JEE Main Mathematics, Feb 25 2021, (Shift-1)

Problem 1 of 30

Let $f, g : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(n + 1) = f(n) + f(1) \forall n \in \mathbb{N}$ and $g$ be any arbitrary function. Which of the following statements is $NOT$ true?

Solution:

$f(n+1) = f(n) + 1$

$\therefore f(2) = f(1) + f(1) = 2 f(1)$

$f(3) = f(2) + f(1) = 2 f(1) + f(1) = 3 f(1)$

$f(4) = f(3) + f(1) = 3 f(1) + f(1) = 4 f(1)$

$.$

$f(n) = n f(1)$

$\because f: \mathbb{N} \rightarrow \mathbb{N}, f(1) \neq 0$

$\therefore f(n)$ is one-one, since one-one functions are functions in which each input has a single corresponding output.

Since $n$ is a positive integer and $f(n)$ is a positive integer such that $f(n) = n f(1)$, $f(n)$ increases with $n$

$f(2) = 2 f(1) $

$\therefore f(2) - f(1) = 2f(1) - f(1) = f(1)(1)$

If $f(1) \neq 1$ then $f(2) - f(1) > 1$

$\Rightarrow f(2) > f(1) + 1$

Then $f(1) + 1$ has no corresponding value in the domain.

Therefore, for $f(n)$ to be onto, $f(1) = 1$ and $f(n) = n$

Problem 2 of 30

Let the lines $(2 - i)z = (2 + i)\overline{z}$ and $(2 + i)z + (i - 2)\overline{z} - 4i = 0$, (here $i^2 = -1$) be normal to a circle $C$. If the line $iz + \overline{z} + 1 + i = 0$ is tangent to this circle $C$, then its radius is:

Solution:

Rewriting the first equation, we get:

$(2 + i)\overline{z} - (2 - i)z = 0$ $...eqn(i)$

Similarly from the second equation, we get:

$(2 + i)z + (i - 2)\overline{z} - 4i = 0$ $...eqn(ii)$

Since, these two lines are normal to a given circle, their intersection point is the center of the circle. Let this point $z_c = a + bi$

Since this is the intersection point, this point satisfies both equations.

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

From $eqn(i)$ we get:

$(2 + i)(a - bi) - (2 - i)(a + bi) = 0$

$\Rightarrow 2a + ai - 2bi - bi^2 - 2a + ai - 2bi + bi^2 = 0$

$\Rightarrow \cancel{2a} + ai - 2bi - \cancel{bi^2} - \cancel{2a} + ai - 2bi + \cancel{bi^2} = 0$

$\Rightarrow 2ai - 4bi = 0$

$\Rightarrow a - 2b = 0$

From $eqn(ii)$ we get:

$(2 + i)(a + bi) + (i - 2)(a - bi) - 4i = 0$

$\Rightarrow \cancel{2a} + ai + 2bi + \cancel{bi^2}+ ai - \cancel{2a} - \cancel{bi^2} + 2bi - 4i = 0$

$\Rightarrow (2a + 4b - 4)i = 0$

$\Rightarrow a + 2b - 2 = 0$ $...eqn(iv)$

Solving $eqn(iii)$ and $eqn(iv)$ we get:

$a = 1$ and $b = \dfrac{1}{2}$

The equation of the tangent, in terms on $x$ and $y$ on the imaginary plane can be written as:

$iz + \overline{z} + 1 + i = 0$

$i(x + yi) + (x - yi) + 1 + i = 0$

$\Rightarrow ix - y + x - yi + 1 + i = 0$

$\Rightarrow (i + 1)x - (i + 1)y + (i + 1) = 0$

$\Rightarrow x - y + 1 = 0$

Therefore, the radius, which is the distance of the center from the tangent, is given by:

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

Problem 3 of 30

The integer $'k'$, for which the inequality $x^2 - 2(3k - 1)x + 8k^2 - 7 > 0$ is valid for every $x$ in $\mathbb{R}$, is:

Solution:

Let $y = x^2 - 2(3k - 1)x + 8k^2 - 7$

To find a value of $k$ that satisfies the given inequality we need the curve of $y$, entirely above the $x$-axis, and since it is a strictly greater than inequality, the curve must not intersect the $x-axis$ at any point, that is the equation $x^2 - 2(3k - 1)x + 8k^2 - 7 = 0$ should not have any real root.

Taking the discriminant, $D < 0$

$\Rightarrow (2(3k-1))^2 - 4(8k^2 - 7) < 0$

$\Rightarrow 4(9k^2 - 6k + 1) - 4(8k^2 - 7) < 0$

$\Rightarrow k^2 - 6k + 8 0$

$\Rightarrow 2<k<4$

$\Rightarrow k = 3$

Problem 4 of 30

If $0 \lt \theta$, $\phi \lt \dfrac{\pi}{2}$, $x = \sum\limits_{n = 0}^{\infty} \cos^{2n} \theta$, $y = \sum\limits_{n = 0}^{\infty} \sin^{2n} \phi$ and $z = \sum\limits_{n = 0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$ then

Solution:

Using the Sum of Infinite Geometric Progression Formula from ,

$x = \sum\limits_{n = 0}^{\infty} \cos^{2n} \theta = 1 + \cos ^2 \theta + \cos ^4 \theta + ... = \dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta}$

$y = \sum\limits_{n = 0}^{\infty} \sin^{2n} \phi = 1 + \sin ^2 \phi + \sin ^4 \phi+ ... = \dfrac{1}{1 - \sin^2 \phi} = \dfrac{1}{\cos ^2 \phi}$

$z = \sum\limits_{n = 0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi = 1 + \sin^2 \phi \cos^2 \theta + \sin^4 \phi \cos^4 \theta + ... $

$= \dfrac{1}{1 - \sin^2 \phi \cos^2 \theta}$

$\therefore z= \dfrac{1}{1 - \left(1 - \dfrac{1}{y}\right)\left( 1 - \dfrac{1}{x}\right)}$

$\Rightarrow z = \dfrac{xy}{xy - (x - 1)(y - 1) }$

$\Rightarrow z = \dfrac{xy}{xy - (x - 1)(y - 1)}$

$\Rightarrow xz + yz - z = xy$

$\Rightarrow xy + z = (x + y) z$

Problem 5 of 30

If Rolle's theorem holds for the function $f(x) = x^3 - ax^2 + bx - 4$, $x \in [1, 2]$ with $f'\left(\dfrac{4}{3}\right) = 0$, then ordered pair $(a, b)$ is equal to:

Solution:

$f(x) = x^3 - ax^2 + bx - 4 $

$f'(x) = 3x^2 - 2ax + b$

$f'(\dfrac{4}{3}) = \dfrac{16}{3} - \dfrac{8}{3}a + b = 0$

$16 - 8a + b = 0$ $...(i)$

Since the function satisfies Rolle's Theorem in the interval $[1,2],$ $f(1) = f(2)$

$\therefore (1)^3 - a(1)^2 + b(1) - 4 = (2)^3 - a(2)^2 + b(2) - 4 $

$1 - a + b - 4 = 8 - 4a + 2b - 4$

$3a - b - 7 = 0$ $...(ii)$

Solving $(i)$ and $(ii)$ we obtain,

$a = 5,\ b = 8$

$\therefore (a,b) = (5,8)$

Problem 6 of 30

$\lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{1 + \dfrac{1}{2} + ... + \dfrac{1}{n}}{n^2}\right)^n$ is equal to:

Solution:

Let $\lim\limits_{n \rightarrow \infty}\dfrac{1 + \dfrac{1}{2} + ... + \dfrac{1}{n}}{n} = L$

Therefore,

$\lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{1 + \dfrac{1}{2} + ... + \dfrac{1}{n}}{n^2}\right)^n = \lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{L}{n}\right)^n$

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

$= \lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{L}{n}\right)^{\frac{n}{L} \times L}$

Let $\dfrac{n}{L} = m$, $\therefore$ as $n \rightarrow \infty$, $m \rightarrow \infty$

Therefore, we get the above limit as:

$\lim\limits_{m \rightarrow \infty}\left(1 + \dfrac{1}{m}\right)^{m \times L}$

$= \left[\lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{1}{m}\right)^{m}\right]^L$

$= e^L$ (using this concept of )

Now we need to see if $L$ exists and is finite or not,

$L = \lim\limits_{n \rightarrow \infty}\dfrac{1 + \dfrac{1}{2} + ... + \dfrac{1}{n}}{n}$

$\Rightarrow L = \lim\limits_{n \rightarrow \infty}\left(\dfrac{1}{n} + \dfrac{1}{2n} + ... + \dfrac{1}{n^2}\right)$

As $n \rightarrow \infty$ each of the terms in the above expression tends to $0$

Therefore, $L = 0$, therefore $L$ is finite and the limit exists.

$\lim\limits_{n \rightarrow \infty}\left(1 + \dfrac{1 + \dfrac{1}{2} + ... + \dfrac{1}{n}}{n^2}\right)^n = e^L = e^0 = 1$

Problem 7 of 30

The value of the integral

$\displaystyle{\int} \dfrac{\sin\theta \cdot \sin 2\theta \left(\sin^{6}\theta + \sin^{4}\theta + \sin^{2}\theta\right) \sqrt{2\sin^{4}\theta + 3\sin^{2}\theta + 6}}{1 - \cos 2\theta}d\theta$ is:

(where $c$ is a constant of integration)

Solution:

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

Let $\sin \theta = t \Rightarrow \cos \theta \ d\theta = dt $

$\displaystyle{\int} \dfrac{\sin\theta \cdot \sin 2\theta \left(\sin^{6}\theta + \sin^{4}\theta + \sin^{2}\theta\right) \sqrt{2\sin^{4}\theta + 3\sin^{2}\theta + 6}}{1 - \cos 2\theta}d\theta$

$= \displaystyle{\int} (t^6 + t^4 + t^2) \sqrt{2t^4 + 3t^2 + 6}\ dt$

$= \displaystyle{\int} (t^5 + t^3 + t) \sqrt{2t^6 + 3t^4 + 6t^2} dt$

Substituting $z = 2t^6 + 3t^4 + 6t^2$

$\Rightarrow \dfrac{dz}{dt} = 12(t^5 + t^3 + t)$

$\Rightarrow dz = 12(t^5 + t^3 + t) dt$

$\displaystyle{\int} (t^5 + t^3 + t) \sqrt{2t^6 + 3t^4 + 6t^2} dt$

$= \displaystyle{\int} \dfrac{1}{12} \dfrac{dz}{dt} \sqrt{z}\ dt$

$= \dfrac{1}{12} \displaystyle \int \sqrt{z}\ dz$

$= \dfrac{1}{18} {\Large{z}^{\large{\frac{3}{2}}}} + c$

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

$= \dfrac{1}{18} (2 \sin^6 \theta + 3 \sin^4 \theta + 6 \sin^2 \theta)^\frac{3}{2} + c$

$= \dfrac{1}{18}[(1 - \cos^2 \theta)(2\cos^4 \theta - 7 \cos^2 \theta + 11)]^\frac{3}{2} + c$

$= \dfrac{1}{18} (-2 \cos^6 \theta + 9 \cos^4 \theta - 18 \cos^2 \theta + 11)^\frac{3}{2} + c$

$= \dfrac{1}{18}\left[11 - 18\cos^{2}\theta + 9\cos^{4}\theta - 2\cos^{6}\theta \right]^{\frac{3}{2}} + c$

Problem 8 of 30

The value of $\displaystyle \int\limits_{-1}^{1} x^2e^{\lfloor x^3 \rfloor} dx$, where $\lfloor t \rfloor$ denotes the greatest integer $\le t$, is:

Solution:

$\lfloor x^3 \rfloor = \left\{\begin{array}{ll} -1 & \text{for } -1 \le x \lt 0 \\ 0 & \text{for } 0 \le x \lt 1 \end{array}\right.$

Therefore, we can split the given integral into two parts, as follows:

$\displaystyle \int\limits_{-1}^{1} x^2e^{\lfloor x^3 \rfloor} dx$

$= \displaystyle \int\limits_{-1}^{0} x^2e^{-1} dx + \int\limits_{0}^{1} x^2e^{0} dx$

$= \displaystyle \dfrac{1}{e}\int\limits_{-1}^{0} x^2 dx + \int\limits_{0}^{1} x^2 dx$

$= \displaystyle \dfrac{1}{e}\left[ \dfrac{x^3}{3}\right]_{-1}^{0} + \left[ \dfrac{x^3}{3} \right]_{0}^{1}$

$= \dfrac{1}{3e} + \dfrac{1}{3} $

$= \dfrac{e + 1}{3e}$

Problem 9 of 30

If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\dfrac{x^2 - 4x + y + 8}{x - 2}$, then this curve also passes through the point:

Solution:

$\dfrac{dy}{dx} = \dfrac{x^2 - 4x + y + 8}{x-2}$

$\Rightarrow \dfrac{dy}{dx} - \dfrac{y}{x-2} = \dfrac{x^2 - 4x + 8}{x-2}$

Using the method we get the factor as $e^{\int -\frac{1}{x-2}dx} = e^{-\log(x-2)} = \dfrac{1}{x-2}$

$\dfrac{dy}{dx} - \dfrac{y}{x-2} = \dfrac{x^2 - 4x + 8}{x-2}$

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

$\Rightarrow \dfrac{y}{x-2} = \displaystyle \int \dfrac{x^2 - 4x + 8}{(x-2)^2} dx$

$\Rightarrow \dfrac{y}{x-2} = \displaystyle \int \dfrac{(x-2)^2 + 4}{(x-2)^2} dx$

$\Rightarrow \dfrac{y}{x-2} = \displaystyle \int 1 dx + \displaystyle \int \dfrac{4}{(x-2)^2} dx$

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

$\Rightarrow y = x(x-2) - 4 + C(x-2)$

$\Rightarrow y = x^2 -2x -4 + C(x-2)$

Since the curve passes through the point $(0,0)$

$0 = -4 - 2C$

$\Rightarrow C =-2$

$\therefore y = x^2 - 2x - 4 -2x +4 = x^2 -4x$

The only point which satisfies the equation of the curve is $(5,5).$

Problem 10 of 30

The image of the point $(3, 5)$ in the line $x - y + 1 = 0$, lies on:

Solution:

The image of a point $(x_1, y_1)$ across a line $ax + by + c = 0$ is given by,

$\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = -2 \left( \dfrac{ax_1 + by_1 + c}{a^2 + b^2} \right)$

$\Rightarrow \dfrac{x - 3}{1} = \dfrac{y - 5}{b} = -2 \left( \dfrac{(1)(3) + (-1)(5) + 1}{(1)^2 + (-1)^2} \right)$

$\Rightarrow \dfrac{x - 3}{1} = \dfrac{y - 5}{-1} = 1$

$\Rightarrow x = 4$ and $y = 4$

The equation that satisfies these values of $x$ and $y$ is $(x - 2)^2 + (y - 4)^2 = 4$

Problem 11 of 30

A tangent is drawn to the parabola $y^2 = 6x$ which is perpendicular to the line $2x + y = 1$. Which of the following points does $NOT$ lie on it?

Solution:

The slope of the line $2x + y = 1$ is $-2$.

Since the tangent is perpendicular to the line, its slope is $\dfrac{1}{2}$

Differentiating both sides of the given equation of the parabola w.r.t $x$, we get:

$2y \dfrac{dy}{dx} = 6$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{3}{y}$

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

$\therefore \dfrac{3}{y} = \dfrac{1}{2}$

$\Rightarrow y = 6$

$\therefore x = \dfrac{y^2}{6} = 6$

Therefore, the equation of the tangent is given by:

$y - 6 = \dfrac{1}{2}(x - 6)$

$\Rightarrow 2y - 12 = x - 6$

$\Rightarrow 2y - x - 6 = 0$

From the given options only the point $(5, 4)$ does not satisfy the equation of the tangent.

Problem 12 of 30

If the curves, $\dfrac{x^2}{a} + \dfrac{y^2}{b} = 1$ and $\dfrac{x^2}{c} + \dfrac{y^2}{d} = 1$ intersect each other at an angle of $90^\circ$, then which of the following relations is $TRUE$?

Solution:

$\dfrac{x^2}{a} + \dfrac{y^2}{b} = 1$ $...(i)$

Differentiating w.r.t $x,$

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

$\Rightarrow \dfrac{dy}{dx} = - \dfrac{bx}{ay}$

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

$\dfrac{x^2}{c} + \dfrac{y^2}{d} = 1$ $...(ii)$

Differentiating w.r.t $x,$

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

$\Rightarrow \dfrac{dy}{dx} = - \dfrac{dx}{cy}$

Since the curves intersect at right angles,

$- \dfrac{dx}{cy} \times - \dfrac{bx}{ay} = -1$

$y^2 = - \dfrac{bd}{ac} x^2$

Substituting $y^2 = - \dfrac{bd}{ac} x^2$ in $(i),$

$\dfrac{x^2}{a} + \dfrac{\dfrac{bd}{-ac}x^2}{b}$

$\Rightarrow x^2 = \dfrac{ac}{c - d}$ $...(iii)$

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

Substituting $y^2 = - \dfrac{bd}{ac} x^2$ in $(ii),$

$\dfrac{x^2}{c} + \dfrac{\dfrac{bd}{-ac}x^2}{d}$

$x^2 = \dfrac{ac}{a - b}$ $...(iv)$

$(iii) = (iv)$

$\therefore \dfrac{ac}{a - b} = \dfrac{ac}{c - d} $

$\Rightarrow a - b = c - d$

Problem 13 of 30

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$. Then the value of $\sin^{4}\alpha + \cos^{4}\alpha$ is:

Solution:

$l^2 + m^2 + n^2 = 1$ $...(i)$

$l^2 + m^2 - n^2 = 0$ $...(ii)$

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

$2n^2 = 1$

$\Rightarrow n = \pm \dfrac{1}{\sqrt{2}}$ $...(iii)$

$l^2 + m^2 = \dfrac{1}{2}$ $...(iv)$

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

$l + m - n = 0 \Rightarrow l + m = n \Rightarrow l + m = \pm \dfrac{1}{\sqrt{2}}$

$\Rightarrow (l + m)^2 = \dfrac{1}{2}$

$\Rightarrow l^2 + 2lm + m^2 = \dfrac{1}{2}$

$\Rightarrow \dfrac{1}{2} + 2lm = \dfrac{1}{2}$

$\Rightarrow 2lm = 0$

$\Rightarrow l = 0, m = \dfrac{1}{\sqrt{2}}$ or $m = 0, l = \dfrac{1}{\sqrt{2}}$

Direction Ratios $= (l, m, n) = \left(\dfrac{1}{\sqrt{2}}, 0, \pm \dfrac{1}{\sqrt{2}}\right)$ or $\left(0, \dfrac{1}{\sqrt{2}}, \pm \dfrac{1}{\sqrt{2}}\right)$

$\Rightarrow \overrightarrow{a} = \dfrac{1}{\sqrt{2}}, 0, \pm \dfrac{1}{\sqrt{2}},\overrightarrow{b} = 0, \dfrac{1}{\sqrt{2}}, \pm \dfrac{1}{\sqrt{2}}$

$\cos \alpha = \dfrac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| \times |\overrightarrow{b}|} = \dfrac{1}{2}$

$\sin \alpha = \pm \dfrac{\sqrt{3}}{2}$

$\cos ^4 \alpha + \sin^4 \alpha = \dfrac{5}{8}$

Problem 14 of 30

The equation of the line through the point $(0, 1, 2)$ and perpendicular to the line $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z-1}{-2}$ is:

Solution:

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

$\Rightarrow x=2\lambda+1,\ y=3\lambda - 1,\ z = -2\lambda + 1$

$((2\lambda+1)\hat{i} + (3\lambda - 1)\hat{j} + (-2\lambda + 1)\hat{k}) \cdot (2 \hat{i} + 3\hat{j} - \hat{k}) = 0$

$\Rightarrow 4\lambda + 2 + 9 \lambda -6 + 4\lambda + 2 = 0$

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

$\Rightarrow 17\lambda = 2$

$\Rightarrow \lambda = \dfrac{2}{17}$

$x=2\lambda + 1 = \dfrac{21}{17}$

$y=3\lambda - 1 = -\dfrac{11}{17}$

$z = -2\lambda + 1 = \dfrac{13}{17}$

$\dfrac{x}{\dfrac{21}{17}-0} = \dfrac{y-1}{-\dfrac{11}{17}-1} = \dfrac{z-2}{\dfrac{13}{17}-2}$

$\Rightarrow \dfrac{x}{\dfrac{21}{17}} = \dfrac{y-1}{-\dfrac{28}{17}} = \dfrac{z-2}{-\dfrac{21}{17}}$

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

Problem 15 of 30

When a missile is fired from a ship, the probability that it is intercepted is $\dfrac{1}{3}$ and the probability that the missile hits the target, given that it is not intercepted, is $\dfrac{3}{4}$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

Solution:

The overall probability that a missile hits a target is:

$\text{(missile does not get intercepted)} \times \text{(missile hits target)}$

$= \dfrac{2}{3} \times \dfrac{3}{4} = \dfrac{1}{2}$

Therefore, the probability that all three missiles hit the target is:

$\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$

Problem 16 of 30

The coefficients $a, b$ and $c$ of the quadratic equation $ax^2 + bx + c = 0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is:

Solution:

The given equation will have equal roots only if the discriminant $b^2 - 4ac = 0$

$\therefore b^2 = 4ac$

We know that $b$ will be an integer between $1$ and $6$, therefore, $a$ and $c$ will have to be such that $4ac$ is a perfect square $\le 36$.

Since $4$ is a perfect square, $ac$ must be a perfect square $\le 9$

The perfect square $\le 9$ are,

$1, 4, 9$

Therefore, the possible values of $(a, c, b)$ are

$(1, 1, 2)$

$(4, 1, 4)$

$(1, 4, 4)$

$(2, 2, 4)$

$(3, 3, 6)$

Therefore, there are $5$ possible cases where the equation will have equal roots, and totally there are $6 \times 6 \times 6 = 216$ equally likely outcomes.

Therefore, the probability is $\dfrac{5}{216}$

Problem 17 of 30

All possible values of $\theta \in [0, 2\pi]$ for which $\sin 2\theta + \tan 2\theta > 0$ lie in:

Solution:

$\sin 2 \theta + \tan 2 \theta > 0$

$\tan 2\theta \cdot (1 + \cos 2 \theta) > 0$

$1 + \cos 2 \theta$ is always non-negative.

If $1 + \cos 2 \theta \ne 0$ then

$\cos 2 \theta \ne -1$

$\Rightarrow 2 \theta \ne (2n + 1)\pi$ where $n \in \mathbb{N_0}$

$\Rightarrow \theta \ne \dfrac{(2n + 1)\pi}{2}$ $...(i)$

Since $\tan 2 \theta \gt 0$ and if $\theta \in [0, 2 \pi]$, $2\theta \in [0, 4 \pi]$

$\therefore 2 \theta \in \left( 0,\dfrac{\pi}{2} \right] \cup \left(\pi,\dfrac{3\pi}{2}\right] \cup \left( 2 \pi, \dfrac{5 \pi}{2}\right] \cup \left( 3 \pi , \dfrac{7 \pi}{2}\right]$

$\theta \in \left( 0,\dfrac{\pi}{4} \right] \cup \left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right] \cup \left( \pi, \dfrac{5 \pi}{4}\right] \cup \left( \dfrac{3 \pi}{2} , \dfrac{7 \pi}{4}\right]$ $...(ii)$

Combining $(i)$ & $(ii)$

$\theta \in \left( 0,\dfrac{\pi}{4} \right) \cup \left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right) \cup \left( \pi, \dfrac{5 \pi}{4}\right) \cup \left( \dfrac{3 \pi}{2} , \dfrac{7 \pi}{4}\right)$

Problem 18 of 30

The total number of positive integral solutions $(x, y, z)$ such that $xyz = 24$ is:

Solution:

$24 = 2^3 \times 3^1$

We can make the $3$ as a factor of any one of $x, y$ or $z$. There are $3$ ways of doing it.

The three $2$s can be distributed between $x, y, z$ using the method , and there are:

$\xacomb{5}{2} = 10$ ways of doing the same.

Therefore, there are $10 \times 3 = 30$ ways to distribute all the factors among $x, y$ and $z$ each giving us a unique $3$-tuple.

Problem 19 of 30

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point $A$, with uniform speed. At the point, angle of depression of the boat with the man's eye is $30^\circ$ (Ignore the man's height). After sailing for $20$ seconds, towards the base of the tower (which is at the level of water), the boat has reached a point $B$, where the angle of depression is $45^\circ$. Then the time taken (in seconds) by the boat to reach the base of the tower is:

Solution:

We will use the following diagram to solve this problem:

Let the foot of the tower be at $P$ and the man's eye be at $M$.

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

Let the height of the tower, $(MP)$, be $a$

$\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{MP}{AP}$

$\Rightarrow AP = \sqrt{3}a $

$\tan 45^\circ = 1 = \dfrac{MP}{BP}$

$\Rightarrow BP = a$

$AB = (\sqrt{3} - 1)a$

$ v = \dfrac{(\sqrt{3} - 1) a}{20} $

Time taken to cover $BP = \dfrac{20}{(\sqrt{3} - 1) a} \times a = 10(\sqrt{3} + 1)$

Problem 20 of 30

The statement $A \rightarrow (B \rightarrow A)$ is equivalent to:

Solution:

Let's see for what input values the statement $A \rightarrow (B \rightarrow A) = 0.$

If $A \rightarrow (B \rightarrow A) = 0, A = 1, B \rightarrow A = 0.$

If $B \rightarrow A = 0, B = 1, A = 0.$

This is a contradiction as $A$ can't be both $0$ and $1,$ therefore, the given statement is a tautology.

Out of the given options only $A \rightarrow (A \lor B)$ is a tautology. Since both are tautologies, for all input combinations of $A$ and $B$ the output is $1$, therefore, they have the same truth table and are equivalent.

Problem 21 of 30

If $A = \begin{bmatrix} 0 & -\tan\left(\dfrac{\theta}{2}\right) \\ \tan\left(\dfrac{\theta}{2}\right) & 0\end{bmatrix}$ and $(I_2 + A)(I_2 - A)^{-1} = \begin{bmatrix} a & -b \\ b & a\end{bmatrix}$,

then $13(a^2 + b^2)$ is equal to

Solution:

$(I_2 - A)^{-1} = \dfrac{1}{1 + \tan^2 \left( \dfrac{\theta}{2} \right)} \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ - \tan\dfrac{\theta}{2} & 1 \end{bmatrix}$

$(I_2 + A) = \begin{bmatrix} 1 & - \tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix}$

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

$\therefore (I_2 + A)(I_2 - A)^{-1}$

$= \dfrac{1}{1 + \tan^2 \left( \dfrac{\theta}{2} \right)} \begin{bmatrix} 1 & - \tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ - \tan\dfrac{\theta}{2} & 1 \end{bmatrix}$

(using )

$= \dfrac{1}{1 + \tan ^2 \left( \dfrac{\theta}{2} \right)} \begin{bmatrix} 1 + \tan^2 \left( \dfrac{\theta}{2} \right) & \tan \left( \dfrac{\theta}{2} \right) - \tan \left( \dfrac{\theta}{2} \right) \\ \tan \left( \dfrac{\theta}{2} \right) - \tan \left( \dfrac{\theta}{2} \right) & 1 + \tan^2 \left( \dfrac{\theta}{2} \right) \end{bmatrix}$

$= \dfrac{1}{1 + \tan ^2 \left( \dfrac{\theta}{2} \right)} \begin{bmatrix} 1 + \tan^2 \left( \dfrac{\theta}{2} \right) & 0 \\ 0 & 1 + \tan^2 \left( \dfrac{\theta}{2} \right) \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$

$\Rightarrow a = 1$ and $b = 0$

$13(a^2 + b^2) = 13(1^2 + 0^2) = 13$

Problem 22 of 30

Let $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y\end{bmatrix}$, where $x, y$ and $z$ are real numbers such that $x + y + z \gt 0$ and $xyz = 2$. If $A^2 = I_3$, then the value of $x^3 + y^3 + z^3$ is ______.

Solution:

$A^2 = I_3$

$\Rightarrow |A|^2 = 1$

$\Rightarrow (x^3 + y^3 + z^3 - 3xyz)^2 = 1$

$\Rightarrow x^3 + y^3 + z^3 - 3xyz = 1$ or $-1$

By $AM - GM$ inequality for $x^3, y^3$ and $z^3,$

$\dfrac{x^3 + y^3 + z^3}{3} \geq \displaystyle \sqrt[3]{x^3y^3z^3}$

$\Rightarrow x^3 + y^3 + z^3 - 3xyz \geq 0$

$\therefore x^3 + y^3 + z^3 - 3xyz = 1$

$\Rightarrow x^3 + y^3 + z^3 - 6 = 1$

$\Rightarrow x^3 + y^3 + z^3 = 7$

Problem 23 of 30

If the system of equations

$\phantom{0000} kx + y + 2z = 1$

$\phantom{0000} 3x - y - 2z = 2$

$\phantom{0000} -2x - 2y - 4z = 3$

has infinitely many solutions, then $k$ is equal to ______.

Solution:

$\begin{bmatrix} k & 1 & 2 \\ 3 & -1 & -2 \\ -2 & -2 & -4\end{bmatrix} \begin{bmatrix}x \\ y \\ z \end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$

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

$B = \begin{bmatrix}x \\ y \\ z \end{bmatrix}$

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

For the given system of equations to have infinte solutions $|A| = 0$ and $adj(A)\times B = O$

$|A| = 0$

Using the method to find ,

$adj(A) = \begin{bmatrix} 0 & 0 & 0 \\ 16 & -4k + 4 & 2k + 6 \\ -8 & 2k - 2 & -k - 3\end{bmatrix}$

$\Rightarrow adj(A) \times B = \begin{bmatrix} 0 & 0 & 0 \\ 16 & -4k + 4 & 2k + 6 \\ -8 & 2k - 2 & -k - 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix} $

$\Rightarrow adj(A) \times B = \begin{bmatrix} 0 \\ -2k + 42 \\ k - 21 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\Rightarrow k = 21$

Problem 24 of 30

The total number of numbers, lying between $100$ and $1000$ that can be formed with the digits $1, 2, 3, 4, 5$, if the repetition of digits is not allowed and the numbers are divisible by either $3$ or $5$, is ______.

Solution:

The numbers between $100$ and $1000$ that can be formed using the digits $1, 2, 3, 4$ and $5$ (without repetition) have to be three-digit numbers.

Numbers divisible by $5:$

Since the last digit of number has to be $5,$ the first digit has $4$ options and the second digit has $3$ options,

$\therefore$ No. of numbers divisible by $5 = 4 \times 3 = 12$

Numbers divisible by $3:$

For the number to be divisible by $3,$ the sum of the digits has to be divisible by $3.$ The possible combination of digits are $(1,3,5), (1,2,3), (2,3,4), (3,4,5).$ Each of these combinations can form $6$ numbers each.

$\therefore$ No. of numbers divisible by $3 = 6 \times 4 = 24$

Numbers divisible by both $3$ and $5:$

Numbers divisible by $3$ and $5$ are $135, 315, 345, 435.$

$\therefore$ No. of numbers divisible by $3$ and $5 = 4$

$\therefore$ No. of numbers divisible by $3$ or $5 = 12 + 24 - 4 = 32$

Problem 25 of 30

Let $A_1, A_2, A_3, ...$ be squares such that for each $n \geqslant 1$, the length of the side of $A_n$ equals the length of diagonal of $A_{n+1}$. If the length of $A_1$ is $12$ cm, then the smallest value of $n$ for which area of $A_n$ is less than one, is ______.

Solution:

Let the area be $a_n$ for each $A_n$

Therefore, $a_{n+1} = \dfrac{1}{2} a_n$

$a_1 = 144, a_2 = 72, ...$

Let $a_t$ be the first term to be less than one in the given G.P. series.

$a_t = 144 \cdot \dfrac{1}{2^{t - 1}} \lt 1$

$\Rightarrow 144 \lt 2^{t-1}$

$\Rightarrow 288 \lt 2^{t}$

$\because 2^8 \lt 288 \lt 2^{9}$

$t = 9$

Problem 26 of 30

The number of points, at which the function $f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$, $x \in \mathbb{R}$ is not differentiable, is ______.

Solution:

We can inspect each of the functions inside the absolute symbol and find the possible points of intersection that may not be differentiable:

Let:

$u(x) = |2x + 1|$

$\Rightarrow u(x) = \left\{\begin{array}{ll} -2x - 1 & \text{for } x \lt -\dfrac{1}{2} \\ 2x + 1 & \text{for } x \ge -\dfrac{1}{2} \end{array} \right.$

Let $v(x) = 3|x + 2|$

$\Rightarrow v(x) = \left\{\begin{array}{ll} -3(x + 2) & \text{for } x \lt -2 \\ 3(x + 2) & \text{for } x \ge -2 \end{array} \right.$

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

Let $w(x) = |x^2 + x - 2|$

$\Rightarrow w(x) = \left\{\begin{array}{ll} -x^2 - x + 2 & \text{for } -2 \le x \le 1 \\ x^2 + x - 2 & \text{otherwise} \end{array} \right.$

Now, we can have the function definition in the ranges $(-\infty, -2)$, $[-2, -\dfrac{1}{2}]$, $(-\dfrac{1}{2}, 1]$ and $(1, +\infty)$

We can define $f(x)$ as:

$f(x) = \left\{\begin{array}{ll}-(2x+1) + 3(x+2) + (x^2 + x - 2) & \text{for } x \in (-\infty, -2) \\ -(2x+1) - 3(x+2) - (x^2 + x - 2) & \text{for } x \in (-2, - \dfrac{1}{2}) \\ (2x+1) - 3(x+2) - (x^2 + x - 2) & \text{for } x \in (-\dfrac{1}{2}, 1) \\ (2x+1) - 3(x+2) + (x^2 + x - 2) & \text{for } x \in (1, \infty) \end{array} \right.$

$= \left\{\begin{array}{ll}x^2 + 2x + 3 & \text{for } x \in (-\infty, -2) \\ -x^2 - 6x - 5 & \text{for } x \in (-2, - \dfrac{1}{2}) \\ -x^2 - 2x - 3 & \text{for } x \in (-\dfrac{1}{2}, 1) \\ x^2 - 7 & \text{for } x \in (1, \infty) \end{array} \right.$

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

Differentiating $f(x)$ in these ranges, we get:

$f'(x) = \left\{\begin{array}{ll}2x + 2 & \text{for } x \in (-\infty, -2) \\ -2x - 6 & \text{for } x \in (-2, - \dfrac{1}{2}) \\ -2x - 2 & \text{for } x \in (-\dfrac{1}{2}, 1) \\ 2x & \text{for } x \in (1, \infty) \end{array} \right.$

Left side derivative and right side derivative at $x = -2$, are $-2$ and $-2$ respectively. Therefore $f'(-2)$ exists, and $f(x)$ is differentiable at $x = -2$.

Left side derivative and right side derivative at $x = -\dfrac{1}{2}$, are $-5$ and $-1$ respectively. Therefore $f'(-\dfrac{1}{2})$ does not exist.

Left side derivative and right side derivative at $x = 1$, are $-4$ and $+2$ respectively. Therefore $f'(1)$ does not exist.

Therefore, the number of points where $f(x)$ is not differentiable for $x \in \mathbb{R}$ is $2$

Problem 27 of 30

Let $f(x)$ be a polynomial of degree $6$ in $x$, in which the coefficient of $x^6$ is unity and it has extrema at $x = -1$ and $x = 1$. If $\lim\limits_{x \rightarrow 0} \dfrac{f(x)}{x^3} = 1$, then $5\cdot f(2)$ is equal to ______.

Solution:

Since the limit $\lim\limits_{x \rightarrow 0} \dfrac{f(x)}{x^3}$ exists, the coefficients of the terms $x^2$ to $x^0$ must be $0$.

Therefore, the polynomial is of the form $x^6 + ax^5 + bx^4 + cx^3$

$\lim\limits_{x \rightarrow 0} \dfrac{f(x)}{x^3} = x^3 + ax^2 + bx + c = 1$

$\Rightarrow c = 1$

$f'(x) = \dfrac{d }{dx}f(x) = 6x^5 + 5ax^4 + 4bx^3 + 3x^2$

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

$f'(-1) = 0$

$\Rightarrow 6(-1)^5 + 5a(-1)^4 + 4b(-1)^3 + 3(-1)^2 = 0$

$\Rightarrow -6 + 5a - 4b + 3 = 0$

$\Rightarrow 5a - 4b = 3$ $...eqn(i)$

$f'(1) = 0$

$\Rightarrow 6(1)^5 + 5a(1)^4 + 4b(1)^3 + 3(1)^2 = 0$

$\Rightarrow 6 + 5a + 4b + 3 = 0$

$\Rightarrow 5a + 4b = -9$ $...eqn(ii)$

$eqn(ii) + eqn(i)$ gives:

$10a = -6 \Rightarrow a = -\dfrac{3}{5}$

$eqn(ii) - eqn(i)$ gives:

$8b = -12 \Rightarrow b = -\dfrac{3}{2}$

Therefore,

$f(x) = x^6 - \dfrac{3}{5}x^5 - \dfrac{3}{2}x^4 + x^3$

$\therefore f(2) = 2^6 - \dfrac{3}{5}2^5 - \dfrac{3}{2}2^4 + 2^3$

$\Rightarrow f(2) = 64 - \dfrac{3}{5} \times 32 - 24 + 8 = 48 - \dfrac{3}{5} \times 32$

Therefore,

$5\cdot f(2) = 5\left(48 - \dfrac{3}{5} \times 32 \right) = 16 (15 - 6) = 144$

Problem 28 of 30

The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area $A$.

Then $A^4$ is equal to ______.

Solution:

The intersection points of $\sin x$ and $\cos x$ are

$\dfrac{\pi}{4}$ and $\dfrac{5\pi}{4}$

The area enclosed by the two curves between these two points is:

$\displaystyle \int\limits_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin x dx - \int\limits_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos x dx$

$= -\cos x - \sin x \biggr\rvert_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$

$= -\cos \left( \dfrac{5 \pi}{4} \right) - \sin \left( \dfrac{5 \pi}{4} \right) + \cos \left( \dfrac{\pi}{4} \right) + \sin \left( \dfrac{\pi}{4} \right)$

$= -\cos \left( \pi + \dfrac{\pi}{4} \right) - \sin \left(\pi + \dfrac{\pi}{4} \right) + \cos \left( \dfrac{\pi}{4} \right) + \sin \left( \dfrac{\pi}{4} \right)$

$= \cos \left( \dfrac{\pi}{4} \right) + \sin \left( \dfrac{\pi}{4} \right) + \cos \left( \dfrac{\pi}{4} \right) + \sin \left( \dfrac{\pi}{4} \right)$

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

$\therefore A^4 = 64$

Problem 29 of 30

The locus of the point of intersection of the lines $\sqrt{3}kx + ky - 4\sqrt{3} = 0$ and $\sqrt{3}x - y - 4(\sqrt{3})k = 0$ is a conic, whose eccentricity is ______.

Solution:

$\sqrt{3}kx + ky - 4\sqrt{3} = 0 ...(i)$

$\sqrt{3}kx -ky= 4\sqrt{3} k ...(ii)$

Adding equation $(i)$ & $(ii)$

$2\sqrt{3} kx = 4\sqrt{3}(k + 1)$

$\Rightarrow x = 2 \times \dfrac{k + 1}{k} ...(iii)$

$\Rightarrow 1 + \dfrac{1}{k} = \dfrac{x}{2}$

$\Rightarrow \dfrac{1}{k} = \dfrac{x}{2} - 1$

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

Subtracting equation $(ii)$ from $(i)$

$y = 2\sqrt{3} \times \dfrac{1 - k}{k} ...(iv)$

$\dfrac{1}{k} - 1 = \dfrac{y}{2 \sqrt{3}}$

$\dfrac{1}{k} = \dfrac{y}{2 \sqrt{3}} + 1$

Equating the values of $\dfrac{1}{k}$, we get:

$\dfrac{1}{k} = \dfrac{x}{2} - 1 = \dfrac{y}{2 \sqrt{3}} + 1$

$\Rightarrow \dfrac{x}{2} - 1 = \dfrac{y}{2 \sqrt{3}} + 1$

$\Rightarrow \dfrac{x^2}{4} - x + 1 = \dfrac{y^2}{12} - \dfrac{y}{\sqrt{3}} + 1$

$\Rightarrow \dfrac{x^2}{4} – \dfrac{y^2}{12} = x - \dfrac{y}{\sqrt{3}}$

$\Rightarrow \dfrac{x^2}{4} – \dfrac{y^2}{12} = 2 \left( 1 + \dfrac{1}{k} \right) - 2 \left(\dfrac{1}{k} - 1 \right)$

$\Rightarrow \dfrac{x^2}{4} – \dfrac{y^2}{12} = 4$

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

$\Rightarrow \dfrac{x^2}{16} - \dfrac{y^2}{48} = 1$

This is the standard form for the equation of a , with eccentricity $e$ as:

$e = \sqrt{1 + \dfrac{48}{16}}$

$\Rightarrow e = 2$

Problem 30 of 30

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\overrightarrow{b} = \hat{i} - \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$, then $\overrightarrow{r} \cdot \overrightarrow{a}$ is equal to ______.

Solution:

$\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$

$\Rightarrow \overrightarrow{r} \times \overrightarrow{a} - \overrightarrow{c} \times \overrightarrow{a} = 0$

$\Rightarrow (\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{a} = 0$

$\therefore \overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{a}$

$\Rightarrow \overrightarrow{r} = \lambda \overrightarrow{a} + \overrightarrow{c}$

$\overrightarrow{r} \cdot \overrightarrow{b} = \lambda \overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{c} \cdot \overrightarrow{b} = 0$

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

$\Rightarrow \lambda = 2$

$\overrightarrow{r} = 2 \overrightarrow{a} + \overrightarrow{c}$

$\overrightarrow{r} \cdot \overrightarrow{a} = 2 |\overrightarrow{a}|^2 + \overrightarrow{a}\cdot\overrightarrow{c} = 2(1 + 4 + 1) + (1 - 2 + 1) = 12$