }{\frac{x^2(sin^2x)}{x^2}}\)

Given:

x2/3 + y2/3 = a2/3 - (i)

The tangent is drawn at (a, 0)

Concept:

The equation of the tangent at point (a, b) on a curve is given by : (y - b) = m(x - a) where m = \((\frac{dy}{dx})_{(a, b)}\)

Solution:

Differentiating (i) both the sides -

⇒ \(\frac{2}{3}\)x-1/3 + \(\frac{2}{3}\)y-1/3\((\frac{dy}{dx}) \)= 0

⇒ \(\frac{dy}{dx} \) = - \(^3\sqrt{\frac{y}{x}}\)

⇒ \(\frac{dy}{dx} \) (at x = a, y = 0) = m = 0

∴ Equation of the tangent : (y - 0) = m(x - a)

⇒ y = 0 which is the x-axis

Clearly, only point (2, 0) lies on the x-axis

Given:

• \(y = f(\frac{x+e^x}{e^x})\)
• f'(1) = 2

Formula:

\(d(\frac{u}{v})= \frac{vdu - u dv}{v^2}\)

Solution:

\(\frac{dy}{dx}\) = \( f'(\frac{x+e^x}{e^x})\) × \(\frac{e^x(1 + e^x)-(x + e^x)e^x}{e^{2x}}\)

⇒ \(\frac{dy}{dx}\)(at x = 0) = f'(1) × 1

⇒ \(\frac{dy}{dx}\)(at x = 0) = 2

Solution:

A - \(\int\limits_0^{\frac{π }{2}} {\frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx} \)

Dividing N^r and D^r by cos⁴x -

= \(\int\limits_0^{\frac{π }{2}} {\frac{{2tanxsec^2x}}{{{{\tan }^4}x + 1}}dx} \)

Putting tan²x = t

⇒ 2tanx sec²x dx = dt

= \(\int\limits_0^{∞} {\frac{dt}{t^2 + 1}}\)

= ₀|tan^-1t|^∞ 

= \(\frac{π }{2}\)

B - I =  \(\int\limits_0^{\frac{π }{2}} {\frac{{\sin x}}{{\sin x + \cos x}}dx} \) - (i)

Applying property \(\int_{a}^{b}f(x) dx = \int_{a}^{b} f(a + b -x) dx \)

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

Adding (i) and (ii) -

⇒ 2I = \(\int_{0}^{\frac{π}{2}} dx\) = \(\frac{π }{2}\)

⇒ I = \(\frac{π }{4}\)

C - I = \(\int\limits_0^{2π } {\frac{1}{{1 + {e^{\sin x}}}}dx} \) - (i)

Applying property \(\int_{a}^{b}f(x) dx = \int_{a}^{b} f(a + b -x) dx \)

⇒ I = \(\int\limits_0^{2π } {\frac{1}{{1 + {e^{-\sin x}}}}dx} \)

⇒ I = \(\int\limits_0^{2π } {\frac{e^{\sin x}}{{1 + {e^{\sin x}}}}dx} \) - (ii)

Adding (i) and (ii) - 

⇒ 2I = \(\int_{0}^{2π} dx\) = 2π 

⇒ I = \(π \)

D - I = \(\int\limits_0^{\frac{π }{2}} {\frac{1}{{9{{\sin }^2}x + 4{{\cos }^2}x}}dx} \)

Dividing Nr and Dr by cos²x -

⇒ I = \(\int\limits_0^{\frac{π }{2}} {\frac{sec^2x}{{5{{\tan }^2}x + 4sec^2x}}dx} \)

⇒ I = \(\int\limits_0^{\frac{π }{2}} {\frac{sec^2x}{{9{{\tan }^2}x + 4}}dx} \)
Putting tanx = t, sec²x dx = dt

⇒ I = \(\int_0^{∞} {\frac{dt}{9{t^2} + 4}}\)

⇒ I = \(\frac{1}{6}\)₀|tan^-1(\(\frac{3t}{2}\))|^∞ 

=  \(\frac{π }{{12}}\)

Given:

x = a(cos θ + θ sin θ)

y = a(sin θ - θ cos θ)

Solution:

\(\frac{dx}{dθ}\) = a(-sinθ + sinθ + θ cosθ)

= aθcosθ - (i)

\(\frac{dy}{dθ}\) = a(cosθ - cosθ + θ sinθ)

= aθsinθ - (ii)

Dividing (ii) by (i) -

\(\frac{dy}{dx}\) = tanθ 

⇒ \(\frac{dy}{dx}\) at (\(\theta = \frac{\pi }{4}\)) = 1

Given:

f(x + y) = f(x)f(y)

f(5) = 2

f'(0) = 3

Solution:

Putting x = y = 0 in f(x + y) = f(x)f(y) -

⇒ f(0) = [f(0)]2

⇒ f(0) = 0 or 1

Differentiating partially both sides wrt x and wrt y respectively -

f'(x + y) = f(y)f'(x) - (i)

f'(x + y) = f(x)f'(y) - (ii)

Dividing (i) and (ii) -

⇒ \(\frac{f'(x)}{f(x)} =\frac{f'(y)}{f(y)} = k (say) \)

⇒ \(\frac{f'(x)}{f(x)} = k \)      - (iii)         (as f'(0) = 3, ∴ f(0) = 0 rejected)

Integrating both sides -

⇒ ln|f(x)| = kx + c - (iv)

Putting f(5) = 2 and f'(0) = 3 -

⇒ k = 3 and c = \(ln2 - 15\)

Putting x = 5 in (iii)-

⇒ f'(5) = 3f(5)

⇒ f'(5) = 6

Concept:

• Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

• Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

• Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

• Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: A = {1, 2, 3} and relation R on A is defined as: R = {(1, 1), (1, 2), (2, 3), (2, 2), (3, 3), (3, 1)}

Reflexive:

As we can see that (a, a) ∈ R, ∀ a ∈ A i.e (1, 1), (2, 2) and (3, 3) ∈ R ⇒ R is reflexive.

Symmetric:

As we can see that for the given relation R, (1, 2) ∈ R but (2, 1) ∉ R and similarly, (3, 1) ∈ R but (1, 3) ∉ R.

So, there is atleast one (a, b) ∈ R but (b, a) ∉ R ⇒ R is not symmetric.

Transitive:

As we can see that for the given relation R, (1, 2) and (2, 3) ∈ R but (1, 3) ∉ R ⇒ R is not transitive.

So, the given relation R is reflexive but neither symmetric nor transitive.

Concept:

The optimal value of an objective function occurs at 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 cost of a table and a chair is 2500 and 500 Rs respectively.

• The maximum number of chairs and tables that can be used for that party is 60.
• Ashok can spend a maximum of money 60000 Rs.
• The profits are 25 Rs per chair and 75 Rs per table.

• Let, Ashok buys x chairs and y tables for the maximum profit.

Then objective function z = 25x + 75y

The constraints are : 

500x + 2500y ≤ 60000 or x + 5y ≤ 120

x + y ≤ 60

x, y ≥ 0

Explanation:

Making the graph of the constraints -

• The corner points of the feasible region are (0 , 0) , (0 , 24) , (60 , 0) and (45 , 15)
• Values at these corner points for z = 25x + 75y are 0 , 1800 , 1500 and 2250 (maximum)
• Hence maximum profit is Rs 2250.

Concept:

• The system of n linear equations AX = B in n variables is -
• consistent and has a unique solution if |A|≠ 0.
• consistent and has infinitely many solutions if  |A| = 0 and (adj A)B =O. 
• inconsistent and has no solution if |A| = 0 and (adj A)B ≠ O.

Calculation:

Given:

\(A=\begin{pmatrix} 1&1 &1 \\ 1&2 &3 \\ 1&4 & λ \end{pmatrix}\) and \(B=\begin{pmatrix} 4\\ 7\\\mu \end{pmatrix}\)

The determinant of the matrix A is given by,

\(|A|=\begin{vmatrix} 1 &1 &1 \\ 1&2 &3 \\ 1&4 &λ \end{vmatrix}\)

⇒ \(|A|=\begin{vmatrix} 1 &1 &1 \\ 0&1 &2 \\ 0&3 &λ-1 \end{vmatrix}\)  Applying R₂→ R₁ - R₂ and R3→ R1 - R3

⇒|A| = (λ -1) -6+0+0 = λ  -7 

• For unique solution,

|A|≠ 0 

⇒ λ ≠ 7

• Therefore option 2 is correct.

Concept: 

• In order to find the global maximum and minimum of f in [a,b], we find the stationary points (where the first derivative vanishes) in [a,b].
• Further, we evaluate the value of the function at those stationary points, say f(c1),f(c2),..., and f(cn).
• Then the absolute maximum M  = max {f(a),f(c1),f(c2),..., f(cn),f(b)}
• The absolute minimum m = min {f(a),f(c1),f(c2),..., f(cn),f(b)}

Calculation:

Given:

\(f(x)=\sin^2x+\cos x\)

• Differentiating with respect to x,

f'(x)= 2sin x cos x - sin x 

• For stationary points,

f'(x) = 0

⇒ 2 sin x cos x - sin x = 0

⇒ sin x(2cos x - 1) = 0

⇒ sin x = 0, cos x =1/2

For  sin x = 0,

⇒ x = 0 and x = π  on  [0, π ]

For cos x =1/2,

⇒ x = π/3  on  [0, π ]

Combing all the values,

⇒ x = 0, π/3 and x = π  on  [0, π]

• The value of the function at these points,

f(0) = 0 + 1 = 1

f(π) = 0 -1 = -1 

f(π/3) = 3/4 + 1/2 = 5/4

The least value of the function m = -1 

• The greatest value of the function M =  5/4

∴ M - m = 5/4 - (-1) = 9/4

⇒  \(\frac{a-4}{4}\) = 9/4

⇒  a - 4 = 13

⇒ a = 13

• Therefore option 4 is correct.

Concept:

• In order to find the global maximum and minimum of f in [a,b], we find the stationary points (where the first derivative vanishes) in [a,b]. Further, we evaluate the value of the function at those stationary points, say f(c₁),f(c₂),..., and f(c_n).
• Then the absolute maximum M  = max {f(a),f(c1),f(c2),..., f(cn),f(b)}
• The absolute minimum m = min {f(a),f(c1),f(c2),..., f(cn),f(b)}

Calculation:

Given:

\(f(x)=2x^3-9x^2+12x+6\)

• Differentiating with respect to x,

\(f'(x)=6x^2-18x+12\)

• For stationary points,

\(f'(x)=0\)

⇒ \(6x^2-18x+12=0\)

⇒ \(x^2-3x+2=0\)

⇒ \((x-1)(x-2)=0\)

⇒ \(x=1,x=2\)

f(1) = 2(1)³ - 9(1)² + 12(1) + 6 =2 - 9 + 12 + 6 = 11

f(2) = 2(2)3 - 9(2)2 +12(2) + 6 =16 - 36 + 24 + 6 = 10

f(3) = 2(3)3 - 9(3)2 + 12(3) + 6 = 54 - 81 + 36 + 6 = 15

• Thus, the absolute minimum is 10 in 1,3] and the absolute maximum is 15 in [1,3].
• Therefore option 1 is correct.

Concept:

• A relation is an equivalence relation if it is reflexive, symmetric, and transitive.
• If aRmb if and only if a ≡  b mod.
• So, a and b have the same "remainder" when they are divided by m.
• We can also write this as,

a = k m + b      ...(1)

or m|(a - b)      ...(2) which means m divides (a-b)

Calculation:

Given: a Rmb if and only if a ≡  b mod.

• Since m|(a-a) = 0, we have a ≡ a mod m
• So  Rm is reflexive.
• if m|(a-b), then

​⇒ m|(-1)(a-b).

⇒ m|(b-a).

​⇒ So, if a ≡  b mod m​​ ⇒ b ≡  a mod m.

• Therefore, Rm is symmetric.
• If aR_mb and bR_mc 

​⇒ m|(a-b) and m|(b - c).

​⇒ m|[(a-b) + (b-c)] 

⇒ m|(a-c) ⇒  a Rmc

• Therefore Rm is transitive.
• Hence the relation is an equivalence relation.
• Therefore option 4 is correct.

Concept:

• For finding the absolute maximum or absolute minimum of a function over the interval, we first find the stationary points (where the first derivative of the function vanishes i.e. f'(x) =0).

Calculation: 

Given:

\(f(x)=\sin x-√{3}\cos x-x\)

• The first derivative of the function is given by,

\(f'(x)=\cos x+√{3}\sin x-1\)

• For stationary points,

f'(x) = 0

⇒ cos x + √3 sin x -1 =0

⇒ cos x + √3 sin x =1

\( ⇒ \frac{1}{2}\cos x+\frac{√{3}}{2}\sin x=\frac{1}{2}\)

\(\Rightarrow \cos\frac{π}{3}\cos x+\sin \frac{π}{3}\sin x=\cos\frac{{π}}3{}\)

⇒ \(cos\left(x-\frac{π}{3}\right)=cos\frac{π}{3}\)

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

∴ \(x=\frac{2π}{3}\) and \(x=2π\)

• So, x = 2π/3 for [0,π]
• Therefore option 1 is correct. 

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: 

X + 2y ≥ 2        ....(1),

5x + 4y ≤ 20        ....(2)

2x –y ≤  4        ....(3)

x ≥ 0, y ≥ 0

After drawing the given constraints, we can see that (3,2)  is one of the feasible solutions for LPP with constraints.

Concept: 

• The system of n linear equations AX = B in n variables is -
• consistent and has unique solution if |A| ≠ 0
• consistent and has infinitely many solutions if  |A| =0  and (adj A)B = 0. 
• inconsistent and has no solution if  |A| =0 and (adjA)B = O.

Calculation:

Calculation:

Given:  The equations are -

4x - 2y = 3

6x + α y = 5

So, \(A=\begin{pmatrix} 4 & -2\\ 6& \alpha \end{pmatrix} \) and \(B=\begin{pmatrix} 3\\5 \end{pmatrix}\)

The determinant of the matrix A,

\(⇒|A|=\begin{vmatrix} 4 &-2 \\ 6&\alpha \end{vmatrix}\)

⇒ |A|=4(α)- 6 (- 2) = 4 α +12

For unique solutions, 

⇒ |A| ≠ 0

⇒ 4 α +12 ≠ 0

⇒ α ​ ≠ - 3

• Therefore option 2 is correct.

Concept:

Equation of plane passing through \(\left( {{{\rm{x}}_1},{{\rm{y}}_1},{{\rm{z}}_1}} \right),{\rm{\;}}\left( {{{\rm{x}}_2},{{\rm{y}}_2},{{\rm{z}}_2}} \right){\rm{\;and\;}}\left( {{{\rm{x}}_3},{{\rm{y}}_3},{{\rm{z}}_3}} \right)\) is given by:

\(\left| {\begin{array}{*{20}{c}} {{\rm{x}} - {{\rm{x}}_1}}&{{\rm{y}} - {{\rm{y}}_1}}&{{\rm{z}} - {{\rm{z}}_1}}\\ {{{\rm{x}}_2} - {{\rm{x}}_1}}&{{{\rm{y}}_2} - {{\rm{y}}_1}}&{{{\rm{z}}_2} - {{\rm{z}}_1}}\\ {{{\rm{x}}_3} - {{\rm{x}}_2}}&{{{\rm{y}}_3} - {{\rm{y}}_2}}&{{{\rm{z}}_3} - {{\rm{z}}_2}} \end{array}} \right| = 0\)

Calculation:

Here, Plane passing through points A (2, 2, 1), B(3, 4, 2) and C (7, 0, 6)

Equation of plane = \(\left| {\begin{array}{*{20}{c}} {{\rm{x}} - 2}&{{\rm{y}} - 2}&{{\rm{z}} - 1}\\ {3 - 2}&{4 - 2}&{2 - 1}\\ {7 - 3}&{0 - 4}&{6 - 2} \end{array}} \right| = 0\)

⇒ (x - 2)(8 – (-4)) – (y - 2)(4 - 4) + (z - 1)(-4 - 8) = 0

⇒ 12x – 24 - 12z + 12 = 0

⇒ x – z - 1 = 0

Now, to find point which lies on plane lets check from options, As if the point lie on the plane then must satisfy the equation of plane. 

So put the value of given points in the equation of plane.

(1, 0, 0):  x – z - 1 = 0 ⇒ 1 – 0 - 1 = 0

This point satisfies the equation

Concept:

General equation: a(x − x₁) + b(y − y­₁) + c(z – z₁) = 0, Where a, b, c are direction ratios of normal to plane

Calculation:

We have, equation of plane: x – z - 1 = 0

This can be written as, 1(x - 0) + 0 (y - 0) + (-1) (z - 0) – 1 = 0

∴ Direction ratios of normal: <1, 0, -1>

Given:

The objective function z = 4x + 6y

The constraints are -

3x + 2y ≤ 12,

x + y ≥ 2,

and x, y ≥ 0

Concept:

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

Solution:

Making the graph of the constraints -

We see that the feasible region has corner points (0, 2), (2, 0), (4, 0) and (0, 6)

The values of the objective function z = 4x + 6y at these points respectively are 12, 8, 14 and 36

Hence maximum value occurs at (0, 6) and is equal to 36

Calculation:

Since, α = β = γ 

⇒ cos² α + cos² α + cos² α = 1

\( \Rightarrow α = {\cos ^{ - 1}}\left( { \pm \frac{1}{{\sqrt 3 }}} \right)\)

Possible direction cosines are,

\(\left( {\pm \frac{1}{{\sqrt 3 }},\pm \frac{1}{{\sqrt 3 }},\pm \frac{{ - 1}}{{\sqrt 3 }}} \right)\)

So, there are four lines whose direction cosines are,

\(\left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right),\left( {\frac{{ - 1}}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right),\left( {\frac{1}{{\sqrt 3 }},\frac{{ - 1}}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)\), \(\left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{{ - 1}}{{\sqrt 3 }}} \right)\)

Thus, four lines are equally inclined to axes.

Solution:

A - \(f\left( x \right) = \frac{1}{{\sqrt {x - 2} }} + \frac{1}{{\sqrt {3 - x} }}\)

Clearly f(x) is defined for (x - 2) > 0 and (3 - x) > 0

For (x - 2) > 0, x > 2 - (i)

For (3 - x) > 0, x < 3 - (ii)

Taking intersection of (i) and (ii) -

Domain f(x) = (2, 3)

B - \(f\left( x \right) = \sqrt {\frac{{x - 2}}{{3 - x}}} \)

Clearly f(x) is defined for \({\frac{{x - 2}}{{3 - x}}} \) ≥ 0 and 3 - x ≠ 0

Making wavy curve for  \({\frac{{x - 2}}{{3 - x}}} \)  -

⇒ Domain f(x) = [2, 3)

C - \(f\left( x \right) = \sqrt {\frac{{3 - x}}{{x - 2}}} \)

Clearly f(x) is defined for \({\frac{{3-x}}{{x-2}}} \) ≥ 0 and x - 2 ≠ 0

Making wavy curve for  \({\frac{{3-x}}{{x-2}}} \)  -

⇒ Domain f(x) = (2, 3]

D - \(f\left( x \right) = \sqrt {\left( {x - 2} \right)\left( {3 - x} \right)} \)

Clearly f(x) is defined for (x - 2)(3 - x) ≥ 0 

Making wavy curve for (x - 2)(3 - x) - 

⇒ Domain f(x) = [2, 3]

Given:

∫ \(\frac{√{x}}{√{1-x^3}} dx\) = \(\frac{2}{3}\)gof(x) + C

Solution:

Putting x3 = (x3/2)2 and then putting x3/2 = t, √xdx = \(\frac{2}{3}\)dt

⇒ I = ∫ \(\frac{√{x}}{√{1-x^3}} dx\) 

After substitution -

I = \(\frac{2}{3}\)∫ \(\frac{1}{\sqrt{1-t^2}} dt\)

⇒ I = \(\frac{2}{3}\)sin-1(t) + C

⇒ I = \(\frac{2}{3}\)sin-1(x3/2) + C

On comparing with \(\frac{2}{3}\)gof(x) + C

⇒ g(x) = sin-1(x) and f(x) = x3/2

Concept:

• If the equation of a plane is ax + by + cz = d, then (a, b, c) are the drs of the normal to the plane.
• The equation of a line passing through (x₁, y₁, z₁) and having drs (a, b, c) is given by \(\frac{x - x_1}{a}=\frac{y-y_1}{b}= \frac{z-z_1}{c}\)
• The nearest points between two lines are those points on the lines that on joining become perpendicular to both the lines.

Calculation:

The drs of the normal to the plane x - 3y + 2z = k are (1, -3, 2)

∵ The line L is normal to the plane, its drs will be parallel to the normal to the plane

∴ drs of the line L are (1, -3, 2)

⇒ Equation of line L passing through (1, 2, -2) is \(\frac{x - 1}{1}=\frac{y-2}{-3}= \frac{z+2}{2}\)

Any point A on the line L can be taken as (λ +1, -3λ +2, 2λ -2)

and any point B on the y-axis can be taken as (0, μ, 0)

For the shortest distance AB must be perpendicular to both L and y-axis

DRS of AB = (λ + 1, -3λ +2 - μ, 2λ -2)

Applying perpendicularity condition with the line L : (1)(λ + 1) + (-3)(-3λ +2 - μ) + (2)(2λ -2) = 0

⇒ 14λ + 3μ = 9 - (i)

Applying perpendicularity condition with the y-axis : (0)(λ + 1) + (1)(-3λ +2 - μ) + (0)(2λ -2) = 0

⇒ 3λ + μ = 2 - (ii)

Solving (i) and (ii) -

λ = 3/5 and μ = 1/5

∴ The point on the line L nearest to y-axis is (λ +1, -3λ +2, 2λ -2)

= (8/5, 1/5, -4/5)

Concept:

The equation of the plane passing through the intersection of two planes is P1 + λP2 = 0

Solution:

Given:

P1 : x + y + z - 6 = 0 and P2 : 2x + 3y + 4z + 5 = 0

Calculation:

The equation of the plane passing through the intersection of two planes is P1 + λP2 = 0

⇒ (x + y + z - 6) + λ(2x + 3y + 4z + 5) = 0

As it contains origin, hence (0, 0, 0) must satisfy it.

⇒ -6 + 5λ = 0

⇒ λ = 6/5

Hence equation of lane P is 17x + 23y + 29z = 0 

Concept:

The reflection of a point in a plane can be calculated by -

• Finding equation of line through the given point and perpendicular to the plane
• Taking a general point on the line and satisfying it with the equation of the plane
• The point on the plane so obtained will act as mid-point for the given point and its reflection.

Solution:

DRS of plane P₂ : 2x + 3y + 4z + 5 = 0 are (2, 3, 4)

Equation of line passing through (0, 0, 0) and having drs 2, 3, 3 is given by :

\(\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = k\)

Any general point on it will be (2k ,3k ,4k)

Satisfying it on the plane -

⇒ 2(2k) + 3(3k) + 4(4k) + 5 = 0

⇒ 29k = -5

or k = -5/29

∴ the point on the plane will be (-10/29, -15/29, -20/29)

This point will act as mid-point, so using mid-point formula : 

⇒ -10/29 = (x + 0)/2

or x = -20/29

⇒ -15/29 = (y + 0)/2

or y = -30/29

⇒ -20/29 = (z + 0)/2

or z = -40/29

Concept:

• Unit vector perpendicular to vectors \(\vec{a}\) and \(\vec{b}\) is given by : \(± \frac{\vec{a}× \vec{b}}{|\vec{a}×\vec{b}|}\)

• If \(\vec{a} = a_x{\hat{i}} + a_y{\hat{j}} + a_z{\hat{k}}\) and \(\vec{b} = b_x{\hat{i}} + b_y{\hat{j}} + b_z{\hat{k}}\) then,

​\(\vec{a}\times\vec{b} = \begin{vmatrix} {\hat{i}} &{\hat{j}} &{\hat{k}}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{vmatrix}\)

Calculation:

Let \(\vec{a}\) = \(\widehat i +\widehat j + \widehat k\space \) be the vector normal to the plane P₁

And,let  \(\vec{b}\) = \( \widehat i -3\widehat j +2\widehat k\) be the vector in the direction of line L

then \(\vec{a}\times\vec{b} = \begin{vmatrix} {̂{i}} &{̂{j}} &{̂{k}}\\1& 1& 1\\1&-3&2\end{vmatrix}\)

= 5î - ĵ - 4k̂ 

⇒ \(|\vec{a}\times\vec{b}|\) = √42
∴ The unit vector perpendicular to both the vectors is \(± \frac{\vec{a}× \vec{b}}{|\vec{a}×\vec{b}|}\)

= ±  \(\frac{{( 5 \widehat i - \widehat j -4 \widehat k)}}{{\sqrt {42} }}\)

Calculation:

Given

Probability of A to fail is, P(A) = 0.2, 

Probability of A to pass is, P(A') = 0.8

Probability of B to fail is, P(B) = 0.3

Probability of B to pass is, P(B') = 0.7

Case 1: A Fails B pass

P(A ∩ B’) = P(A)P(B’) = (0.2)(0.7) = 0.14

Case 2 : A Pass B fails

P(A’ ∩ B) = P(A’)P(B) = 0.8 (0.3) = 0.24

Probability that either A or B fails is

Concept:

• Since direction ratios of the line are also the direction ratios of normal to the plane.
• So the direction of the line joining these two points (1, 2, 0) and (4, 13, 5) is given by,

= (4 -1), (13 - 2), (5 - 0) i.e., (3, 11, 5)

Explanation:

∴ Required coefficients of x, y, and z will be the direction ratios of the normal to the plane which will be,

= (4 -1), (13 - 2), (5 - 0) i.e., (3, 11, 5).

Concept:

To find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.

Calculation:

Required area = 

\(= \rm \int _0^\frac{\pi}{4}(cos x-sinx)dx\\ =[sinx+cosx]_0^\frac{\pi}{4}\\ =[\frac{1}{\sqrt2}+\frac{1}{\sqrt2}-1]\\ =\sqrt2-1\)

Concept:

Corner point method:

•  As per the method, the coordinates of all corner points of the feasible region are determined and the value of the objective function at these points is computed since the optimal value of the objective function lies at one of the corner points as per the LP theory.
• The first step to finding the solution to an LP problem is to determine the feasible region. It is obtained by drawing the X-Y coordinates and plotting all the constraints.
• The common region on the graph obtained that satisfies all the constraints will be the feasible region.
• If no feasible region is obtained via plotting the constraints, then the linear program is infeasible.
• All the possibilities are summarized below: 

• In the case of a feasible region problem, if the value of the objective function is different at all points, then the point which gives the maximum and minimum value will be the optimal solution.
• In the case of a feasible region problem, if the optimal value of the objective function is the same at more than one point, then the solution will the entire line joining those points at which the optimal value of the objective function is the same.

Draw the constraints to find the feasible region:

• First, draw the equation form of the inequalities to draw 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 following problem has a feasible region.
• The inequality constraints are 2x + 2y ≤ 5 and 4x - 3y ≥ 3.
• Corner points are the origin, A, B, and C.
• A is the point where line 2x + 2y = 5 intersects the Y-axis.
• A is calculated by putting x = 0 in the constraint, hence A is (0, \(\frac{5}{2}\)).
• Similarly, B is calculated by putting y = 0 in 4x - 3y = 3, since B is the point where line 4x - 3y = 3 intersects the X-axis.
• Hence B = (\(\frac{3}{4}\), 0).
• C is the intersection of two inequality constraints, hence C is obtained by solving 2x + 2y = 5 and 4x - 3y = 3 simultaneously.   
• Solving the two constraints, we get C = (\(\frac{3}{2}\), 1).
• Below is the table showing the corner points and the value of the objective functions at those points.

• The table indicates that the objective function has a maximum value of \(\frac{5}{2}\) at points A and C.
• Hence, the maximum value of the function lies on any point on the line joining points A and C.
• Hence the correct answer is option 3.

Concept:

• Perpetuity: A perpetuity is an annuity where payments continue forever.
• We can not calculate the future value of the Perpetuity where payments continue forever but we can calculate the present value of the perpetuity.
• Amount of a Perpetuity: The amount of perpetuity is undefined since it increases beyond all bounds as time progresses.
• Present value of Perpetuity: We consider two types of perpetuity which are as follows:
• Present value of a perpetuity of Rs. R payable a the end of each period is given by the following equation,

P = \(\frac{R}{i}\)      ....(1)

• Where R = size of each payment at the end of each period
• i = interest rate for one period
• Perpetuity of Rs. R payable at the beginning of each period is given by the following equation,

P = R + \(\frac{R}{i}\)       ....(2)

Explanation:

Given:​

• P (when payment is done at beginning of each period) = R + \(\frac{R}{i}\)
• Here is the present value of the perpetuity,

\(\frac{R}{i}\) =  50,000, and R = Rs 1,000

• Hence the present value of the perpetuity when payments are done at the beginning of each period = Rs 50,000 + Rs 1,000 = Rs 51,000.
• Hence the correct answer is option 2.

Concept:

• The point that lies in a region bounded by any lines will satisfy the inequation of all the lines.
• The point does not lie in the bounded region if any of the inequations are unsatisfied.
• For example, (1, 2) lies in the region bounded by the lines, x + y ≤ 3 and  -x + y ≥ 0, since putting 1 in place of x and 2 in place of y in the inequations will give 3 ≤ 3 and 1 ≥ 0, which are true.

Calculation:

Given: The pair of lines are 7x + y ≥ 40 and 2x + 3y ≤ 25.

• The first point is (0, 0). Putting x = 0 and y = 0 in 2x  + 3y ≤ 25 gives 0 ≤ 25 which is true. Hence the point (0, 0) lies in 2x  + 3y ≤ 25.
• However, putting x = 0 and y = 0 in 7x  + y ≥ 40 gives 0 ≥ 40, which is not true. Hence the point does not lie in 7x + y ≥ 40.
• Overall, the point (0, 0) does not lie in the region bounded by 7x + y ≥ 40 and 2x + 3y ≤ 25.
• Similarly, for point (8, 5), putting x = 8 and y = 5 in both inequations give 61 ≥ 40 and 31 ≤ 25, out of which the latter is not true. Hence (8, 5) does not lie in the bounded region.   
• For point (3, 7), putting x = 3 and y = 7 in both inequations give 28 ≥ 40 and 27 ≤ 25, which are not true. Hence (3, 7) does not lie in the bounded region.
• For point (6, 4), putting x = 6 and y = 4 in both inequations give 46 ≥ 40 and 24 ≤ 25. Both the conditions are true, hence (6, 4) lie in the bounded region.
• So, the correct answer is option 4.

Concept:

Vector along the line joining A, B,
\(\vec {AB} = (x_2 - x_1 ) \hat i + (y_2 - y_1)\hat j + (z_2 - z_1)) \hat k \)

Calculation:

The vector along the line joining A and B will have the projection s along the x, y, and z-axis respectively will be,

Here, x₂ - x₁ = 6,

y₂ - y₁ = 6,

z₂ - z₁ = 4

∴ Projections of line AB on co-ordinate axes are 6, 6, and 4 respectively.

Concept:

• In an event, a discrete random variable X whose possible finite values x₁, x₂, x_3,…, x_n occur with the probabilities  p₁, p₂,p_3,…, p_n such that ∑p_i = 1.
• The mean or the mathematical expectation or expected value is the weighted average of all possible values of X which is given by:
• E(X) = x₁p₁ + x₂p₂ + x₃p₃ +…+ x_np_n = \(\sum_{i = 1}^{n}x_{i}p_{i}\)

Calculation:

Given:

• Coin is tossed three times.
• Hence all the possible outcomes = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}
• The total number of outcomes = 8.
• x_i will be equal to the number of tails in each of the 8 possible outcomes.  
• To calculate the mathematical expectation, the calculations are done and summarized in the table below:

• Thus mathematical expectation can be given by:

E(x) = \(\frac{3}{8}\) + \(\frac{6}{8}\) + \(\frac{3}{8}\) = \(\frac{3}{2}\).

• Hence, the correct answer is option 4. 

Concept:

• A sinking fund is a sum of money created by saving small amounts over a period of time and is completely separate from the savings account or the emergency fund.
• Its purpose is to pay off a financial obligation at some future designated date.
• A sinking fund can be used to pay for home repairs, save for a new car, pay for vacation, or cover large medical bills.
• The sinking fund is calculated by the formula mentioned below:

A = R \(\times\) Sn|i, ….(1)

• Where, Sn|i = \(\frac{(1+i)^{n} - 1}{i}\)
• n is the total number of time periods
• i is the interest rate.
• R is the amount of periodic payment done.

Calculation:

Given:

• This is a sinking fund problem.
• Total college expenses = Rs 4,00,000.
• Due to 50% of the subsidy, the actual amount to be accumulated by Nisha will be A = 0.5 × 4,00,000 = 2,00,000
• Annual interest rate, i = 8%.
• Time period after which the amount of the fee needs to be accumulated, n = 4 years
• According to the formula for sinking fund, we can write,

A = R \(\frac{(1 + i)^{n}-1}{i}\)

⇒ 2,00,000 =  R × \(\frac{1.08^{4}-1}{0.08}\)

⇒ 2,00,000 =  R × 4.5

⇒ R = \(\frac{2,00,000}{4.5}\) = 44,444

• Hence the correct answer is option 4.

Concept:

• Value of any objective function like Z = ax + by at any point can be found by putting the values of x and y in the equation of the objective function.
• For example value of Z = 3x + 4y at (2, 5) is Z = 3 × 2 + 4 × 5 = 6 + 20 = 26.

Calculation:

Given:

• The objective function is Z = ax + by has its maximum value at points (9, 3) and (5, 7).
• Hence we can write the following,

a \(\times \) 9 + b \(\times \) 3 = a \(\times \) 5 + b \(\times \) 7

\(\Rightarrow\) 9a + 3b = 5a + 7b

\(\Rightarrow\) 4a - 4b = 0 ....(1)

• Since a, b ≥ 0 and ab = 16, we can rewrite equation (1) as,

\(\Rightarrow\) 4a - 4 \(\times\) \(\frac{16}{a}\) = 0

\(\Rightarrow\) 4a - \(\frac{64}{a}\) = 0

\(\Rightarrow\) 4a² - 64 = 0

\(\Rightarrow\) a² - 16 = 0

\(\Rightarrow\) a = 4

• a = 4 gives b = 4, since ab = 16.
• Hence, the objective function is 4x + 4y
• Maximum value of the objective function is given by,

Z_max = 4 \(\times\) 9 + 4 \(\times\) 3 = 36 + 12 = 48.

• Hence the correct answer is option 1.

Concept:

• Any company or business has to spend certain costs for running its business. These costs are classified as fixed costs and variable costs. 
• Fixed Cost: These are the expenses that are independent of the production output units. Even with zero production, this cost has to be paid regularly. 
• Variable Cost: These expenses are dependent on the production output and will increase with an increase in production. Some examples are the cost of raw material, packaging costs, energy consumption per unit cost, etc.

• Cost Function or Total cost (C(x)): It is represented in terms of the sum of variable cost V(x) for producing ‘x’ units and fixed cost k as below:

C(x) = V(x) + k ….(1)  

• Marginal Cost: It is the increase in total cost due to a unit increase in output. It is calculated by deriving the partial derivative of the cost function with respect to output. 

Calculation: 

Given:

• Number of battery units produced = 50 units, k = Rs 2000
• Variable Cost, V(x) = 3x² + 3500
• From equation (1),
• Total Cost Function,

C(x) = Fixed Cost + Variable Cost

⇒ C(x) = 2000 + 3x ² + 3500 = 5500 + 3x²

• C(x) at 50 units = 5500 + 3 × 50² = 5500 + 7500 = 13000
• Marginal Cost = \(\frac{\mathrm{d} C(x)}{\mathrm{d} x}\) 

\(⇒ \frac{\mathrm{d} C(x)}{\mathrm{d} x} = \frac{\mathrm{d} (5500 + 3x^{2})}{\mathrm{d} x}= 6x\)

• Marginal Cost of producing 50 units = 6*50 = 300
• Hence the correct answer is an option (3).

Concept:

Corner point method:

• As per the method, the coordinates of all corner points of the feasible region are determined and the value of the objective function at these points is computed since the optimal value of the objective function lies at one of the corner points as per the LP theory.
• The first step to finding the solution to an LP problem is to determine the feasible region. It is obtained by drawing the X-Y coordinates and plotting all the constraints.
• The common region on the graph obtained that satisfies all the constraints will be the feasible region.
• If no feasible region is obtained via plotting the constraints, then the linear program is infeasible.
• All the possibilities are summarized below: 

• In the case of a feasible region problem, if the value of the objective function is different at all points, then the point which gives the maximum and minimum value will be the optimal solution.
• In the case of a feasible region problem, if the optimal value of the objective function is the same at more than one point, then the solution will have an infinite number of solutions on the line joining those points.

Draw the constraints to find the feasible region:

• First, draw the equation form of the inequalities to draw 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:

• Graph of the constraints in xy-plane needed to be drawn.
• The two constraints are given by 3x + 4y ≤ 60 and 5x - 3y ≤  20.
• To draw 3x + 4y ≤ 60, by putting x = 0, we get y = 15. Subsequently putting y = 0 gives x = 20.
• Two points obtained are (0, 15) and (20, 0). Joining these points and extending the line on both sides will give the line 3x + 4y ≤ 60.
• Similarly, 5x - 3y ≤ 20 will be drawn.
• Based on the inequalities, the feasible region is obtained which is shown in red color in the figure below:   

• Now the corner points need to be determined.
• First corner point will be the origin (0, 0).
• Second corner point is the point where 3x + 4y ≤ 60 intersects the y-axis. Putting x = 0, we get the first corner point as (0, 15).  
• Third corner point is the point where 5x - 3y ≤ 20 intersects the x-axis. Putting y = 0, we get the first corner point as (4, 0).  
• Fourth corner point will be the intersection of two inequalities. Solving them we get, y = \(\frac{240}{29}\) and x = \(\frac{260}{29}\).
• Below is the table showing the corner points and the value of the objective functions at those points.

• The table indicates that the objective function has a maximum value at (4, 0) and (\(\frac{260}{29}\), \(\frac{240}{29}\)).
• Hence, the maximum value of the function lies on any point on the line joining the points (4, 0) and (\(\frac{260}{29}\), \(\frac{240}{29}\)).
• There will infinite number of points on the line where the value of the objective function will be maximum.
• Hence the answer is option 4.

Concept:

The area between the curves y1 = f(x) and y2 = g(x) is given by:

Area enclosed = \(\rm \left|\int_{x_1}^{x_2}(y_1-y_2)dx\right|\)

Where x1 and x2 are the intersections of curves y1 and y2 

Calculation:

Given

Curve 1: y = 16x2 = f(x) (say)

Curve 2: y - 9 = 0

⇒ y = 9 = g(x) (say)

To find the intersections (or limits of the area) putting value of y from curve 1

⇒ 16x2 = 9

⇒ 16x2 - 9 = 0

⇒ (4x - 3)(4x + 3) = 0

⇒ x1 = \(-3\over 4\) , x2 = \(3\over 4\)

Now the required area (A) is

A = \(\rm \left|\int_{x_1}^{x_2}[f(x)-g(x)]dx\right|\)

⇒ A = \(\rm \left|\int_{-3\over4}^{3\over4}[16x^2-9]dx\right|\)

We know that,

\(\because \ \int_{-a}^{a}[f(x)]dx = \int_{0}^{a}[f(x)+f(-x)]dx\)

If f(x) = f(- x) then the function f(x) will be the even function then,

\(\Rightarrow \ \int_{-a}^{a}[f(x)]dx =2 \int_{0}^{a}[f(x)]dx\;\;\;\;\; \ldots \left( 1 \right)\)

If f(x) = - f(- x) then the function f(x) will be the odd function then,

\(\Rightarrow \ \int_{-a}^{a}[f(x)]dx =0\)

Since the given function is an even function, then using equation 1,

\(\Rightarrow A = 2\int_{0}^{3\over4}[16x^2-9]dx\)

⇒  \(A = 2 \left|\left[{16x^3\over3}-9x\right]_{0}^{3\over 4}\right|\)

⇒ A = \(A= 2\left|\left\{{16\over3}\times{\left(3\over 4\right)^3}-9\times{3\over 4}\right\}-0\right| \)

⇒ A = 9

Additional Information

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Concept:

• Statistical significance signifies that a set of observed data are not the result of luck but can instead be attributed to a specific reason. 
• When analyzing a data set to find whether one or more variables have an effect on an outcome, strong statistical significance claim that the results are real and not caused by luck.
• For example, statistical significance can be used to indicate that an increased number of mobile towers has reduced the population of birds in a region and the phenomenon is not just a mere coincidence. 
• Statistical inference is the process of extracting conclusions about populations or scientific details from the data and it could be reached by either conventional hypothesis testing or confidence intervals.
  • Null Hypothesis (H₀) : It proposes that no statistical significance exists in a set of given observations. According to it, any kind of difference between the chosen characteristics observed in a set of data is due to chance. In other words, it states that there is no difference between a parameter or statistic or between two parameters.
  • Alternative Hypothesis (H1) : It proposes that statistical significance exists in a set of given observations. Any difference between the chosen characteristics observed in a set of data is due to genuine factors. In other words, it states that there is a difference between a parameter or statistic or between two parameters  It basically contradicts everything that the null hypothesis says. 
• To achieve the condition of statistical significance, it is mandatory to fully reject the null hypothesis.

Explanation:

• It has been discussed that the alternative hypothesis is exactly the opposite of the null hypothesis.
• Vice versa, the null hypothesis is exactly the opposite of the alternative hypothesis.
• As per the question, the alternative hypothesis will indicate that Average sunlight received in February > Annual average sunlight received.
• The null hypothesis will be the exact opposite of the same: Average sunlight received in February <= Annual average sunlight received. 
• Hence the correct answer will be option (4).

Concept:

If two matrices A and B are said to be equal if the following conditions hold true:

• Order of matrix A = Order of matrix B
• Corresponding element of matrix A = Corresponding element of matrix B

According to the question

AB = 0

\(\begin{bmatrix} 6 & -3 \\ 4 & -2 \end{bmatrix}\) × \(\begin{bmatrix} 2 & y\\ x & 7 \end{bmatrix}\) = \(\begin{bmatrix} 0&0\\0&0\end{bmatrix}\)

\(\begin{bmatrix} 12-3x &6y-21\\ 8-2x & 4y-14 \end{bmatrix}\) = \(\begin{bmatrix} 0& 0\\ 0& 0\end{bmatrix}\) 

⇒ 12 - 3x = 0

⇒ 3x = 12

⇒ x = 4

Also 4y - 14 = 0

⇒ 4y = 14

⇒ y = 3.5

Concept:

• Revenue Function (R(x)): It is represented in terms of the product of per unit selling price p and the number of units sold x as below:

R(x) = px ….(1)  

• Marginal Revenue: It is the increase in revenue due to a unit increase in output. It is calculated by deriving the partial derivative of the revenue function with respect to output.  
• Cost Function or Total cost (C(x)): It is represented in terms of the sum of variable cost V(x) for producing ‘x’ units and fixed cost k as below:

C(x) = V(x) + k ….(2)  

• Marginal Cost: It is the increase in total cost due to a unit increase in output. It is calculated by deriving the partial derivative of the cost function with respect to output.
• Marginal Profit = Marginal Revenue - Marginal Cost ….(3)
• Total profit reaches its maximum value when marginal revenue becomes equal to marginal cost, ie Marginal Cost = Marginal Revenue

Explanation: 

Given:

• Marginal cost indicates an increase in the cost when production is increased by 1 unit.
• Marginal revenue indicates an increase in the revenue when selling is increased by 1 unit.
• Below is the example of a curve where marginal revenue and marginal cost are plotted:  

• The marginal revenue line (orange one) shows that a product is always sold for Rs 50, no matter how many units are sold.
• The marginal cost line (grey one) shows that marginal cost will reduce up to from starting to produce 3 units, and after that, it will gradually increase. For example, from the graph, the cost of producing the 5th unit is Rs 21, while for the 6th unit, it is Rs 35.  
• At one point, the marginal cost curve will intersect the marginal revenue line, and after that marginal revenue will be lower than the marginal cost.  
• From the graph, in the region where the marginal cost curve is above the marginal revenue line, marginal cost > marginal revenue, an increase in cost by producing 1 more unit will be greater than the increase in revenue by selling 1 more unit. In this case, the company has already incurred some losses and should not produce more units.
• From the graph, in the region where the marginal cost curve is below the marginal revenue line, marginal cost < marginal revenue, an increase in cost by producing 1 more unit will be lower than the increase in revenue by selling 1 more unit. In this case, more units need to be produced to increase the overall profit until the marginal cost becomes greater than the marginal revenue.
• In the region ​where the marginal cost curve intersects the marginal revenue line, marginal cost = marginal revenue.  Up to this point, the company has sold its products to make all the profit that can be made, since the marginal revenue was greater than the marginal cost (up to 7 units for this case). After that, production needs to be stopped, since the marginal revenue will become lower than the marginal cost.
• Hence, the company makes maximum profit until the point where marginal cost becomes equal to marginal revenue.   

• Hence, the correct answer is an option (2).

Additional Information 

• Any company or business has to spend certain costs for running its business. These costs are classified as fixed costs and variable costs. 
• Fixed Cost: These are the expenses that are independent of the production output units. Even with zero production, this cost has to be paid regularly. 
• Variable Cost: These expenses are dependent on the production output and will increase with an increase in production. Some examples are the cost of raw material, packaging costs, energy consumption per unit cost, etc.

Given : 

The total allocation of funds in a five-year plan is 36000 (in crore of rupees)

Formula used : 

The angle subtended = (Given data/Total data) × 360° 

Calculation : 

Funds allocated to agriculture = \(\frac{90^\circ}{360^\circ}× 36000 = 9000\)

Funds allocated to industry = \(\frac{45^\circ}{360^\circ}× 36000 = 4500\)

Funds allocated to employment = \(\frac{120^\circ}{360^\circ}× 36000 = 12000\)

Funds allocated to miscellaneous = \(\frac{75^\circ}{360^\circ}× 36000 = 7500\)

∴ The employment head is allocated maximum funds.

Given : 

The total allocation of funds in a five-year plan is 36000 (in crore of rupees)

Formula used : 

The angle subtended = (Given data/Total data) × 360° 

Calculations :

Money allocated to Education = \(\frac{30^\circ}{360^\circ}\times 36000 = 3000\)

∴ The fund allocated to Education is 3000.

Given : 

The total allocation of funds in a five-year plan is 36000 (in crore of rupees)

Formula used : 

 The angle subtended = (Given data/Total data) × 360°  

Calculations : 

Money allocated to agriculture = \(\frac{90^\circ}{360^\circ}\times 36000 = 9000\)

Money allocated to employment = \(\frac{120^\circ}{360^\circ}\times 36000 = 12000\)

⇒ Money allocated to agriculture + employment = 9000 + 12000

∴ The money allocated to both agriculture and employment is 21000.

Given : 

Total allocation of funds in a five-year plan is 36000 (in crore of rupees)

Formula used : 

The angle subtended = (Given data/Total data) × 360° 

Calculations :

Money allocated to Miscellaneous = \(\frac{75^\circ}{360^\circ}\times 36000 = 7500\)

Money allocated to education = \(\frac{30^\circ}{360^\circ}\times 36000 = 3000\)

Excess money allocated to miscellaneous over education = 7500 - 3000

∴ Excess money(in crore) allocated to miscellaneous over education is 4500. 

Given:

Three out of the four vertex of the parallelogram -

A = (1, 2, 3)

B = (-1, -2, -1)

C = (2, 3, 2)

Concept:

The diagonals of a paralleogram bisect each other.

Formula:

The mid-point of the line segment joining A (x1, y1, z1) and B (x2, y2, z2) is given by -

M (x, y, z) where \(x = \frac{x_1 + x_2}{2}\), \(y = \frac{y_1 + y_2}{2}\) and \(z = \frac{z_1 + z_2}{2}\)

Solution:

Let the midpoint of AC be M, then it is also the midpoint of BD

Now, M = (\(\frac{1+2}{2}\), \(\frac{2+3}{2}\), \(\frac{3+2}{2}\))

= (3/2, 5/2, 5/2)

Let D = (a, b, c)

Then, the midpoint of BD is given by (\(\frac{a-1}{2}\), \(\frac{b-2}{2}\), \(\frac{c-1}{2}\))

Equating it to M, we get a = 4, b = 7, c = 6

Hence D = (4, 7, 6)