Created By - Thunder

Concept:

|A'| = |A|, where A' is the transpose of A.

|A–1| = \(\frac{1}{|A|}\), where A–1 is the inverse of A. 

Calculation:

Given, A = \(\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]\)

⇒ |A| = 1 + tan²x = sec²x 

∴ |A' A–1| 

= |A'||A–1|

= |A'| |A|–1

= \(\frac{|A|}{|A|}\)

= 1

∴ The value of |A' A–1| is 1.

The correct answer is Option 4.

Explanation -

Required area = area of the circle – area bounded by given line and parabola

Required area = πr2 –

\(\int_0^1\left(y-y^2\right) d y\)

Area = \(2 \pi-\left(\frac{y^2}{2}-\frac{y^3}{3}\right)_0^1\)

= 2π – (1/6)

= (1/6)(12π – 1) 

Hence Option (2) is correct.

Concept -

(i) A diagonal matrix is a square matrix where all the elements outside the main diagonal are zero. The main diagonal is the set of elements where the row index equals the column index.

In other words, a diagonal matrix is a matrix in which all the off-diagonal elements are zero.

(ii) Identity matrix is \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \)

Explanation -

We have a 3 x 3 diagonal matrix:

\(\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \)

In this matrix, all the elements off the main diagonal (i.e., non-diagonal elements) are zero. and diagonal elements are no zero.

Hence Option(4) is true.

Calculation:

Given, \(\rm D=\begin{vmatrix}y+z&z&y\\\ z&z+x&x\\\ y&x&x+y\end{vmatrix}\)

C₁ → C₁ + C₂ + C₃

⇒ \(\rm D=\begin{vmatrix}2(y+z)&z&y\\\ 2(y+z)&z+x&x\\\ 2(y+z)&x&x+y\end{vmatrix}\)

⇒ \(\rm D=2\begin{vmatrix}y+z&z&y\\\ y+z&z+x&x\\\ y+z&x&x+y\end{vmatrix}\)

C1 → C1 - C2 - C3

⇒ \(\rm D=2\begin{vmatrix}0&z&y\\\ -x&z+x&x\\\ -x&x&x+y\end{vmatrix}\)

⇒ \(\rm D=-2x\begin{vmatrix}0&z&y\\\ 1&z+x&x\\\ 1&x&x+y\end{vmatrix}\)

R2 → R2 - R3

⇒ \(\rm D=-2x\begin{vmatrix}0&z&y\\\ 0&z&-y\\\ 1&x&x+y\end{vmatrix}\) = - 2x(-yz - yz) = 4xyz = 8

⇒ xyz = 2

Possible cases are:

{(2, 1, 1), (1, 2, 1), (1, 1, 2), (−2, −1, 1), (−2, 1, −1), (−1, −2, 1), (−1, 2, −1), (1, −2, −1), (1, −1, −2), (−1, −1, 2), (−1, 1, −2), (1, −1, 2)}

∴ The number of solutions are 12.

The correct answer is Option 3.

Calculation:

Given, x2 + y2 = 36…(i) which is outside the parabola y2 = 9x…(ii)

Using (ii) in (i), we get x2 + 9x = 36

⇒ x2 + 12x - 3x - 36 = 0

⇒ x(x + 12) - 3(x + 12) = 0

⇒ (x + 12)(x - 3) = 0

⇒ x = 3[As they meet on the +ve side x-axis]

⇒ y = ± 3√3

∴ The curves intersect at point (3, ± 3√3)

∴ Required area

= \(\pi r^2-2\left[\int_0^3 \sqrt{9 x} d x+\int_3^6 \sqrt{36-x^2} d x\right]\)

= \(36 \pi-12 \sqrt{3}-2\left(\frac{x}{2} \sqrt{36-x^2}+18 \sin ^{-1}\left(\frac{x}{6}\right)\right)_3^6\)

= \(36 \pi-12 \sqrt{3}-2\left(9-\left(\frac{9 \sqrt{3}}{2}+3 \pi\right)\right)\)

= 24π – 3√3

∴ The required area is 24π – 3√3.

The correct answer is Option 4.

Formula Used:

The determinant of a 3x3 matrix  A  is given by the following formula:

\( \text{det}(A) = a_{11} \cdot \text{det}(A_{11}) - a_{12} \cdot \text{det}(A_{12}) + a_{13} \cdot \text{det}(A_{13}) \)

where  A_ij is the submatrix obtained by removing the i^th row and j^th column, and a_ij is the element in the  ith row and jth column of matrix  A.

Explanation:

Given Matrix \(A = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 22 \\ 0 & \frac{1}{7} & π \end{array}\right] \)

\( \text{det}(A) =1 \cdot \text{det}(A_{11}) - 0 \cdot \text{det}(A_{12}) +0 \cdot \text{det}(A_{13}) \)

⇒ \( \text{det}(A) = \text{det}(A_{11}) = \left[\begin{array}{cc} 1 & 22 \\ \frac{1}{7} & π \end{array}\right] \)

⇒ det(A) = π - 22/7

As \(\frac{22}{7} = 3.\overline{142857}\) and π = 3.14159256 . . . 

Thus, π - 22/7 < 0 < 1

Also π - 22/7 is an irrational number as well.

Thus, det(A) is an irrational number less than 1.

Calculation:

Given, f(x) = e3x + 3x + 2

Now, f(0) = 1 + 0 + 2 = 3

and, f(\(\frac{1}{3}\)) = e + 1 + 2 = e + 3

We know, \(\displaystyle \int_a^bf(x) dx + \int_{f(a)}^{f(b)}f^{-1}(x)dx\) = bf(b) - af(a)

Putting a = 0 and b = \(\frac{1}{3}\), we get:

\(\displaystyle \int_0^\frac{1}{3}f(x) dx + \int_{f(0)}^{f(\frac{1}{3})}f^{-1}(x)dx \) = \(\frac{1}{3}\) f(\(\frac{1}{3}\)) - 0f(0) = \(\frac{1}{3}\)(e + 3)

⇒ \(\displaystyle \int_0^\frac{1}{3} (e^{3x}+3x+2)dx + \int_3^{e+3}f^{-1}(x)dx\) = \(\frac{1}{3}\)(e + 3)

⇒ \(\frac{1}{6}\)(2e + 3) + \(\displaystyle \int_3^{e+3}f^{-1}(x)dx\) = \(\frac{1}{3}\)(e + 3)

⇒ \(\displaystyle \int_3^{e+3}f^{-1}(x)dx\) = 1/2

∴ The value of \(\displaystyle \int_3^{e+3}f^{-1}(x)dx\) is 1/2.

The correct answer is Option 2.

Given:

 y = 2x + x log x

Concept:

Use formula

\(\rm \frac{d}{dx}(a^x)=a^x\ loga\)

\(\rm \frac{d}{dx}[f(x)g(x)]=\frac{d}{dx}[f(x)]g(x)+f(x)\frac{d}{dx}[g(x)]\)

Calculation:

 y = 2x + x log x

Differentiate with respect to x

\(\rm \frac{dy}{dx}=2^x\ log 2 + 1\cdot log x + x\cdot\frac{1}{x}\)

\(\rm \frac{dy}{dx}\)= 2x log 2 + log x + 1

Hence the option (4) is correct.

Concept Used:

The integrating factor (IF) for a first-order linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\) is given by \(e^{\int P(x) dx}\).

Calculation:

Given:

The differential equation is \(x \frac{dy}{dx} - y = x^2\).

⇒ \(\frac{dy}{dx} - \frac{1}{x}y = x\)

Here, \(P(x) = -\frac{1}{x}\) and \(Q(x) = x\).

The integrating factor is:

⇒ \(IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx}\)

⇒ \( = e^{-\ln|x|} = e^{\ln|x^{-1}|} = x^{-1} = \frac{1}{x}\)

Hence option 2 is correct

Concept:

\(P\left(\frac{A}{B}\right)=P\left(\frac{P(A\bigcap B)}{P(B)}\right)\)

Calculation:

Given, P(A') = 0.3, P(B) = 0.4 and P (A ∩ B') = 0.5

∴ \(\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A} \cup \mathrm{B}^{\prime}}\right)\)

= \(\frac{P(\mathrm{B}\bigcap (\mathrm{A} \cup \mathrm{B}^{\prime}))}{P (\mathrm{A} \cup \mathrm{B}^{\prime})}\)

= \(\frac{P(\mathrm{B}\bigcap (\mathrm{A} \cup \mathrm{B}^{\prime}))}{P (\mathrm{A})+ P( \mathrm{B}^{\prime})-P(\mathrm{A}\bigcap \mathrm{B}^{\prime})}\)

= \(\frac{P(\mathrm{A}\bigcap \mathrm{B})}{P (\mathrm{A})+ P( \mathrm{B}^{\prime})-P(\mathrm{A}\bigcap \mathrm{B}^{\prime})}\)

= \(\frac{P (\mathrm{A})-P(\mathrm{A}\bigcap \mathrm{B}^{\prime})}{P (\mathrm{A})+ P( \mathrm{B}^{\prime})-P(\mathrm{A}\bigcap \mathrm{B}^{\prime})}\)

= \(\frac{0.7-0.5}{0.7+0.6-0.5}\)

= \(\frac{0.2}{0.8}\)

= \(\frac{1}{4}\)

∴ The required probability is \(\frac{1}{4}\)

The correct answer is Option 1.

Calculation:

Given, f(x) = e2x + 4x - 3

Now, f(0) = 1 + 0 - 3 = -2

and, f(\(\frac{1}{2}\)) = e + 2 - 3 = e - 1

We know, \(\displaystyle \int_a^bf(x) dx + \int_{f(a)}^{f(b)}f^{-1}(x)dx\) = bf(b) - af(a)

Putting a = 0 and b = \(\frac{1}{2}\), we get:

\(\displaystyle \int_0^\frac{1}{2}f(x) dx + \int_{f(0)}^{f(\frac{1}{2})}f^{-1}(x)dx\) = \(\frac{1}{2}\)f(\(\frac{1}{2}\)) - 0f(0) = \(\frac{1}{2}\)(e - 1)

⇒ \(\displaystyle \int_0^\frac{1}{2} e^{2x}+4x-3dx + \int_{-2}^{e-1}f^{-1}(x)dx\) = \(\frac{1}{2}\)(e - 1)

⇒ \(\frac{1}{2}\)(e - 3) + \(\displaystyle \int_{-2}^{e-1}f^{-1}(x)dx\) = \(\frac{1}{2}\)(e - 1)

⇒ \(\displaystyle \int_{-2}^{e-1}f^{-1}(x)dx\) = 1

∴ The value of \(\displaystyle \int_{-2}^{e-1}f^{-1}(x)dx\) is 1.

The correct answer is Option 1.

Calculation:

Given, x + y + z = 4,

2x + 5y + 5z = 17,

x + 2y + mz = n

∴ D = \(\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & \mathrm{~m} \end{array}\right|\) = 0 

⇒ 1(5m - 10) - 1(2m - 5) + 1(4 - 5) = 2 

⇒ m = 2

Also, D_z = \(\left|\begin{array}{ccc} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{array}\right|\) = 0 

⇒ 1(5n - 34) - 1(2n - 17) + 4(4 - 5) = 0

⇒ n = 7

∴ m2 + n2 – mn = 22 + 42 – 14 = 4 + 49 – 14 = 39

∴ m and n satisfy the equation m2 + n2 – mn = 39.

The correct answer is Option 4.

Calculation:

Let A_i denote the event that the number i appears on the dice and let E denote the event that only white balls are drawn.

∴ \(P(A_i) = \frac{1}{6} \quad \text{for} \quad i = 1, 2, \dots, 6 \)

and, \(P(E / A_i) = \frac{^6C_i}{^{10} C_i}, \quad i = 2, \dots, 6 \)

∴ P(E)

⇒ \(\sum_{i=1}^{6} P(E \cap A_i) \)

⇒ \(\sum_{i=1}^{6} P(A_i) P(E \big| A_i) \)

⇒ \(\frac{1}{6} \left[ \frac{6}{10} + \frac{15}{45} + \frac{20}{120} + \frac{15}{210} + \frac{6}{252} + \frac{1}{210} \right] \)

⇒ \(\frac{1}{5}\)

∴ The probability that the balls chosen are white is \(\frac{1}{5}\).

The correct answer is Option 1.

Calculation:

Given: Z = ax + 3ay is maximum at points (1, 3) and (3, 9).

• If the objective function is maximum at points (1, 3) and (3, 9), we can write  the following:

⇒ 1 × a + 3 × 3 = 3 × a + 9 × 3

⇒ a + 9 = 3a + 27

⇒ a = - 9

• The objective function is given by -9x + 3y.
• Putting any of the points out of (1, 3) and (3, 9) will give the maximum value.
• Hence the maximum value will be given by,

⇒ Z_max  = - 9 × 3 + 9 × 3

⇒ Zmax  = -27 + 27 = 0

• So, the correct answer is option 1.

CONCEPT:

The inverse of a matrix: The Inverse of an n × n matrix is given by:

\({A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\) where adj(A) is called an adjoint matrix.

Adjoint Matrix: If Bn× n is a cofactor matrix of matrix An× n then the adjoint matrix of An× n is denoted by adj(A) and is defined as BT. So, adj(A) = BT.

CALCULATION:

Given: A-1 = x A + y I

\(\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 5}&1 \end{array}} \right]\therefore {A^{ - 1}} = \frac{{adjA}}{{\left| A \right|}} =\frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 5&1 \end{array}} \right]\\ \end{array}\)

\( ⇒ \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 1&-2\\ { -5}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} x&{2x}\\ { - 5x}&x \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} y&0\\ 0&y \end{array}} \right]\)

\( ⇒ x + y = \frac{1}{{11}},2x = \frac{{ - 2}}{{11}} \).

\(⇒ x = \frac{{ - 1}}{{11}},y = \frac{2}{{11}}\)

Calculation:

Given, I = \(\int_{0}^{π}\frac{dx}{1+e^{\cos x}}\) … (i)

Putting t = π + 0 - x ⇔ dt = - dx

At x = 0, t = π

At x = π, t = 0

∴ I = \(\int_{π}^{0}\frac{-dt}{1+e^{\cos (π - t)}}\)

= \(\int_{0}^{π}\frac{dt}{1+e^{-\cos t}}\)

= \(\int_{0}^{π}\frac{e^{\cos t}}{1+e^{\cos t}}dt\) = \(\int_{0}^{π}\frac{e^{\cos x}}{1+e^{\cos x}}dx\) … (ii)

Addign (i) and (ii), we get:

2I = \(\int_{0}^{π}dx\)

⇒ I = π/2.

∴ The value of the integral is π/2.

The correct answer is Option 2.

Concept:

Let y = f(x) be a function. Then,

The function is continuous if it satisfies the following conditions:

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}^{-}}{}}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a^+}}{ }}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right){\rm{}}\)

Calculation:

 f(x) = \(\rm \left\{\begin{matrix} \rm 12x + 3 λ, \, x\ne1 \\ \rm 0, \, x=1 \end {matrix} \right.\)

Since, f(x) is continuous at x  =1 

⇒  lim_x→112x + 3λ = 0

⇒ 12(1) + 3λ = 0 ⇒ 12 + 3λ = 0 

⇒ λ = -4

∴ Option 1 is correct

Given:

Set \( A = \{1, 2, 3\} \)

We need to find the number of non-empty equivalence relations on set \( A \).

Concept:

The number of equivalence relations on a set is equal to the number of partitions of the set. The number of partitions of a set with \( n \) elements is given by the Bell number, \( B_n \).

Formula Used:

The Bell number, \( B_n \), for \( n = 3 \) is calculated using:

\( B_n = \sum_{k=0}^{n} S(n, k) \), where \( S(n, k) \) is the Stirling number of the second kind.

Calculation:

The partitions of set \( A = \{1, 2, 3\} \) are:

1. One subset: \( \{\{1, 2, 3\}\} \)

2. Two subsets: \( \{\{1\}, \{2, 3\}\}, \{\{2\}, \{1, 3\}\}, \{\{3\}, \{1, 2\}\} \)

3. Three subsets: \( \{\{1\}, \{2\}, \{3\}\} \)

Total number of partitions = \( B_3 = 1 + 3 + 1 = 5 \).

\( \therefore \) The number of non-empty equivalence relations on set \( A \) is 5.

Concept:

• Limit of f(x) at x = a is said to exist if the function approaches the same value from both sides.
• \(\rm \displaystyle\lim_{x\rightarrow a}\) f(x) = exist if LHL = RHL at x = a
• L'Hôpital's rule : \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) if the limit on the right exists. L'Hôpital's rule can be used when we encounter an indeterminate form like \((\frac{0}{0}) or (\frac{\infty}{\infty})\)

Calculation:

Consider the limit:

\(\rm \displaystyle\lim_{x\rightarrow0}\frac{\sin(3x)-3\sin(x)}{x^3}\)

We have an indeterminate form  \((\frac{0}{0})\) 

Apply L'Hôpital's rule:

⇒ \(\lim_{x \to 0} \frac{\sin 3x - 3 \sin x}{x^3} = \lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2} \)

We still have an indeterminate form  \((\frac{0}{0})\) as both cos 3x and cos x approach 1 as \(x \to 0 \).

So, let's apply L'Hôpital's rule again.

⇒ \( \lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2} = \lim_{x \to 0} \frac{-3 \sin 3x + \sin x}{2x} \)

We still get the indeterminate form \((\frac{0}{0})\), so we need to apply L'Hôpital's rule one more time.

⇒ \( \lim_{x \to 0} \frac{-3 \sin 3x + \sin x}{2x} = \lim_{x \to 0} \frac{-9 \cos 3x + \cos x}{2} \)

⇒ \(\lim_{x \to 0} \frac{-9 \cos 3x + \cos x}{2}\) = \(\frac{-9 \cdot 1 + 1}{2}\) = \(\frac{-9 + 1}{2}\) =  -4 

So, the value of the limit is: -4

∴ Option 2 is correct

Calculation:

Given y² = 4x

⇒ a = 1

∴ A = 4t(12 - t²) = 48t - 4t³  

For maximum area, \(\frac{dA}{dt}\) = 0

⇒ 48 - 12t² = 

⇒ t² = 4

⇒ t = 2, - 2

Now,  \(\frac{d^2A}{dt^2}\) = - 24t

At t = 2, \(\frac{d^2A}{dt^2}\) < 0

⇒ t = 2 is the point of maxima.

∴ A_max = 4×2(12 - 2²) = 8(12 - 4) = 64 sq. units

∴ The maximum area is 64.

The correct answer is Option 1.

Calculation

\(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}\)

\(\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)\)

⇒ cos^–1 (cosx cosα + sinx sinα)

⇒ cos^–1 (cos(x – α)) 

⇒ x – α because x – α ∈ \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

⇒ \(x-\tan ^{-1} \frac{5}{12}\)

Hence option 2 is correct

Concept:

• \({d \over dx} (p(x) q(x)) = p'(x) q(x) + p(x)q'(x)\)
• \({d \over dx} (f(g(x)) = f'(g(x))g'(x)\)

Calculation:

If y = (xx)x, 

Taking log both sides,

⇒ log y = log (xx)x, 

⇒ log y = x log (xx), 

⇒ log y = x.x log x,

⇒ log y = x² log x,

Now differentiating both sides with respect to x,

⇒ \({1 \over y}{dy \over dx} = 2x \log x + x^2 .{1 \over x}\)

⇒ \({dy \over dx} = 2xy \log x + xy\)

⇒ ​\({dy \over dx} - xy (1 + 2\ln x) = 0\)​

∴ The correct answer is option (2).

Given:

Two matrices A and B of same order is given.

Concept Used:

A' = A, then A will be a symmetric matrix,

A' = - A, then A will be a skew- symmetric matrix.

Solution:

We have, 

(AB' - BA') if we calculate (AB' - BA')'

⇒ (AB' - BA')' = (AB')' - (BA')'

⇒ (AB' - BA')' = (B')' A' - (A')' B'  {(AB)' = B'A'}

⇒ (AB' - BA')" = B A' - A B'    {(A')' = A}

⇒ (AB' - BA')' = - (AB' - BA')

Hence (AB' - BA') is skew-symmetric matrix.

\(\therefore\) Option 1 is correct.

Calculation:

Given, x = 2t2 - 4t + 3 and y = t3 - 4t² + 6  

∴ \(\frac{dx}{dt}\) = 4t - 4

⇒ \(\frac{dx}{dt} \bigg|_{t=2}\) = 4(2) - 4 = 4

Also, \(\frac{dy}{dt}\) = 3t² - 8t

⇒ \(\frac{dy}{dt} \bigg|_{t=2}\) = 3(2)2 - 8(2) = - 4

∴ \(\frac{dy}{dx}\) = \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) = - 4/4 = - 1

∴ The slope of the tangent to the curve x = 2t2 - 4t + 3 and y = t3 - 4t² + 6 at t = 2 is

- 1.

The correct answer is Option 3.

Calculation

Given

\(\vec{a}=-5 \hat{i}+j-3 \hat{k}\)

\(\vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}\)

\((\vec{a} \times \vec{b}) \times \hat{i}=(\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}\)  =  \(-5 \vec{b}-\vec{a}\)

 \(\rm \vec{c}\) = \((((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i})\)

⇒ \(((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i}\)

⇒ \((11 \hat{k}+23 \hat{j}) \times \hat{i}\)

⇒ \((11 \hat{j}-23 \hat{k})\)

\(\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=11-23\) = -12

∴ Option 1 is correct

Concept:

Codomain - A codomain is the group of possible values that the dependent variable can take.

Range - The range is all the elements from set B that have the corresponding pre-image in set A.

Calculation:

Given, \(f(x)=\frac{2 x+3}{3 x+5}, x \in A\)

For f(x) to be defined, 3x + 5 ≠ 0

⇒ x = \(-\frac{5}{3}\)

∴ A = \(R \backslash\left\{-\frac{5}{3}\right\}\)

Now, y = f(x) is onto

⇒ y = \(\frac{2 x+3}{3 x+5}\)

⇒ 3xy + 5y = 2x + 3

⇒ 3xy - 2x = 3 - 5y

⇒ x(3y - 2) = 3 - 5y

⇒ x = \(\frac{3-5y}{3y-2}\)

For x to be defined for \(R \backslash\left\{\frac{2}{3}\right\}\)

Since, for onto functions co-domain = range

∴ B = \(R \backslash\left\{\frac{2}{3}\right\}\)

∴ \(A=R \backslash\left\{-\frac{5}{3}\right\} \text { and } B=R \backslash\left\{\frac{2}{3}\right\}\)

Calculation:

Given, f(x) = x³ + 6x² - 36x + 7

⇒ f'(x) = 3x² + 12x - 36

= 3(x² + 4x - 12)

For f(x) to be increasing,

f'(x) > 0

⇒ 3 (x² + 4x - 12) > 0

⇒ x² + 4x - 12 > 0

⇒ (x + 6) (x - 2) > 0

⇒ x ∈ (- ∞, −6) ∪ (2, ∞)

∴ The range of x is (- ∞, −6) ∪ (2, ∞).

The correct answer is Option 1.

Calculation

Given \(f(x)=|\sin 4x|+|\cos 2x|\)

\(\Rightarrow\sqrt { \sin ^{ 2 }{ 4x } } +\sqrt { \cos ^{ 2 }{ 2x } } \)

\(\Rightarrow\sqrt { \cfrac { 1-\cos { 8x } }{ 2 } } +\sqrt { \cfrac { 1+\cos { 4x } }{ 2 } } \)

Now, period of \( \sqrt { \cfrac { 1-\cos { 8x } }{ 2 } } \) is \(\cfrac{\pi}{4}\)

Period of \(\sqrt { \cfrac { 1+\cos { 4x } }{ 2 } } \) is \( \cfrac{\pi}{2}\).

LCM of \( \cfrac{\pi}{4}\) and \(\cfrac{\pi}{2}\) is \( \cfrac{\pi}{2}\).

Hence, period of \(f(x)\) is \(\displaystyle \cfrac{\pi}{2}\).

Hence, option 3 is correct

Concept:

For y = sin–1 x:

• Domain = [– 1, 1]
• Range = \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

For y = cos–1 x:

• Domain = [– 1, 1]
• Range = \(\left[0,\pi\right]\)

Calculation:

For sin–1 (x – 1) to be defined, – 1 ≤ x – 1 ≤ 1

⇒ 0 ≤ x ≤ 2

For cos–1 (x – 3) to be defined, – 1 ≤ x – 3 ≤ 1

⇒ 2 ≤ x ≤ 4

∴ Possible value of x is x = 2

∴ sin–1 (x – 1) + cos–1 (x – 3) + tan–1 \(\left(\frac{x}{2-x^2}\right)\) = cos–1 k + π

⇒ sin–1 (2 – 1) + cos–1 (x – 2) + tan–1 \(\left(\frac{2}{2-2^2}\right)\) = cos–1 k + π

⇒ sin–1 (1) + cos–1 (– 1) + tan–1 (– 1) = cos–1 k + π

⇒ π/2 + π – π/4 = cos–1 k + π

⇒ cos–1 k = π/4

⇒ k = cos π/4 = 1/\(\sqrt2\)

∴ The require dvalue of k is 1/\(\sqrt2\)

Calculation

y = (x –2)², y² = 8(x – 2)

y = x² , y² = – 8x 

= \(\frac{16 \mathrm{ab}}{3}=\frac{16 \times \frac{1}{4} \times 2}{3}=\frac{8}{3}\)

Hence option 1 is correct

Concept:

The equation of line passing through (x1, y1, z1) and having direction ratios (a, b, c) is given by \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)

Calcuation:

Given, \(\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z\) …(1)

⇒ - 9μ - 4λ - 1 + 7 = 0

⇒ 4λ + 9μ = 6

∴ The value of 4λ + 9μ is 6.

The correct answer is Option 4.

Explanation:

\(\rm \frac{dy}{dx}=1+x^2+y^2+x^2y^2\)

\(\rm \frac{dy}{dx}=1(1+x^2)+y^2(1+x^2)\)

\(\rm \frac{dy}{dx}=(1+x^2)(1+y^2)\)

\(\rm \frac{dy}{1+y^2}=(1+x^2)dx\)

Integrating both sides

\(\rm \tan^{-1}y=x+\frac{x^3}{3}+C\) where C is integrating constant.

Option (2) is true.

Concept:

1. The vectors which are parallel to the same plane , or lie on the same plane are called coplanar vectors.

2. Three vectors are coplanar if their scalar triple product is zero. \(⃗{a} \cdot (⃗{b} \times ⃗{c}) = 0 \)

Calculation

Let \(\vec{a} \) = 2i – j + k

\(\vec{b} \) =i + 2j – 3k

\(\vec{c} \)= 3i + a j + 5k

\( ⃗{b} \times ⃗{c} = \begin{vmatrix} \hat{i} &\hat{j}& \hat{k}\\ 1 & 2& -3\\ 3& a& 5\\\end{vmatrix} \)

⇒ \(\hat{i}(10+3a)-\hat{j}(5+9)+\hat{k}(a-6) \)

⇒ \((10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k} \)

⇒ \(\vec {a}\cdot(\vec{b}\times \vec {c})=(2\hat{i}-\hat{j}+\hat{k})\cdot((10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k}) \)

⇒ 2(10 + 3a) - (-14) + (a - 6)

⇒ 20 + 6a + 14 + a - 6

⇒ 7a + 28

Since 7a + 28 = 0

⇒ a = -4

Hence Option(4) is correct.

Calculation:

Given, \(16^{\sin ^{2} x}+16^{\cos ^{2} x}=10\)

⇒ \(16^{\sin ^{2} x}+16^{1-\sin ^{2} x}=10\)

⇒ \(16^{\sin ^{2} x}+\frac{16}{16^{\sin ^{2} x}}=10\)

Let \(16^{\sin ^{2} x}\) = t

Now, \(t+\frac{16}{t}=10\)

⇒ t² - 10t + 16 = 0

⇒ (t - 2)(t - 8) = 0

⇒ t = 2 or t = 8

∴ \(16^{\sin ^{2} x}\) = 2 or \(16^{\sin ^{2} x}\) = 8

⇒ \(2^{4\sin ^{2} x}\) = 2¹ or \(2^{4\sin ^{2} x}\) = 2³ 

⇒ 4sin²x = 1 or 4sin²x = 3

⇒ sin x = ±\(\frac{1}{2}\) or sin x = ±\(\frac{\sqrt{3}}{2}\)

⇒ x = \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), \(\frac{7\pi}{6}\), \(\frac{11\pi}{6}\) or x = \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(\frac{4\pi}{3}\), \(\frac{5\pi}{3}\)

∴ The number of solutions is 8.

The correct answer is Option 4.

Concept:

If a plane is parallel to a line then, the sum of the product of direction ratios of the line and the plane is zero.

Calculation:

Given equation of a Line is

\(\rm \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)

Direction Ratio's of line = 3, 4, 5.

By checking the options,

Option 1: Dr's of 2x + 2y + z - 1 = 0 are 2, 2, 1

Product of direction ratios = 3×2 + 4×2 + 5×1 = 6 + 8 + 5 = 19

Option 2: Dr's of 2x - y - 2z + 5 = 0 are 2, -1, -2

Product of direction ratios = 3×2 + 4×-1 + 5×-2 = 6 - 4 - 10 = -8

Option 3: Dr's of 2x + 2y - 2z + 1 = 0 are 2, 2, -2

Product of direction ratios = 3×2 + 4×2 + 5×-2 = 6 + 8 - 10 = 4

Option 4: Dr's of x - 2y + z - 1 = 0 are 1, -2, 1

Product of direction ratios = 3 × 1+ 4 × -2 + 5 × 1 = 3 - 8 + 5 = 0

Hence, option 4 is correct.

Concept:

\(\tan^{-1}a+\tan^{-1}b=\left\{\begin{matrix} \tan^{-1}\left(\frac{a+b}{1-ab}\right ), ab<1 \\ \pi +\tan^{-1}\left(\frac{a+b}{1-ab} \right ),ab>1 \end{matrix}\right.\)

Calculation:

We can observe that, \(\left(\frac{x}{1-x^2}\right) \times \frac{1}{x^3}=\left(\frac{1}{1-x^2}\right) \frac{1}{x^2}>1\) for x ∈ (0, 1)

∴ \(\tan^{-1}\left(\frac{x}{1-x^2} \right )+\tan^{-1}\left(\frac{1}{x^3} \right )=\frac{3\pi}{4}\)

⇒ \(\pi+\tan ^{-1}\left(\frac{\frac{x}{1-x^2}+\frac{1}{x^3}}{1-\frac{x}{x^3\left(1-x^2\right)}}\right)=\frac{3\pi}{4}\)   

⇒ \(\pi+\tan^{-1}\left(\frac{x^4+1-x^2}{(x^2-x^4-1)x} \right )=\frac{3\pi}{4} \)

⇒ \(\tan^{-1}\left(\frac{-1}{x} \right )=\frac{-\pi}{4} \)

⇒ \(-\tan^{-1}\left(\frac{1}{x} \right )=\frac{-\pi}{4}\)

⇒ \(\frac{1}{x}=\tan \frac{\pi}{4}\) = 1

⇒ x = 1 ∉ (0, 1)

∴ The number of solutions of the given equation lying in the interval (0, 1) is zero.

The correct answer is Option 1.

Calculation

Length of \(\mathrm{OC}=\frac{\sqrt{136}}{3}=\frac{2 \sqrt{34}}{3}\)

Hence option 2 is correct

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: 

Z = 3x1 – x2

Constraints,

2x1 – x2 ≤ 1

x1 ≤ 3

• The corner points of the feasible region are (0, 0) (0.5, 0) and (3,5)

Z(0, 0) = 0

Z(0.5, 0) = 1.5

Z(3, 5) = 3(3) – 5 = 4

∴ Z(3, 5) = 4

⇒ optimal solution

• So, the correct answer is option 2.

Explanation:

Let \(\vec{r}=\hat i\)

Then \(\vec{r} \cdot \hat{i}=\hat{i} \cdot \hat{i}=1\)

\(\vec{r} \times \hat{i}=\hat{i} \times \hat{i}\) = 0

\(\vec{r} \cdot \hat{j}=\hat{i} \cdot \hat{j}\) = 0

\(\vec{r} \cdot \hat{k}=\hat{i} \cdot \hat{k}\) = 0

So,

\((\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})+(\vec{r} \cdot \hat{j})\)\((\vec{r} \times \hat{j})+(\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})\)

= 0 + 0 + 0 = 0

Option (1) is true.

Calculation:

Given, y2 = 8x = 4⋅2x

General point is P(at², 2at) = P(2t², 4t)

Now, y2 = 8x 

⇒ 2yy'= 8

⇒ y' = \(\frac{4}{y} \bigg|_{(2t^2,4t)}\) = \(\frac{1}{t}\)

∴ Normal at P is give by:

(y - 4t) = - t(x - 2t²)

Since it passes through the centre, (0 - 4t) = - t(5 - 2t²)

⇒ 4 = 5 - 2t²

⇒ 2t² = 1

⇒ t = \(\frac{1}{\sqrt{2}}\)

⇒ P = (2t2, 4t) = (1, 2√2)

∴ The point is (1, 2√2).

The correct answer is Option 2.

Calculation:

Given:

A = \(\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\)

A² = A × A = \(\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\) × \(\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\)

⇒ A² = \(\begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2\cos \theta \sin \theta \\ -2\cos \theta \sin \theta & \cos^2 \theta - \sin^2 \theta \end{bmatrix}\)

⇒ A² = \(\begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}\)

A³ = A² × A = \(\begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}\) × \(\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\)

⇒ A³ = \(\begin{bmatrix} \cos 2\theta \cos \theta - \sin 2\theta \sin \theta & \cos 2\theta \sin \theta + \sin 2\theta \cos \theta \\ -\sin 2\theta \cos \theta - \cos 2\theta \sin \theta & -\sin 2\theta \sin \theta + \cos 2\theta \cos \theta \end{bmatrix}\)

⇒ A³ = \(\begin{bmatrix} \cos (2\theta + \theta) & \sin (2\theta + \theta) \\ -\sin (2\theta + \theta) & \cos (2\theta + \theta) \end{bmatrix}\)

⇒ A³ = \(\begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix}\)

Hence option 3 is correct

Concept:

Since the initial position is P with the final position S
Vector addition, the net displacement is SP = ( R² + R² + 2RRCosθ )^1/2

Calculation:

Cosθ = 90 = 0

Displacement = R√2= 2√2 = √8 km 

∴ The correct option is 2)

Concept Used:

Conditional Probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

Calculation

Total Possible cases

{H, TH, TTH}

P(H) = 1/2

P(TH) = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

P(TTH) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

P(coin is tossed thrice) = P(TTH) + P(TTT) = 1/4

P(coin is tossed thrice | first toss not head) = \(\frac{1/4}{1/2} = \frac{1}{2}\)

Hence option 1 is correct

Concept:

Draw the constraints to find the feasible region:

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

Calculation:

Given:

7x1 + 12x2 ≥ 84

• Now, check all the given options by satisfying the points on this constraint.
• We have to choose the points which do not satisfy the constraints.
• The point (2,8) is lying in the lower plane of the given constraint so this is not satisfying the region of interest of the given constraints.

 7 × 2 + 12 × 3 ≥ 84

⇒ 50 ≥ 84 (False)

• So,  the correct answer is option 1

Concept:

The points where the f'(x) = 0 are known as critical values.

At critical point, if f ''(x) > 0 then function has minima.

If f ''(x) < 0 then function has maxima.

Calculation:

Given, f (x) = \(\left\{\begin{array}{cc} x^3+x^2+10 x, & x<0 \\ -3 \sin x, & x ≥ 0 \end{array}\right.\)

For x ≥ 0

f(x) = − 3sin x

⇒ f '(x) = − 3cos x

⇒ f ′(0) = − 3

For x < 0, f (x) = x³ + x² + 10x

⇒ f ′(x) = 3x² + 2x + 10

⇒ f ′(0) = 10

⇒ f ′(x) > 0 for x < 0 and f ′(x) < 0 for x ≥ 0

⇒ f (x) has maxima at x = 0.

∴ At x = 0, there is a point of maximum.

The correct answer is Option 1.

Calculation:

Let the length and breadth of the rectangular field be x and y respectively.

⇒ 2(x + y) = 200

⇒ y = 100 − x   …(i)

Now,

Area,

A = xy

= x(100 − x)

= 100x − x²   …(ii)

Differentiate equation (ii) w.r.t  x, we get

\(\rm\frac{dA}{dx}\) = 100 − 2x

&

\(\rm\frac{d^2A}{dx^2}\) = −2 < 0 gives the maximum value of the function.

Thus, putting x = 50 in A = 100 x − x², we get,

⇒ A = 100 × 50 − 2500 = 2500 m²

Hence, the required maximum area = 2500 m².

The correct answer is option "4"

Calculation:

Given, f'(x) = f'(2 – x)

On integrating both sides, we get

⇒ f(x) = - f(2 – x) + c

Putting x = 0, we get:

f(0) + f(2) = c

⇒ c = 1+ e²

⇒ f(x) + f(2 – x) = 1 + e²

∴ I = \(\rm \int_0^2 f(x) d x\)

= \(\int_0^1f(x)dx+\int_1^2f(x) d x\)

Putting x = 2 - t, in the second integral, we get

= \(\rm \int_0^1f(x)dx+\int_1^0f(2-t) (-d t)\)

= \(\rm \int_0^1(f(x)+f(2-x)) d x\) 

= 1 + e2 

∴ The value of \(\int_0^2 f(x) d x\) is 1 + e2.

The correct answer is Option 1.

Calculation

Required probability = 

⇒ \(\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . .\)

\(⇒\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \)

Hence option 3 is correct

Formula Used:

cos(3π/4) = \(\frac{-1}{\sqrt{2}}\)

cosec(π/6) = 2

Calculation:

\(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)+\operatorname{cosec}^{-1}(2)\)

⇒ \(\cos ^{-1}\left(cos\frac{3π}{{4}}\right)+\operatorname{cosec}^{-1}cosec(\frac{π}{6})\)

⇒ 3π/4 + π/6

⇒ \(\frac{11 \pi}{12}\)

∴ Correct answer is  \(\frac{11 \pi}{12}\).

Explanation:

A fair coin is tossed till two heads occur in succession.

Then,

P (the number of tosses required is less than 6) =  P(HH) + P(THH) + P(TTHH) + P(TTTHH)

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

= \(\frac{(\frac{1}{2})^2[ 1 -( \frac{1}{2})^4}{1-\frac{1}{2}}\)

= \(\frac{\frac{1}{4}[1-\frac{1}{16}]}{\frac{1}{2}}\)

= \(\frac{1}{4}\times\frac{15}{16}\times\frac{2}{1} = \frac{15}{32}\)

∴ Option (b) is correct.