Problem 1 of 30
If $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular then $\overrightarrow{a} \times \left(\overrightarrow{a} \times \left(\overrightarrow{a} \times \left(\overrightarrow{a} \times \overrightarrow{b}\right)\right)\right)$ is equal to:Solution:
Let the unit vectors in the direction of $\overrightarrow{a}$ and $\overrightarrow{b}$, be $\hat{a}$ and $\hat{b}$ respectively.Also, let the unit vector in the direction of $\hat{a} \times \hat{b}$ be $\hat{c}$.
Therefore, $\overrightarrow{a} = |\overrightarrow{a}| \hat{a}$ and $\overrightarrow{b} = |\overrightarrow{b}| \hat{b}$
Therefore,
$\overrightarrow{a} \times \left(\overrightarrow{a} \times \left(\overrightarrow{a} \times \left(\overrightarrow{a} \times \overrightarrow{b}\right)\right)\right)$
$= |\overrightarrow{a}| \hat{a} \times \left(|\overrightarrow{a}| \hat{a} \times \left(|\overrightarrow{a}| \hat{a} \times \left(|\overrightarrow{a}| \hat{a} \times |\overrightarrow{b}| \hat{b}\right)\right)\right)$
$= |\overrightarrow{a}|^4|\overrightarrow{b}| \left[\hat{a} \times \left(\hat{a} \times \left(\hat{a} \times \left(\hat{a} \times \hat{b}\right)\right)\right) \right]$
$= |\overrightarrow{a}|^4|\overrightarrow{b}| \left[\hat{a} \times \left(\hat{a} \times \left(\hat{a} \times \hat{c}\right)\right) \right]$
$= |\overrightarrow{a}|^4|\overrightarrow{b}| \left[\hat{a} \times \left(\hat{a} \times \left(- \hat{b} \right)\right) \right]$
$= |\overrightarrow{a}|^4|\overrightarrow{b}| \left[\hat{a} \times \left(-\hat{c}\right) \right]$
$= |\overrightarrow{a}|^4|\overrightarrow{b}| \left( \hat{b} \right)$
$= |\overrightarrow{a}|^4\overrightarrow{b}$
Problem 2 of 30
If $(1, 5, 35)$, $(7, 5, 5)$, $(1, \lambda, 7)$ and $(2\lambda, 1, 2)$ are coplanar, then the sum of all possible values of $\lambda$ is:Solution:
Let $P = (1, 5, 35)$, $Q = (7, 5, 5)$, $R = (1, \lambda, 7)$ and $S = (2\lambda, 1, 2)$
Since $P, Q, R$ and $S$ are coplanar,
$\begin{vmatrix} \overrightarrow{PQ}_x & \overrightarrow{PQ}_y & \overrightarrow{PQ}_z \\ \overrightarrow{PR}_x & \overrightarrow{PR}_y & \overrightarrow{PR}_z \\ \overrightarrow{PS}_x & \overrightarrow{PS}_y & \overrightarrow{PS}_z \end{vmatrix} = 0$
(Here $\overrightarrow{AB}_x$ denotes the $x$ -component of the vector $\overrightarrow{AB}$)
-----------book page break-----------
$\Rightarrow \begin{vmatrix} 6 & 0 & -30 \\ 0 & \lambda - 5 & -28 \\ 2 \lambda - 1 & -4 & -33 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} 1 & 0 & -5 \\ 0 & \lambda - 5 & -28 \\ 2 \lambda - 1 & -4 & -33 \end{vmatrix} = 0$
$\Rightarrow 1[(-33)(\lambda - 5) - (-28)(-4)] + (-5)[(0)(-4) - (\lambda - 5)(2 \lambda - 1)] = 0$
$\Rightarrow -33 \lambda + 165 - 112 + 10\lambda ^2 - 55 \lambda + 25 = 0$
$\Rightarrow 10 \lambda^2 - 88 \lambda + 78 = 0$
$\Rightarrow 5 \lambda^2 - 44 \lambda + 39 = 0$
$\Rightarrow (5 \lambda - 39)(\lambda - 1) = 0$
$\Rightarrow \lambda = \dfrac{39}{5}$ or $1$
Therefore, the sum of all possible values of $\lambda = \dfrac{39}{5} + 1 = \dfrac{44}{5}$
Problem 3 of 30
The intersection of three lines $x - y = 0$, $x + 2y = 3$ and $2x + y = 6$ is a :Solution:
No two lines have slopes which have a product of $-1,$ therefore the triangle formed is not a right-angled triangle. The intersection of the lines $x - y = 0$ and $x + 2y = 3, A$ is $(1,1).$
The intersection of the lines $x - y = 0$ and $2x + y = 6, B$ is $(2,2).$
The intersection of the lines $x +2y = 3$ and $2x + y = 6, C$ is $(3,0).$
$AB = \sqrt{2}$
$BC = \sqrt{5}$
$AC = \sqrt{5}$
Since $BC = AC \neq AB,$ $\triangle ABC$ is an isosceles triangle.
Problem 4 of 30
The value of $\begin{vmatrix} (a + 1)(a + 2) & a + 2 & 1 \\ (a + 2)(a + 3) & a + 3 & 1 \\ (a + 3)(a + 4) & a + 4 & 1 \end{vmatrix}$ is:Solution:
Using the of finding determinants:
$= (a+1)(a+2)\{(a+3) - (a+4)\} - (a+2)\{(a+2)(a+3)$
$- (a+3)(a+4)\} + (a+2)(a+3)(a+4) - (a+3)(a+3)(a+4)$
$= -(a+1)(a+2) + 2(a+2)(a+3) - (a+3)(a+4)$
$= -(a+1)(a+2) + (a+3)(a)$
$= -a^2 - 3a - 2 + a^2 + 3a$
$= -2$
Problem 5 of 30
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is $1000$ at initial time $t = 0$. The number of bacteria is increased by $20\%$ in $2$ hours. If the population of bacteria is $2000$ after $\dfrac{k}{\log_e\left(\dfrac{6}{5}\right)}$ hours, then $\left(\dfrac{k}{\log_e2}\right)^2$ is equal to:Solution:
Let $y$ be the number of bacteria, therefore $y = f(t)$
Given that:
$\dfrac{dy}{dt} \propto y$$\Rightarrow \dfrac{dy}{dt}= ky$
-----------book page break-----------
$\Rightarrow \dfrac{dy}{y} = k dt$
$\Rightarrow \log_e y = kt + C $
At $t=0$
$\log_e y_0 = C $
$\therefore \log_e y = kt + \log_e y_0$
$\Rightarrow kt = \log _e \dfrac{y}{y_0}$
As the number of bacteria increases by $20\%$ in $2$ hours,
$y = \dfrac{120}{100} y_0$
$\Rightarrow \dfrac{y}{y_0} = \dfrac{6}{5}$
$ kt = \log _e \dfrac{y}{y_0}$
$\Rightarrow 2k = \log _e \dfrac{6}{5}$
$\Rightarrow k = \dfrac{1}{2} \log _e \dfrac{6}{5}$
-----------book page break-----------
$\Rightarrow t = \dfrac{2 \log _e \dfrac{y}{y_0}}{\log _e \dfrac{6}{5}}$
$\Rightarrow \dfrac{k}{\log _e \dfrac{6}{5}} = \dfrac{2 \log _e \dfrac{2000}{1000}}{\log _e \dfrac{6}{5}}$
$\Rightarrow \log _e 2 = \dfrac{k}{2}$
$\left(\dfrac{k}{\log_e 2}\right)^2$
$= \left(\dfrac{k}{\dfrac{k}{2}}\right)^2$
$= (2)^2$
$= 4$
Problem 6 of 30
In the circle given below, let $OA = 1$ unit, $OB = 13$ units and $PQ \perp OB$. Then, the area of the area of the triangle $PQB$ (in square units) is: 
Solution:
[Although the problem does not mention anything about the $y$-axis being a tangent to the circle, from the given diagram we will need to conclude the same to be able to solve this problem.]
$AB = OB - OA = 13 - 1 = 12$
If we draw $\triangle OPB$ (not shown in the figure), then $\angle OPB = 90^\circ$
(If $y$-axis is a tangent, then $OB$ is a diameter of the circle)
$\therefore PA^2 = OA \times AB = 12$
$\Rightarrow PA = \sqrt{12} = 2\sqrt{3}$
$[\triangle PAB] = \dfrac{1}{2} PA \times AB = \dfrac{1}{2} 2\sqrt{3} \times 12 = 12\sqrt{3}$
$\triangle PAB \cong \triangle QAB$,
$\therefore [\triangle PQB] = [\triangle PAB] + [\triangle QAB] = 2 \times [\triangle PAB] = 24\sqrt{3}$
(In the above solution $[XYZ]$ denotes the area of the closed polygon $XYZ$)
Problem 7 of 30
The value of $\lim\limits_{h \rightarrow 0} 2\left\{\dfrac{\sqrt{3}\sin\left(\dfrac{\pi}{6} + h\right) - \cos\left(\dfrac{\pi}{6} + h\right)}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)}\right\}$Solution:
$\lim\limits_{h \rightarrow 0} 2\left\{\dfrac{\sqrt{3}\sin\left(\dfrac{\pi}{6} + h\right) - \cos\left(\dfrac{\pi}{6} + h\right)}{\sqrt{3}h\left(\sqrt{3}\cos h - \sin h\right)}\right\}$
$= \lim\limits_{h \rightarrow 0} 2\left\{\dfrac{\dfrac{\sqrt{3}}{2}\sin\left(\dfrac{\pi}{6} + h\right) - \dfrac{1}{2}\cos\left(\dfrac{\pi}{6} + h\right)}{\sqrt{3}h\left(\dfrac{\sqrt{3}}{2}\cos h - \dfrac{1}{2}\sin h\right)}\right\} \phantom{00} \left(\begin{array}{l} \text{by dividing both} \\ \text{numerator and} \\ \text{denominator by 2}\end{array}\right)$
-----------book page break-----------
$= \lim\limits_{h \rightarrow 0} 2\left\{\dfrac{\cos{\dfrac{\pi}{6}}\sin\left(\dfrac{\pi}{6} + h\right) - \sin{\dfrac{\pi}{6}}\cos\left(\dfrac{\pi}{6} + h\right)}{\sqrt{3}h\left(\cos{\dfrac{\pi}{6}}\cos h - \sin{\dfrac{\pi}{6}}\sin h\right)}\right\} \phantom{00} \left(\begin{array}{l} \text{by substituting } \\ \dfrac{\sqrt{3}}{2} = \cos{\dfrac{\pi}{6}} \\ \text{and } \dfrac{1}{2} = \sin{\dfrac{\pi}{6}} \end{array}\right)$
$=\lim\limits_{h \rightarrow 0} 2\left\{ \dfrac{\sin{\left(\dfrac{\pi}{6} + h - \dfrac{\pi}{6} \right)}}{\sqrt{3}h \cos\left(\dfrac{\pi}{6} + h\right)} \right\}$
$=\lim\limits_{h \rightarrow 0} 2\left\{ \dfrac{\sin{h}}{\sqrt{3}h \cos\left(\dfrac{\pi}{6} + h\right)} \right\}$
$=\lim\limits_{h \rightarrow 0} 2\left\{ \dfrac{\sin{h}}{h} \right\} \times \lim\limits_{h \rightarrow 0} \left\{ \dfrac{1}{\sqrt{3} \cos\left(\dfrac{\pi}{6} + h\right)} \right\}$
$=2 \times \left\{ \dfrac{1}{\sqrt{3} \cos\left(\dfrac{\pi}{6}\right)} \right\}$
$=2 \times \left\{ \dfrac{1}{\sqrt{3} \times \dfrac{\sqrt{3}}{2}}\right\}$
$= \dfrac{4}{3}$
Problem 8 of 30
The maximum slope of the curve $y = \dfrac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$ occurs at the point:Solution:
The slope of the given curve, $y = \dfrac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$ is:$s = \dfrac{dy}{dx} = 2x^3 - 15x^2 + 36x - 19$
The of the slope occurs at:
$\dfrac{ds}{dx} = 0$
$\Rightarrow 6x^2 - 30x + 36 = 0$
$\Rightarrow x^2 - 5x + 6 = 0$
$\Rightarrow x = 2$ and $x = 3$
To find which of these is the maxima and which is the minima, we need the second derivative of $s$,
$f''(s) = \dfrac{d^2s}{dx^2} = 2x - 5$
$f''(3) = 1$, which means this is the minima, and $f''(2) = -1$ therefore, this is the maxima.
The value of $y$ at $x = 2$ is:
$y = \dfrac{1}{2}(2)^4 - 5(2)^3 + 18(2)^2 - 19(2)$
$\Rightarrow y = 8 - 40 + 72 - 38 = 2$
Therefore, the maximum slope of the given curve occurs at $(2, 2)$
Problem 9 of 30
The value of $\displaystyle{\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}} \dfrac{\cos^2 x}{1 + 3^x} dx$ is:Solution:
Let $I = \displaystyle \int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 3^x} \ dx$
$\Rightarrow I = \displaystyle \int \limits_{-\frac{\pi}{2}}^0 \dfrac{\cos^2 x}{1 + 3^{x}} \ dx + \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x}\ dx$
-----------book page break-----------
Substituting $x = -z$ in the first part of the integral we get:
$I = \displaystyle \int \limits_{\frac{\pi}{2}}^0 \dfrac{\cos^2 (-z) }{1 + 3^{-z}} (-dz) + \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x} \ dx$
$\Rightarrow I = \displaystyle \int \limits_0^{\frac{\pi}{2}} \dfrac{\cos^2 (-z) }{1 + 3^{-z}} \ dz + \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x} \ dx$
$\Rightarrow I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 z}{1 + 3^{-z}} \ dz+ \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x} \ dx$
Ignoring the variable name and replacing $z$ with $x$, we get:
$I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^{-x}} \ dx+ \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x} \ dx$
$\Rightarrow I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{1 + 3^x} + \dfrac{\cos^2 x}{1 + 3^{-x}} dx$
$\Rightarrow I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \cos ^2 x dx$
-----------book page break-----------
$\Rightarrow I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos 2x + 1}{2} dx$
$\Rightarrow I = \left[\dfrac{1}{2} \times \dfrac{\sin 2x}{2} + \dfrac{x}{2} \right]_0^{\frac{\pi}{2}}$
$= \dfrac{\pi}{4}$
Problem 10 of 30
The number of seven digit integers with sum of the digits equal to $10$ and formed by using the digits $1, 2$ and $3$ only is:Solution:
This problem can be done by brute force searching, but here we will see a more mathematical way of solving it.
Let there be $a$ $1$s, $b$ $2$s and $c$ $3$s.
Therefore,
$a + b + c = 7 \Rightarrow a = 7 - b - c$ and
$a + 2b + 3c = 10$
$\Rightarrow 7 - b - c + 2b + 3c = 10$
$\Rightarrow b + 2c = 3$
The only possible integer solutions for $(b, c)$ are $(3, 0)$ and $(1, 1)$
Therefore, the possible values of $(a, b, c)$ are:
$(4, 3, 0)$ and $(5, 1, 1)$
The number of permutations for each of these cases are:
$\dfrac{7!}{4!3!}$ and $\dfrac{7!}{5!}$, that is:
$35$ and $42$.
Therefore, there are $77$ possible numbers satisfying the given conditions.
Problem 11 of 30
The maximum value of the term independent of $'t'$ in the expansion of $\left(\large{tx}^{\frac{1}{5}} + \dfrac{(1-x)^{\frac{1}{10}}}{t}\right)^{10}$ where $x \in (0, 1)$ is:Solution:
$\large T_{r+1}$
$= \large \xacomb{10}{r} \cdot ({tx}^{\frac{1}{5}})^{10 - r} \cdot \left( \dfrac{{(1 - x)^{\frac{1}{10}}}}{t} \right) ^r$
$= \large \xacomb{10}{r} \cdot t^{10 - 2r} \cdot x ^\frac{10 - r}{5} \cdot (1 - x)^{\frac{r}{10}}$
Since we need $10 - 2r = 0 \Rightarrow r = 5$
$T_6 = \xacomb{10}{5} \cdot x \cdot \sqrt{1 - x}$
$\dfrac{d (T_6)}{dx} = \xacomb{10}{5} \left( \sqrt{1 - x} - \dfrac{x}{2\sqrt{1 - x}} \right) = 0$
$\Rightarrow 1 - x =\dfrac{x}{2}$
$\Rightarrow x = \dfrac{2}{3}$
Evaluating $T_6$ for $x = \dfrac{2}{3},$ we obtain $T = \dfrac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$
Problem 12 of 30
Let $R = \{(P, Q) | P$ and $Q$ are at the same distance from the origin$\}$ be a relation, then the equivalence class of $(1, -1)$ is the set:Solution:
Since $P$ and $Q$ are equidistant from the origin, they lie on a circle with centre $(0,0).$$S = \{(x, y) | x^2 + y^2 = k\}$
Since the equation must satisfy $(1,-1)$
$k = x^2 + y^2$
$\Rightarrow k = (1)^2 + (-1)^2$
$\Rightarrow k = 2$
$\therefore S = \{(x, y) | x^2 + y^2 = 2\}$
Problem 13 of 30
Let $A$ be a symmetric matrix of order $2$ with integer entries. If the sum of the diagonal elements of $A^2$ is $1$, then the possible number of such matrices is:Solution:
$A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$
$\therefore A^2 = \begin{bmatrix} a^2 + b^2 & ab + bc \\ ab + bc & b^2 + c^2 \end{bmatrix}$
Given that:
$a^2 + 2b^2 + c^2 = 1$
Possible integer solutions for $a, b , c$ are:
$a = \pm 1, b = 0, c = 0$
or
$a = 0, b = 0, c = \pm 1$
$\therefore $ There are $4$ possible symmetric matrices $A$ such that $A^2 = 1$
Problem 14 of 30
Let $f$ be any function defined on $\mathbb{R}$ and let it satisfy the condition:$\phantom{00} |f(x) - f(y)| \le |(x - y)^2|, \forall (x, y) \in \mathbb{R}$
If $f(0) = 1$, then:
Solution:
$|f (x) - f (y)| \leq |(x - y)^2|, \forall \ x, y \in \mathbb{R}$
$\Rightarrow \left|\dfrac{f (x) - f (y)}{(x - y)}\right| \leq |x - y|$
Taking the limit $x \rightarrow y$ on both sides we obtain,
$\lim\limits_{x \rightarrow y} \left|\dfrac{f (x) - f (y)}{(x - y)}\right| \leq 0$
$\Rightarrow |f’ (y)| ≤ 0$
-----------book page break-----------
Since, $|f'(y)| \nless 0,$
$\therefore f’ (y) = 0$
$\Rightarrow f (y) = c$
Since $f (0) = 1, f (y) = 1, \forall \ y \in R$
Therefore, $f(x) \gt 0, \forall x \in \mathbb{R}$
Problem 15 of 30
A fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to probability of getting $9$ heads, then the probability of getting $2$ heads is:Solution:
Let the number of times the coin has been tossed be $n.$
$P( 7 \ heads ) = \dfrac{\xacomb{n}{7}}{2^n}$$P( 9 \ heads ) = \dfrac{\xacomb{n}{9}}{2^n}$
Given: $P(7 \ heads) = P(9\ heads)$
$\therefore \dfrac{\xacomb{n}{7}}{2^n} = \dfrac{\xacomb{n}{9}}{2^n}$
$\Rightarrow \xacomb{n}{7} = \xacomb{n}{9}$
$\Rightarrow n= 16$
$P(2 \ heads) $
$= \dfrac{\xacomb{16}{2}}{2^{16}}$
$= \dfrac{16 \times 15}{2 \times 2^{16}}$
$= \dfrac{15}{2^{13}}$
Problem 16 of 30
If $\dfrac{\sin^{-1}x}{a} = \dfrac{\cos^{-1}x}{b} = \dfrac{\tan^{-1}y}{c}; 0 \lt x \lt 1$, then the value of $\cos\left(\dfrac{\pi c}{a + b}\right)$ is:Solution:
Let,
$\dfrac{\sin^{-1}x}{a} = \dfrac{\cos^{-1}x}{b} = \dfrac{\tan^{-1}y}{c} = k$
Therefore,
$x = \sin \left(ka\right)$, $x = \cos \left(kb\right)$ and $y = \tan \left(kc\right)$
-----------book page break-----------
Therefore,
$\sin \left(ka\right) = \cos \left(kb\right)$
$\Rightarrow \sin \left(ka\right) = \sin \left(\dfrac{\pi}{2} - kb\right)$
Therefore, $ka = \dfrac{\pi}{2} - kb$
$\Rightarrow ka + kb = \dfrac{\pi}{2}$
$\Rightarrow a + b = \dfrac{\pi}{2k}$
Therefore,
$\cos\left(\dfrac{\pi c}{a + b}\right) = \cos\left(\dfrac{\pi c}{\frac{\pi}{2k}}\right) = \cos (2kc)$
$= 2 \cos^2 (kc) - 1 = \dfrac{2}{\sec^2 (kc)} - 1 = \dfrac{2}{1 + \tan^2 (kc)} - 1$
$= \dfrac{1 - \tan^2 (kc)}{1 + \tan^2 (kc)} = \dfrac{1 - y^2}{1 + y^2}$
Problem 17 of 30
In an increasing geometric series, the sum of the second and the sixth term is $\dfrac{25}{2}$ and the product of the third and the fifth term is $25$, then the sum of $4\xasuper{th}$, $6\xasuper{th}$ and $8\xasuper{th}$ terms is equal to:Solution:
Let the first term and the common ratio be $a$ and $r$ respectively, where $r \gt 1$.
Therefore, based on the given conditions:
$ar + ar^5 = \dfrac{25}{2}$ and
$ar^2 \times ar^4 = 25$
$\Rightarrow a^2r^6= 25$
$\Rightarrow ar^3 = 5$
$\Rightarrow a = \dfrac{5}{r^3}$
-----------book page break-----------
Therefore,
$\dfrac{5}{r^3} \times r + \dfrac{5}{r^3} \times r^5 = \dfrac{25}{2}$
$\Rightarrow \dfrac{5}{r^2} + 5r^2 = \dfrac{25}{2}$
$\Rightarrow \dfrac{1}{r^2} + r^2 = \dfrac{5}{2}$
$\Rightarrow 2 + 2r^4 = 5r^2$
$\Rightarrow 2r^4 -5r^2 + 2= 0$
$\Rightarrow 2r^4 -4r^2 - r^2 + 2= 0$
$\Rightarrow (2r^2 - 1)(r^2 - 2) = 0$
$\therefore r = \pm \dfrac{1}{\sqrt{2}}$ or $r = \pm \sqrt{2}$
Since $r \gt 1 \Rightarrow r = \sqrt 2$
The sum of the given terms is:
$ar^3 + ar^5 + ar^7$
$= \dfrac{5}{r^3}r^3 + \dfrac{5}{r^3}r^5 + \dfrac{5}{r^3}r^7$
$= 5 + 5r^2 + 5r^4$
$= 5 + 10 + 20 = 35$
Problem 18 of 30
The sum of the infinite series $1 + \dfrac{2}{3} + \dfrac{7}{3^2} + \dfrac{12}{3^3} + \dfrac{17}{3^4} + \dfrac{22}{3^5} + ...$ is equal to:Solution:
Ignoring the first term, $1$, we get the rest of the series as:
$\dfrac{2}{3} + \dfrac{7}{3^2} + \dfrac{12}{3^3} + \dfrac{17}{3^4} + \dfrac{22}{3^5} +...$$= 2 \times \dfrac{1}{3} + 7 \times \dfrac{1}{3^2} + 12 \times \dfrac{1}{3^3} + ...$
$= \sum \limits_{i = 1}^{\infty} \{2 + (i - 1)5 \} \times \dfrac{1}{3} \times \left(\dfrac{1}{3}\right)^{i-1}$
-----------book page break-----------
Using the concept of Arithmetic-Geometric progression explained , we get
$a = 2, d = 5, b = \dfrac{1}{3}, r = \dfrac{1}{3}$
$k_1 = ab - bd = -1$, and $k_2 = db = \dfrac{5}{3}$
Sum of Arithmetic-Geometric Progression $= \dfrac{k_1}{1 - r} + \dfrac{k_2}{(1-r)^2} = \dfrac{-1}{\dfrac{2}{3}} + \dfrac{\dfrac{5}{3}}{(\dfrac{2}{3})^2} = -\dfrac{3}{2} + \dfrac{15}{4} = \dfrac{9}{4}$
Adding the initial $1$ back to the sum, we get:
$S = 1 + \dfrac{9}{4}$
$= \dfrac{13}{4}$
Problem 19 of 30
Consider the three planes$\phantom{0000} 3x + 15y + 21z = 9$
$\phantom{0000} x - 3y - z = 5$
$\phantom{0000} 2x + 10y + 14z = 5$
Then, which of the following is true?
Solution:
Since the direction ratios of the normals to $P_1 \ (3, 15, 21)$ and $P_3 \ (2, 10 , 14)$ are proportional, the normals are parallel and therefore the planes are parallel.
Problem 20 of 30
The value of $\displaystyle{\sum\limits_{n = 1}^{100} \int\limits_{n-1}^{n}} \large{e^{x - \lfloor x \rfloor}}dx$, where $\lfloor x \rfloor$ is the greatest integer $\le x$, is:Solution:
We can start by evaluating just one interval of the expression, let's say for $x \in [m, m+1)$:
We can write $x$ as $m + f$ where $m = \lfloor x \rfloor$ (the integer part) and $f = m - \lfloor x \rfloor$ (the fraction part)
Therefore, any single interval can be written as:
$\displaystyle \int\limits_{m}^{m + 1} e^{x - \lfloor x \rfloor} dx$
$= \displaystyle \int\limits_{0}^{1} e^{z} dz$
-----------book page break-----------
$= e^z {\biggr |}_{0}^{1}$
$= e^1 - e^0 = e - 1$
Therefore,
$\displaystyle{\sum\limits_{n = 1}^{100} \int\limits_{n-1}^{n}} e^{x - \lfloor x \rfloor}dx$
$= \displaystyle \sum\limits_{n = 1}^{100} (e - 1)$
$= 100(e - 1)$
Problem 21 of 30
The difference between degree and order of a differential equation that represents the family of curves $y^2 = a\left(x + \dfrac{\sqrt{a}}{2}\right), a \gt 0$ is ______.Solution:
$y^2 = a \left(x + \left(\dfrac{\sqrt{a}}{2}\right)\right)$
Differentiating w.r.t. $x$
$2y\times y’ = a$
$\therefore y^2 = 2y\times y’ \left[x + \left(\dfrac{\sqrt{2y \times y’}}{2} \right) \right]$
$\Rightarrow y = 2y’ \left[\left(x + \dfrac{\sqrt{2y \times y’}}{2}\right)\right]$
$\Rightarrow y - 2xy’ = \sqrt{2y \times y'} \times y’$
$\Rightarrow \left[y - 2x \left(\dfrac{dy}{dx}\right)\right]^2 = 2y \left(\dfrac{dy}{dx}\right)^3$
Degree $= 3$ and Order $= 1$
Degree - Order $= 3 - 1 = 2$
Problem 22 of 30
The sum of $162\xasuper{th}$ power of the roots of the equation $x^3 - 2x^2 + 2x - 1 = 0$ is ______.Solution:
We can see from the given cubic equation that $x = 1$ is a root of the equation, therefore $x - 1$ is a factor of the given polynomial.Now let us factorise the $LHS$ of the given equation:
$x^3 - 2x^2 + 2x - 1 = 0$
$\Rightarrow x^3 - x^2 - x^2 + x + x - 1 = 0$
$\Rightarrow x^2(x - 1) - x(x - 1) + 1(x - 1) = 0$
The quadratic equation $x^2 - x + 1 = 0$, has complex roots of the form $\dfrac{1 \pm \sqrt{3}i}{2}$ , which we know from , are the complex cube roots of of $-1$.
We can mark each of complex roots as $z_1$ and $z_2$, where $(z_1)^3 = -1$ and $(z_2)^3 = -1$
Therefore the sum of $162\xasuper{th}$ power of the roots are:
$(1)^{162} + (z_1)^{162} + (z_2)^{162}$$
$= 1 + ({z_1}^3)^{54} + ({z_2}^3)^{54}$
$= 1 + (-1)^{54} + (-1)^{54}$
$= 1 + 1 + 1 = 3$
Problem 23 of 30
The area bounded by the lines $y = ||x - 1| - 2|$ is ______.Solution:
The equation $y = ||x - 1| - 2|$ can represent the following four lines -
$y = x - 3$ $...(i)$
$y = 3 - x$ $...(ii)$
$y = x + 1$ $...(iii)$
$y = -x - 1$ $...(iv)$
Since $(i)$ and $(iii)$ have a slope of $1$and $(ii)$ and $(iv)$ have a slope of $-1$, the lines form a parallelogram.
Let $m_1$ be the slope of lines $(i)$ and $(iii)$ and let $m_2$ be the slope of the lines $(ii)$ and $(iv).$
Let $c_1, c_2, d_1$ and $d_2$ be the constants of the lines $(i), (iii), (ii)$ and $(iv)$ respectively.
-----------book page break-----------
$\therefore c_1 = 1, c_2 = -3, d_1 = 3, d_2 = -1, m_1 = 1, m_2 = -1$
Area of parallelogram
$= \left| \dfrac{(c_1 - c_2)(d_1 - d_2)}{(m_1 - m_2)} \right|$
$= \left| \dfrac{(1 - (-3))( 3 -(-1)}{(1 - (-1)} \right|$
$= \left| \dfrac{4 \times 4}{2} \right|$
$= 8$ sq. units
Problem 24 of 30
If $y = y(x)$ is the solution of the equation $e^{\sin y}\cos y \dfrac{dy}{dx} + e^{\sin y}\cos x = \cos x$, $y(0) = 0;$ then $1 + y\left( \dfrac{\pi}{6} \right) + \dfrac{\sqrt{3}}{2}y\left( \dfrac{\pi}{3} \right) + \dfrac{1}{\sqrt{2}}y\left( \dfrac{\pi}{4} \right)$ is equal to ______.Solution:
$e^{\sin y} \cos y \left(\dfrac{dy}{dx}\right) + e^{\sin y} \cos x = \cos x$
Substituting $e^{\sin y} = t$
$e^{\sin y } \times \cos y \left(\dfrac{dy}{dx} \right) = \dfrac{dt}{dx}$
Then, $\dfrac{dt}{dx} + t \cos x = \cos x$
Using the integrating factor method from , we get:
IF $= e^{\int \cos x \ dx} = e^{\sin x}$
-----------book page break-----------
Therefore the solution of the differential equation
$ t e^{\sin x} =\displaystyle \int e^{\sin x} \cos x dx$
$\Rightarrow e^{\sin y} . e^{\sin x} = e^{\sin x} + c$
At $x = 0, y = 0$
$1 = 1 + c \Rightarrow c = 0$
$\sin y + \sin x = \sin x$
$y = 0$
$y \dfrac{\pi}{6} = 0, y \dfrac{\pi}{3} = 0, y \dfrac{\pi}{4} = 0$
The required answer is $1 + 0 + 0 + 0 = 1.$
Problem 25 of 30
The number of solutions of the equation $\log_4(x - 1) = \log_2(x - 3)$ is ______.Solution:
$\log_4(x - 1) = \log_2(x - 3)$
$\Rightarrow \log_2(x - 1) \times \log_4 2 = \log_2(x - 3)$
$\Rightarrow \log_2(x - 1) \times \dfrac{1}{2} = \log_2(x - 3)$
$\Rightarrow \log_2(x - 1) = 2\log_2(x - 3)$
$\Rightarrow \log_2(x - 1) = \log_2(x - 3)^2$
$\Rightarrow (x - 1) = (x - 3)^2$
$\Rightarrow (x - 1) = x^2 - 6x + 9$
$\Rightarrow x^2 - 7x + 10 = 0$
$x = 5$ or $x = 2$
If $x = 2$, then $\log_2(x - 3) = \log_2 (-1)$ which is not defined.
Therefore, $x = 5$ and there is only one solution of the given equation.
Problem 26 of 30
The number of integral values of $'k'$ for which the equation $3\sin x + 4\cos x = k + 1$ has a solution, $k \in \mathbb{R}$ is ______.Solution:
$3 \sin x + 4 \cos x = k + 1$
The extrema will be at,
$\dfrac{d(3 \sin x + 4 \cos x)}{dx} = 0$
$\Rightarrow 3 \cos x - 4 \sin x = 0 $
$\Rightarrow 3 \cos x = 4 \sin x$
$\Rightarrow \dfrac{\sin x}{\cos x} = \dfrac{3}{4}$
-----------book page break-----------
$\Rightarrow \tan x = \dfrac{3}{4}$
$\Rightarrow \sin x = \pm \dfrac{3}{5}, \cos x = \pm\dfrac{4}{5}$
The maximum value of $f(x) = 3 \sin x + 4 \cos x$ is $f(x) = 3 \left( \dfrac{3}{5} \right) + 4 \left( \dfrac{4}{5} \right) = 5$
Similarly the minimum value of $f(x)$ is at $f(x) = 3 \left(- \dfrac{3}{5} \right) + 4 \left( -\dfrac{4}{5} \right) = -5$
Since $3 \sin x + 4 \cos x$ is a continuous function, it will assume all the integral values in the range $[-5, 5]$ for some real values of $x$.
$\therefore −5 \leq k + 1 \leq 5$
$\Rightarrow −6 \leq k \leq 4$
$-6,\ -5,\ -4,\ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4$ satisfy the given equation.
$\therefore 11$ values of $k$ satisfy the given equation.
Problem 27 of 30
Let $(\lambda, 2, 1)$ be a point on the plane which passes through the point $(4, -2, 2)$. If the plane is perpendicular to the line joining the points $(-2, -21, 29)$ and $(-1, -16, 23)$, then$\left(\dfrac{\lambda}{11}\right)^2 - \dfrac{4\lambda}{11} - 4$ is equal to ______.
Solution:
Let $A = (-2, -21, 29)$ and $B = (-1, -16, 23)$Therefore, the direction ratios of $AB$ are $(1, 5, -6)$
Let $P$ be the plane perpendicular to $AB$ and passing through $(4, -2, 2).$
Since $P$ is perpendicular to $AB$ the equation of $P$ is,
$P: 1x + 5y - 6z + k = 0$
As it passes through the point $(4, -2, 2)$
$1(4) + 5(-2) - 6(2) + k = 0$
$\Rightarrow k = 18$
$\Rightarrow P: 1x + 5y - 6z + 18 = 0$
As it passes throuht $(\lambda, 2, 1)$
$\lambda + 10 - 6 + 18 = 0$
$\Rightarrow \lambda = -22$
$\therefore \left( \dfrac{\lambda}{11} \right)^2 - \dfrac{4\lambda}{11} - 4 = \left( \dfrac{-22}{11} \right)^2 - \dfrac{4\times -22}{11} - 4 = 4 + 8 - 4 = 8 $
Problem 28 of 30
The value of the integral $\displaystyle{\int\limits_{0}^{\pi}}|\sin 2x|dx$ is ______.Solution:
Substituting $2x = z$, we get $dx = \dfrac{dz}{2}$ and the range $x = [0, \pi]$ $z = [0, 2 \pi]$, and we get the integral as:
$A = \displaystyle \int\limits_{0}^{2 \pi} \dfrac{1}{2} |\sin z| dz$
Since $0 \le \sin z \le 1$ for $0 \le z \le \pi$, and $0 \gt \sin z \ge -1$ for $\pi \lt z \le 2 \pi$ we can split the integration as:
$A = \displaystyle \int\limits_{0}^{\pi} \dfrac{1}{2} \sin z dz + \int\limits_{\pi}^{2 \pi} - \dfrac{1}{2} \sin z dz$
$= \left[ -\dfrac{1}{2}\cos z\right]_{0}^{\pi} - \left[ -\dfrac{1}{2}\cos z\right]_{\pi}^{2\pi}$
$= -\dfrac{1}{2}(-1 - 1) + \dfrac{1}{2}(1 - (-1))$
$= 2$
Problem 29 of 30
If $\sqrt{3} (\cos^{2}x) = (\sqrt{3} - 1)\cos x + 1$, the number of solutions of the given equation when $x \in \left[0, \dfrac{\pi}{2}\right]$ is ______.Solution:
The given equation is a quadratic equation in $\cos x$, and can be written as:
$\sqrt{3} (\cos^{2}x) - (\sqrt{3} - 1)\cos x - 1 = 0$
Therefore, $\cos x = \dfrac{(\sqrt{3} - 1) \pm \sqrt{(\sqrt{3} - 1)^2 + 4\sqrt{3}}}{2\sqrt{3}}$
$= \dfrac{(\sqrt{3} - 1) \pm \sqrt{(\sqrt{3} + 1)^2}}{2\sqrt{3}}$
$= \dfrac{(\sqrt{3} - 1) \pm (\sqrt{3} + 1)}{2\sqrt{3}}$
$= 1$ or $-\dfrac{1}{\sqrt{3}}$
$\because x \in \left[0, \dfrac{\pi}{2}\right]$
$\cos x \ne -\dfrac{1}{\sqrt{3}} \therefore \cos x = 1$
$\therefore x = 0$
Therefore, the given equation has only $one$ solution.
Problem 30 of 30
Let $m, n \in \mathbb{N}$ and $gcd(2, n) = 1$. If $\displaystyle{30{30\choose 0} + 29{30\choose 1}} + ... + 2{30\choose 28} + 1{30\choose 29} = n \cdot 2^m$, then $n + m$ is equal to ______.$\left(\text{Here} \displaystyle{n \choose k} = \xacomb{n}{k}\right)$
Solution:
We can write the given series as:
$\displaystyle{30{30\choose 0} + 29{30\choose 1}} + ... + 2{30\choose 28} + 1{30\choose 29}$
$= \displaystyle{30{30\choose 0} + 29{30\choose 1}} + ... + 2{30\choose 28} + 1{30\choose 29} + 0 {30 \choose 30}$
We can see that the $i\xasuper{th}$ term of the series is,
$\displaystyle (30 - i) \times {30 \choose {i}}$ where $i \longrightarrow 0...30$
-----------book page break-----------
For each $i$, (except $i = 15$),
$T_i + T_{30 - i} = \displaystyle (30 - i) \times {30 \choose {i}} + (i) \times {30 \choose {30 - i}}$
$= \displaystyle (30 - i) \times {30 \choose {i}} + (i) \times {30 \choose {i}}$
$= \displaystyle (30 - i + i)\times {30 \choose {i}}$
$= \displaystyle 30 \times {30 \choose i} = 15 \times {30 \choose i} + 15 \times {30 \choose {30-i}}$
For $i = 15$
$T_{15} = 15 \times \xacomb{30}{15}$
Therefore the given series can be rewritten as:
$\displaystyle 15{30\choose 0} + 15{30\choose 1} + ... + 15{30\choose 28} + 15{30\choose 29} + 15 {30 \choose 30}$
$= \displaystyle 15 \left\{{30\choose 0} + {30\choose 1} + ... + {30\choose 28} + {30\choose 29} + {30 \choose 30}\right\}$
$= 15 \times 2^{30}$ (because $\sum \limits_{r = 0}^{k} \xacomb{k}{r} = 2^k$)
Therefore, $n + m = 15 + 30 = 45$