Mathematics Mock Test - 10

Given:

A is a 2 × 2 matrix such that

1. A + AT = I
2. ATA = I​

Concept:

• Matrix multiplication is row into column-wise.
• Transpose of a matrix A represented by AT is obtained by interchanging rows into columns and vice-versa.

​Solution:

Let A = \(\begin {bmatrix} a & b \\ c & d \end {bmatrix} \)

then AT = \(\begin {bmatrix} a & c \\ b & d \end {bmatrix} \)

∵ A + AT = I

⇒ \(\begin {bmatrix} 2a & b+c \\ b+c & 2d \end {bmatrix} \) = \(\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix} \)

On equating the elements -

2a = 1

⇒ a = 1/2    - (i)

b + c = 0

⇒ b = -c    - (ii)

2d = 1

⇒ d = 1/2    - (iii)

Also, 

ATA = I

⇒ \(\begin {bmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end {bmatrix} \) = \(\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix} \)

Putting values of a, b, c from (i), (ii) and (iii) -

We see a2 + c2 = 1

⇒ c2 = 3/4

⇒ c = ±\(\frac{\sqrt{3}}{2}\) 

Similarly, we can see, b = ±\(\frac{\sqrt{3}}{2}\) 

So, possible matrices are \(\begin {bmatrix} 1/2 & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & 1 \end {bmatrix}\) and \(\begin {bmatrix} 1/2 & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1 \end {bmatrix} \)

So, the total number of matrices = 2

Given:

f(x) = (x - 1)(x - 2)(x - 3)

Concept:

For maximum value of a differentiable function f(x) -

• f'(x) = 0
• f''(x) < 0 at x where f'(x) = 0

Solution:

f(x) = (x - 1)(x - 2)(x - 3)

⇒ f(x) = x3 - 6x2 + 11x - 6

⇒ f'(x) = 3x2 - 12x + 11

Putting f'(x) = 0

⇒ 3x2 - 12x + 11 = 0

⇒ x = 2 ± √(1/3)

⇒ x ≈ 1.423 , 2.577

Now, f''(x) = 6x - 12

f"(1.423) < 0

Hence, the maximum value occur at x = 2 - √(1/3)

⇒ f(2 ± √(1/3)) = (1 - √(1/3))(-√(1/3))(-1 - √(1/3))

= \(\frac{2}{3\sqrt{3}}\)

Concept:

• Conditional Probability is the possibility or probability of an event B happening, given that another event A has already occurred.  
• Event B has some sort of relationship with event A in which the occurrence of one event affects the happening of another event.

• The above figure shows a sample space S, in the form of a rectangle and two events, A and B.
• Normally, the probability of event B occurrence will be given by the ratio of the area covered by event B and the area covered by total sample space.
• When we try to find the probability of event B occurrence when A has already occurred, the sample space area will be reduced to A.
• It is the fraction of probability B that intersects with A.
• In the green portion in the above figure, A ∩ B represents the portion of the probability of event B and event A occurring simultaneously.
• The formula of the conditional probability of A given B has already occurred is given by,

P(A | B) = \(\frac{P(A \cap B)}{P(B)}\) ….(1)

Explanation:

•  Based on the above explanation, the white-colored area will represent the probability of events A and B B and A occurring simultaneously.
• Hence the correct answer is option 2.  

Concept:

Reflexive:- A relation of elements of a set such that each element of the set is related to itself.

Symmetric:- A relation of elements of a set such that if the first element is related to the second element, then the second element is also related to the first element as defined by the relation.

Transitive:- A relation of elements of a set such that if the first element is related to the second element, and the second element is related to the third element, then the first element must be related to the third element.

Calculation:

Given:

The function is defined by aRb if |a - b| ≤ 1

Since, |a - a| = 0 ≤ 1

So,  aRb, ∀ a ∈ R

⇒  R is reflexive.

Now, aRb ⇒  |a - b| ≤ 1

⇒ |b - a| ≤ 1

⇒ bRa

⇒  R is symmetric.

Now, 1R2, 2R3 but 1R3,

⇒ |1 - 3| = 2 > 1

⇒ R is not transitive.

∴ Then R is reflexive and symmetric.

Given:

\(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \)

Formula Used:

2cos^-1x = cos^-1(2x² - 1)

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

Calculation:

We have,

\(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \)

⇒ \(sin \left[cos^{-1}\left( 2(\frac{\sqrt5}{3})^2 - 1\right) \right]\)

⇒ \(sin\left[ cos^{-1} \left(2 \times \frac{5}{9} - 1 \right) \right]\)

⇒ \(sin \left[ cos^{-1} \frac{1}{9}\right]\)

⇒ \(sin\left[sin^{-1} \sqrt{1 - (\frac{1}{9})^2 } \ \right]\)

⇒ \(sin \left[ sin^{-1} \sqrt{\frac{80}{81}} \right] \)

⇒ \(\frac{\sqrt{80}}{9}\)

⇒ \(\frac{4\sqrt5}{9}\)

∴ The value of \(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \) is \(\frac{4\sqrt5}{9}\)

Given:

y = (1 + x)(1 + x2)(1 + x4)........(1 + x2n)

Concept:

For functions where many terms are being multiplied, it is better to take log on both sides and then differentiate.

Formula:

\(\frac{d(logx)}{dx}=\frac{1}{x}\)

log(a.b.c...) = loga + logb + logc + .....

Calculation:

y (at x = 0) = 1

Taking log both sides -

⇒ log(y) = log(1 + x) + log(1 + x²) + ........... + log(1 + x2n)

Differentiating both sides wrt x -

\(⇒ \frac{1}{y}\frac{dy}{dx} = \frac{1}{1+x} + \frac{2x }{1+x^2} + ....... +\frac{2nx^{2n-1}}{1+x^{2n}}\)

⇒ \(\frac{dy}{dx} = y(\frac{1}{1+x} + \frac{2x }{1+x^2} + ....... +\frac{2nx^{2n-1}}{1+x^{2n}})\)

∴ dy/dx (at x = 0) = (1)(1 + 0 + 0 + .....)

⇒ dy/dx (at x = 0) = 1

Given:

Linear equations are:-

x + 2ay + az = 0

x + 3by + bz = 0 

x + 4cy + cz = 0

Concept:

For a homogeneous system of equations to have non zero solution, Δ = 0

Calculation:

We have, linear equations 

⇒ \(\begin{array}| {{1 \ 2a \ 0 \\\\ 1 \ 3b \ b \\\\ 1 \ 4c \ c }} \end{array} \)= 0 

C₂ → C₂ - 2C₃

⇒ \(\begin{array}|{1 \ 0 \ a \\ 1 \ b \ b \\ 1 \ 2c \ c} \end{array} \) = 0 

R₃ → R₃ - R₂, R₂ → R₂ - R₁

⇒ \(\begin{array}|{1 \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ a \\ 0 \ \ \ \ \ \ \ b \ \ \ \ \ \ \ \ \ \ b-a \\ 0 \ \ {2c - b} \ \ \ \ \ {c- b} } \end{array}\) = 0

⇒ b(c - b) - (b - a)(2c - b) = 0

⇒ bc - b² - (2bc - b² - 2ac + ab) = 0

⇒ - bc + 2ac - ab = 0

⇒ 2ac = ab + bc

⇒ 2ac = b(a + c)

⇒ \(b = \frac{2ac}{a + c}\)

∴ The solution is \(b = \frac{2ac}{a + c}\)

Concept:

Componendo and Dividendo Rule:-  A theorem on proportions that is used to perform calculations and reduce the number of steps.

According to the rule:-

\(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a + b}{a - b} = \frac{c + d}{c - d}\)

Calculation:

Given:

\(f\left( x \right) = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}} \)

Let f(x) = y

⇒ \(y = \frac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}\)

⇒ \(\frac{y + 1}{y - 1} = \frac{{{{10}^x} - {{10}^{ - x}} + 10^x + 10^{-x}}}{{{{10}^x} + {{10}^{ - x}} - 10^x - 10 ^{-x}}}\)   [ by using componendo and dividendo rule]

⇒ \(\frac{y + 1}{y - 1} = \frac{10^x}{-10^{-x}}\)

⇒ \(\frac{y + 1}{y - 1} = -10^{2x}\)

⇒ \(\frac{1 + y}{1 - y} = 10^{2x}\)

Taking log₁₀ on both sides, we get,

⇒ \(2x = log_{10}\left( \frac{1 +y}{1 - y}\right)\)

⇒ \(x = \frac{1}{2}log_{10}\left(\frac{1 + y}{1 - y}\right)\)

⇒ \(f^{-1}(y) = \frac{1}{2}log_{10}\left(\frac{1+y}{1-y}\right)\)

Now, replacing y with x, we get inverse,

⇒ \(f^{-1}(x) = \frac{1}{2}log_{10}\left(\frac{1+x}{1-x}\right)\)

∴ The inverse of the function is \(\frac{1}{2}log_{10}\left(\frac{1+x}{1-x}\right)\)

Given:

f(x) = sin x - cos x - ax + b

Formula Used:

sin(A + B) = sin A cos B + cos A sin B

Calculation:

We have,

⇒ f(x) = sin x - cos x - ax + b

⇒ f'(x) = cos x + sin x - a

Divide and multiply by \(\sqrt2\)

⇒ f'(x) = \(\sqrt2\left[ \frac{1}{\sqrt2}cosx + \frac{1}{\sqrt2}sinx \right] - a\)

⇒ f'(x) = \(\sqrt2\left[ sin\frac{\pi}{4} cosx + cos\frac{\pi}{4} sinx \right] - a\)

⇒ f'(x) = \(\sqrt2 \left[ sin\left(\frac{\pi}{4} + x \right) \right] - a\)

If f(x) decreases for all x , f'(x) is negative.

⇒ a ≥ max of \(\sqrt2sin\left(\frac{\pi}{4} + x \right)\)

∴ The value of a ≥ \(\sqrt2\)  

Given: 

f(x) = log_x (log x)

Calculation:

We have, 

f(x) = log_x(logx)

⇒ \(f(x) = \frac{log(logx)}{logx}\)

On differentiating both the sides, with respect to 'x'

⇒ \(f'(x) = \frac{(logx)\frac{d}{dx}log(logx) - log(logx)\frac{d}{dx}logx}{(logx)^2}\)

⇒ \(f'(x) = \frac{logx(\frac{1}{logx}.\frac{1}{x}) - log(logx).\frac{1}{x}}{(logx)^2}\)

⇒ \(f'(x) = \frac{\frac{1}{x} - \left[\frac{log(logx)}{x}\right]}{(logx)^2}\)

Now, Put x = e, 

⇒ \(f'(e) = \frac{\frac{1}{e} - \frac{(log\ log\ e)}{e}}{(log\ e)^2}\)

⇒ \(f'(e) = \frac{1}{e}\)    [∵ log e = 1]

∴ The value of function at x = e is \(\frac{1}{e}\)

Given:

\(f(α ) = \left[ \begin{array}{l} \cos α \,\sin α \\ - \sin α \,\cos α \end{array} \right] \)

α, β, and γ are the angles of a triangle.

Calculation:

\(f(α ) = \left[ \begin{array}{l} \cos α \,\sin α \\ - \sin α \,\cos α \end{array} \right] \)

So, the determinant of f(α),

⇒ |f(α)| = cosα. cosα + sinα.sinα 

⇒ |f(α)| = cos²α + sin²α 

⇒ |f(α)| = 1

\(f(β ) = \left[ \begin{array}{l} \cos β \,\sin β \\ - \sin β \,\cos β \end{array} \right] \)

⇒ |f(β)| = cosβ.cosβ + sinβ.sinβ   

⇒ |f(β)| = cos²β + sin²β

⇒ |f(β)| = 1 

\(f(γ ) = \left[ \begin{array}{l} \cos γ \,\sin γ \\ - \sin γ \,\cos γ \end{array} \right] \)

⇒ |f(γ)| = cos²γ + sin²γ 

⇒ |f(γ)| = 1

∵  |f(α). f(β). f(γ)|³ = |f(α)|³. |f(β)|³. |f(γ)|³

⇒ |f(α). f(β). f(γ)|³  = 1³ × 13 × 13 

⇒ |f(α). f(β). f(γ)|3 = 1

⇒ x³ = 1

Here, the above equation is representing a cube root of unity.

⇒ x = 1, ω, ω2

Given:

\(\int\limits_0^{\pi /2} {\left| {\sin x - \cos x} \right|dx} \)

Calculation:

Let I = \(\int\limits_0^{\pi /2} {\left| {\sin x - \cos x} \right|dx} \)

⇒ I = \(-\int\limits_0^{\pi /4} {\left( {\sin x - \cos x} \right)dx} +\int\limits_{\pi/4}^{\pi /2} {\left( {\sin x - \cos x} \right)dx} \)

⇒ I = \(- \left| - cosx - sinx \right|_0^{\frac{\pi}{4}} + \left|-cosx - sinx \right|_\frac{\pi}{4}^{\frac{\pi}{2}}\)

⇒ I = \(- \left( - cos\frac{\pi}{4} - sin \frac{\pi}{4} - \left(- cos0 - sin0\right) \right) + \left(- cos\frac{\pi}{2} - sin \frac{\pi}{2} - \left(-cos \frac{\pi}{4} - sin\frac{\pi}{4}\right)\right)\)

⇒ I = \(- \left(- \frac{1}{\sqrt2} - \frac{1}{\sqrt2} - (-1 - 0)\right) + \left(0 - 1 - \left(- \frac{1}{\sqrt2} - \frac{1}{\sqrt2} \right)\right)\)

⇒ I = \(- (- \sqrt2 + 1) + (-1 + \sqrt2)\)

⇒ I = \((\sqrt2 - 1) + (\sqrt2 - 1)\)

⇒ I = \(2(\sqrt2 - 1)\)

∴ The value of the integral is \(2(\sqrt2 - 1)\)

Given:

x² + y² = π²

y = sin x

Calculation:

We have,

⇒ x² + y² = π² is a circle of radius π and center at the origin.

Required area = Area of the circle (1st quadrant) - \(\int\limits_0^\pi sinx dx\)

⇒ Required Area = \(\frac{\pi.\pi^2}{4} - \left.[-cosx\right] _0^\pi\)  

⇒ Required Area = \(\frac{\pi^3}{4} + (cos\pi - cos0)\)

⇒ Required Area = \(\frac{\pi^3}{4} + (-1 -1)\)

⇒ Required Area = \(\frac{\pi^3}{4} - 2\)

⇒ Required Area = \(\frac{\pi^3 - 8}{4}\) sq. units

∴ The required area is \(\frac{\pi^3 - 8}{4}\)sq. units.

Given:

\(\frac{{dy}}{{dx}} + 2y\,\tan x = \sin x\)

Calculation:

Differential equation is of the  form

\(\frac{dy}{dx} + Py = Q\)

⇒ \(\frac{dy}{dx} + 2ytan x = sinx\) 

Where, P = 2 tan x and Q = sin x

⇒ I.F. = \(e^{\int P dx}\)

⇒ I.F. = \(e^{\int 2 tanx dx}\)

⇒ I.F. = \(e^{2log\ secx}\)

⇒ I.F. =  \(e^{log\ sec^2x}\)

⇒ I.F. = \(sec^2x\)

Solution is,

⇒ y(I.F.) = ∫(Q × IF)dx + c

⇒ y(sec²x) = ∫sin x sec²x dx + c

⇒ \(y\ sec^2x = ∫sinx \times \frac{1}{cos^2x}dx + c\)

⇒ \(y\ sec^2x = \int \frac{sinx}{cosx} \times \frac{1}{cosx} dx + c\)

⇒ y sec²x = ∫tan x sec x dx + c

⇒ y sec²x = sec x + c

∴ The solution of the given equation is y sec2x = sec x + c

Given:

\(\vec a = \hat i + \hat j + 2 \hat k\)

\(\vec b = \hat i + 2 \hat j + \hat k\)

\(\vec c = \hat i + \hat j + \hat k\)

\(\vec n = x \vec a + y \vec b\)

Concept:

A unit vector coplanar to the vector:- \(\vec n = x \vec a + y \vec b\)

Calculation:

We have,

⇒ \(\vec n. \vec c = 0 = x( \vec a . \vec c) + y (\vec b . \vec c)\)

⇒ \(x(\vec a .\vec c) + y (\vec b . \vec c) = 0\)

⇒ x(4) + y (4) = 0

⇒ 4x + 4y = 0

⇒ y = - x

⇒ \(\vec n = x \vec a - x \vec b\)

⇒ \(\vec n = x ( \vec a - \vec b)\)

⇒ \(\vec n = x ( \hat i + \hat j + 2 \hat k - \hat i - 2\hat j - \hat k)\)

⇒ \(\vec n = x ( - \hat j + \hat k)\)

⇒ \(|\vec n| = 1\)

⇒ \(\sqrt{x ^2 + x^2} = 1\)

⇒ \(\sqrt{2x^2}= 1\)

⇒ \(2x^2 = 1\)

⇒ \(x^2 = \frac{1}{2}\)

⇒ \(x = ± \frac{1}{\sqrt2}\)

⇒ \(\vec n = x( - \hat j + \hat k)\)

⇒ \(\vec n = ± \frac{1}{\sqrt2} (- \hat j + \hat k)\)

∴ The unit vector is \(\vec n = ± \frac{1}{\sqrt2} (- \hat j + \hat k)\)

Formula:

∫dx = x + c

∫(dx/x) = ln|x| + c

Solution:

I = ∫\(\frac{sinx}{sinx - cosx}dx\)

⇒ I = \(\frac{1}{2}\)∫\(\frac{2sinx}{(sinx - cosx)}dx\)

⇒ I = \(\frac{1}{2}\)∫\(\frac{(sinx - cosx) + (sinx + cosx)}{(sinx - cosx)}dx\)

⇒ I = \(\frac{1}{2}\)∫[1 + \(\frac{ (sinx + cosx)}{(sinx - cosx)}\)]dx

⇒ I = \(\frac{1}{2}\)∫dx + \(\frac{1}{2}\)∫\(\frac{ (sinx + cosx)}{(sinx - cosx)}\)dx

Putting sinx - cosx = t in the second integration -

⇒ sinx + cosx = dt 

Second integration becomes ∫(dt/t) = ln|t| + c

Putting t = sinx - cosx back , we get ∫\(\frac{ (sinx + cosx)}{(sinx - cosx)}\)dx = ln|sinx - cosx| + c

⇒ I = \(\frac{x}{2}\) + \(\frac{1}{2}\)ln|sinx - cosx| + c

Given:

A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6)

Concept:

Hypotenuse² = Perpendicular² + Base²

Formula Used:

Distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2 + (z_2 - z_1)^2} \)

Calculation:

AB = \(\sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2}\)

⇒ AB = \(\sqrt{(-1)^2 + (-1)^2 +(-4)^2}\)

⇒ AB = \(\sqrt{1 + 1 + 16}\)

⇒ AB = \(\sqrt{18}\)

⇒ AB = \(3\sqrt2\)

BC = \(\sqrt{(-4 + 1)^2 + (9 - 6)^2 + (6 - 6)^2}\)

⇒ BC = \(\sqrt{(-3)^2 + (3)^2 + (0)^2}\)

⇒ BC = \(\sqrt{ 9 + 9}\)

⇒ BC = \(\sqrt{18}\)

⇒ BC = \(3\sqrt2\)

CA = \(\sqrt{(0 + 4)^2 + (7 - 9)^2 + (10 - 6)^2}\)

⇒ CA = \(\sqrt{ (4)^2 + (-2)^2 + (4)^2}\)

⇒ CA = \(\sqrt{16 + 4 + 16}\)

⇒ CA = \(\sqrt{36}\)

⇒ CA = 6

Now, AB² + BC² = \((3\sqrt2)^2 + (3\sqrt2)^2 \) = 18 + 18 = 36 = AC²

AB² + BC² = AC²

So, By Pythagoras's Theorem, ABC is a right-angled triangle. 

⇒ AB = BC

So, ABC is also an isosceles triangle.

∴ ABC is a right angle isosceles triangle.

Solution:

Given:

Equation of the line : \(\frac{{x - 1}}{2} = \frac{{y + 2}}{{ - 3}} = \frac{{z - 3}}{4}\)

Equation of the plane : x - y + az + b = 0 

Also, the line lies in the plane.

Concept:

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

So if the line lies in the plane then the direction ratios of the line will b perpendicular to the direction ratios of the normal to the plane i. e. a , b, c

⇒ The sum of the product of respective direction cosines will be 0.

2. Point through which the line passes must also satisfy the equation of the plane.

Calculation:

The direction ratios of the line are 2, -3, 4.

⇒ (2)(1) + (-3)(-1) + (4)(a) = 0

⇒ a = - 5/4

Also,

Line passes through (1, -2 , 3)

Hence (1, -2 , 3) must satisfy plane -

⇒ (1) - (-2) + (a)(3) + b = 0

Using a = -5/4, we get b = 3/4

∴ b - a = 2

Given:

Equation of plane P₁ : x + 2y + 3z - 2 = 0

Equation of plane P₂ : x - y + z - 3 = 0 

The plane through the intersection of above planes is at a distance \(\frac{2}{{\sqrt 3 }}\)from the point (3, 1, -1)

Concept:

Equation of plane passing through the intersection of two planes P1 and P2 can be assumed as :

P1 + λP2 = 0

Formula:

The distance of a point (p, q, r) from a plane ax + by + cz = d is given by :

\(D = |\frac{ap + bq + rc -d}{\sqrt{a^2 + b^2 + c^2 }}|\)

Calculation:

Using P1 + λP2 = 0,

Equation of plane through the intersection of above planes is -

(x + 2y + 3z - 2) + λ(x - y + z - 3) = 0      ----      (i)

⇒ (1 + λ)x + (2 - λ)y + (3 + λ)z + (-2 - 3λ) = 0

As this plane is at a distance \(\frac{2}{{\sqrt 3 }}\)from the point (3, 1, -1)

⇒ \(|\frac{(1 + λ)(3) + (2 - λ )(1) + (3 + λ)(-1) +(-2 -3λ)}{ \sqrt {(1 + λ )^{2} + (2 - λ )^{2} + (3 + λ)^{2}} }|\) = \(\frac{2}{{\sqrt 3 }}\)

On solving, λ = -7/2

Putting λ = -7/2 in equation (i), we get equation of plane as -

-5x + 11y - z + 17 = 0

⇒ 5x - 11y + z - 17 = 0

∴ a + b + c - 17 = - 22

Concept:

• if P(A) ≠ 0 and P(B) ≠ 0 and if A is independent of B, then B is also independent of A.
• If A and B are independent events, then P(A/B) = P(A), P(B) ≠ 0 and P(B/A) = P(B), P(A) ≠ 0
• If A and B are independent events, then P(A ∩ B) = P(A)P(B)
• If A and B are independent events, then A' and B' are also independent events.
• If A and B are independent events, then (A and B') and (A' and B) are also independent events.
• P(A/B) = P(A ∩ B)/P(B), P(B) ≠ 0 
• P(B/A) = P(A ∩ B)/P(A), P(A) ≠ 0
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given:

P(A) = P(A|B) = \(\frac{1}{4}\) 

and P(B|A) = \(\frac{1}{2}\).

Explanation:

As P(A) = P(A|B), it means A is independent of B

By concept (i), B is also independent of A or A and B are independent events.

Now, P(B/A) = P(B) = 1/2 as B is an independent event.

⇒ P(B̅) = 1 - (1/2) = 1/2

Also, P(A̅) = 1 - P(A)

⇒ P(A̅) = 1 - (1/4) = 3/4 

Option (i) - 

P(A̅|B) = P(A̅)

⇒ P(A̅|B) = 3/4           ------      TRUE

Option (ii) - 

P(B̅|A̅) = P(B̅ ∩ A̅)/P(A̅)

⇒  P(B̅|A̅) = P(B̅)    ----    (by concept (iv))

⇒  P(B̅|A̅) = 1/2        --------      TRUE

Option (iv) - 

P(A ∩ B) = P(A)P(B)     (by concept (iii))

⇒ P(A ∩ B) = (1/4)(1/2) = 1/8      ------     TRUE

Option (iii) -

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A ∪ B) = 1/4 + 1/2 - 1/8

= 5/8        --------      FALSE

Explanation:

Given :

P(A) = 1/3, P(A') = 2/3

P(B) = 3/4, P(B') = 1/4

Where A and B are the events that 'A' and 'B' speak the truth 

Calculation:

Let T be the event that both speak truth and T' be the event that both do not speak the truth.

Let G be the event that they both agree (When one speaks the truth then the other will also speak the truth and vice versa)

∴ P(T) = P(A)P(B) = 1/4

∴ P(T') = P(A')P(B') = 1/6

Here, P(G/T) = 1, (As both are speaking the truth so they will agree to all conditions)

Similarly, P(G/T') = 1

So, if A and B agree on a certain statement then the probability that the statement is true = The probability of speaking the truth when both get agree with each other

 \(\Rightarrow P(\frac{T}{G})= \frac{P(T)P(\frac{G}{T})}{P(T)P(\frac{G}{T}) + P(T')P(\frac{G}{T'})}\)

= [(1/4) × 1]/[(1/4) × 1 + (1/6) × 1]

= 3/5

Given:

\(\left| {\frac{{{x^2}}}{{x - 1}}} \right| \le 1\)

Concept:

(i) \(|\frac{a}{b}|= \frac{|a|}{|b|}\)

(ii) |x - a| = (x - a) if x ≥ a and -(x - a) if x < a

Calculation:

\(\left| {\frac{{{x^2}}}{{x - 1}}} \right| \le 1\)

⇒ \( \frac{|x^2|}{|x - 1|} \le 1\)

⇒ |x²| ≤ |x - 1|

⇒ x2 ≤ |x - 1| as x2 is always positive for real x

(i) Now for x > 1, |x - 1| = (x - 1)

∴ above inequality will be x2 ≤ x - 1

⇒ x2 - x + 1 ≤ 0

Which gives no solution as for x2 - x + 1, (a > 0 and D < 0) which makes it always positive.

Hence no solution for x > 1

(ii) Now for x ≤ 1, |x - 1| = - (x - 1)

∴ above inequality will be x2 ≤ -(x - 1)

⇒ x2 + x - 1 ≤ 0

which gives \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)

Taking the intersection of  x ≤ 1 and \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)

We get \(\frac{-1-\sqrt5}{2} ≤ x≤\frac{-1+\sqrt5}{2}\)

Given:

f: R→ R such that f(x + y) = f(x).f(y) for all x, y ∈ R and f(x) ≠ 0.

f(x) is differentiable at x = 0 and f'(0) = 2

f'(x) = λ. f(x)

Calculation:

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

f(0) = [f(0)]²

⇒ f(0)[1 - f(0)] = 0

⇒ f(0) = 1 as f(x) ≠ 0

Also, f'(x) = λ. f(x)

Putting x = 0, we get f'(0) = λ. f(0)

⇒ λ = 2

Given:

y = f(x) = |cos x - sin x|

Concept:

To find derivative of piecewise functions such as GIF, modulus, etc.; it is better to check the Left-Hand Derivative and Right-Hand Derivative at the point of concern.

Formula:

\(LHD \space (at\space x = a) = \lim_{h \to 0}\frac{f(a - h) - f (a)}{-h}\)

\(RHD \space (at\space x = a) = \lim_{h \to 0}\frac{f(a + h) - f (a)}{h}\)

Also, |x| = x if x ≥ 0 and -x if x < 0 

Calculation:

LHD (at x = π/4) = \( \lim_{h \to 0}\frac{f(π/4 - h) - f (π/4)}{-h}\)

f(π/4) = 0

Here, for x → (π/4)⁺ , sin x > cos x so the mod will open as negative because the inner function will be negative.

So, f(x) = - (cos x - sin x) 

Here, for x → (π/4)^- , sin x < cos x so the mod will open as positive because the inner function will be positive.

So, f(x) =  (cos x - sin x) 

⇒ LHD = f'(x → (π/4)^-) = ​\( \lim_{h \to 0}\frac{ cos(π/4 - h) -sin(π/4 - h) ​​​​}{-h} → \frac{0}{0}\)

So, we can use  L' Hospital,

⇒ LHD = \(\)-√2

RHD = f'(x → (π/4)⁺) = \( \lim_{h \to 0}\frac{f(π/4 + h) - f (π/4)}{h}\)

⇒ RHD = \( \lim_{h \to 0}\frac{sin(π/4 + h) - cos(π/4 + h) ​​​​}{h} → \frac{0}{0}\)

⇒ RHD = \( \lim_{h \to 0}\frac{cos(π/4 + h) +sin(π/4 + h) ​​​​}{1} \)

Using L'Hospital, RHD = √2

∴ We can see LHD ≠  RHD, hence derivative is not defined at x = π/4.

Concept:

• People spend money for specific purposes like on housing, gadgets, holiday trips, or buying a car.
• Some of them borrow loans for such activities.
• The loan is borrowed within a defined length of time.
• EMI (Equated Monthly Installment) is a monthly payment at a fixed date done towards a loan that has been borrowed by the borrower from the lender.
• One of the methods to calculate the EMI is the Flat Rate Method.
• Each interest charged is calculated on the original loan amount.
• The EMI is given by the formula below,

EMI = \(\frac{P+i}{n}\) ....(1)

• Where, P = Principal amount of the loan,  
• i = Interest on the loan amount, 
• n = Number of months in the loan period            

Calculation:

Given:

• Vikas is paying 50% of the EMI.
• Rest 50% of EMI must be paid from some other source.
• Since Vikas is paying Rs 800, the total EMI will be equal to Rs 800 × 2 = Rs 1,600.
• Number of months, n = 5 × 12 = 60.
• Let i be the interest amount and P be the principal amount taken.
• From the information above, the total interest amount can be written as,

i = P × \(\frac{10}{100}\) × 5 = 0.5P.  

The EMI will be given by,

EMI = \(\frac{P + i}{n}\)

\(\Rightarrow\) 1,600 = P + 0.5P/60 \(\frac{P + 0.5P}{n}\)

\(\Rightarrow\) 1,600 = \(\frac{1.5P}{60}\)

\(\Rightarrow\) P = \(\frac{96,000}{1.5}\)

\(\Rightarrow\) P = 64,000

The principal amount taken is equal to Rs 64,000.

Hence the correct answer is option 3.

Given:

The curve : 5x5 - 10x3 + x + 2y + 6 = 0

Normal is at point A (0 , -3) and it meets the curve again at B and C

Concept:

The normal to a curve at a point can be found by evaluating the slope of the normal at that point and then writing the equation of the normal using the point-slope form of the line.

Formula:

The slope of the normal at a point P (x₁ , y₁) on the curve is  \((\frac{-1}{\frac{dy}{dx}})_{(x_1,y_1)}\)

The distance between two points is given by,

d = \(√ {(x_2 - x_1)^2 + (y_2 - y_1)^{2}}\)

Calculation:

Differentiating the curve 5x5 - 10x3 + x + 2y + 6 = 0, we get-

25x⁴ - 30x² + 1 + 2\(\frac{dy}{dx}\) = 0

⇒ \(\frac{dy}{dx}\) =  (30x2 - 25x4 - 1)/2

∴ The slope of the normal at a point P (0, -3) on the curve is \((\frac{-1}{\frac{dy}{dx}})_{(0,-3)}\)

⇒ \((\frac{dy}{dx})_{(0,-3)}\) = -1/2

∴ \((\frac{-1}{\frac{dy}{dx}})_{(0,-3)}\) = 2 (SLOPE OF THE NORMAL)

Now equation of the normal = (y - (-3)) = 2(x - 0)

⇒ 2x - y = 3 

Putting y = 2x - 3 in the curve equation -

We get  5x5 - 10x3 + 5x = 0

Which gives,

⇒ x = 0, 1, -1

Corresponding y are : -3, -1, -5

∴ B = (1 , -1) and C = (-1 , -5)

Distance between them d = \(√ {((-1) - 1)^2 + ((-5) - (-1))^{2}} \)

⇒ d = √20 or 2√5

Given:

The lines

L1 : x = py + q and z = ry + s

L2 ​: x = p'y + q' and z = r'y + s'​

Concept:

If two lines are perpendicular, the sum of the product of their drs is equal to zero.

Solution:

For line L1 - 

\(\frac{x-q}{p} = \frac{y}{1}\) and \(\frac{z-s}{r} = \frac{y}{1}\)

Hence, L1 : \(\frac{x-q}{p} = \frac{y}{1} =\frac{z-s}{r}\)

For line L2 - 

\(\frac{x-q'}{p'} = \frac{y}{1}\) and \(\frac{z-s'}{r'} = \frac{y}{1}\)

Hence, L1 : \(\frac{x-q'}{p'} = \frac{y}{1} =\frac{z-s'}{r'}\)

Equating the sum of the product of their drs = 0

⇒ pp' + 1 + rr' = 0

Hence pp' + rr' + 1 = 0

Given:

\(∫ {\frac{{dx}}{{x - {x^3}}} = A.} \)In \(\left| {\frac{{{x^2}}}{{1 - {x^2}}}} \right| + c\)

Formula:

\(\int \frac{dx}{x} = ln|x| + c\)

\(\int \frac{dx}{(ax+b)} = \frac{1}{a}ln|(ax+b)| + c\)

Calculation:

I = \(∫ {\frac{{dx}}{{x - {x^3}}} } \)

⇒ I = \(∫ {\frac{{dx}}{{x(1 - {x^2 )}}} } \)

Multiplying the numerator and denominator by x, we have -

⇒ I = \(∫ {\frac{{xdx}}{{x^2(1 - {x^2 )}}} } \)

Substituting x² = t,

⇒ 2xdx = dt

Now the integral becomes -

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

⇒ I = \(\frac{1}{2}\)∫ \(\frac{(1-t+t)dt}{t(1-t)}\)

⇒ I = \(\frac{1}{2}\)∫\(\frac{dt}{t}\) + \(\frac{1}{2}\)∫\(\frac{dt}{(1-t)}\)  

⇒ I = \(\frac{1}{2}\) ln |t| - \(\frac{1}{2}\)ln |(1 - t)| + c

Putting t = x² back, we get -

I = \(\frac{1}{2}\)\(ln|\frac{x^2}{1-x^2}| + c\)

∴ On comparing A = 1/2

Concept:

Greatest Integer Function (GIF) gives the greatest integer less than or equal to the given value of x.

Formula:

\(\int x^ndx = \frac{x^{n+1}}{n+1} + c, n\neq -1\)

If I  ≤ x ≤ (I + 1), then [x] = I where I is an integer

Calculation:

\(\int\limits_{ - 1}^1 {f\left( x \right).dx} \) = \(\int\limits_{ - 1}^0 {f\left( x \right).dx} \) + \(\int\limits_{ 0}^1 {f\left( x \right).dx} \)

⇒ \(\int\limits_{ - 1}^1 {(x-[x])dx} \) = \(\int\limits_{ - 1}^0 {(x-[x])dx} \) + \(\int\limits_{ 0}^1 {(x-[x])dx} \)

Now as [x] = -1 for -1 ≤ x ≤ 0 and [x] = 0 for 0 ≤ x ≤ 1

⇒ \(\int\limits_{ - 1}^1 {(x-[x])dx} \) = \(\int\limits_{ - 1}^0 {(x+1)dx} \) + \(\int\limits_{ 0}^1 {(x)dx} \)

⇒ \(\int\limits_{ - 1}^1 {(x-[x])dx} \) = 1/2 + 1/2 = 1

Given:

f(x) = sin x

and g(x) = cos x

Concept:

The area bounded by the curves f(x) and g(x) between x = a and x = b is given by \(\int_{a}^{b} |(f(x) - g(x)|dx\)

Solution:

Referring to the combination of sinx and cosx, the shaded region will be repeated an infinite number of times.

Hence we can see there are infinite regions each of the same area.

Since, sinx ≥ cosx on the interval  [π/4 , 5π/4]

∴ Area = \(\int_{\pi/4}^{5\pi/4} |(sin(x) - cos(x)|dx\)

⇒ Area = \(\int_{\pi/4}^{5\pi/4} (sin(x) - cos(x))dx\)      as sinx > cosx in given domain of x.

Which on integration gives Area = \(2\sqrt 2 \)

Concept:  

• Binomial distribution summarizes the number of trials when each trial has the same probability of attaining one particular value.
• Only two outcomes are possible in a binomial distribution, either success or failure.
• Some of the underlying assumptions of the binomial distribution are:
• It consists of multiple numbers of trials, for example, tossing a coin multiple times.
• Each trial has the same probability of each outcome.
• Each trial is mutually exclusive or independent of one another.
• The expected value or the mean of a binomial distribution is calculated by multiplying the number of trials (n) by the probability of success (p).
• Mean = np
• The binomial distribution formula is given by,

P(x:n,p) = \(\frac{n!}{r!(n-r)!}p^{r}(1-p)^{n-r}\) ....(1)

• Where, n = the number of trials.
• p is the probability of a successful outcome in a trial.
• 1 - p is the probability of an unsuccessful outcome in a trial.
• r is the number of successful trials.
• The binomial distribution with n trials and success probability p is denoted by B(n, p).

Calculation:

Given: A fair dice is thrown six times, hence the probability of each outcome of a trial will remain the same.

n = 6, The number of successful trials, r = 4

• The trials are independent, hence this is a case of a binomial distribution problem.
• Factors of 6 are 1, 2, 3, and 6.
• Hence probability of achieving a successful outcome for this problem,

p = \(\frac{4}{6}\) = \(\frac{2}{3}\).

• Hence probability of unsuccessful outcome = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\).
• According to the binomial theorem from equation (1), the probability of getting the factors of 6 four times will be equal to,

P(x: n,p) = \(\frac{n!}{r!(n-r)!}p^{r}(1-p)^{n-r}\) 

P(x: 6, \(\frac{2}{3}\)) = \(\frac{6!}{4!(6-4)!}(\frac{2}{3})^{4}(1-\frac{2}{3})^{6-4}\) = \(\frac{6!}{4!(6-4)!}(\frac{2}{3})^{4}(\frac{1}{3})^{2}\)

P(x: 6, \(\frac{2}{3}\)) = 15 × \(\frac{16}{81}\) × \(\frac{1}{9}\) = \(\frac{80}{243}\)

• Hence, the correct answer is option 3.

Given:

\(\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{f\left( {\frac{y}{x}} \right)}}{{f'\left( {\frac{y}{x}} \right)}}\)

Concept:

To solve such type of differential equations, simply put y = vx.

Calculation:

Putting y = vx

⇒ \(\frac{dy}{dx} = v + x\frac{dv}{dx} \) 

Now, \(\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{f\left( {\frac{y}{x}} \right)}}{{f'\left( {\frac{y}{x}} \right)}}\)

⇒ \( v + x\frac{dv}{dx} \) = v + \(\frac{f(v)}{f'(v)}\)

⇒ \(\frac{f'(v)}{f(v)}dv\) = \(\frac{dx}{x}\)

Integrating both sides -

⇒ ln|f(v)| = ln|x| + lnC

⇒ f(v) = Cx

putting v = y/x back,

⇒ f(y/x) = Cx

Given:

 \(\vec{a}\) = cxî - 6ĵ + 3k̂

 \(\vec{b}\) = xî + 2ĵ +2cxk̂ 

Vectors make an obtuse angle with each other.

Concept:

The cosine of an obtuse angle is negative.

Formula:

If the angle between two vectors \(\vec{a}\) and \(\vec{b}\) is θ, then,

cosθ = \(\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\)

Calculation:

\(\vec{a}\).\(\vec{b}\) = (cxî - 6ĵ + 3k̂).(xî + 2ĵ + 2cxk̂)

⇒  \(\vec a .\vec b\) = cx² - 12 + 6cx 

Also |\(\vec{a}\)| and |\(\vec{b}\)| are always positive

Hence if cosθ < 0 

⇒ ​\(\vec{a}\).​​\(\vec{b}\) < 0

⇒ cx² - 12 + 6cx < 0​

Now, since x is real so the above inequality is only possible when the above parabola will be opening downward with no real root of x.

⇒ c < 0 and D < 0

⇒ 36c² + 48c < 0

⇒ c(3c + 4) < 0

⇒ -4/3 < c < 0

Taking the common range of c, 

⇒ -4/3 < c < 0

Concept:

The range of a function is the set of all the elements which are obtained by mapping all the elements in the domain of the function.

Solution:

(i) y = \(f(x) = \frac{x}{{1 + {x^2}}}\)

We can easily see that the domain of a function is x ∈ R

For range, express x as a function of y

⇒ x²y - x + y = 0

⇒ x = \({-1 ± √{1-4y^2} \over 2y}\)

We can see for real x, (1 - 4y²) ≥ 0 

⇒ -1/2 ≤ y ≤ 1/2 

Hence solution set is \({R_f} = \left[ { - \frac{1}{2},\frac{1}{2}} \right]\)

(ii) y = \(f(x) = \frac{3}{{2 - {x^2}}}\)

We can easily see that the domain of a function is x ∈ R - {-√2, √2}

For range, express x as a function of y

⇒ x = ±

We can see for real x, \(\frac{2y - 3}{y}\) ≥ 0 

⇒ y < 0 , y ≥ 3/2

Hence solution set Rf = (-∞, 0) ∪ \(\left[ {\frac{3}{2},∞ } \right]\)

(iii) y = \(f(x) = \frac{1}{{√ {x - \left[ x \right]} }}\) 

We know that x - [x] ≥ 0 ,therefore function is defined for all non-integral values of x as on integer values x - [x] = 0 which makes denominator 0.

Also 0 < x - [x] < 1 for non-integer x

⇒ 0 < √(x - [x]) < 1 for non-integer x

or 1 < \( \frac{1}{{√ {x - \left[ x \right]} }}\) < ∞ 

Hence solution set is Rf = (1, ∞)

(iv) \(f\left( x \right) = \frac{1}{{2 - \sin 3x}}\)

We can easily see it is defined for all real x as denominator can never be 0 as -1 ≤ sin3x ≤ 1

Now, ∵ -1 ≤ sin3x ≤ 1

⇒ -1 ≤  -sin3x ≤ 1

⇒ 1 ≤ 2 - sin3x ≤  3

⇒ \(​​\frac{1}{3} \leq \frac{1}{{2 - \sin 3x}}\leq 1\)

Hence solution set \({R_f} = \left[ {\frac{1}{3},1} \right]\)

Hence correct option is (ii).

Concept: 

There are three main components of linear programming:

• Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
• The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
• Constraints: These represent real-life limitations such as money, time, labor, or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.
• For example, suppose x and y are the decision variables. The objective function will be given by:

Z = ax + by ….(1)

Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

• The equation looks very simple since there are various assumptions involved while forming a linear programming example. These are mentioned below:
• Parameters such as resources available, profit contribution of unit decision variable, and resources used by unit decision variable need to be known.
• Decision variables are continuous. Hence the outputs can be an integer or a fraction.
• The contribution of each decision variable in the objective function is directly proportional to the objective function.

Calculation:

Given:

• From the above problem, Durgesh aims to maximize his earnings. x and y represent the packets of dishes A and B respectively. These represent the decision variables for the problem.
• Rs 50 can be earned by selling a packet of dish A while Rs 70 can be earned by selling a packet of dish B. 
• Hence, the objective function is given by,

Z = 50x + 70y

• Hence, the correct answer will be option (3).

Concept:

In the yz-plane, the x coordinate is always 0.

To find the ratio, we assume it λ : 1.

On solving with the given conditions, if λ is positive then ratio is internal otherwise external.

Formula:

If the join of two points \((x_1,y_1,z_1) \) and \((x_2,y_2,z_2) \) is divided in the ratio m : n by a point P (x , y , z), then the coordinates of the point P will be -

\(x = \frac{mx_2 + nx_1}{m + n}\) , \(y = \frac{my_2 + ny_1}{m + n}\)  , \(z = \frac{mz_2 + nz_1}{m + n}\)

Calculation:

Given:

Points A(4, 8, 10) and B(6, 10, -8)

By evaluating x-coordinate,

⇒ \(0 = \frac{4 + 6λ}{λ + 1}\)

⇒ 4 + 6λ = 0 

⇒ λ = -2/3   (negative sign means externally)

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 (adj A)B = O.

Calculation:

Given: \(adj A=\begin{pmatrix} 1 &-4 &1 \\ 2&8 &2 \\ 3&-12 &3 \end{pmatrix}\)

• A is the matrix formed by the coefficient of x, y, and z of the given system of linear equations

\(A=\begin{pmatrix} 1 &-2 &1 \\ 1&1 &-1 \\ 3 &6 &-5 \end{pmatrix}\) and \(B=\begin{pmatrix} 5\\6 \\7 \end{pmatrix}\)

• The determinant of the matrix A is given by,

\(|A|=\begin{vmatrix} 1 &-2 &1\\ 1&1 &-1 \\ 3&6 &-5 \end{vmatrix}\)

• Applying R2 →  R2 - R1 and R3→  R3 -3R1

⇒\(|A|=\begin{vmatrix} 1 &-2 &1 \\ 0&3 &-2 \\ 0&12 &-8 \end{vmatrix}\) 

⇒ |A| = (-24 + 24) + 0 + 0 = 0

• Then the system has not a unique solution.
• Let us evaluate the (adj A) B of the matrix.

\(\Rightarrow (adj A)B=\begin{pmatrix} 1 &-4 &1 \\ 2&8 &2 \\ 3&-12 &3 \end{pmatrix}\begin{pmatrix} 5\\6 \\7 \end{pmatrix}=\begin{pmatrix} -12\\72 \\-36 \end{pmatrix}\ne O\)

• Thus the system is inconsistent and has no solution.
• Therefore option 2 is correct.

Concept:

•  Linear Regression is used to draw a line or find a linear equation that best fits a given set of data.
• This method calculates the parameters such that the sum of squares of the difference between actual and computed values is minimized. 
• The data needs to be brought in the form of,

y = a + bx ....(1),

• For the purpose of plotting the best-fitted line for trend analysis, the real values of constants ‘a’ and ‘b’ are estimated by solving the following two equations:

Σy = n a + b Σx --------------------- (ii)
Σxy = a Σ x + b Σ x² -------------------- (iii)

• Where ‘n’ = number r of years given in the data.
• Remember that the time unit is usually of successive uniform duration. Therefore, when the middle time period is taken as the point of origin, it reduces the sum of the time variable x to zero which means that by taking the mid-point of the time as the origin, we get

Σx = 0

⇒ b = \(\frac{\sum xy}{\sum x^2}\)

Also, a = \(\frac{\sum y}{n}\), = The average of the dependent variable.

Where y is the estimated value of the data and x is time. 

• a and b are calculated from actual values of x and y.

Calculation:

Given:

• ∑xy = 6 and ∑x² = 5.
• ‘b’ will be calculated by the formula,

b =  \(\frac{\sum xy}{\sum x^2}\) = \(\frac{6}{5}\) = 1.2

• Hence the correct answer is option 3. 

Given:

\(f(x) = \sin \left( {x + \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \)

Formula Used:

sin(A + B) = sin A cos B + cos A sin B

Calculation:

We have, 

\(f(x) = \sin \left( {x + \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \)

Multiply and divide the given function by \(\sqrt2\)

⇒ \(f(x) = \frac{\sqrt2}{\sqrt2}\left[\sin \left( {x + \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \right] \)

⇒ \(f(x) = \sqrt2\left[\frac{1}{\sqrt2}\sin \left( {x + \frac{\pi }{6}} \right) + \frac{1}{\sqrt2}\cos \left( {x + \frac{\pi }{6}} \right) \right]\)

⇒ \(f(x) = \sqrt2 \left[cos\frac{\pi}{4}sin(x + \frac{\pi}{6}) + sin\frac{\pi}{4}cos(x + \frac{\pi}{6})\right]\)

⇒ \(f(x) = \sqrt2\left[ sin(\frac{\pi}{4} + x + \frac{\pi}{6})\right]\)

⇒ \(f(x) = \sqrt2 \ sin(x + \frac{5\pi}{12})\)

For maximum value:-

⇒ \(x + \frac{5\pi}{12} = \frac{\pi}{2}\)

⇒ \(x = \frac{\pi}{2} - \frac{5\pi}{12}\)

⇒ \(x = \frac{\pi}{12}\)

∴ The maximum value of the given function is \(\frac{\pi}{12}\)

Given:

f(x) = \(2tan^{-1}x + sin^{-1}\frac{2x}{1 + x^2}\)

Calculation:

We know that,

2tan^-1x = \( sin^{-1}x(\frac{2x}{1 + x^2}), where \ \ \ 1 ≤ x ≤ 1\)

2tan^-1x = \(- π - sin^{-1}(\frac{2x}{1 + x^2}), \ where \ x < - 1\)

2tan^-1x = \(π - sin^{-1}x (\frac{2x}{1 + x^2}), \ where \ x > 1\)

So, according to conditions,

⇒ f(x) = 2tan^-1x + (π - 2tan^-1x)

⇒ f(x) = π 

∴ The value of f(x) is π.

Given:

\(f(x) = [x+1] - [x] +\frac{1}{x^2-1}\)

Concept:

A continuous function f(x) is decreasing in an interval [a, b] if f'(x) < 0 in the interval [a, b]

Formula:

[x + I] = [x] + I, where I is some integer and [x] is the greatest integer less than or equal to x.

Calculation:

By above formula, [x + 1] = [x] + 1

∴ [x + 1] - [x] = [x] + 1 - [x] 

⇒ [x + 1] - [x] = 1

∴ \(f(x) = 1 + \frac{1}{x^2-1}\)

⇒ \(f(x) = \frac{x^2}{x^2-1}\)

Now, f'(x) = \(\frac{-2x}{(x^2-1)^2}\) , x ≠ ±1 as f'(x) is not defined there.

For decreasing function, f'(x) < 0

⇒ \(\frac{-2x}{(x^2-1)^2}\) < 0

⇒ -2x < 0 as (x² - 1)² is always positive

or x > 0

Taking the intersection of x > 0 and x ≠ ±1, f(x) decreases in the interval (0, 1) ∪ (1, ∞)

Concept:

• For local maximum or local minimum of a continuous and differentiable function, f'(x) = 0
• If f''(x) < 0 where f'(x) = 0, then it is a point of local maximum.
• If f''(x) > 0 where f'(x) = 0, then it is a point of local minimum.

Calculation:

f'(x) = \(\frac{-2x}{(x^2-1)^2}\) 

For maximum or minimum f'(x) = 0

⇒ \(\frac{-2x}{(x^2-1)^2}\) = 0

⇒ x = 0

Now, f''(x) = \(\frac{6x^2+2}{(x^2-1)^3}\)

f''(x) at (x = 0) = -2 < 0, hence maxima

∴ We can see that f'(x) = 0 at only x = 0 and maxima is there.

Concept:

Critical points of a function are those values of x in the interval where -

• f(x) does not exist.
• f'(x) does not exist
• f'(x) = 0

Calculation:

\(f(x) = \frac{x^2}{x^2-1}\)

⇒ f(x) is not defined at x = ± 1

f'(x) = \(\frac{-2x}{(x^2-1)^2}\)  

⇒ f'(x) is not defined at x = ± 1 as (x² - 1 = 0) there.

Also, f'(x) = 0 at x = 0

Hence critical points are x = -1, 0, +1

Given:

The tangent to the curve f(x) is at P(2, \(\frac{4}{3}\))

Concept:

• The slope of the tangent at point P(a, b) to a curve y = f(x) is given by : m = f'(x)_{|at (a, b)}
• _​The equation of the tangent at point P(a, b) to a curve y = f(x) is (y - b) = m(x - a) where m = f'(x)|at (a, b) 

Calculation:

f'(x) = \(\frac{-2x}{(x^2-1)^2}\)

⇒ f'(x) (at P) = \(\frac{-4}{9}\)

∴ m = \(\frac{-4}{9}\)

Now, the equation of tangent : (y - \(\frac{4}{3}\)) = \(\frac{-4}{9}\)(x - 2)

⇒ 4x + 9y = 20

It cuts x-axis at (5, 0) and y-axis at (0, 20/9)

∴ Area = \(\frac{1}{2} \times 5\times \frac{20}{9}\)

= 50/9 sq. units

Explanation:

From above, it has only one local maximum and that is at x = 0

Hence local maximum value at x = 0 -

f(0) = 0

Concept:

If the value of the objective function maximizes at two corner points then it will maximize at all the infinite points in the line joining those two corner points.

Calculation:

Given:

The objective function : z = 3x + 4y

The constraints are :

4x + 3y ≤ 24,

3x + 4y ≤ 24,

and x, y ≥ 0

The graph of the constraints is :

• The corner points are (0 , 0) , (6 , 0) , (0 , 6) and (24/7 , 24/7)
• The values at these points are 0, 18, 24, and 24 respectively.

∴ We can see that at two points (0  6) and (24/7, 24/7) maximum value occurs.

Concept:

• A random variable is a variable that can take multiple values based on the outcome of an event like throwing dice or tossing a coin.
• If outcomes are infinite (like the weight of a person can range from 60.4 kg to 75.3 kg)), then the random variable is continuous.   
• The description of the probability of each possible value of the random variable that can occur is called its probability distribution.
• When the random variable is continuous, the probability distribution is called the probability density function.
• For continuous random variables, the probability of some given value x is equal to zero.
• Probability of X lying in an interval (a, b) is calculated which is given by the following formula:

P(a ≤ X ≤ b) = P(a < X < b) = \(\int_{a}^{b}f(x)dx\) ....(1)  

• Here, P(a ≤ X ≤ b) = P(a < X < b), since X is continuous, hence endpoints of intervals while finding probabilities can be ignored.​

Calculation:

Given:

• This is a continuous random variable-based problem.

\(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} { - \;e^x,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;if\ \;-1\leq x < 0}\\ {\frac{1}{1 +x^2},\;\;\;\;if\ \;0 \leq x < 1}\\ {e^{-1}-\frac{\pi}{4},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;else} \end{array}} \right.\) 

• P(-1 < x < 1) can be calculated by probability density function formula from equation (1).

\(\Rightarrow\) P(-1 < x < 1) = \(\int_{-1}^{1}f(x)dx\)  

\(\Rightarrow\) P(-1 < x < 1) = \(\int_{-1}^{0}f(x)dx\)  + \(\int_{0}^{1}f(x)dx\)

\(\Rightarrow\) P(-1 < x < 1) = \(\int_{-1}^{0}(e^x)dx\)  + \(\int_{0}^{1}(\frac{1}{1+x^2})dx\)

• Since, \(\int_{}^{}(e^x)dx\) = e^x and \(\int_{}^{}(\frac{1}{1+x^2})dx\) = tan^-1x

\(\Rightarrow\) P(-1 < x < 1) = e⁰ - e^-1 + tan-1(1) - tan-1(0)​

\(\Rightarrow\) P(-1 < x < 1) = 1 -  e-1 + \(\frac{\pi}{4}\) - 0

​\(\Rightarrow\) P(-1 < x < 1) = ​1 - \(\frac{1}{e}\) + \(\frac{\pi}{4}\)

\(\Rightarrow\) P(-1 < x < 1) =  \(\frac{4e-4+e\pi}{4e}\)

• Hence the correct answer is option 1.

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)

Calculation:

Given: Payment of perpetuity is done on a quarterly basis at the end of each period.

• The regular perpetuity payment (R) = 40,000 Rs
• Present value, P = Rs 10,00,000.
• Annual interest rate = 100r% = r
• Quarterly interest rate = \(\frac{r}{4}\)
• If R is the periodic payment, then from the formula of the present value of a perpetuity, the following can be written,

\(\frac{r}{4}\) = \(\frac{40,000}{10,00,000}\)

⇒ r = 0.16

• Hence the quarterly periodic payment = Rs 40,000 
• Hence the correct answer is option 3.

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. 
• 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.  

Calculation: 

Given:

• Cost of each ball = Rs 500
• An increase in revenue by selling 1 more unit of the ball will always be fixed at Rs 500. 
• The marginal cost is Rs 500.      
• Hence the correct answer is an option (2).

Concept:

(X+Y) mod M = X mod M + Y mod M

Calculation:

Given: X = 48, Y=15 and X mod Y = (X + kY) mod Y

∴ X mod Y = 48 mod 15 = 3

(48 + k 15) mod 15 = 48 mod 15 + 15k mod 15

⇒ (48 + k15) mod 15 = 3 + 0 = 3 for all integers k

• Thus k can be any integer.
• Therefore the option 4 is correct.