JEE Main Mathematics, Feb 24 2021, Shift-2

Problem 1 of 30

Let $a, b \in \mathbb{R}$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\dfrac{x - 3}{7} = \dfrac{y - 2}{5} = \dfrac{z - 1}{-9}$ is $(20, b, -a - 9)$ then $|a + b|$ is equal to:

Solution:

Let $N$ be foot of the perpendicular from $P$ to the given line.

Therefore, $N$ satisfies:

$\dfrac{x - 3}{7} = \dfrac{y - 2}{5} = \dfrac{z - 1}{-9} = \lambda$ for some constant $\lambda$

$P = (7\lambda + 3, 5\lambda + 2, -9\lambda + 1)$

Since $N$ is the midpoint of $P$ and its mirror image,

$\dfrac{20 + a }{2} = 7\lambda + 3$ (using the $x$ coordinate)

$\Rightarrow 20 + a = 14 \lambda + 6$ $...(i)$

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$\dfrac{-a - 9 +9}{2} = - 9\lambda + 1 $ $...(ii)$ (using the $z$ coordinate)

Solving equations $(i)$ and $(ii),$

$\lambda = -3$

and

$a = -56$

Using the $y$ coordinate:

$\dfrac{b+6}{2} = 5 \lambda + 2$

$\Rightarrow b = -32$

$\therefore | a+b | =| - 56 + -32 | = 88 $

Problem 2 of 30

Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c), (2, b)$ and $(a, b)$ be $\left(\dfrac{10}{3}, \dfrac{7}{3}\right)$. If $\alpha , \beta$ are the roots of the equation $ax^2 + bx + 1 = 0$, then the value of $\alpha^2 + \beta^2 - \alpha \beta$ is:

Solution:

Let $x_1 = a, x_2 = 2, x_3 = a, y_1 = c, y_2 = b$ and $y_3 = b.$

Since $a, b$ and $c$ are in A.P, $b = a + d$ and $c = a + 2d.$

The centroid of a triangle with vertices $(x_1,y_1), (x_2,y_2)$ and $(x_3,y_3)$ is given by, $\left( \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$

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Therefore,

$\dfrac{10}{3} = \dfrac{x_1 + x_2 + x_3}{3}$

$\Rightarrow 10 = x_1 + x_2 + x_3$

$\Rightarrow 10 = a + 2 + a$

$\Rightarrow 2a + 2 = 10$

$\Rightarrow a = 4$

Similarly,

$\dfrac{7}{3} = \dfrac{y_1 + y_2 + y_3}{3}$

$\Rightarrow 7 = y_1 + y_2 + y_3$

$\Rightarrow 7 = c + b + b$

$\Rightarrow 7 = (a + 2d) + (a + d) + (a + d)$

$\Rightarrow 7 = 3a + 4d$

$\Rightarrow 7 = 12 + 4d$

$\Rightarrow 4d = -5$

$\Rightarrow d = \dfrac{-5}{4}$

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$b = a + d \Rightarrow b = 4 + \left(\dfrac{-5}{4}\right) = \dfrac{11}{4}$

Therefore,

$ax^2 + bx + 1 = 0$

$\Rightarrow 4x^2 + \dfrac{11}{4}x + 1 = 0$

If $\alpha$ and $\beta$ are the roots of the equation above then $\alpha + \beta = -\dfrac{11}{16}$ and $\alpha\beta = \dfrac{1}{4}$

$\alpha^2 + \beta^2 - \alpha\beta$

$= (\alpha + \beta)^2 -3\alpha\beta$

$= \left( -\dfrac{11}{16} \right)^2 - \dfrac{3}{4}$

$= \dfrac{121}{256} - \dfrac{3}{4}$

$= \dfrac{121 - 192}{256}$

$= - \dfrac{71}{256}$

Problem 3 of 30

If $P$ is a point on the parabola $y = x^2 + 4$ which is closest to the straight line $y = 4x - 1$, then the coordinates of $P$ are:

Solution:

If $P$ is the point on the given parabola closest to the given line, then, the slope of the the tangent to the parabola at the given point is parallel to this line.

The slope of the tangent at any point is given by:

$\dfrac{dy}{dx} = 2x$

The slope of the line $4x - y - 1 = 0$ is $4$.

Therefore, if $P = (x_1, y_1)$,

$2x_1 = 4$

$\Rightarrow x_1 = 2$

$\therefore y_1 = {x_1}^2 + 4 = 8$

Therefore, $P = (2, 8)$

Problem 4 of 30

For which of the following curves, the line $x + \sqrt{3}y = 2\sqrt{3}$ is the tangent at the point $\dfrac{3\sqrt{3}}{2}, \dfrac{1}{2}$?

Solution:

If a given line is a tangent to a curve, at any point, then it satisfies two conditions:

- The slope of the line is equal to the slope of the curve at that point.

- The curve passes through the point.

Also, observe that for any given line and a point on the line, there can be an infinite number of curves to which the line is tangent at that point, therefore, it is impossible to arrive at a unique solution without eliminating all the incorrect options in the given set of options.

The slope of the given line is $s = -\dfrac{a}{b} = - \dfrac{1}{\sqrt{3}}$

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All of the curves given in the options pass through the given point $P = \left(\dfrac{3\sqrt{3}}{2}, \dfrac{1}{2}\right)$

The slope of the curve $2x^2 - 18y^2 = 9$ can be obtained by differentiating both sides of the equation w.r.t., as follows:

$4x - 36y\dfrac{dy}{dx} = 0$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{x}{9y}$

Therefore, the slope of the curve at $P$ is $\dfrac{1}{3\sqrt{3}}$ which is not equal to the slope of the given line.

Likewise, out of the remaining options, for $x^2 + 9y^2 = 9$, we can see that:

$2x + 18y \dfrac{dy}{dx} = 0$

$\Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{9y}$

The slope at $\left(\dfrac{3\sqrt{3}}{2}, \dfrac{1}{2}\right)$ for this curve is, $- \dfrac{\dfrac{3\sqrt{3}}{2}}{9 \times \dfrac{1}{2}}$

$= -\dfrac{3\sqrt{3}}{9} = -\dfrac{1}{\sqrt{3}}$, which is equal to the slope of the given line.

Since this curve has a slope equal to the given line, and also satisfies the given point, $x^2 + 9y^2 = 9$ is the correct option.

Problem 5 of 30

The value of the integral $\displaystyle{\int\limits_{1}^{3}}\lfloor x^2 - 2x - 2 \rfloor dx$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ is:

Solution:

$I = \displaystyle \int \limits_1^3 \lfloor x^2 - 2x - 2 \rfloor dx$

$\Rightarrow I = \displaystyle \int \limits_1^3 -3 dx + \displaystyle \int \limits_1^3 \lfloor (x - 1)^2 \rfloor dx$

$\Rightarrow I = -6 + \displaystyle \int \limits_1^3 \lfloor (x - 1)^2 \rfloor dx$

Substituting $t = x - 1$, we get:

$I = -6 + \displaystyle \int \limits_0^2 \lfloor t^2 \rfloor dt$

$\Rightarrow I = -6 + \displaystyle \int \limits_0^1 0 \ dt + \displaystyle \int \limits_1^\sqrt 2 1 \ dt+ \displaystyle \int \limits_\sqrt 2 ^\sqrt 3 2 \ dt + \displaystyle \int \limits_\sqrt 3 ^2 3 \ dt$

$\Rightarrow I = -6 + \sqrt 2 - 1 + 2 \sqrt 3 - 2 \sqrt 2 + 6 - 3 \sqrt 3$

$\Rightarrow I = - \sqrt 2 - \sqrt 3 - 1$

Problem 6 of 30

The negation of the statement

$\lnot p \land (p \lor q)$ is:

Solution:

The negation of the given statement is:

$\lnot(\lnot p \land (p \lor q))$

$ = \lnot(\lnot p)\lor \lnot(p \lor q))$

$ = p \lor (\lnot p \land \lnot q))$

$ = (p \lor \lnot p) \land(p \lor \lnot q))$

$ = p \lor \lnot q$

Problem 7 of 30

Let $A$ and $B$ be $3 \times 3$ real matrices such that $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix. Then the system of linear equations $(A^2B^2 - B^2A^2)X = O$, where $X$ is a $3 \times 1$ column matrix of unknown variables and $O$ is a $3 \times 1$ null matrix, has:

Solution:

$(A^2B^2 - B^2A^2)X = O$

Since $A$ is a symmetric matrix,

$A^T = A$

Since $B$ is a skew-symmetric matrix,

$B^T = - B$

Let $A^2B^2 - B^2A^2 = P$

$P^T =(A^2B^2 - B^2A^2)^T = (A^2B^2)^T - (B^2A^2)^T$

$= (B^2)^T(A^2)^T - (A^2)^T(B^2)^T = B^2A^2 - A^2B^2 = - P$

(The square of a skew - symmetric matrix is a symmetric matrix and therefore, $(A^2)^T = A^2,$ if $A$ is a skew-symmetric matrix)

$\therefore P$ is a skew-symmetric matrix.

Since $P$ is a skew-symmetric matrix, $|P| = 0$

Given that,

$(A^2B^2 - B^2A^2)X = O$

$\Rightarrow P \times X = O$

Since, $PX = O$ and $|P| = 0$ the given system of equations has an infinite number of solutions.

Problem 8 of 30

Let $f(x)$ be a differentiable function defined on $[0, 2]$ such that $f'(x) = f'(2 - x)$ for all $x \in (0, 2), f(0) = 1$ and $f(2) = e^2$.

Then the value of $\displaystyle{\int\limits_{0}^{2}}f(x)dx$ is:

Solution:

$f'(x) = f'(2 - x)$

Integrating w.r.t $x$ on both sides, we obtain,

$f(x) = -f(2 - x) + c$

$\Rightarrow c = f(x) + f(2 - x)$

Substituting $x = 2$,

$c = f(2) + f(0) = e^2 + 1$

$\therefore f(x) = -f(2 - x) + e^2 + 1$

$\Rightarrow f(x) + f(2 - x) = e^2 + 1$ $...(i)$

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Let $\displaystyle{I = \int\limits_{0}^{2}}f(x)dx$

Substituting $x = 2 - z$ and $dx = -dz$

$I = \displaystyle \int \limits_{2}^{0} f(2 - z) (-dz)$

$I = \displaystyle \int \limits_{0}^{2} f(2 - z) dz$

Ignoring variable names, we can write:

$I = \displaystyle \int\limits_{0}^{2}f(x)dx = \int \limits_{0}^{2} f(2 - x) dx$

Integrating both sides of $eqn\ (i)$ w.r.t. $x$ from $0$ to $2$, we get:

$\displaystyle \int \limits_{0}^{2} f(x)dx + \int \limits_{0}^{2} f(2 - x)dx = \int \limits_{0}^{2}[e^2 + 1]dx$

$I + I = e^2x + x {\Large{\vert}}_0^2$

$\Rightarrow 2I = 2e^2 + 2$

$\Rightarrow I = e^2 + 1$

Problem 9 of 30

Let $f$ be a twice differentiable function defined on $\mathbb{R}$ such that $f(0) = 1, f'(0) = 2$ and $f'(x) \ne 0$ for all $x \in \mathbb{R}$. If $\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$, for all $x \in \mathbb{R}$, then the value of $f(1)$ lies in the interval:

Solution:

$\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$

$\Rightarrow f(x) \times f''(x) - (f'(x))^2 = 0$ $...eqn(i)$

Differentiating $\dfrac{f'(x)}{f(x)}$ using the quotient rule we get,

$\dfrac{d}{dx} \dfrac{f'(x)}{f(x)} = f(x) \times f''(x) - (f'(x))^2$

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From $eqn(i)$, $f(x) \times f''(x) - (f'(x))^2 = 0$

$\therefore \dfrac{d}{dx} \dfrac{f'(x)}{f(x)} = 0$

$\Rightarrow \dfrac{f'(x)}{f(x)} = \lambda$ where $\lambda$ is a constant.

$\therefore \displaystyle \int \limits_0^t \dfrac{f'(x)}{f(x)} dx = \displaystyle \int \limits_0^t \lambda dx$

$\Rightarrow \displaystyle \left[ \ln |f(x)|\right]_0^t = \displaystyle \left[ \lambda x \right]_0^t$

$\Rightarrow \ln |f(t)| - \ln |f(0)| = \lambda t $

$\Rightarrow \ln |f(t)| = \lambda t$

$\Rightarrow f(t) = e^{\lambda t}$

$\Rightarrow f'(t) = \lambda e^{\lambda t}$

Given that $f'(0) = 2$

$\therefore \lambda = 2$

$\Rightarrow f(x) = e^{2x}$

$\therefore f(1) = e^2 = (2.7...)^2 \approx 7.29$

$\therefore f(1) \in (6,9)$

Problem 10 of 30

If the curve $y = ax^2 + bx + c, x \in \mathbb{R}$, passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$, then the possible values of $a, b, c$ are:

Solution:

Since the curve $y = ax^2 + bx + c$ passes through the point $(1, 2)$,

$a + b + c = 2$

The curve also passes through the origin, therefore,

$c = 0$

The slope of the curve is given by $2ax + b$ at any $x$. Therefore at $x = 0$, slope $= b$ which is equal to the slope of the line $y = x$ which is $1$.

Therefore, $b = 1$, and

$a = 2 - 1 = 1$

Therefore, $a = 1, b = 1, c = 0$

Problem 11 of 30

If $n \ge 2$ is a positive integer, then the sum of the series

$\xacomb{n+1}{2} + 2\left(\xacomb{2}{2} + \xacomb{3}{2} + \xacomb{4}{2} + ... + \xacomb{n}{2}\right)$ is:

Solution:

$\xacomb{k}{2} = \dfrac{k!}{2! \times (k-2)!} = \dfrac{(k)(k-1)}{2}$

$\Rightarrow 2 \times \xacomb{k}{2} = (k-1)(k)$

$\xacomb{n+1}{2} + 2(\xacomb{2}{2}+\xacomb{3}{2}+\xacomb{4}{2} + ... + \xacomb{n}{2})$

$= \dfrac{(n)(n+1)}{2} + (1)(2) + (2)(3) + (3)(4) + ... + (n-1)(n) $

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$= \dfrac{(n)(n+1)}{2} + (1)(1+1) + (2)(2+1) + (3)(3+1) + ...$

$+ (n-1)[(n-1) + 1] $

$= \dfrac{(n)(n+1)}{2} + 1^2 + 1 + 2^2 + 2 + 3^2 + 3 + ... (n-1)^2 + (n-1)$

$= \dfrac{(n)(n+1)}{2} + [1 + 2+ 3 + ... + (n-1)] + [1^2 + 2^2 + 3^2 + ... + (n-1)^2]$

$= \dfrac{(n)(n+1)}{2} + \dfrac{(n -1)(n)}{2} + \dfrac{(n-1)(n)(2n-1)}{6}$

$= n^2 + \dfrac{(n-1)(n)(2n-1)}{6}$

$= \dfrac{6n^2 + (n-1)(n)(2n-1)}{6}$

$= \dfrac{n((n-1)(2n-1) + 6n)}{6}$

$= \dfrac{n((n^2 - 3n + 1) + 6n)}{6}$

$= \dfrac{n(n^2 + 3n + 1)}{6}$

$= \dfrac{n(n+1)(2n+1)}{6}$

Problem 12 of 30

For the system of linear equations:

$x - 2y = 1$, $x - y + kz = -2$, $ky + 4z = 6$, $k \in \mathbb{R}$ consider the following statements:

$(A)$ The system has unique solution if $k \ne 2, k \ne -2$

$(B)$ The system has unique solution if $k = -2$

$(C)$ The system has unique solution if $k = 2$

$(D)$ The system has no-solution if $k = 2$

$(E)$ The system has infinite number of solutions if $k \ne 2$

Which of the following statements are correct?

Solution:

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$\begin{bmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\6\end{bmatrix}$

$A = \begin{bmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{bmatrix}$

$B = \begin{bmatrix} 1 \\ -2 \\ 6 \end{bmatrix}$

For the given system of equations to have a unique solution $|A| \neq 0 $

$|A| \neq 0$

$\Rightarrow 1(-4 - k^2) + 2(4) \neq 0$

$\Rightarrow 4 - k^2 \neq 0$

$\Rightarrow k^2 \neq 4$

$\Rightarrow k \neq 2$ and $k \neq -2$

The given system of equations has a unique solution for $k \neq 2$ and $k \neq -2.$

$\therefore$ Option $(A)$ is correct and options $(B), (C)$ and $(E)$ are incorrect.

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To check whether option $(D)$ is correct, we need to eliminate the possibility that the given system of equation has infinite solutions which will satisfy the condition:

$adj(A)\times B = O$ for $k = 2$

$\begin{bmatrix} -8 & 8 & -4 \\ -4 & 4 & -2 \\ 2 & -2 & 1\end{bmatrix}\begin{bmatrix} 1 \\ -2 \\ 6 \end{bmatrix} = \begin{bmatrix} -48 \\ -24 \\ 12 \end{bmatrix} \neq O$

Therefore, the given system of equation does not have infinite number of solutions.

$\therefore$ Option $(D)$ is also correct.

Therefore options $(A)$ and $(D)$ are correct options.

Problem 13 of 30

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined as

$f(x) = \left\{\begin{array}{ll} -55x, & \text{ if } x < -5 \\ 2x^3 - 3x^2 -120x, & \text{ if } -5 \le x \le 4 \\ 2x^3 - 3x^2 - 36x -336, & \text{ if } x \gt 4\end{array}\right.$

Let $A = \{x \in R:f$ is increasing$\}$. Then $A$ is equal to:

Solution:

$f(x) = \left\{\begin{array}{ll} -55x & \text{ if } x < -5 \\ 2x^3 - 3x^2 -120x & \text{ if } -5 \le x \le 4 \\ 2x^3 - 3x^2 - 36x -336 & \text{ if } x \gt 4\end{array}\right.$

$f'(x) = \left\{\begin{array}{ll} -55 & \text{ if } x < -5 \\ 6(x^2 - x -20) & \text{ if } -5 \le x \le 4 \\ 6(x^2 - x - 6) & \text{ if } x \gt 4\end{array}\right.$

$f'(x) = \left\{\begin{array}{ll} -55 & \text{ if } x < -5 \\ 6(x-5)(x+4) & \text{ if } -5 \le x \le 4 \\ 6(x-3)(x+2) & \text{ if } x \gt 4\end{array}\right.$

$\therefore f(x)$ increases in $(-5,-4) \cup (4, \infty)$

Problem 14 of 30

A possible value of $\tan\left(\dfrac{1}{4}\sin^{-1}\dfrac{\sqrt{63}}{8}\right)$ is:

Solution:

Let $\sin \theta = \dfrac{\sqrt{63}}{8}$,

then, $\tan \left ( \dfrac{1}{4} \sin^{-1} \dfrac{\sqrt{63}}{8}\right) = \tan \dfrac{\theta}{4}$

Since $\sin \theta = \dfrac{\sqrt{63}}{8},$

$\cos \theta = \dfrac{1}{8}$

$\tan \dfrac{\theta}{2} = \dfrac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}} = \sqrt{\dfrac{7}{9}}$

$\therefore \cos \dfrac{\theta}{2} = \dfrac{3}{4}$

$\tan \dfrac{\theta}{4} = \sqrt{\dfrac{1 - \cos \dfrac{\theta}{2} }{ 1 + \cos \dfrac{\theta}{2}}} = \sqrt{\dfrac{1 - \dfrac{3}{4}}{ 1 + \dfrac{3}{4}}} = \dfrac{1}{\sqrt{7}}$

Therefore,

$\tan \left ( \dfrac{1}{4} \sin^{-1} \dfrac{\sqrt{63}}{8}\right ) = \dfrac{1}{\sqrt{7}}$

Problem 15 of 30

The angle of elevation of a jet plane from a point $A$ on the ground is $60^\circ$. After a flight of $20$ seconds at the speed of $432$ km/hour, the angle of elevation changes to $30^\circ$. If the jet plane is flying at a constant height of, then the height is:

Solution:

Let $h$ be the height of the plane in metres. And the initial horizontal distance be $b$.

$\therefore \dfrac{h}{b} = \tan 60^\circ = \sqrt{3}$

$\Rightarrow b = \dfrac{h}{\sqrt{3}}$

In $20$ sec the plane will travel, $\dfrac{432000 \times 20}{3600} = 2400$ m

Therefore, $\dfrac{h}{b + 2400} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{h}{\dfrac{h}{\sqrt{3}} + 2400} = \dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{\sqrt{3}h}{h + 2400\sqrt{3}} = \dfrac{1}{\sqrt{3}}$

$\Rightarrow h + 2400\sqrt{3} = 3h$

$\Rightarrow h = 1200 \sqrt{3}$

Problem 16 of 30

The vector equation of the plane passing through the intersection of the planes

$\overrightarrow{r} \cdot \left(\hat{i} + \hat{j} + \hat{k}\right) = 1$ and $\overrightarrow{r} \cdot \left(\hat{i} - 2\hat{j}\right) = -2$, and the point $(1, 0, 2)$ is:

Solution:

The plane passing through the intersection of the given planes is,

$\overrightarrow{r} \cdot \left(\hat{i} + \hat{j} + \hat{k}\right) + \lambda \left( \overrightarrow{r} \cdot \left(\hat{i} - 2\hat{j}\right) \right) = 0$

and it passes through $\hat{i} + 2\hat{k}.$

Therefore,

$\left( \hat{i} + 2\hat{k} \right) \cdot \left(\hat{i} + \hat{j} + \hat{k}\right) + \lambda \left( \left( \hat{i} + 2\hat{k} \right) \cdot \left(\hat{i} - 2\hat{j}\right) \right) = 0$

$\Rightarrow (3 - 1) + \lambda(1 + 2) = 0$

$\Rightarrow \lambda = - \dfrac{2}{3}$

Therefore, the equation of the plane is,

$\overrightarrow{r} \cdot \left(\hat{i} + \hat{j} + \hat{k}\right) - \dfrac{2}{3} \left( \overrightarrow{r} \cdot \left(\hat{i} - 2\hat{j}\right) \right) = 0$

$\Rightarrow 3\overrightarrow{r} \cdot \left(\hat{i} + \hat{j} + \hat{k}\right) - 2 \left( \overrightarrow{r} \cdot \left(\hat{i} - 2\hat{j}\right) \right) = 0$

$\Rightarrow \overrightarrow{r} \cdot \left(\hat{i} + 7\hat{j} + 3\hat{k}\right) = 7$

Problem 17 of 30

For the statements $p$ and $q$, consider the following compound statements:

$(a) \phantom{000} (\lnot q \land (p \rightarrow q)) \rightarrow \lnot p$

$(b) \phantom{000} ((p \lor q) \land \lnot p) \rightarrow q$

Solution:

We will solve this problem using two approaches, as follows.

Approach 1 - Using Logical Deduction:

In this approach we will find at least one case where either of the expressions evaluates to $0$, which will show that the respective expression is not a tautology.

If statement $(a)$ is not a tautology, then for some value of $p, q$ $(\lnot q \land (p \rightarrow q)) \rightarrow \lnot p = 0$, which means $\lnot p = 0$ and $(\lnot q \land (p \rightarrow q)) = 1$

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$\lnot p = 0$

$\Rightarrow p = 1$

$(\lnot q \land (p \rightarrow q)) = 1$

$\Rightarrow \lnot q \land( 1 \Rightarrow q) = 1$

$\therefore \lnot q = 1$ and $ 1 \rightarrow q = 1$

If $\lnot q = 1$

$\Rightarrow q = 0$

If $1 \rightarrow q = 1$

$\therefore q = 1$

This is a contradiction, therefore there is no case where the overall expression evaluates to $0$ and this is a tautology.

If statement $(b)$ is not a tautology, then there is some tuple $(p, q)$ for which $((p \lor q) \land \lnot p) \rightarrow q = 0$

Therefore, $((p \lor q) \land \lnot p) = 1$ and $q = 0$

$((p \lor 0) \land \lnot p) = 1$

$\therefore \lnot p = 1$

$\Rightarrow p = 0$

Putting $p = 0$ and $q = 0$, we get the overall expression as:

$((p \lor q) \land \lnot p) \rightarrow q$

$= ((0 \lor 0) \land \lnot 0) \rightarrow 0$

$= (0 \land 1) \rightarrow 0$

$= 0 \rightarrow 1$

$= 1$

Therefore, there is no tuple $(p, q)$ for which this expression yields a result of $0$, hence, this is a tautology.

Therefore, $(a)$ and $(b)$ both are tautologies.

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Approach 2 - Using Truth Tables:

Here we will construct the truth table for each of the given expressions to see if one or both are tautologies or not.

$(\lnot q \land (p \rightarrow q)) \rightarrow \lnot p$

$p$	$q$	$p \rightarrow q$	$\lnot q \land (p \rightarrow q)$	$(\lnot q \land (p \rightarrow q)) \rightarrow \lnot p$
$0$	$0$	$1$	$1$	$1$
$0$	$1$	$1$	$0$	$1$
$1$	$0$	$0$	$0$	$1$
$1$	$1$	$1$	$0$	$1$

$((p \lor q) \land \lnot p) \rightarrow q$

$p$	$q$	$(p \lor q) \land \lnot p$	$((p \lor q) \land \lnot p) \rightarrow q$
$0$	$0$	$0$	$1$
$0$	$1$	$1$	$1$
$1$	$0$	$0$	$1$
$1$	$1$	$1$	$1$

From the above tables we can conclude that none of the statements have any set of inputs, that will evaluate to $0$. Therefore, both statements are tautologies.

Problem 18 of 30

If a curve $y = f(x)$ passes through the point $(1, 2)$ and satisfies the $x\dfrac{dy}{dx} + y = bx^4$, then for what value of $b$, $\displaystyle{\int\limits_{1}^{2}}f(x)dx = \dfrac{62}{5}$?

Solution:

$x \dfrac{dy}{dx} + y = bx^4$

$\Rightarrow \dfrac{dy}{dx} + \dfrac{y}{x} = bx^3$

Using the method and the I.F. as:

$e^{\int \frac{1}{x} dx} = e^{\log_e x} = x$

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$xy = \int bx^4dx$

$\Rightarrow xy = \dfrac{bx^5}{5} + C$

$\Rightarrow y = \dfrac{bx^4}{5} + \dfrac{C}{x}$

Since the curve passes through the point $(1,2),$

$2 = \dfrac{b(1)^4}{5} + \dfrac{C}{1}$

$\Rightarrow C = \dfrac{10-b}{5}$

$\therefore y = \dfrac{bx^4}{5} + \dfrac{10-b}{5x}$

$\displaystyle {\int_{1}^2 f(x)dx}$

$= \displaystyle {\int_{1}^2 \dfrac{bx^4}{5} + \dfrac{10-b}{5x}dx}$

$= \dfrac{b\times2^5}{25} + \dfrac{10-b}{5} \log_e 2 - \dfrac{bx}{25} + \dfrac{10-b}{5} \log_e 1$

$= \dfrac{31b}{25} + \dfrac{10-b}{5} \log_e 2$

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$\displaystyle {\int_{1}^2 f(x)dx} = \dfrac{62}{5}$

$\Rightarrow \dfrac{31b}{25} + \dfrac{10-b}{5} \log_e 2 = \dfrac{62}{5}$

$\Rightarrow b = 10$

Problem 19 of 30

The probability that two randomly selected subsets of the set $\{1, 2, 3, 4, 5\}$ have exactly two elements in their intersection, is:

Solution:

The possible number of subsets, including the null subset, of the the given set is $2^5$.

Therefore, the two subsets can be selected in $2^5 \times 2^5 = 2^{10}$ ways.

For an intersection of exactly two elements, we can select the two elements in $\xacomb{5}{2}$ ways, and assign each of the remaining $3$ either to the first set, or to the second set or to neither, which can be done in $3^3$ ways. Therefore, the total number of ways to select these sets is:

$\xacomb{5}{2} \times 3^3$

$= 270$ ways

The required probability is $\dfrac{270}{2^{10}} = \dfrac{135}{2^9}$

Problem 20 of 30

The area of the region $R = \{(x, y) : 5x^2 \le y \le 2x^2 + 9\}$ is:

Solution:

We can start by finding the intersection points of the two given curves, by solving the equation:

$5x^2 = 2x^2 + 9$

$\Rightarrow 3x^2 = 9$

$\Rightarrow x = \pm \sqrt{3}$

Therefore, the region bounded by the two given curves is given by:

$\displaystyle \int\limits_{-\sqrt{3}}^{\sqrt{3}} (2x^2 + 9) dx - \int\limits_{-\sqrt{3}}^{\sqrt{3}} 5x^2dx$

$= \displaystyle \int\limits_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2) dx$

$= \displaystyle (9x - x^3) \biggr\rvert_{-\sqrt{3}}^{\sqrt{3}}$

$= 9\sqrt{3} - 3\sqrt{3} + 9\sqrt{3} - 3\sqrt{3}$

$= 18\sqrt{3} - 6\sqrt{3} = 12\sqrt{3}$

Problem 21 of 30

If the variance of $10$ natural number $1, 1, 1, ..., 1, k$ is less than $10$, then the maximum possible value of $k$ is ______.

Solution:

$\sigma^2 = \dfrac{\sum x^2}{n} - \left( \dfrac{\sum x}{n} \right)^2 = \dfrac{9 + k^2}{10} - \left ( \dfrac{9 + k}{10} \right ) ^2 < 10$

$\Rightarrow \dfrac{90 + 10k^2 - (k^2 + 18k + 81)}{100} < 10$

$\Rightarrow 90 + 10k^2 - (k^2 + 18k + 81) < 1000$

$\Rightarrow 9k^2 - 18k + 9 < 1000$

$\Rightarrow 9(k^2 - 2k + 1) < 1000$

$\Rightarrow (k - 1)^2 < \dfrac{1000}{9}$

$\Rightarrow k - 1 < \dfrac{10 \sqrt{10}}{3}$

$\Rightarrow k < \dfrac{10 \sqrt{10}}{3} + 1$

$\Rightarrow k_{max} = 11$

Problem 22 of 30

Let a point $P$ be such that its distance from the point $(5, 0)$ is thrice the distance of $P$ from the point $(-5, 0)$. If the locus of the point $P$ is a circle of radius $r$, then $4r^2$ is closest to the integer ______.

Solution:

Based on the given condition, the locus of $P$ is given by:

$\sqrt{(x - 5)^2 + (y - 0)^2} = 3 \sqrt{(x + 5)^2 + (y - 0)^2}$

$\Rightarrow (x - 5)^2 + y^2 = 9 \{(x + 5)^2 + y^2\}$

$\Rightarrow x^2 - 10x + 25 + y^2 = 9x^2 + 90x + 225 + 9y^2$

$\Rightarrow 8x^2 + 100x + 200 + 8y^2 = 0$

$\Rightarrow x^2 + \dfrac{25}{2}x + y^2 + 25 = 0$

$\Rightarrow x^2 + 2 \dfrac{25}{4}x + \dfrac{625}{16} + y^2 = \dfrac{625}{16} - 25$

$\Rightarrow \left(x + \dfrac{25}{4}\right)^2 + y^2 = \dfrac{225}{16}$

Therefore, the radius of the circle $r^2 = \dfrac{225}{16} \Rightarrow 4r^2 = \dfrac{225}{4} = 56\dfrac{1}{4}$

The closest integer to $4r^2$ is $56$

This question, as originally published, read $\unicode{0x201C}$ Find the value of $4r^2 \unicode{0x201D}$. Since $4r^2$ is a fraction, we changed the wordings of the question to clarify that the nearest integer to $4r^2$ is required.

Problem 23 of 30

If the area of the triangle formed by the $x$-axis, and the normal and tangent to the circle $(x - 2)^2 + (y - 3)^2 = 25$ at the point $(5, 7)$ is $A$, then $24A$ is equal to ______.

Solution:

The normal to the given circle at $(5, 7)$ will pass through the centre $(2, 3)$ and the given point.

Therefore, the equation of the normal is:

$\dfrac{y - 3}{x - 2} = \dfrac{7 - 3}{5 - 2}$

$\Rightarrow \dfrac{y - 3}{x - 2} = \dfrac{4}{3}$

$\Rightarrow 3y - 9 = 4x - 8$

$\Rightarrow 4x - 3y + 1 = 0$

This line has a slope of $\dfrac{4}{3}$.

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Therefore, the tangent at $(5, 7)$ will have a slope of $-\dfrac{3}{4}$

Therefore, the equation of the tangent is:

$y - 7 = -\dfrac{3}{4}(x - 5)$

$\Rightarrow 3x + 4y - 43 = 0$

If the normal intersects the $x$-axis at $x_1$, then

$4x_1 - 3\times 0 + 1 = 0$

$\Rightarrow x_1 = -\dfrac{1}{4}$

Similarly, if the tangent intersects the $x$-axis at $x_2$, then

$3x_2 + 4 \times 0 - 43 = 0$

$\Rightarrow x_2 = \dfrac{43}{3}$

Therefore, the hypotenuse of the given triangle is of length $x_2 - x_1 = \dfrac{43}{3} - (-\dfrac{1}{4}) = \dfrac{175}{12}$

The altitude from the vertex $(5, 7)$ to the hypotenuse is of length $7$.

Therefore, the area is:

$A = \dfrac{1}{2} \times 7 \times \dfrac{175}{12}$

$\Rightarrow 24A = 24 \times \dfrac{1}{2} \times 7 \times \dfrac{175}{12} = 1225$

This question, as published originally, read $\text{positive x-axis}$. Since one of the intersection points is on the negative $x-axis$, it was unsolvable. We changed the text $\text{positive x-axis}$ to $\text{x-axis}$, which makes the problem solvable.

Problem 24 of 30

Let $i = \sqrt{-1}$. If $\dfrac{(-1 + i\sqrt{3})^{21}}{(1 - i)^{24}} + \dfrac{(1 + i\sqrt{3})^{21}}{(1 + i)^{24}} = k$ and $n = \lfloor |k| \rfloor$ be the greatest integral part of $|k|$. Then $\sum\limits_{j = 0}^{n+5}(j + 5)^2 - \sum\limits_{j = 0}^{n+5}(j + 5)$ is equal to ______.

Solution:

We know from , that the complex cube roots of $1$ are $\dfrac{-1 \pm \sqrt{3}i}{2}$ that of $-1$ are $\dfrac{1 \pm \sqrt{3}i}{2}$

Therefore,

$(-1 + \sqrt{3}i)^{21} = 2^{21}\left(\dfrac{-1 + \sqrt{3}i}{2}\right)^{21}$

$= 2^{21}\left\{\left(\dfrac{-1 + \sqrt{3}i}{2}\right)^{3}\right\}^7$

$= 2^{21} (1)^7 = 2^{21}$

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Similarly,

$(1 + i\sqrt{3})^{21} = -2^{21}$

Therefore,

$\dfrac{(-1 + i\sqrt{3})^{21}}{(1 - i)^{24}} + \dfrac{(1 + i\sqrt{3})^{21}}{(1 + i)^{24}} = k$

$\Rightarrow k = \dfrac{2^{21}}{(1 - i)^{24}} - \dfrac{2^{21}}{(1 + i)^{24}}$

Expanding $(1 - i)^8$ we get:

$(1 - i)^8$

$= [(1 - i)^2]^4$

$= [(1 - 2i + i^2)]^4$

$= [-2i]^4$

$= 16$

Similarly, we will get $(1 - i)^8 = 16$

$\therefore k = \dfrac{2^{21}}{16^3} - \dfrac{2^{21}}{16^3}$

$\Rightarrow k = 0$

$\therefore n = \lfloor |0| \rfloor = 0$

$\therefore \sum\limits_{j = 0}^{n+5}(j + 5)^2 - \sum\limits_{j = 0}^{n+5}(j + 5)$

$= \sum\limits_{j = 0}^{5}(j + 5)^2 - \sum\limits_{j = 0}^{5}(j + 5)$

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$= \sum\limits_{j = 0}^{5}\left\{(j + 5)^2 - (j+5)\right\}$

$= \sum\limits_{j = 0}^{5}(j + 5)(j + 4)$

$= \sum\limits_{j = 0}^{5}(j^2 + 9j + 20)$

$= \sum\limits_{j = 0}^{5}j^2 + 9 \times \sum\limits_{j = 0}^{5}j + 6 \times 20$

$= \dfrac{(5)(6)(11)}{6} + 9 \times \dfrac{5(6)}{2} + 120$

$= 55 + 9 \times 15 + 120$

$= 55 + 135 + 120$

$= 190 + 120$

$= 310$

Problem 25 of 30

The students $S_1, S_2, S_3, ..., S_{10}$ are to be divided into $3$ groups $A$, $B$ and $C$ such that each group has at least one student and the group $C$ has at most $3$ students. Then the total number of possibilities of forming such groups is ______.

Solution:

We can approach this problem by assigning members of group $C$ first, and then dividing the rest between groups $A$ and $B$.

We can assign the members of group $C$ in $\xacomb{10}{1}$ or $\xacomb{10}{2}$ or $\xacomb{10}{3}$ ways.

For each of the above selections, we can distribute the remaining $r$ students into $A$ and $B$ in $2^r$ ways without any restrictions.

These $2^r$ will contain one combination where all members went to $A$ and one where all remaining member went to $B$.

Therefore, with the restriction of minimum one per group, there are $2^r - 2$ ways of assigning the remaining $r$ student to $A$ and $B$.

Therefore, the total number of ways of assigning the students is:

$\xacomb{10}{1} (2^9 - 2) + \xacomb{10}{2} (2^8 - 2) + \xacomb{10}{3} (2^7 - 2)$

$= 20 (2^8 - 1) + 90 (2^7 - 1) + 240 \times (2^6 - 1)$

$= 20 \times 255 + 90 \times 127 + 240 \times 63$

$= 5100 + 11430 + 15120$

$= 31650$

Problem 26 of 30

For integers $n$ and $r$ let $\left(\begin{matrix} n \\ r \end{matrix}\right) = \left\{\begin{array}{ll} \xacomb{n}{r}, & \text{ if } n \ge r \ge 0 \\ 0, & \text{ otherwise} \end{array}\right.$

The maximum value of $k$ for which the sum

$\displaystyle{\sum\limits_{i = 0}^{k} {10 \choose i}{15 \choose {k-i}}} + \displaystyle{\sum\limits_{i = 0}^{k+1} {12 \choose i}{13 \choose {k+1-i}}}$ exists, is equal to ______.

This question, as published, was incorrect and was declared a bonus question.

Do not attempt this question, just enter $0$ as answer.

Solution:

This question, as published, was incorrect and was declared a bonus question.

The problem cannot be solved because for $r \gt n$ the function is defined as $0$, therefore, $r$ can be infinitely large.

Problem 27 of 30

If $a + \alpha = 1, b + \beta = 2$ and $a f(x) + \alpha f\left(\dfrac{1}{x}\right) = bx + \dfrac{\beta}{x}, x \ne 0$, then the value of the expression

$\dfrac{f(x) + \left(\dfrac{1}{x}\right)}{x + \dfrac{1}{x}}$ is ______.

Solution:

$a f(x) + \alpha f\left(\dfrac{1}{x}\right) = bx + \dfrac{\beta}{x}$ $...(i)$

Interchanging $x$ and $\dfrac{1}{x}$

$a f\left(\dfrac{1}{x}\right) + \alpha f(x) = \dfrac{b}{x} + \beta x$ $...(ii)$

Adding $(i)$ and $(ii),$

$a f(x) + \alpha f\left(\dfrac{1}{x}\right) + a f\left(\dfrac{1}{x}\right) + \alpha f(x) = bx + \dfrac{\beta}{x} + \dfrac{b}{x} + \beta x$

$\Rightarrow (a + \alpha) \left( f(x) + f\left(\dfrac{1}{x}\right) \right) = (b + \beta) \left( x + \dfrac{1}{x} \right)$

$\Rightarrow \dfrac{f(x) + f\left(\dfrac{1}{x}\right)}{ x + \dfrac{1}{x}} = \dfrac{b + \beta}{a + \alpha} = \dfrac{2}{1} = 2$

Problem 28 of 30

Let $\lambda$ be an integer. If the shortest distance between the lines $x - \lambda = 2y - 1 = -2z$ and $x = y + 2\lambda = z - \lambda$ is $\dfrac{\sqrt{7}}{2\sqrt{2}}$, then the value of $| \lambda |$ is ______.

Solution:

From the equation of the first line:

$x - \lambda = 2y - 1 = - 2z$

$\Rightarrow \dfrac{x - \lambda}{1} = \dfrac{ y - \dfrac{1}{2}}{\dfrac{1}{2}} = \dfrac{z}{- \dfrac{1}{2}}$

$\Rightarrow \dfrac{x - \lambda}{2} = \dfrac{y - 1}{2} = \dfrac{z}{-1}$

$\therefore \left( \lambda, \dfrac{1}{2}, 0 \right)$ is lies on this line

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Similarly, using the equation of the second line:

$\dfrac{x}{1} = \dfrac{y + 2 \lambda}{1} = \dfrac{z - \lambda}{1} $

$\therefore (0, -2 \lambda, \lambda)$ lies on this line

Distance between skew lines $= \dfrac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}| }$

$= \dfrac{\begin{vmatrix}\lambda & 2 + \dfrac{1}{2} & - \lambda \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{vmatrix}}{\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{vmatrix}}$

$= \dfrac{| -5 \lambda - \dfrac{3}{2}|}{\sqrt{14}}$

Given, $\dfrac{| -5 \lambda - \dfrac{3}{2}|}{\sqrt{14}} = \dfrac{\sqrt{7}}{2 \sqrt{2}}$

$\Rightarrow 10 \lambda + 3 = 7 $

$\Rightarrow \lambda = -1$

$\therefore |\lambda| = 1$

Problem 29 of 30

The sum of the first four terms of a geometric progression (G.P.) is $\dfrac{65}{12}$ and the sum of their respective reciprocals is $\dfrac{65}{18}$. If the product of first three terms of the G.P. is $1$, and the third term is $\alpha$, then $2\alpha$ is ______.

Solution:

Let the first four terms be $a, ar, ar^2, ar^3$

Their reciprocals are $\dfrac{1}{a}, \dfrac{1}{ar}, \dfrac{1}{ar^2}, \dfrac{1}{ar^3}$

Therefore,

$\dfrac{a(r^4 - 1)}{r - 1} = \dfrac{65}{12}$ $...eqn(i)$

and

$\dfrac{\frac{1}{a}(\frac{1}{r^4} - 1)}{\frac{1}{r} - 1} = \dfrac{65}{18}$

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$\Rightarrow \dfrac{(1 - r^4)r}{ar^4(1 - r)} = \dfrac{65}{18}$

$\Rightarrow \dfrac{r^4 - 1}{ar^3(r-1)} = \dfrac{65}{18}$ $...eqn(ii)$

$(i) \div (ii)$ gives us:

$a^2r^3 = \dfrac{3}{2}$

Product of first $3$ terms = $a^3r^3 = 1$

$\therefore a = \dfrac{a^3r^3}{a^2r^3} = \dfrac{2}{3}$

and

$r = \dfrac{3}{2}$

The third term $\alpha = ar^2$

$\therefore 2 \times \alpha = 2ar^2 = 2 \times \dfrac{2}{3} \times \dfrac{3}{2} \times \dfrac{3}{2} = 3$

Problem 30 of 30

The number of the real roots of the equation $(x + 1)^2 + |x - 5| = \dfrac{27}{4}$ is ______.

Solution:

The given equation can be broken down into two different quadratic equations:

$(x + 1)^2 + (x - 5) = \dfrac{27}{4}$ (for $x \ge 5$) $...eqn(i)$

and

$(x + 1)^2 - (x - 5) = \dfrac{27}{4}$ (for $x \lt 5$) $...eqn(ii)$

Solving equation $(i)$,

$x^2 + 3x - 4 = \dfrac{27}{4}$

$\Rightarrow 4x^2 + 12x - 43 = 0$

The discriminant of the above equation is:

$\sqrt{144 + 4\cdot 4 \cdot 43} = 4\sqrt{52}$

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The two roots are

$\dfrac{-12 \pm 4\sqrt{52}}{2}$

$= -6 \pm 2\sqrt{52}$

Performing an inequality check, we get:

$\sqrt{52} \lt \sqrt{64}$

$\Rightarrow 2\sqrt{52} \lt 2 \times 8$

$\Rightarrow -12 + 2\sqrt{52} \lt -12 + 16$

$\Rightarrow -12 + 2\sqrt{52} \lt 4 \lt 5$

$\therefore -12 - 2\sqrt{52} \lt 5$

These two roots do not satisfy the given condition $x \ge 5$, hence are not valid roots of the given function.

Similarly, expanding equation $(ii)$,

$x^2 + x + 6 = \dfrac{27}{4}$

$\Rightarrow 4x^2 + 4x - 3 = 0$

The discriminant of this equation is:

$\sqrt{16 + 48} = \sqrt{64}$

Therefore, the two roots are:

$\dfrac{-4 \pm \sqrt{64}}{2}$

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$= \dfrac{-4 \pm 8}{2}$

$= -2 \pm 4$

$= -6$ and $2$

Both these roots satisfy the condition $x \lt 5$ and, therefore, are valid roots.

Therefore, the given equation has a total of $2$ real roots.