JEE Main Mathematics, Feb 24 2021, (Shift-1)

Problem 1 of 30

Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be defined as $f(x) = 2x - 1$ and $g:\mathbb{R - \{1\}} \longrightarrow \mathbb{R}$ be defined as $g(x) = \dfrac{x - \dfrac{1}{2}}{x - 1}$.

Then the composition function $f(g(x))$ is:

Solution:

$f(x) = 2x - 1, g(x) = \dfrac{x - \dfrac{1}{2}}{x - 1}$

$f(g(x)) = 2 g(x) - 1 = 2 \left( \dfrac{x - \dfrac{1}{2}}{x - 1} \right) - 1 = \dfrac{x}{x - 1} - 1$

$\therefore f(g(x))$ is a hyperbola.

Therefore, $f(g(x))$ is one-one but not onto

Problem 2 of 30

Let $p$ and $q$ be two positive numbers such that $p + q = 2$ and $p^4 + q^4 = 272$. Then $p$ and $q$ are the roots of the equation:

Solution:

$p+q = 2$

$\Rightarrow (p+q)^4 = 2^4 = 16$

$\Rightarrow p^4 + 4p^3q + 6p^2q^2 + 4pq^3 + q^4 = 16$

$\Rightarrow 4p^3q + 6p^2q^2 + 4pq^3 = 16 - 272 = -256$

$\Rightarrow 2p^3q + 3p^2q^2 + 2pq^3 = -128$

$\Rightarrow pq(2p^2 + 3pq + 2q^2) = -128$

$\Rightarrow pq\{2(p+q)^2 - pq\} = -128$

$\Rightarrow pq(8 - pq) = -128$

-----------book page break-----------

Let $pq = x$

$\therefore x(8 - x) = -128$

$\Rightarrow 8x - x^2 = -128$

$\Rightarrow x^2 - 8x -128 = 0$

$\Rightarrow (x-16)(x+8)=0$

$\Rightarrow x = 16$ or $-8$

$\therefore pq = 16$ or $-8$

An equation with roots $p$ and $q$ is of the form $x^2 -(p+q)x + pq = 0$

Therefore, the equations are, $x^2 - 2x + 16 = 0$ and $x^2 - 2x - 8 = 0$

Since the latter isn't an option, the answer is $x^2 - 2x + 16 = 0$

Problem 3 of 30

The system of linear equations:

$3x - 2y - kz = 10$

$2x - 4y - 2z = 6$

$x + 2y - z = 5m$

is inconsistent if:

Solution:

For system of equations to be inconsistent $|A| = 0$ and $adj(A)B \neq O$

$A = \begin{bmatrix} 3 & -2 & -k\\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{bmatrix}$ $B = \begin{bmatrix} 10 \\ 6 \\ 5m \end{bmatrix}$

-----------book page break-----------

$|A| = \begin{vmatrix} 3 & -2 & -k\\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{vmatrix}$

$= 3[(-4)(-1) - (-2)(2)] - (-2)[(2)(-1) - (-2)(1)]$

$+ (-k)[(2)(2) - (-4)(1)]$

$= 8(3 - k)$

$\because |A| = 0, \ k = 3$

$\therefore A = \begin{bmatrix} 3 & -2 & -3\\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{bmatrix}$

$\Rightarrow adj(A) = \begin{bmatrix} 8 & -8 & -8\\ 0 & 0 & 0 \\ 8 & -8 & -8 \end{bmatrix}$

For the system of equations to be inconsistent,

$adj(A) B \neq O$

$\Rightarrow \begin{bmatrix} 8 & -8 & -8\\ 0 & 0 & 0 \\ 8 & -8 & -8 \end{bmatrix} \begin{bmatrix} 10 \\ 6 \\ 5m \end{bmatrix} \neq O$

$\Rightarrow 8(10) + (-8)(6) + (-8)(5m) \neq 0$

$\Rightarrow 80 - 48 - 40m \neq 0$

-----------book page break-----------

$\Rightarrow 32 - 40 m \neq 0$

$\Rightarrow 40 m \neq 32$

$\Rightarrow m \neq \dfrac{4}{5}$

$\therefore$ For the given system of linear equations to be inconsistent $k = 3, m \neq \dfrac{4}{5}$

Problem 4 of 30

The value of

$-\xacomb{15}{1} + 2\cdot \xacomb{15}{2} - 3\cdot \xacomb{15}{3} +... -15\cdot \xacomb{15}{15} + \xacomb{14}{1} + \xacomb{14}{3} + \xacomb{14}{5} + ...$

$+ \xacomb{14}{11}$ is

Solution:

Let the two parts of the given series be $S_1$ and $S_2$.

Therefore,

$S_1 = -\xacomb{15}{1} + 2\cdot \xacomb{15}{2} - 3\cdot \xacomb{15}{3} +... -15\cdot \xacomb{15}{15}$

$= \sum\limits_1^{15} (-1)^r \times r \times \xacomb{15}{r}$

$= 15 (\sum\limits_1^{15} (-1)^r \times r \times \xacomb{14}{r-1})$

$= 15 (- \xacomb{14}{0} + \xacomb{14}{1} - \xacomb{14}{2} + \xacomb{14}{3} ... - \xacomb{14}{14})$

-----------book page break-----------

$\therefore S_1 = 15 (-\xacomb{14}{0} + (\xacomb{13}{0} + \xacomb{13}{1}) - (\xacomb{13}{1} + \xacomb{13}{2}) + (\xacomb{13}{2} + \xacomb{13}{3})...$

$- (\xacomb{13}{12} + \xacomb{13}{13}) - \xacomb{14}{14})$

$\Rightarrow S_1 = 15(-\xacomb{14}{0} + \xacomb{13}{0} + \xacomb{13}{13} - \xacomb{14}{14})$

$\Rightarrow S_1 = 15(-1 + 1 + 1 -1)$

$\Rightarrow S_1 = 15(0) = 0$

Similarly,

$S_2 = \xacomb{14}{1} + \xacomb{14}{3} + \xacomb{14}{5}...\xacomb{14}{11}$

$= (\xacomb{14}{1} + \xacomb{14}{3} + \xacomb{14}{5}...\xacomb{14}{11} + \xacomb{14}{13}) - \xacomb{14}{13}$

$= (\xacomb{13}{0} + \xacomb{13}{1}) + (\xacomb{13}{2} + \xacomb{13}{3}) + (\xacomb{13}{4} + \xacomb{13}{5})...$

$+ (\xacomb{13}{10} + \xacomb{13}{11}) + (\xacomb{13}{12} + \xacomb{13}{13}) - \xacomb{14}{13}$

$= 2^{13} - 14$

Therefore,

$-\xacomb{15}{1} + 2\cdot \xacomb{15}{2} - 3\cdot \xacomb{15}{3} +... -15\cdot \xacomb{15}{15}$

$+ \xacomb{14}{1} + \xacomb{14}{3} + \xacomb{14}{5} + ... + \xacomb{14}{11}$

$= S_1 + S_2$

$= 0 + 2^{13} - 14$

$= 2^{13} - 14$

Problem 5 of 30

If $e^{(\cos^2x + \cos^4x + \cos^6x + ... \infty)log_e2}$ satisfies the equation $t^2 - 9t + 8 = 0$, then the value of

$\dfrac{2\sin x}{\sin x + \sqrt{3}\cos x}\ \left(0 \lt x \lt \dfrac{\pi}{2}\right)$ is:

Solution:

We can evaluate the series in the exponent first,

$\cos^2x + \cos^4x + \cos^6x + ... \infty$

This is an infinite G.P. series with common ratio $\lt 1$, therefore, the sum is finite.

The first term is $\cos^2x$ and the common ratio is also $\cos^2 x$, therefore the sum is:

$\dfrac{\cos^2x}{1 - \cos^2x} = \dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

-----------book page break-----------

Therefore,

$e^{(\cos^2x + \cos^4x + \cos^6x + ... \infty)\log_e2}$

$= e^{(\cot^2 x)\log_e2} = \left(e^{\log_e2} \right)^{\cot^2 x} = 2^{\cot^2 x}$

The roots of the equation $t^2 - 9t + 8 = 0$ are $8$ and $1$

Therefore,

$2^{\cot^2 x} = 1 = 2^0 \Rightarrow \cot^2 x = 0 \Rightarrow x = \dfrac{\pi}{2}$ This is outside the given range.

Therefore,

$2^{\cot^2 x} = 8 = 2^3 \Rightarrow \cot^2 x = 3 \Rightarrow \cot x = \sqrt{3}$

It is possible to evaluate the given expression without actually finding the value of $x$:

$\cot x = \sqrt{3} \Rightarrow \cos x = \sqrt{3} \sin x$

Therefore,

$\dfrac{2\sin x}{\sin x + \sqrt{3}\cos x}$

$= \dfrac{2\sin x}{\sin x + 3\sin x}$

$= \dfrac{2}{4} = \dfrac{1}{2}$

Problem 6 of 30

$\displaystyle{\lim\limits_{x \rightarrow 0}}\dfrac{\int\limits_0^{x^2}\left(\sin \sqrt{t}\right)dt}{x^3}$ is equal to:

Solution:

We will first evaluate the function in the numerator using integration.

Let,

$I = \displaystyle \int\left(\sin \sqrt{t}\right)dt$

Substituting $\sqrt{t} = z$, we get:

$z^2 = t \Rightarrow 2zdz = dt$

Therefore,

$I = \displaystyle \int 2z \left(\sin z\right) dz = \displaystyle 2\int z \left(\sin z\right) dz$

-----------book page break-----------

We can find this integral, using integration by parts, where we can take $u(z) = z$ and $v(z) = \sin z$

Therefore,

$I = \displaystyle 2\left(z \int \sin z dz - \int \left[\dfrac{d}{dz}z \right] \left[\int \sin z dz \right]dz\right)$

$= \displaystyle 2\left(- z \cos z + \int \cos z dz\right)$

$= 2 (\sin z - z \cos z)$

Substituting $z = \sqrt{t}$, we get:

$I = 2 (\sin \sqrt{t} - \sqrt{t} \cos \sqrt{t})$

Therefore,

$2 (\sin \sqrt{t} - \sqrt{t} \cos \sqrt{t}) \biggr\rvert_{0}^{x^2}$

$= 2 \sin x - 2x cos x - 2 \sin 0 - 2 \times 0 \cos 0$

$= 2 \sin x - 2x cos x$

Therefore,

$\lim\limits_{x \rightarrow 0}\dfrac{\int\limits_0^{x^2}\left(\sin \sqrt{t}\right)dt}{x^3}$

$= \lim\limits_{x \rightarrow 0}\dfrac{2 \sin x - 2x cos x}{x^3}$

-----------book page break-----------

The above limit is of the indeterminate form, $\dfrac{0}{0}$, therefore, can be solved using .

Let $f(x) = 2 \sin x - 2x \cos x$ and $g(x) = x^3$

Therefore,

$\dfrac{f'(x)}{g'(x)} = \dfrac{2 \cos x - 2\cos x + 2 x \sin x}{3x^2} = \dfrac{2 x \sin x}{3x^2} = \dfrac {2 \sin x}{3x}$

Therefore,

$\lim\limits_{x \rightarrow 0} \dfrac{f(x)}{g(x)}$

$= \lim\limits_{x \rightarrow 0} \dfrac{f'(x)}{g'(x)}$

$= \lim\limits_{x \rightarrow 0} \dfrac{2 \sin x}{3x}$

$= \dfrac{2}{3}\lim\limits_{x \rightarrow 0} \dfrac{\sin x}{x}$

$= \dfrac{2}{3}$ (using the standard forms of limits from )

Problem 7 of 30

The function $f(x) = \dfrac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x$:

Solution:

$f(x) = \dfrac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1)\cos x $

$f'(x) = \dfrac{12x^2}{6} - \dfrac{6x}{6} - 2 \cos x - (2x - 1) \sin x + 2 \cos x$

$= 2x^2 - x - (2x - 1) \sin x$

$= (2x - 1)x - (2x - 1) \sin x$

$= (2x -1) (x - \sin x)$

Clearly, $x - \sin x$ is greater than $0$ for $x \in \left(1, \infty \right)$.

To find its behaviour for $0 \le x \le 1$, we can differentiate $x - \sin x$

$\dfrac{d}{dx}(x - \sin x) = 1 - \cos x$

$1 - \cos x = 0$ for $x = 0$ and for $0 \lt x \le 1$, $1 - \cos x \gt 0$

For $x = 0$, $x - \sin x = 0$ and it increases in the interval $0 \lt x \le 1$

Therefore, for $0 \lt x \le 1$, $x - \sin x$ is positive.

$\Rightarrow f'(x) \geq 0$ for $x \in \left[\dfrac{1}{2}, \infty \right)$ as $(x - \sin x)$ is positive $[0, \infty)$ and $2x -1$ is positive in $x \in \left[\dfrac{1}{2}, \infty \right)$

Problem 8 of 30

A scientific committee is to be formed from $6$ Indians and $8$ foreigners, which includes at least $2$ Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is :

Solution:

The ways to form the committee based on the given conditions are:

$2$ Indians, $4$ foreigners $= \xacomb{6}{2} \times \xacomb{8}{4} = 15 \times 70$

$3$ Indians, $6$ foreigners $= \xacomb{6}{3} \times \xacomb{8}{6} = 20 \times 28$

$4$ Indians, $8$ foreigners $= \xacomb{6}{4} \times \xacomb{8}{8} = 15 \times 1$

The total number of ways to form the committee is:

$15 \times 70 + 20 \times 28 + 15$

$= 5(3 \times 70 + 4 \times 28 + 3)$

$= 5(210 + 112 + 3)$

$= 5 \times 325 = 1625$

Problem 9 of 30

If $f : \mathbb{R} \longrightarrow \mathbb{R}$ is defined by $f(x) = \lfloor x - 1\rfloor \cdot \cos\left(\dfrac{2x - 1}{2}\right)\pi$ where $\lfloor \cdot \rfloor$ denotes the greatest integer function, the $f$ is:

Solution:

Possible points of discontinuity are at $x = n, n \in \mathbb{Z}$

$f(n) = \lfloor n - 1 \rfloor \cdot \cos \dfrac{2n - 1}{2} \pi = 0$

$\lim \limits_{x \rightarrow n^-} f(x) = \lim \limits_{x \rightarrow n^-} \left ( \lfloor x - 1 \rfloor \cdot \cos \dfrac{2x - 1}{2} \pi \right ) = ( n - 2 ) \cdot \cos \dfrac{2n - 1}{2} \pi = 0$

$\lim \limits_{x \rightarrow 1^+} f(x) = \lim \limits_{x \rightarrow 1^+} \left ( \lfloor x - 1 \rfloor \cdot \cos \dfrac{2x - 1}{2} \right ) = ( n - 1 ) \cdot \cos \dfrac{2n - 1}{2} \pi = 0$

$\therefore f(x)$ is continuous at every real $x$

Problem 10 of 30

If $\displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}dx = a\sin^{-1}\left(\dfrac{\sin x + \cos x}{b}\right) + c$, where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to:

Solution:

$\displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - 1 - \sin 2x}}dx$

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - 1 - 2\sin x \cos x}}dx$

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - (1 + 2\sin x \cos x)}}dx$

-----------book page break-----------

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - (\sin^2x + \cos^2x + 2\sin x \cos x)}}dx$

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}}dx$

$u = \sin x + \cos x$

$\Rightarrow \dfrac{du}{dx} = \cos x - \sin x$

$\Rightarrow dx = \dfrac{du}{\cos x - \sin x}$

$\therefore \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}}dx$

$= \displaystyle{\int} \dfrac{\cos x - \sin x}{\sqrt{9 - u^2}}\dfrac{du}{\cos x - \sin x}$

$= \displaystyle{\int} \dfrac{1}{\sqrt{9-u^2}}du$

$= \sin^{-1} \dfrac{u}{3} + C$

$= \sin^{-1} \dfrac{\cos x - \sin x}{3} + C$

$a \sin^{-1} \dfrac{\cos x - \sin x}{b} + C = \sin^{-1} \dfrac{\cos x - \sin x}{3} + C$

-----------book page break-----------

Therefore, $a = 1$ and $b = 3$

$\Rightarrow (a,b)=(1,3)$

Problem 11 of 30

The area (in sq. units) of the part of the circle $x^2 + y^2 = 36,$ which is outside the parabola $y^2 = 9x,$ is :

Solution:

Solving the two equations we get:

$x^2 + 9x - 36 = 0$

$\Rightarrow (x + 12)(x - 3) = 0$

$\because x \ge 0$, $x = 3$

Therefore, the parabola and the circle intersect at $x = 3$

-----------book page break-----------

Now we can use the following diagram:

The area inside the circle and outside the parabola is given by:

$\text{Area of the circle} - 2 \times (\text{Area shown in green} + \text{Area shown in red})$

$\pi r^2 - 2 \left [ \displaystyle \int \limits_0^3 \sqrt{9x} dx + \displaystyle \int \limits_3^6 \sqrt{36 - x^2} dx \right ]$

$= 36 \pi - 12 \sqrt{3}- 2 \left[ \dfrac{x}{2} \cdot \sqrt{36 -x^2} + 18 \sin^{-1} \left( \dfrac{x}{6} \right) \right]$

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

$= 24 \pi - 3 \sqrt{3}$

Problem 12 of 30

The population $P = P(t)$ at time $'t'$ of a certain species follows the differential equation $\dfrac{dP}{dt} = 0.5P - 450$. If $P(0) = 850$, then the time at which population becomes zero is:

Solution:

$\dfrac{dP}{dt} = 0.5P - 450$

$\Rightarrow \dfrac{dP}{dt} - 0.5P = -450$

This is a linear differential equation and can be solved using the method of integrating factor described .

-----------book page break-----------

Integrating factor $= e^{\int -0.5 dt}= e^{-\frac{t}{2}}$

$e^{-\frac{t}{2}}P = \int (-450) e^{-\frac{t}{2}} dt$

$\Rightarrow e^{-\frac{t}{2}}P = -450 \int e^{-\frac{t}{2}} dt$

$u=-\dfrac{t}{2}$

$\dfrac{du}{dt}=-\dfrac{1}{2}$

$dt = -2 du$

$Pe^{-\frac{t}{2}} = - 450 \int e^{-\frac{t}{2}} dt$

$\Rightarrow Pe^{-\frac{t}{2}} = - 450 \int e^u -2 du$

$\Rightarrow Pe^{-\frac{t}{2}} =900 e^u + C$

$\Rightarrow Pe^{-\frac{t}{2}} =900 e^{-\frac{t}{2}} + C$

At $t=0$

$850 = 900 + C$

$\Rightarrow C = -50$

$Pe^{-\frac{t}{2}} =900 e^{-\frac{t}{2}} - 50$

$\Rightarrow P = 900 - 50 e^{\frac{t}{2}}$

-----------book page break-----------

$\Rightarrow 0 = 900 - 50 e^{\frac{t}{2}}$

$\Rightarrow 50 e^{\frac{t}{2}}= 900$

$\Rightarrow e^{\frac{t}{2}}= 18$

$\Rightarrow \dfrac{t}{2} = \log_e 18$

$\Rightarrow t = 2 \log_e 18$

Therefore, the time at which the population becomes zero at $t = 2 \log_e 18.$

Problem 13 of 30

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\dfrac{1}{4}.$ Three stones $A,\ B$ and $C$ are placed at the points $(1,1),\ (2,2)$ and $(4,4)$ respectively. Then which of these stones is/are on the path of the man?

Solution:

Let the $x$ and $y$ intercepts be $I_x$ and $I_y$ respectively and let the coordinates of the stone be $(h,k).$

The equation of the line, formed using the intercepts, is given by

$\dfrac{x}{I_x} + \dfrac{y}{I_y} = 1$

$\Rightarrow \dfrac{h}{I_x} + \dfrac{k}{I_y} = 1$ $... eqn (i)$

-----------book page break-----------

The question states,

$\dfrac{\dfrac{1}{I_x} + \dfrac{1}{I_y}}{2} = \dfrac{1}{4}$

$\Rightarrow \dfrac{1}{I_x} + \dfrac{1}{I_y} = \dfrac{1}{2}$

$\Rightarrow \dfrac{2}{I_x} + \dfrac{2}{I_y} = 1$ $... eqn (ii)$

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

$\dfrac{h}{I_x} + \dfrac{k}{I_y} = \dfrac{2}{I_x} + \dfrac{2}{I_y}$

$\Rightarrow h = 2$ and $y = 2$

The coordinates of the stone on the man's path is $(2,2)$ and the stone is $B.$

Problem 14 of 30

The locus of the mid-point of the line segment joining the focus of the parabola $y^2 = 4ax$ to a moving point of the parabola, is another parabola whose directrix is :

Solution:

A parabola with equation $y^2 = 4ax$ has its focus at $(a, 0)$, the directrix is $x = -a$ and the vertex is at $(0,0).$

The locus of the mid-point of the line segment joining the focus and the parabola is another parabola whose focus remains at $(a, 0),$ but the vertex changes to $(\dfrac{a}{2}, 0).$

Since the vertex is equidistant from the focus and directrix, the new directrix is at $x = 0.$

Problem 15 of 30

If the tangent to the curve $y = x^3$ at the point $P(t, t^3)$ meets the curve again at $Q$, then the ordinate of the point which divides $PQ$ internally in the ratio $1:2$ is:

Solution:

$y=x^3$

$\Rightarrow \dfrac{dy}{dx} = 3x^2$

Slope of the tangent at $(t,t^3)$ $= 3t^2$

The equation of the tangent at $(t, t^3)$ is

$(y-t^3) = 3t^2 (x - t)$

-----------book page break-----------

The intersection point of the curve and the tangent satisfies $y=x^3$

$(x^3-t^3) = 3t^2 (x - t)$

$\Rightarrow (x-t)(x^2 + xt + t^2) = 3t^2 (x-t)$ ($\because P$ and $Q$ are distinct points, $x \ne t$)

$\therefore x^2 + xt + t^2 = 3t^2$

$\Rightarrow x^2 + xt - 2t^2 = 0$

The roots of the above equation are $x = t$ and $x = -2t$

For $x=-2t,$ $y= -8t^3$

The ordinate ($y$ value) of the point that divides the line joining $(t,t^3)$ and $(-2t, -8t^2)$ is

Therefore, the ordinate value of the point dividing $PQ$ in the ratio $1:2$

$= \dfrac{(2)t^3 + (1)(-8t^3)}{3} = -2t^3$

Problem 16 of 30

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$, is:

Solution:

Let the given planes represented by the equations $3x + y - 2z = 5$ and $2x - 5y - z = 7$ be $P_1$ and $P_2$ respectively.

Let the direction ratios of the normal to the line of intersection of $P_1$ and $P_2$ be $a, b, c.$

Let $P_3$ be the plane perpendicular to both $P_1$ and $P_2$, and passing through the given point.

The normal to $P_3$ will be parallel to the line of intersection of $P_1$ and $P_2$ and therefore, will have the same direction ratios.

-----------book page break-----------

The line of intersection of the two planes will be perpendicular to the normals to these planes. Therefore, we have,

$3a + 1b - 2c = 0$ $...(i)$

$2a - 5b -1c = 0$ $...(ii)$

Solving $(i)$ and $(ii)$, we get the direction ratios of the line of intersection.

$\dfrac{a}{-11} =\dfrac{b}{-1}=\dfrac{c}{-17}$

$\dfrac{a}{11} =\dfrac{b}{1}=\dfrac{c}{17}$

Therefore, equation of the plane, $P_3$, is,

$11x + y + 17z + k = 0$

Since the plane passes through$(1, 2, -3),$

$11(1) + 2 + 17(-3) + k = 0$

$k = 38$

Therefore, equation of the plane is,

$11x + y + 17z + 38 = 0$

Problem 17 of 30

The distance of the point $(1, 1, 9)$ from the point of intersection of the line

$\dfrac{x-3}{1} = \dfrac{y - 4}{2} = \dfrac{z - 5}{2}$ and the plane $x + y + z = 17$ is:

Solution:

Let point of intersection of the given line and the plane be represented by:

$\dfrac{x-3}{1} = \dfrac{y - 4}{2} = \dfrac{z - 5}{2} = \lambda$

$x = \lambda + 3,\ y = 2\lambda + 4,\ z = 2\lambda + 5$

Therefore,

$x + y + z = 17$

$\Rightarrow (\lambda + 3) + (2\lambda + 4) + (2\lambda + 5) = 17$

$\Rightarrow 5 \lambda + 12 = 17$

$\Rightarrow \lambda = 1$

Therefore, the coordinates of the intersection point are:

$x = 4,\ y = 6, \ z = 7$

Required distance $= \sqrt{(1-4)^2 + (1-6)^2 + (9-7)^2}$

$= \sqrt{(-3)^2 + (-5)^2 + (2)^2} = \sqrt{38}$

Problem 18 of 30

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number $2$ times is equal to the probability of getting an even number $3$ times, then the probability of getting an odd number odd number of times is:

Solution:

Let there be $n$ roll of the dice.

For any ordinary dice the probability of getting an odd number is equal to the probability of getting an even number $= \dfrac{1}{2}$

The probability of getting an odd number $2$ times is:

$\xacomb{n}{2} \left(\dfrac{1}{2}\right)^2 \left(\dfrac{1}{2}\right)^{n-2}$

The probability of getting an even number $3$ times is:

$\xacomb{n}{3} \left(\dfrac{1}{2}\right)^3 \left(\dfrac{1}{2}\right)^{n-3}$

-----------book page break-----------

Since they are equal,

$\xacomb{n}{2} \left(\dfrac{1}{2}\right)^2 \left(\dfrac{1}{2}\right)^{n-2} = \xacomb{n}{3} \left(\dfrac{1}{2}\right)^3 \left(\dfrac{1}{2}\right)^{n-3}$

$\Rightarrow \xacomb{n}{2} \left(\dfrac{1}{2}\right)^n = \xacomb{n}{3} \left(\dfrac{1}{2}\right)^n$

$\Rightarrow \xacomb{n}{2} = \xacomb{n}{3}$

$\therefore n = 5$

The probability of getting an odd number odd number of times, that is, $1, 3, 5$ is:

$\xacomb{5}{1} \left(\dfrac{1}{2}\right)^1 \left(\dfrac{1}{2}\right)^{4} + \xacomb{5}{3} \left(\dfrac{1}{2}\right)^3 \left(\dfrac{1}{2}\right)^{2} + \xacomb{5}{5} \left(\dfrac{1}{2}\right)^5 \left(\dfrac{1}{2}\right)^{0}$

$= \left(\dfrac{1}{2}\right)^5 \left(\xacomb{5}{1} + \xacomb{5}{3} + \xacomb{5}{5}\right)$

$= \left(\dfrac{1}{2}\right)^5 (5 + 10 + 1)$

$= \left(\dfrac{1}{2}\right)^5 (16)$

$= \dfrac{1}{2}$

Problem 19 of 30

Two vertical poles are $150$ m apart and the height of one is three times the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Solution:

Let the height of the two poles be $h$ and $3h$ with angles of elevation as $\theta$ and $90 - \theta$ respectively.

The man is standing at the midpoint, therefore, is $75$ m from each of the poles.

Therefore,

$\tan \theta = \dfrac{h}{75}$ and

$\tan (90 - \theta) = \dfrac{3h}{75}$

$\Rightarrow \cot \theta = \dfrac{h}{25}$

$\Rightarrow \tan \theta = \dfrac{25}{h}$

Therefore,

$\dfrac{h}{75} = \dfrac{25}{h}$

$\Rightarrow h = 25 \sqrt{3}$

Problem 20 of 30

The statement among the following that is a tautology is:

Solution:

We will solve this problem using two different approaches.

Approach 1 - Using Truth Table:

Using truth table. This would involve creating the truth tables for the given options, till you come across a case which is a tautology.

Here, for the sake of brevity, we will show the truth table just for the correct option, which is $[A \land (A \rightarrow B)] \rightarrow B$

$A$	$B$	$A \rightarrow B$	$A \land (A \rightarrow B)$	$[A \land (A \rightarrow B)] \rightarrow B$
$0$	$0$	$1$	$0$	$1$
$0$	$1$	$1$	$0$	$1$
$1$	$0$	$0$	$0$	$1$
$1$	$1$	$1$	$1$	$1$

-----------book page break-----------

Since for all input combinations for this expression evaluates to $1$, this is a tautology.

If you evaluate all the other options, for each you will find at least one set of input for which the expression evaluates to $0$

Approach 2 - Using Logical Deduction:

Here we will see the logical approach to solve this problem. For each of the given options we will try to find a set of input, which will lead to a $0$ value of expression. Not finding such a value will show that the given option is a tautology.

$A \land (A \lor B) = 0$

We can see that $A = 0$ will lead to the whole expression being $0$, no matter what the value of $B$ is.

Therefore, this is not a tautology.

$A \lor (A \land B)$

Here,

$A = 0$

Therefore, $A \land B = 0$ no matter what the value of $B$ is. Therefore, this is not a tautology.

$B \rightarrow [A \land (A \rightarrow B)]$

This expression will yield a result of $0$ only if $B = 1$ and $A \land (A \rightarrow B) = 0$

If $A \land (A \rightarrow B) = 0$ then one of the possible values of $A$ is $0$

Putting $A = 0$ and $B = 1$, we get:

$B \rightarrow [A \land (A \rightarrow B)]$

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

$= 1 \rightarrow [0 \land 1]$

$= 1 \rightarrow 0$

$= 0$

Since we have a $0$ value of the expression, this cannot be a tautology.

-----------book page break-----------

$[A \land (A \rightarrow B)] \rightarrow B$

For this expression to be $0$, the only possible condition is $[A \land (A \rightarrow B)] = 1$ and $B = 0$

If $[A \land (A \rightarrow B)] = 1$, then $A = 1$

Putting $A = 1$ and $B = 0$, we get:

$[A \land (A \rightarrow B)] \rightarrow B$

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

$= [1 \land 0] \rightarrow 0$

$= 0 \rightarrow 0$

$= 1$

Therefore, there is no input set which will lead to a result of $0$ for the expression, hence, this is a tautology.

It may appear that the second method is longer than the first method, however, given that we have evaluated only one option in the first approach, and all the four options in the second approach, the second approach is much more efficient.

Problem 21 of 30

If the least and the largest real values of $\alpha$, for which the equation $z + \alpha|z-1| + 2i = 0$ $z \in \mathbb{C}$ and $i = \sqrt{-1}$ has a solution, are $p$ and $q$ respectively then $4(p^2 + q^2)$ is equal to:

Solution:

Let $z = x + yi$, where $x,y \in \mathbb{R}$

$\therefore |z - 1| = |x + yi - 1| = \sqrt{(x-1)^2 + y^2}$

Therefore,

$z + \alpha|z-1| + 2i = 0$

$\Rightarrow x + yi + \alpha \sqrt{(x-1)^2 + y^2} + 2i = 0$

$\Rightarrow x + yi + 2i = - \alpha \sqrt{(x-1)^2 + y^2}$

$\Rightarrow x + (y + 2)i = - \alpha \sqrt{(x-1)^2 + y^2} + 0i$

-----------book page break-----------

Equating the real and imaginary parts of the above equation,

$y + 2 = 0 \Rightarrow y = -2$

$x = -\alpha \sqrt{(x-1)^2 + y^2}$

$\Rightarrow x = -\alpha \sqrt{(x-1)^2 + 4}$

$\Rightarrow x^2 = \alpha^2 \{(x-1)^2 + 4\}$

$\Rightarrow x^2 - \alpha^2 x^2 + 2\alpha^2 x - 5\alpha^2 = 0$

$\Rightarrow (1 - \alpha^2)x^2 + 2\alpha^2 x - 5\alpha^2 = 0$

For $x$ to be real, the discriminant of the above equation must be $ \ge 0$

$\therefore (2\alpha^2)^2 + 20(1 - \alpha^2)\alpha^2 \ge 0$

$\Rightarrow \alpha^4 + 5(1 - \alpha^2)\alpha^2 \ge 0$

$\Rightarrow \alpha^4 + 5\alpha^2 - 5\alpha^4 \ge 0$

$\Rightarrow 5\alpha^2 - 4\alpha^4 \ge 0$

$\Rightarrow \alpha^2 (5 - 4\alpha^2) \ge 0$

$\Rightarrow 5 - 4\alpha^2 \ge 0$

$\Rightarrow 4 \alpha^2 \le 5$

$\Rightarrow -\dfrac{\sqrt{5}}{2} \le \alpha \le \dfrac{\sqrt{5}}{2}$

-----------book page break-----------

Therefore, $p = -\dfrac{\sqrt{5}}{2}$ and $q = \dfrac{\sqrt{5}}{2}$

$\therefore 4(p^2 + q^2) = 10$

Problem 22 of 30

Let $B_i (i = 1, 2, 3)$ be three independent events in a sample space. The probability that only $B_1$ occurs is $\alpha$, only $B_2$ occurs is $\beta$ and only $B_3$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_i$ occurs and these $4$ probabilities satisfy the equation $(\alpha - 2\beta)p = \alpha \beta$ and $(\beta - 3\gamma)p = 2\beta \gamma$ (All probabilities are assumed to lie in the interval $(0, 1)$). Then $\dfrac{P(B_1)}{P(B_3)}$ is equal to ___________.

Solution:

Let $x,y,z$ be probability of $B1, B2, B3$ respectively.

Using the given conditions of only the $i \xasuper{th}$ event happening for each $i$,

$x(1 – y) (1 – z) = \alpha$

$y(1 – x)(1 – z) = \beta$

$z(1 – x)(1 – y ) = \gamma$

Given that the probability of none of the events happening is $p$,

$\therefore (1 – x)(1 – y)(1 – z) = p$

-----------book page break-----------

GIven that $(\alpha - 2\beta)p = \alpha \beta$

$\Rightarrow (x(1–y)(1–z)-2y(1-x)(1–z)) (1–x)(1–y)(1–z) = xy(1–x)(1–y) (1–z)2$

$\Rightarrow x+ xy - 2y = xy$

$ \therefore x = 2y$ $⋯(i)$

Similarly, using $(\beta–3\gamma) p = 2 \beta \gamma$ we get:

$ y = 3z$ $⋯(ii)$

From $(i)$ and $(ii)$

$\Rightarrow x = 6z$

Hence $\dfrac{x}{z} = \dfrac{P(B_1)}{P(B_3)} = 6$

Problem 23 of 30

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in \mathbb{R}$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some $k \in \mathbb{R}$. If $q_{23} = -\dfrac{k}{8}$ and $|Q| = \dfrac{k^2}{2}$ then $\alpha^2 + k^2$ is equal to ____________.

Solution:

$PQ = kI$

$\Rightarrow Q = kP^{-1} I$

$\Rightarrow Q = \dfrac{k}{|P|}\times adj(P) \times I$

$|P| = 20 + 12 \alpha$

For our purpose we do not need to find the complete $adj(P)$, all we need is $adj(P)_{2, 3}$

-----------book page break-----------

$adj(P)_{2, 3} = - $Cofactor$(P_{3, 2}) = - \begin{vmatrix} 3 & -2 \\ 2 & \alpha\end{vmatrix} = -3 \alpha - 4$

$Q = \dfrac{k}{20 + 12 \alpha} \begin{bmatrix} - & - & - \\ - & - & -3 \alpha - 4 \\ - & - & - \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\because q_{23} = - \dfrac{k}{8}$

$\Rightarrow \dfrac{k}{20 + 12 \alpha}(-3 \alpha - 4) = - \dfrac{k}{8}$

$\Rightarrow \alpha = -1$

$|Q| = \dfrac{k^3 |I|}{|P|}$

$\Rightarrow \dfrac{k^2}{2} = \dfrac{k^3}{20 + 12 \alpha}$

$\Rightarrow 2k =20 + 12 \alpha $

$\Rightarrow 2k = 8$

$\Rightarrow k = 4$

$\therefore \alpha^2 + k^2 = (-1)^2 + (4)^2 = 17$

Problem 24 of 30

Let $M$ be any $3 \times 3$ matrix with entries from the set $\{0, 1, 2\}$. The maximum number of such matrices, for which the sum of the diagonal elements of $M^TM$ is seven is ______________.

Solution:

Let $M = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$

Then $M^T = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}$

Since we are only concerned with the elements of the diagonal elements of $M^T$ we will be computing only those.

The first diagonal element will be $(a_1)(a_1) + (b_1)(b_2) + (c_2)(c_2)$ or $a_1^2+ b_1^2 + c_1^2.$.

-----------book page break-----------

Similarly, the second and third diagonal elements will be $a_2^2+ b_2^2 + c_2^2$ and $a_3^2+ b_3^2 + c_3^2$ respectively.

$a_1^2+ b_1^2 + c_1^2 + a_2^2+ b_2^2 + c_2^2 + a_3^2+ b_3^2 + c_3^2 = 7$

The possible ways this can be accomplished in the following ways,

Case I:

Any seven of the terms are $1$ and the rest are $0.$

Number of possible matrices $= \xacomb{9}{7} = 36$

Case II:

Any one term is $2$ and three are $1.$

Number of possible matrices $= \xacomb{9}{1} \times \xacomb{8}{3} = 504$

$\therefore$ Maximum number of possible matrices= $36 + 504 = 540$

Problem 25 of 30

Let $A = \{n \in \mathbb{N} : n$ is a $3$-digit number$\}$

$B = \{9k + 2 : k \in \mathbb{N}\}$

$C = \{9k + l : k \in \mathbb{N}\}$ for some $l (0 < l < 9)$

If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$, then $l$ is equal to _______.

Solution:

Set $A$ contains all $3$-digits numbers, therefore,

$A = \{100, 101, ..., 999\}$

Set $B$ contains numbers of the form $9k + 2$.

Therefore,

$A \cap B$ = $\{101, 110, 119, ..., 992\}$

$\therefore$ Sum of all the elements in $A \cap B = 101 + 110 + 119... + 992$

$= \left(\dfrac{(992 - 101)}{9} + 1\right)\left(\dfrac{101 + 992}{2}\right)$

$= 100 \times \dfrac{1093}{2}$

-----------book page break-----------

$A \cap C$ contains all $3$-digit numbers of the form $9k + l$

If $l = 2$ then $B = C$, and the sum of elements in $A \cup (B \cap C) = A \cup B$ which is not true.

Therefore, $l \ne 2$ and $B$ and $C$ are mutually exclusive sets.

Given that $l \gt 0$, therefore, the least and the greatest element in $A \cap C$ are $99 + l$ and $990 + l$ respectively.

Therefore, sum of $3$- digit numbers of the form $9k + l$ is:

$\dfrac{100}{2} \times [(99 + l) + (990 + l)] $

Since $B$ and $C$ are mutually exclusive:

$\left(\dfrac{100}{2} \times [(99 + l) + (990 + l)]\right) + \left(\dfrac{100}{2} \times 1093\right) = 400 \times 274$

$\Rightarrow \dfrac{100}{2} \times [(99 + l) + (990 + l)] = 400 \times 274 - \dfrac{100}{2} \times 1093$

$\Rightarrow \dfrac{100}{2} \times [(99 + l) + (990 + l)] = \dfrac{100}{2} (8 \times 274 - 1093)$

$\Rightarrow 2l = 2192 - 1093 - 1089 = 10$

$\Rightarrow l = 5$

Problem 26 of 30

The minimum value of $\alpha$ for which the equation $\dfrac{4}{\sin x} + \dfrac{1}{1 - \sin x} = \alpha$ has at least one solution in $\left(0, \dfrac{\pi}{2}\right)$ is ___________.

Solution:

The given equation:

$\dfrac{4}{\sin x} + \dfrac{1}{1 - \sin x} = \alpha$

$\Rightarrow \dfrac{4 - 3 \sin x}{\sin x - \sin^2 x} = \alpha$ $...eqn(i)$

$\Rightarrow 4 - 3 \sin x = \alpha\sin x - \alpha\sin^2 x$

$\Rightarrow \alpha\sin^2 x - (3 + \alpha)\sin x + 4 = 0$

$\Rightarrow \sin x = \dfrac{(3 + \alpha) \pm \sqrt{(3 + \alpha)^2 - 16\alpha}}{2 \alpha}$

-----------book page break-----------

For the above equation to have real solution/s:

$(3 + \alpha)^2 \ge 16\alpha$

$\Rightarrow 9 - 10\alpha + \alpha^2 \ge 0$

$\Rightarrow (\alpha - 9)(\alpha - 1) \ge 0$

$\therefore \alpha \le 1$ or $\alpha \ge 9$

We can observe from the given equation that for the given range $\left(0, \dfrac{\pi}{2}\right)$,

$\dfrac{4}{\sin x} \gt 4$ and $\dfrac{1}{1 - \sin x} \gt 1$, therefore, $\alpha \gt 5$.

Therefore, the minimum value of $\alpha$ is $9$

Problem 27 of 30

If $\displaystyle{\int\limits_{-a}^{a}} (|x| + |x - 2|)dx = 22, (a \gt 22)$ and $\lfloor x \rfloor$ denotes the greatest integer $\le x$ then $\displaystyle{\int\limits_{a}^{-a}} (x + \lfloor x \rfloor)dx$ is equal to ________.

Solution:

$\displaystyle \int \limits_{-a}^a |x| + |x-2| dx = 22$

$\Rightarrow \displaystyle \int \limits_{-a}^0 (-2x + 2) dx + \displaystyle \int \limits_{0}^2 2 \ dx + \displaystyle \int \limits_{2}^a (2x - 2) dx = 22$

$\Rightarrow [-x^2 + 2x]_{-a}^0 + [2x]_0^2 + \displaystyle [x^2 - 2x]_2^a = 22$

$\Rightarrow a^2 + 4 + a^2 = 22$

-----------book page break-----------

$\Rightarrow 2a^2 = 18$

$\Rightarrow a = 3$

$\displaystyle{\int\limits_{a}^{-a}} (x + \lfloor x \rfloor)dx$

$=\displaystyle{\int\limits_{3}^{-3}} (x + \lfloor x \rfloor)dx$

$= - \displaystyle{\int\limits_{-3}^{3}} (x + \lfloor x \rfloor)dx$

$= -\left[ \displaystyle \int\limits_{-3}^{-2} (x + \lfloor x \rfloor)dx + \int\limits_{-2}^{-1} (x + \lfloor x \rfloor)dx + \int\limits_{-1}^{0} (x + \lfloor x \rfloor)dx \right.$

$\left.\displaystyle + \int\limits_{0}^{1} (x + \lfloor x \rfloor)dx + \int\limits_{1}^{2} (x + \lfloor x \rfloor)dx + \int\limits_{2}^{3} (x + \lfloor x \rfloor)dx \right]$

-----------book page break-----------

$= - \left[ \displaystyle \int\limits_{-3}^{-2} (x + (- 3))dx + \int\limits_{-2}^{-1} (x + (- 2))dx + \int\limits_{-1}^{0} (x + (-1))dx \right.$

$\left.\displaystyle + \int\limits_{0}^{1} (x + 0)dx + \int\limits_{1}^{2} (x + 1)dx + \int\limits_{2}^{3} (x + 2)dx \right]$

$= -\sum\limits_{n = -3}^{2} \left( \dfrac{(n + 1)^2}{2} - \dfrac{n^2}{2} + n \right)$

$= -\sum\limits_{n = -3}^{2} \left( \dfrac{4n + 1}{2} \right)$

$= -\sum\limits_{n = -3}^{2} 2n + 6 \times \dfrac{1}{2}$

$= -(-6 + 3) = 3$

Problem 28 of 30

If one of the diameters of the circle $x^2 + y^2 – 2x – 6y + 6 = 0$ is a chord of another circle $'C'$, whose centre is at $(2, 1),$ then its radius is __________

Solution:

We can rewrite the equation of the given circle as:

$(x - 1)^2 + (y - 3)^2 = 2^2$

This circle is centered at $(1, 3)$ with a radius of $2$. Let $r_c$ be the radius of the circle $'C'$

We can draw the diagram for this problem as follows:

The distance between the two centres is:

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

$= \sqrt{5}$

Therefore, we have ${r_c}^2 = (\sqrt{5})^2 + 2^2$

$\Rightarrow r_c = 3$

Problem 29 of 30

Let three vectors $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ be such that $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$, $\overrightarrow{a} \cdot \overrightarrow{c} = 7$ and $\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$, where $\overrightarrow{a} = -\hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + \hat{k}$, then the value of $2|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^2$ is _______.

Solution:

Using the conditions of vector coplanarity,

$\overrightarrow{c} = \lambda(\overrightarrow{b} \times (\overrightarrow{a} \times \overrightarrow{b}))$

$\Rightarrow \overrightarrow{c} = \lambda (\overrightarrow{b} \cdot \overrightarrow{b})\overrightarrow{a} - (\overrightarrow{a}\cdot\overrightarrow{b})\overrightarrow{b} $

$\Rightarrow \overrightarrow{c} = \lambda (5 (-\hat{i} + \hat{j} + \hat{k}) + 2\hat{i} + \hat{j})$

$\Rightarrow \overrightarrow{c} = \lambda (- 3 \hat{i} + 5 \hat{j} + 6 \hat{k})$

-----------book page break-----------

$\overrightarrow{c} \cdot \overrightarrow{a} = 7$

$\Rightarrow ( \lambda (- 3 \hat{i} + 5 \hat{j} + 6 \hat{k}))\cdot(-\hat{i} + \hat{j} + \hat{k}) = 7$

$\Rightarrow 3\lambda + 5 \lambda + 6\lambda = 7$

$\Rightarrow 14 \lambda = 7$

$ \Rightarrow \lambda = \dfrac{1}{2}$

$\overrightarrow{c} = \lambda (- 3 \hat{i} + 5 \hat{j} + 6 \hat{k}) = -\dfrac{3}{2} \hat{i} + \dfrac{5}{2} \hat{j} + 3 \hat{k}$

Therefore,

$2 |\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^2$

$=2 \left|(-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k}) + \left( -\dfrac{3}{2} \hat{i} + \dfrac{5}{2} \hat{j} + 3 \hat{k} \right) \right|$

$=2 \left|-\dfrac{1}{2}\hat{i} + \dfrac{7}{2}\hat{j} + 5\hat{k}\right|$

$= 2 \left( \sqrt{\dfrac{1}{4} + \dfrac{49}{4} + 25} \right)^2 = 75$

Problem 30 of 30

$\lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n}\tan^{-1}\left(\dfrac{1}{1 + r + r^2}\right)\right\}$ is equal to _______.

Solution:

$\lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n}\tan^{-1}\left(\dfrac{1}{1 + r + r^2}\right)\right\}$

$= \lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n}\tan^{-1}\left(\dfrac{(r + 1) - (r)}{1 + (r)(1 + r)}\right)\right\}$

Substituting $\tan \phi = r + 1$ and $\tan \theta = r$, we get:

$\lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n}\tan^{-1}\left(\dfrac{ \tan \phi - \tan \theta}{1 + \tan \phi \tan \theta } \right ) \right \}$

$= \lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n}\tan^{-1} [\tan (\phi - \theta)] \right\} $

-----------book page break-----------

$= \lim\limits_{n \rightarrow \infty}\tan \left\{\sum\limits_{r=1}^{n} (\phi - \theta) \right\} $

$= \tan \left( \lim \limits_{n \rightarrow \infty} \sum \limits_{r = 1}^n \left[ \tan^{-1}(r + 1) - \tan^{-1}(r) \right] \right) $

$= \tan \left( \lim \limits_{n \rightarrow \infty} \left[ \cancel{\tan^{-1}(2)} - \tan^{-1}(1) + \cancel{\tan^{-1}(3)} - \cancel{\tan^{-1}(2)} \right. \right.$

$\left. \left. + \cancel{\tan^{-1}(4)} - \cancel{\tan^{-1}(3)}... + \tan^{-1}(n+1) - \cancel{\tan^{-1}(n)}\right] \right)$

$= \tan \left( \lim \limits_{n \rightarrow \infty} \left( \tan^{-1} (n+1) - \dfrac{\pi}{4}\right) \right)$

$= \tan \left( \lim \limits_{n \rightarrow \infty} \tan^{-1} (n+1) - \dfrac{\pi}{4} \right)$

$= \tan \left(\dfrac{\pi}{2} - \dfrac {\pi}{4} \right)$

$= \tan \left( \dfrac{\pi}{4} \right)$

$= 1$