Problem 1 of 30
If the mirror image of the point $(1, 3, 5)$ with respect to the plane $4x - 5y + 2z = 8$ is $(\alpha, \beta, \gamma)$, then $5(\alpha, \beta, \gamma)$ equals:Solution:
The direction ratios of the normal to $4x - 5y + 2z - 8 = 0$ are $(4, -5, 2)$The mirror image of the point $(1,3,5)$ satisfies the following condition,
$\dfrac{x - 1}{4} = \dfrac{y - 3}{-5} = \dfrac{z - 5}{2} = -2 \dfrac{4(1) - 5(3) + 2(5) - 8}{\sqrt{4^2 + (-5)^2 + 2^2}}$
$\Rightarrow \dfrac{x - 1}{4} = \dfrac{y - 3}{-5} = \dfrac{z - 5}{2} = \dfrac{2}{5}$
$\Rightarrow x = \dfrac{13}{5},\ y = 1, \ z = \dfrac{29}{5}$
Therefore,
$\alpha = \dfrac{13}{5}, \ \beta = 1, \ \gamma = \dfrac{29}{5}$
$\therefore 5(\alpha + \beta + \gamma) = 5 \left( \dfrac{13}{5} + 1 + \dfrac{29}{5} \right) = 47$
Problem 2 of 30
Let $A = \{1, 2, 3, ..., 10\}$ and $f:A \rightarrow A$ be defined as$f(k) = \left\{\begin{array}{cl} k + 1 & \text{if k is odd} \\ k & \text{if k is even} \end{array}\right.$
Then the number of possible functions $g:A \rightarrow A$ such that $g \circ f = f$ is:
Solution:
$f(1) = f(2) = 2$$f(3) = f(4) = 4$
$f(5) = f(6) =6$
$.$
$.$
$.$
$f(9) = f(10) = 10$
$\therefore g(2) = 2, g(4) = 4, g(6) = 6...g(10) = 10.$ Therefore, each of the even elements can be mapped in exactly way.
However, $g(1), \ g(3),\ g(5),\ g(7), \ g(9)$ can be mapped to any one of the $10$ elements in $A$.
Therefore, the total number of functions satisfying the given conditions = $10 \times 10 \times 10 \times 10 \times 10 = 10^5$
Problem 3 of 30
Let $A_1$ be the area of the region bounded by the curves $y = \sin x$, $y = \cos x$ and the $y$-axis in the first quadrant. Also, let $A_2$ be the area of the region bounded by the curves $y = \sin x$, $y = \cos x$, $x$-axis and $x = \dfrac{\pi}{2}$ in the first quadrant. Then,Solution:
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Refer to the diagram below:
$= [ \sin x]_0^{\frac{\pi}{4}} - [- \cos x]_0^{\frac{\pi}{4}}$
$= \dfrac{1}{\sqrt{2}} - \left( - \dfrac{1}{\sqrt{2}} + 1\right)$
$= \sqrt{2} - 1$
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$A_2 = \displaystyle \int \limits_0^{\frac{\pi}{4}} \cos x dx + \displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x dx$
$= [ \sin x]_0^{\frac{\pi}{4}} + [- \cos x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$= \left(1-\dfrac{1}{\sqrt{2}}\right) + \left( 1 - \dfrac{1}{\sqrt{2}} \right)$
$= 2 - \sqrt{2}$
$A_1 : A_2 = 1:\sqrt{2}$ and $A_1 + A_2 = 1$
Problem 4 of 30
If $0 \lt a, b \lt 1$, and $\tan^{-1}a + \tan^{-1}b = \dfrac{\pi}{4}$, then the value of $a + b - \dfrac{a^2 + b^2}{4} + \dfrac{a^3 + b^3}{3} - \dfrac{a^4 + b^4}{4} + ...$ is:
Solution:
Let $\tan \theta_1 = a,\ \tan \theta_2 = b$
$\tan^{-1} a + \tan^{-1} b = \dfrac{\pi}{4}$
$\Rightarrow \theta_1 + \theta_2 = \dfrac{\pi}{4}$
$\Rightarrow \tan (\theta_1 + \theta_2) = \tan \dfrac{\pi}{4} = 1$
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$\therefore \dfrac{\tan \theta_1 + \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} = 1$
$\Rightarrow \dfrac{a + b}{1 - ab} = 1$
$\Rightarrow a + b = 1 - ab$
$\Rightarrow (1 + a)(1 + b) = 2$
$(a + b) - \left ( \dfrac{a^2 + b^2}{2} \right ) + \left ( \dfrac{a^3 + b^3}{3} \right )...\infty$
$= \left (a - \dfrac{a^2}{2} + \dfrac{a^3}{3}... \right ) + \left (b - \dfrac{b^2}{2} + \dfrac{b^3}{3}... \right ) $
$= \ln (1 + a) + \ln (1 + b)$
$= \ln (1 + a)(1 + b)$
$= \ln 2$
Problem 5 of 30
Let slope of the tangent line to a curve at any point $P(x, y)$ be given by $\dfrac{xy^2 + y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$, then the value of $y$, for which the point $(3, y)$ lies on the curve, is:Solution:
Slope of the tangent $= \dfrac{xy^2 + y}{x}$$\Rightarrow \dfrac{dy}{dx} = \dfrac{xy^2 + y}{x}$
$\Rightarrow \dfrac{dy}{dx} = \dfrac{y}{x} + y^2$
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$\Rightarrow \dfrac{dy}{dx} - \dfrac{y}{x} = y^2$
$\Rightarrow y^{-2}\dfrac{dy}{dx} - \dfrac{y^{-1}}{x} = 1$
We will use the method to solve this problem, by using the substitution
$z=y^{-1}$ and $1-n=-1$
and the integrating factor as $x = e^{(1-n) \int -\frac{1}{x} dx}$
$zx = (1-n) \int x dx$
$\Rightarrow zx = \dfrac{-x^2}{2} + C$
$\Rightarrow z = \dfrac{-x}{2} + \dfrac{C}{x}$
$\Rightarrow \dfrac{1}{y} = \dfrac{-x}{2} + \dfrac{C}{x}$
Since the curve passes through the point $(-2,3)$
$\dfrac{1}{3} = \dfrac{-(-2)}{2} + \dfrac{C}{-2}$
$\Rightarrow \dfrac{1}{3} = 1 - \dfrac{C}{2}$
$\Rightarrow C = \dfrac{4}{3}$
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$\therefore \dfrac{1}{y} = \dfrac{-x}{2} + \dfrac{4}{3x}$
For $x = 3$
$\dfrac{1}{y} = \dfrac{-3}{2} + \dfrac{4}{3(3)}$
$\Rightarrow \dfrac{1}{y} = \dfrac{-3}{2} + \dfrac{4}{9}$
$\Rightarrow \dfrac{1}{y} = - \dfrac{19}{18}$
$\Rightarrow y = -\dfrac{18}{19}$
Problem 6 of 30
The sum of the series $\displaystyle{\sum\limits_{n = 1}^{\infty}} \dfrac{n^2 + 6n + 10}{(2n + 1)!}$ is equal to:Solution:
$\sum \limits_{n=1}^\infty \dfrac{n^2 + 6n + 10}{(2n+1)!}$$r = 2n + 1, \ r = 3, 5, 7$
$\sum \limits_{n = 1}^\infty \dfrac{n^2 + 6n + 1}{(2n+1)!}$
$= \dfrac{\sum \limits_{r = 3, 5...}^\infty \left ( \dfrac{r - 1}{2} \right )^2 +6 \times \dfrac{r - 1}{2} + 10}{r!}$
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$= \dfrac{\sum \limits_{r = 3,5...}^\infty \dfrac{r^2 - 2r + 1}{4} + 3r - 3 + 10}{ r!}$
$= \sum \limits_{r = 3,5...}^\infty \dfrac{r^2 - 2r + 1 + 12r - 12 + 40}{4r!}$
$\sum \limits_{r = 3,5...} \dfrac{r^2 + 10r + 29}{4r!}$
$= \dfrac{1}{4}\sum \limits_{r = 3,5...} \dfrac{(r^2 - r) + 11r + 29}{r!}$
$= \dfrac{1}{4} \sum \limits_{r = 3,5...}^\infty \dfrac{1}{(r - 2)!} + \dfrac{11}{(r - 1)!} + \dfrac{29}{r!}$
$= \dfrac{1}{4} \left [ \left ( \dfrac{1}{1!} + \dfrac{1}{3!} + \dfrac{1}{5!} ... \right) + 11 \left ( \dfrac{1}{2!} + \dfrac{1}{4!} + \dfrac{1}{6!} ... \right) + 29\left ( \dfrac{1}{3!} + \dfrac{1}{5!} + \dfrac{1}{7!} ... \right) \right ]$
$= \dfrac{1}{4} \left [ \left ( \dfrac{e - \dfrac{1}{e}}{2} \right ) + 11 \left ( \dfrac{e + \dfrac{1}{e} - 2}{2} \right ) + 29 \left ( \dfrac{e - \dfrac{1}{e} - 2}{2} \right ) \right ]$ (using the properties of )
$= \dfrac{41}{8}e - \dfrac{19}{8}e^{-1} - 10$
Problem 7 of 30
Let $f(x) = \displaystyle{\int\limits_{0}^{x}} e^t f(t) dt + e^x$ be a differentiable function for all $x \in \mathbb{R}$. Then $f(x)$ equals:Solution:
$f(x) = \displaystyle{\int\limits_{0}^{x}} e^t f(t) dt + e^x$
Substituting $g(t) = e^tf(t)$ we get:
$f(x) = \displaystyle \int \limits_0^x g(t)dt + e^x$
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Using the Leibniz integral rule:
$f'(x) = g(x)\cdot \dfrac{d}{dx}(x) - g(0). \dfrac{d}{dx} (0) + \displaystyle \int \limits_0^x \left(\dfrac{\partial }{\partial x} g(t) \right) dt + \dfrac{d}{dx}e^x$
$= g(x)\cdot \dfrac{d}{dx}(x) - g(0). \dfrac{d}{dx} (0) + \displaystyle \int \limits_0^x 0 dt + e^x$
$= g(x) + e^x$
$= e^x \cdot f(x) + e^x$
$= e^x(f(x) + 1)$
$\therefore \dfrac{f'(x)}{f(x) + 1} = e^x$ $...(i)$
Integrating both sides of $(i)$ w.r.t $x$, we get:
$\displaystyle \int \dfrac{f'(x)}{f(x) + 1} dx = \int e^x dx$
$\Rightarrow \ln(f(x) + 1) = e^x + c$ $...(ii)$
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Given that:
$f(x) = \displaystyle{\int\limits_{0}^{x}} e^t f(t) dt + e^x$
$\Rightarrow f(0) = \displaystyle{\int\limits_{0}^{0}} e^t f(t) dt + e^0 = 0 + 1 = 1$
Substituting $x = 0$ in $eqn(ii)$ we get:
$\ln(f(0) + 1) = e^0 + c$
$\Rightarrow \ln 2 = 1 + c$
$\Rightarrow c = \ln 2 - 1$
Substituting the above value of $c$ in $eqn(ii)$ we get:
$\ln(f(x) + 1) = e^x + \ln 2 - 1$
$\Rightarrow f(x) + 1 = e^{(e^x + \ln 2 - 1)}$
$\Rightarrow f(x) + 1 = e^{(e^x - 1)} \cdot e^{(\ln 2)}$
$\Rightarrow f(x) + 1 = 2 e^{(e^x - 1)}$
$\Rightarrow f(x) = 2 e^{(e^x - 1)} - 1$
Problem 8 of 30
Let $f(x)$ be a differentiable function at $x = a$ with $f'(a) = 2$ and $f(a) = 4$. Then $\lim\limits_{x \rightarrow a} \dfrac{xf(a) - af(x)}{x - a}$ equals:Solution:
Given:
$f'(a) = 2$$\therefore \lim\limits_{h \rightarrow 0} \dfrac{f(a+h) - f(a)}{h} = 2$
$\lim\limits_{h \rightarrow 0} \dfrac{f(a+h) - 4}{h} = 2$
$\lim\limits_{h \rightarrow 0} f(a+h) = \lim\limits_{h \rightarrow 0} 2h + 4$
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$f(x) = f(a + (x-a))$
$\therefore \lim\limits_{x-a \rightarrow 0} f(a+ (x-a)) = \lim\limits_{x-a \rightarrow 0} 2(x-a) + 4$
$= \lim\limits_{x \rightarrow a} 2x - 2a +4$
$\lim\limits_{x \rightarrow a} \dfrac{xf(a) - af(x)}{x - a}$
$=\lim\limits_{x \rightarrow a} \dfrac{ 4x - a(2x - 2a + 4)}{x - a}$
$=\lim\limits_{x \rightarrow a} \dfrac{ 4x - 2ax + 2a^2 - 4a}{x - a}$
$= \lim\limits_{x \rightarrow a} \dfrac{(4 - 2a)(x - a)}{x - a}$
$= 4 - 2a$
Problem 9 of 30
Let $f(x) = \sin^{-1}x$ and $g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim\limits_{x \rightarrow 2} g(x)$, then the domain of the function $f \circ g$ is:Solution:
The domain of $f \circ g$ includes all values of $x$ for which $g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6}$ lies in $[-1,1]$$g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6} = \dfrac{(x - 2)(x + 1)}{(x - 2)(2x + 3)} = \dfrac{x + 1}{2x + 3}$
$-1 \le \dfrac{x + 1}{2x + 3} \le 1$
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$\Rightarrow 0 \le \dfrac{(x+1)^2}{(2x + 3)^2} \le 1$
$\Rightarrow (x + 1)^2 \le (2x + 3)^2$
$\Rightarrow x^2 + 2x + 1 \le 4x^2 + 12x + 9 $
$\Rightarrow 3x^2 + 10x + 8 \ge 0$
$\Rightarrow 3x^2 + 6x + 4x + 8 \ge 0$
$\Rightarrow 3x(x + 2) + 4(x + 2) \ge 0$
$\Rightarrow (3x + 4)(x + 2) \ge 0$
$\Rightarrow x \ge - \dfrac{4}{3}$ or $x \le -2$
It is given that $g(2) = \lim \limits_{x \rightarrow 2} g(x)$
$= \dfrac{2 + 1}{3 \times 2 + 3}$
$= \dfrac{3}{7}$
Therefore, $g(2)$ is defined and is in the required range $[-1, 1]$ and the domain of $f \circ g$ is given by:
$x \in (- \infty, -2] \cup \left[ -\dfrac{4}{3}, \infty \right)$
Problem 10 of 30
Let $A(1, 4)$ and $B(1, -5)$ be two points. Let $P$ be a point on the circle $(x - 1)^2 + (y - 1)^2 = 1$ such that $(PA)^2 + (PB)^2$ have maximum value, then the points $P, A$ and $B$ lie on:Solution:
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Using the following diagram:
The coordinates of $P$ is $(1 + \cos \theta, 1 + \sin \theta)$
$\therefore PA = \sqrt{(1 - (1 + \cos \theta))^2 + (4 - (1 + \sin \theta))^2}$
$= \sqrt{(-\cos \theta)^2 + (3 - \sin \theta)^2}$
Similarly,
$PB^2 = \cos^2 \theta + (\sin \theta + 6)^2 = 37 + 12 \sin \theta$
$PA^2 + PB^2 = 47 + 6 \sin \theta$
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To maximise $PA^2 + PB^2$
$\sin \theta = 1$
$\Rightarrow \theta = \dfrac{\pi}{2}$
$\therefore P, A$ and $B$ lie on a straight line
Problem 11 of 30
If vectors $\overrightarrow{a_1} = x\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{a_2} = \hat{i} + y\hat{j} + z\hat{k}$ are collinear, then a possible unit vector parallel to the vector $x\hat{i} + y\hat{j} + z\hat{k}$ is:Solution:
Since $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ are collinear:$\dfrac{x}{1} = \dfrac{-1}{y} = \dfrac{1}{z} = \lambda$
Unit Vector $\parallel \ x\hat{i} + y\hat{j} + z\hat{k} = \dfrac{\lambda \hat{i} + \dfrac{-1}{\lambda} \hat{j} + \dfrac{1}{\lambda} \hat{k}}{\sqrt{\lambda^2 + \dfrac{2}{\lambda^2}}}$
For $\lambda = 1,$
$ \dfrac{\lambda \hat{i} + \dfrac{-1}{\lambda} \hat{j} + \dfrac{1}{\lambda} \hat{k}}{\sqrt{\lambda^2 + \dfrac{2}{\lambda^2}}} = \dfrac{1}{\sqrt{3}}\left(\hat{i} -\hat{j} + \hat{k}\right)$
Problem 12 of 30
Let $F_1(A, B, C) = (A \land \lnot B) \lor [\lnot C \land (A \lor B)] \lor \lnot A$ and $F_2(A, B) = (A \lor B) \lor (B \rightarrow \lnot A)$ be two logical expressions. Then:Solution:
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We will solve this problem using two different approaches, and then see which one is more efficient.
Approach 1:
In this approach, we will form the truth table for both the given expressions. These are as follows:
$F_1(A, B, C) = (A \land \lnot B) \lor [\lnot C \land (A \lor B)] \lor \lnot A$
| $A$ | $B$ | $C$ | $A \land \lnot B$ | $\lnot C \land (A \lor B)$ | $\lnot A$ | $[\lnot C \land (A \lor B)] \lor \lnot A$ | $(A \land \lnot B) \lor [\lnot C \land (A \lor B)] \lor \lnot A$ |
| $0$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ |
| $0$ | $0$ | $1$ | $0$ | $0$ | $1$ | $1$ | $1$ |
| $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ | $1$ |
| $0$ | $1$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ |
| $1$ | $0$ | $0$ | $1$ | $1$ | $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ | $1$ | $0$ | $0$ | $0$ | $1$ |
| $1$ | $1$ | $1$ | $0$ | $0$ | $0$ | $0$ | $0$ |
| $1$ | $1$ | $0$ | $0$ | $1$ | $0$ | $1$ | $1$ |
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$F_2(A, B) = (A \lor B) \lor (B \rightarrow \lnot A)$
| $A$ | $B$ | $A \lor B$ | $B \rightarrow \lnot A$ | $\lnot A$ | $(A \lor B) \lor (B \rightarrow \lnot A)$ |
| $0$ | $0$ | $0$ | $1$ | $1$ | $1$ |
| $0$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $1$ | $0$ | $1$ | $0$ | $0$ | $1$ |
| $1$ | $1$ | $1$ | $1$ | $0$ | $1$ |
Since $F_1$ has one case where the overall expression evaluates to $0$, it is not a tautology, while all possible combinations of inputs to $F_2$ evaluates to $1$, therefore the statement is a tautology.
Approach 2:
In this approach, we will analyse each expression, and work backwards to find at least one input combination that results into $0$. If we are able to find it, then the expression is not a tautology, but if there is no case where the expression evaluates to $0$, then it is a tautology.
Let us consider the first function:
$F_1(A, B, C) = (A \land \lnot B) \lor [\lnot C \land (A \lor B)] \lor \lnot A$
For this to be false $(0)$,
$\lnot A = 0$
$\Rightarrow A = 1$
$A \land \lnot B = 0$
$1 \land \lnot B = 0$
Therefore, $\lnot B = 0 \Rightarrow B = 1$
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$[\lnot C \land (A \lor B)] = 0$
$\Rightarrow \lnot C \land (1 \lor 0) = 0$
$\Rightarrow \lnot C \land 1 = 0$
$\Rightarrow \lnot C = 0$
$\Rightarrow C = 1$
The input set $A = 1, B = 1, C = 1$ gives a overall value of $0$ for the whole expression. Therefore, this is not a tautology.
For the second function:
$F_2(A, B) = (A \lor B) \lor (B \rightarrow \lnot A)$
$(B \rightarrow \lnot A) = 0$
Therefore,
$B = 1$ and $\lnot A = 0 \Rightarrow \lnot A = 1$
Therefore,
$(A \lor B)$
$= 1 \lor 1 = 1$
Therefore, it is never possible for this expression to evaluate to $0$, and hence, this is a tautology.
We can clearly make out that the second approach is much more efficient in terms of the number of calculations, therefore, is the recommended method.
Problem 13 of 30
A seven digit number is formed using digits $\{3, 3, 4, 4, 4, 5, 5\}$. The probability, that number so formed is divisible by $2$, is:Solution:
The number of possible $7$-digit numbers with the given digits is:$\dfrac{7!}{2!3!2!}$
The number will be divisible by $2$ only if the least significant digit is a $4$. Fixing $4$ as the least significant digit, the possible permutations of the remaining digits is:
$\dfrac{6!}{2!2!2!}$
Therefore, the probability of the number being even is:
$\dfrac{\dfrac{6!}{2!2!2!}}{\dfrac{7!}{2!3!2!}}$
$= \dfrac{6!}{2!2!2!} \times \dfrac{2!3!2!}{7!}$
$= \dfrac{3}{7}$
Problem 14 of 30
Consider the system of equations:$\phantom{0000} x + 2y - 3z = a$
$\phantom{0000} 2x + 6y - 11z = b$
$\phantom{0000} x - 2y + 7z = c$
where $a, b$ and $c$ are real constants. Then the system of equations:
Solution:
Given:
$\begin{bmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7\end{bmatrix}\begin{bmatrix} x\\ y \\ z\end{bmatrix} = \begin{bmatrix} a\\ b \\ c\end{bmatrix}$-----------book page break-----------
$A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7\end{bmatrix}, \ B = \begin{bmatrix} a\\ b \\ c\end{bmatrix}$
$|A|$
$= 1((6)(7) - (-2)(11)) - 2((2)(7) - (1)(-11)) + (-3)((2)(-2) - (1)(6))$
$ = 20 - 50 + 30 = 0$
$\therefore$ The given equation doesn't have a unique solution for any value of $a, b , c.$
$adj(A) = \begin{bmatrix} 20 & -8 & -4 \\ -25 & 10 & 5 \\ -10 & 4 & 2\end{bmatrix}$
$adj(A) \times B = \begin{bmatrix} 20a - 8b - 4c\\ -25a + 10b + 5c \\ -10a + 4b +2c\end{bmatrix}$
$adj(A) \times B = O$ if $5a = 2b + c$
Since, $|A| = 0$ and $adj(A) \times B = O$ if $5a = 2b + c,$ the given system of equations has an infinite number of solutions.
Problem 15 of 30
The triangle of maximum area that can be inscribed in a given circle of radius $'r'$ is:Solution:
Comparing all the given options, we get:
Case A:
An equilateral triangle, inscribed in a circle of radius $r$, will have a side length of $\sqrt{3} r$ and a height of $\dfrac{3r}{2}$. Therefore, the area will be $\dfrac{3\sqrt{3} r^2}{4}$
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Case B:
An isosceles triangle with one side $= 2r$, will be a right angle triangle with the base $= 2r$ and the height as $r$.
Therefore, this will have an area of $r^2$
Case C:
Equilateral triangle with height $= \dfrac{2r}{3}$.
If we have an equilateral triangle with all three of its vertices on the circle, its height will always be $\dfrac{3r}{2}$. Therefore, this equilateral triangle does not exist.
Case D:
A right angle triangle with sides $2r$ and $r$ as its sides, will have $2r$ as the hypotenuse, and $\sqrt{4r^2 - r^2} = \sqrt{3}r$ as the third side.
Therefore, the area will be $\dfrac{\sqrt{3}r^2}{2}$
Comparing the three valid cases, we get:
$\dfrac{3\sqrt{3} r^2}{4} \gt r^2 \gt \dfrac{\sqrt{3}r^2}{2}$
Therefore, Case A, that is the inscribed equilateral triangle, with each side $= \sqrt{3}r$, will have the maximum area.
Problem 16 of 30
Let $L$ be a line obtained from the intersection of two planes $x + 2y + z = 6$ and $y + 2z = 4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3, 2, 1)$ on $L$, then the value of $21(\alpha + \beta + \gamma)$ equals:Solution:
Let $A$ be the point $(3,2,1)$ and let $B$ be the foot of the perpendicular from $A$ to line $L$.
The direction ratios of $L$ are $a, b, c$ where,
$a + 2b + c = 0$
$b + 2c = 0$
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Therefore, the direction ratios of $L$ are $3, -2, 1.$
Then the equation of the $L$ is,
$L: \dfrac{x + 2}{3} = \dfrac{y - 4}{-2} = \dfrac{z}{1}$
The co-ordinates of $B$ are,
$B: \dfrac{x + 2}{3} = \dfrac{y - 4}{-2} = \dfrac{z}{1} = \lambda$
$\Rightarrow B = 3 \lambda - 2, -2 \lambda + 4, \lambda$
The direction ratios of $AB$ are $3 \lambda - 5, -2 \lambda + 2, \lambda - 1$
$3(3 \lambda - 5) - 2 (-2 \lambda + 2) + 1( \lambda - 1) = 0$
$\Rightarrow 9 \lambda - 15 + 4\lambda - 4 + \lambda - 1 = 0$
$\Rightarrow 14\lambda -20 = 0$
$\Rightarrow \lambda = \dfrac{10}{7}$
$\therefore B = 3 \lambda - 2, -2 \lambda + 4, \lambda = \dfrac{16}{7}, \dfrac{8}{7}, \dfrac{10}{7}$
$21(\alpha + \beta + \gamma) = 21 \left( \dfrac{16}{7} + \dfrac{8}{7} + \dfrac{10}{7} \right) = 21 \times \dfrac{34}{7} = 102$
Problem 17 of 30
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined as$f(x) = \left\{\begin{array}{ll}2\sin\left(-\dfrac{\pi x}{2}\right), & \text{ if }x \lt -1 \\ |ax^2 + x + b|, & \text{ if }-1 \le x \le 1 \\ \sin\left(\pi x \right), & \text{ if }x \gt 1\end{array}\right.$
If $f(x)$ is continuous on $\mathbb{R}$, then $a + b$ equals:
Solution:
For the function to be continuous,
$|ax^2 + x + b | = 2\sin\left(-\dfrac{\pi x}{2} \right)$ for $x = -1$
and
$|ax^2 + x + b | = sin(\pi x)$ for $x = 1$
Therefore, substituting $x = -1$
$|a(-1)^2 + (-1) + b | = 2\sin\left(-\dfrac{\pi (-1)}{2} \right)$
$\Rightarrow |a + b - 1| = 2$
$\Rightarrow a +b = 3$ or $a + b = -1$
Again, substituting $x= 1$
$|a(1)^2 + (1) + b | = \sin(\pi (1))$
$\Rightarrow |a + b + 1| = 0$
$\Rightarrow a + b = -1 $
$\therefore a + b = -1$
Problem 18 of 30
If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^2 + y^2 = 1$ is a circle of radius $r$, then $r$ is equal to:Solution:
We know that the of a point dividing a line from a fixed point to a point on a circle in the ratio $m:n$ is given by:
$\left(x - \dfrac{na}{m + n} \right)^2 + \left(y - \dfrac{nb}{m + n} \right)^2 = \left( \dfrac{mr}{m + n} \right)^2$
Since in this problem the given point is the midpoint, $m:n = 1:1$
Also, $a = 3, b = 2$ and $r = 1$
Therefore, the locus of the midpoint is:
$\left(x - \dfrac{3}{2} \right)^2 + \left(y - \dfrac{2}{2} \right)^2 = \left(\dfrac{1}{2}\right)^2$
Therefore, the radius of the locus (circle) is $\dfrac{1}{2}$.
Problem 19 of 30
A natural number has prime factorization given by $n = 2^x3^y5^z$, where $y$ and $z$ are such that $y + z = 5$ and $y^{-1} + z^{-1} = \dfrac{5}{6}$, $y \gt z$. Then the number of odd divisors of $n$, including $1$, is:Solution:
$y+z=5$ $... eq \ (i)$Also,
$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{5}{6}$
$\Rightarrow \dfrac{y+z}{yz} = \dfrac{5}{6}$
$\Rightarrow yz = 6$
$\therefore y-z = \sqrt{(y-z)^2} = \sqrt{(y+z)^2 - 4yz} = \sqrt{(5)^2 - 24} = \sqrt{1} = 1$
$y - z = 1$ $... eq \ (ii)$
Solving equations $(i)$ and $(ii),$ we get $y = 3$ and $z=2.$
The number of odd divisors can be found by finding all factors of $3^y5^z$.
Therefore, the number of odd divisors of $2^x3^y5^z$
$ = (y+1)(z+1) = (3+1)(2+1) = (4)(3) = 12$
Problem 20 of 30
For $x \gt 0$, $f(x) = \displaystyle \int\limits_{1}^{x}\dfrac{\log_{e}t}{(1+t)}dt$, then $f(e) + f\left(\dfrac{1}{e}\right)$ is equal to:Solution:
From the given definition,
$f\left(\dfrac{1}{x}\right) = \displaystyle \int\limits_{1}^{1/x}\dfrac{\log_{e}t}{(1+t)}dt$
Substituting $t = \dfrac{1}{z}$, we get:
$dt = - \dfrac{1}{z^2} dz$, $\ln t = - \ln z$ and $\dfrac{1}{1 + t} = \dfrac{z}{z + 1}$
Therefore, we get:
$f\left(\dfrac{1}{x}\right) = \displaystyle \int\limits_{1}^{x} \dfrac{z (-\ln z)}{(z + 1)(-z^2)} dz = \int\limits_{1}^{x} \dfrac{\ln z}{(z + 1)z} dz$
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Expressing both $f(x)$ and $f \left(\dfrac{1}{x} \right)$, in terms of the same variable, $y$, we get:
$f(x) + f\left(\dfrac{1}{x} \right) = \displaystyle \int\limits_{1}^{x}\dfrac{\ln y}{(1+y)}dy + \int\limits_{1}^{x} \dfrac{ \ln y}{(y + 1)y} dy$
$= \displaystyle \int\limits_{1}^{x} \ln y\left[\dfrac{1}{1 + y} + \dfrac{1}{(1 + y)y} \right]dy$
$= \displaystyle \int\limits_{1}^{x} \ln y\left[\dfrac{(1 + y)}{(1 + y)y} \right]dy$
$= \displaystyle \int\limits_{1}^{x} \dfrac{\ln y}{y} dy$
$= \dfrac{(\ln y)^2}{2} \biggr\rvert_{1}^{x} = \dfrac{(\ln x)^2}{2}$
Therefore,
$f(e) + f\left(\dfrac{1}{e}\right) = \dfrac{(\ln e)^2}{2} = \dfrac{1}{2}$
Problem 21 of 30
If $I_{m, n} = \displaystyle{\int\limits_{0}^{1}} x^{m-1}(1-x)^{n-1} dx,$ for $m,n \geqslant 1,$ and $\displaystyle{\int\limits_{0}^{1}} \dfrac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} dx = \alpha I_{m,n}, \alpha \in R,$ then $\alpha$ equals _________.Solution:
$I_{m, n} = \displaystyle{\int\limits_{0}^{1}} x^{m-1}(1-x)^{n-1} dx$ $...(i)$Substituting $x = \dfrac{1}{y + 1}$
$I_{m, n} = \displaystyle{\int\limits_{\infty}^{0}} \dfrac{y^{n-1}}{(y + 1)^{m + n}} (-1) dy $
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$\Rightarrow I_{m, n} = \displaystyle{\int\limits_{0}^{\infty}} \dfrac{y^{n-1}}{(y + 1)^{m + n}} dy $ $...(ii)$
Substituting $z = 1 - x$ and $dx = -dz$ in $eqn\ (i)$
$I_{m,n} = \displaystyle{\int\limits_{1}^{0}} (1-z)^{m-1}(z)^{n-1} (-dz)$
$I_{m,n} = \displaystyle{\int\limits_{0}^{1}} (z)^{n-1}(1-z)^{m-1} (dz)$
Again, substituting $z = \dfrac{1}{y}$, we get:
$I_{m, n} = \displaystyle{\int\limits_{0}^{\infty}} \dfrac{y^{m-1}}{(y + 1)^{m + n}} dy $ $...(iii)$
Adding equations $(ii)$ and $(iii)$
$2I_{m, n} = \displaystyle{\int\limits_{0}^{\infty}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy $
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{1}^{\infty}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy$
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Substituting $y = \dfrac{1}{z}, \ dy = - \dfrac{1}{z^2} dz$ in the second integral,
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{1}^{\infty}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy$
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{1}^{0}} \dfrac{\left( \dfrac{1}{z} \right) ^{m-1} + \left( \dfrac{1}{z} \right) ^{n - 1}}{\left(\dfrac{1}{z} + 1\right)^{m + n}} \left( - \dfrac{1}{z^2}dz\right)$
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{1}^{0}} \dfrac{z^{m+1} + z^{n + 1}}{(z + 1)^{m + n}} \left( - \dfrac{1}{z^2}dz\right)$
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{1}^{0}} \dfrac{z^{m-1} + z^{n - 1}}{(z + 1)^{m + n}} (- dz)$
$\Rightarrow 2I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy + {\int\limits_{0}^{1}} \dfrac{z^{m-1} + z^{n - 1}}{(z + 1)^{m + n}} dz$
Ignoring variable names in definite integrals, we get:
$2 I_{m, n} = 2 \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy$
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$\Rightarrow I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy$
$\because I_{m, n} = \displaystyle{\int\limits_{0}^{1}} \dfrac{y^{m-1} + y^{n - 1}}{(y + 1)^{m + n}} dy = \alpha I$
$\therefore \alpha = 1$
Problem 22 of 30
Let $z$ be those complex numbers which satisfy $|z+5| \leq 4$ and $z(1+i) + \overline{z}(1-i) \geqslant -10, \ i = \sqrt{-1}.$If the maximum value of $|z+1|^2$ is $\alpha + \beta \sqrt{2}$ then the value of $(\alpha + \beta)$ is ________.
Solution:
Let $z = a + bi$ where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$
Because,
$|z + 5| \le 4$$\Rightarrow \sqrt{(a + 5)^2 + b^2} \le 4$
$\Rightarrow (a + 5)^2 + b^2 \le 16$
Again,
$z(1 + i) + \overline{z}(1 - i) \ge -10$
$\Rightarrow (a + bi)(1 + i) + (a - bi)(1 - i) \ge -10$
$\Rightarrow a + bi + ai + bi^2 + a - bi - ai + bi^2 \ge -10$
$\Rightarrow 2a + 2bi^2 \ge -10$
$\Rightarrow a - b \ge -5$
$\Rightarrow b \le a + 5$
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Substituting $b = a + 5$ in $eqn(i)$
$(a + 5)^2 + (a + 5)^2 \le 16$
$\Rightarrow 2(a + 5)^2 \le 16$
$\Rightarrow (a + 5)^2 \le 8$
$\Rightarrow -2\sqrt{2} \le a + 5 \le 2\sqrt{2}$
$\Rightarrow -2\sqrt{2} - 5 \le a \le 2\sqrt{2} - 5$
and
$\because b \le a + 5$
$-2\sqrt{2} \le b \le 2\sqrt{2}$
We need to find the maximum value of:
$|z+1|^{2}$
$= (a + 1)^2 + b^2$
Since the above is the sum of two positive quantities, maximising each, will give the maximum value of the expression.
We can maximise $(a+1)^2$ by maximising $|a + 1|$
For $a = 2\sqrt{2} - 5$, $|a + 1| = |2\sqrt{2} - 5 + 1| = 4 - 2\sqrt{2}$
For $a = -2\sqrt{2} - 5$, $|a + 1| = |-2\sqrt{2} - 5 + 1| = 4 + 2\sqrt{2}$
For $a = -2\sqrt{2} - 5$, $b \le a + 5 \Rightarrow b \le -2\sqrt{2} - 5 + 5 \Rightarrow b \le -2\sqrt{2}$
$\therefore max[(a + 1)^2 + b^2] = (-4 - 2\sqrt{2})^2 + (-2\sqrt{2})^2$
$= 8 + 16 + 16\sqrt{2} + 8$
$= 32 + 16\sqrt{2}$
Therefore, $\alpha = 32$ and $\beta = 16$ and $\alpha + \beta = 48$
Problem 23 of 30
Let the normals at all points on a given curve pass through a fixed point $(a,b).$ If the curve passes through $(3,-3)$ and $(4, -2 \sqrt{2})$, and given that $a - 2 \sqrt{2} b = 3,$ then $(a^2+b^2+ab)$ is equal to ________.Solution:
Since the normals to all points on a curve pass through $(a, b)$, the curve is a circle with $(a, b)$ as the centre.
The points $(3,-3)$ and $(4, -2 \sqrt{2})$ lie on the circle and forms a chord, therefore, the centre lies on the perpendicular bisector this chord.
The slope of the chord is: $\dfrac{-2\sqrt{2} + 3}{1} = 3 - 2\sqrt{2}$. The line perpendicular to the chord has a slope of $\dfrac{1}{2\sqrt{2} - 3}$
The midpoint of the chord is $\left(\dfrac{7}{2}, \dfrac{-3-2\sqrt{2}}{2}\right)$
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The equation of the diameter perpendicular to this chord is, therefore:
$\dfrac{y - \left(\dfrac{-3-2\sqrt{2}}{2}\right)}{x - \dfrac{7}{2}} = \dfrac{1}{2\sqrt{2} - 3}$
$\Rightarrow \dfrac{2y + 3 + 2\sqrt{2}}{2x - 7} = \dfrac{1}{2\sqrt{2} - 3}$
$\because (a, b)$ lies on the diameter, it satisfies the above equation:
$\therefore \dfrac{2b + 3 + 2\sqrt{2}}{2a - 7} = \dfrac{1}{2\sqrt{2} - 3}$
Substituting $a = 2\sqrt{2}b + 3$ in the above equation:
$\dfrac{2b + 3 + 2\sqrt{2}}{4\sqrt{2}b - 1} = \dfrac{1}{2\sqrt{2} - 3}$
$\Rightarrow 4\sqrt{2}b - 6b + 6\sqrt{2} - 9 + 8 - 6\sqrt{2}= 4\sqrt{2}b - 1$
$\Rightarrow -6b = 0 \Rightarrow b = 0$
$\therefore a = 2\sqrt{2} \times 0 + 3 = 3$
Therefore,
$a^2 + b^2 + ab$
$= 9 + 0 + 0 = 9$
Problem 24 of 30
Let $a$ be an integer such that all real roots of the polynomial $2x^5+5x^4+10x^3+10x^2+10x+10$ lie in the interval $(a,a+1)$ Then $|a|$ is equal to _________.Solution:
The slope $(s)$ of the given polynomial is given by:
$f'(x) = 10x^4 + 20x^3 + 30x^2 + 20x + 10$
$f''(x) = 40x^3 + 60x^2 + 60x + 20$
Assigning $f''(x) = 0$ we get:
$40x^3 + 60x^2 + 60x + 20 = 0$
$\Rightarrow 2x^3 + 3x^2 + 3x + 1 = 0$
$\Rightarrow x^3 + (x + 1)^3 = 0$
$\Rightarrow (2x + 1)(x^2 - x^2 - x + x^2 + 2x + 1) = 0$
$\Rightarrow (2x + 1)(x^2 + x + 1) = 0$
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This equation has only one real root $x = -\dfrac{1}{2}$ and two complex roots.
Therefore, $f'(x)$ has either a minima or a maxima.
$f'''(x) = 120x^2 + 120x + 60$
Putting $x = -\dfrac{1}{2}$ in $f'''(x)$, we get
$f'''\left(-\dfrac{1}{2}\right) = 30 - 60 + 60 = 30$
Therefore, $x = -\dfrac{1}{2}$ is a minima for the function $f'(x)$
Putting $x = -\dfrac{1}{2}$ in $f'(x)$,
$f'\left(-\dfrac{1}{2}\right) = 10\left(-\dfrac{1}{2}\right)^4 + 20\left(-\dfrac{1}{2}\right)^3 + 30\left(-\dfrac{1}{2}\right)^2 + 20\left(-\dfrac{1}{2}\right) + 10$
$\dfrac{10}{16} - \dfrac{20}{8} + \dfrac{30}{4} - \dfrac{20}{2} + 10 = \dfrac{45}{8}$
Since, the minima for $f'(x)$ is $\dfrac{45}{8}$,
$f'(x) \ge \dfrac{45}{8}$ which implies that $f'(x)$ is always positive.
Therefore $f(x)$ has a positive slope for all real $x$, which means it is an increasing function.
Using manual substitution,
$f(-1) = 3$
and
$f(-2) = -34$
Since the function $f$ is increasing for all real values of $x$, the curve intersect the $x$-axis between $x = -2$ and $x = -1$ and nowhere else. That is, it has only one real root, between $-2$ and $-1$.
Therefore, $a = -2$
Therefore, $|a| = 2$
Problem 25 of 30
If $X_1, X_2, ..., X_{18}$ be eighteen observations such that $\displaystyle{\sum\limits_{i = 1}^{18}}(X_i - \alpha) = 36$ and $\displaystyle{\sum\limits_{i = 1}^{18}}(X_i - \beta)^2 = 90$ where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1$, then the value of $|\alpha - \beta|$ is ______.Solution:
$\sum x_i - 18 \alpha = 36$$\Rightarrow \sum x_i = 18(\alpha + 2)$
$\sum\limits_{i = 1}^{18}(X_i - \beta)^2 = 90$
$\Rightarrow \sum (x_i)^2 + 18 \beta^2 - 2\beta \sum x_i = 90$
$\therefore \sum (x_i)^2 + 18 \beta ^2 - 36 \beta (\alpha + 2) = 90$
$\Rightarrow \sum (x_i)^2 = 90 - 18 \beta^2 + 36 \beta (\alpha + 2)$
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$\sigma^2 = 1$
$ \Rightarrow \dfrac{1}{18} \sum (x_i)^2 - \left ( \dfrac{\sum x_i}{18} \right ) ^2 = 1$
$ \Rightarrow \dfrac{1}{18} (90 - \beta ^2 + 36 \alpha \beta + 72 \beta) - \left ( \dfrac{18(\alpha + 2)}{18} \right ) = 1 $
$ \Rightarrow 5 - \beta^2 + 2 \alpha \beta + 4 \beta - \alpha^2 - 4\alpha - 4 = 1 $
$ \Rightarrow - \alpha ^2 - \beta ^2 + 2 \alpha \beta + + 4 \beta - 4 \alpha = 0$
$ \Rightarrow - (\alpha - \beta)^2 - 4(\alpha - \beta) = 0$
$ \Rightarrow - (\alpha - \beta) (\alpha - \beta + 4) = 0$
$ \Rightarrow \alpha - \beta = -4$ or $\alpha - \beta = 0$
$ \Rightarrow \alpha - \beta = - 4$ $\because \alpha \neq \beta$
$ \Rightarrow |\beta - \alpha| = 4$
Problem 26 of 30
If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{bmatrix}$ satisfies the equation $A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix}$ for some real numbers $\alpha$ and $\beta$, then $\beta - \alpha$ is equal to ______.Solution:
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$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{bmatrix}$
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$A^3 =\begin{bmatrix} 1 & 0
& 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2
& 0 \\ 3 & 0 & -1\end{bmatrix} = \begin{bmatrix} 1 & 0
& 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1\end{bmatrix}$
$\cdot$
$\cdot$
$\cdot$
$A^{19} = \begin{bmatrix} 1 & 0
& 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1\end{bmatrix}$
$A^{20} = \begin{bmatrix} 1 & 0
& 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1\end{bmatrix}$
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$A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1 & 0
& 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1\end{bmatrix} + \begin{bmatrix} \alpha & 0
& 0 \\ 0 & 2^{19}\alpha & 0 \\ 3\alpha & 0 & -\alpha\end{bmatrix} + \begin{bmatrix} \beta & 0 & 0 \\ 0 & 2\beta & 0 \\ 3\beta & 0 & -\beta\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$\Rightarrow 3\alpha + 3\beta = 0$
$\Rightarrow \alpha = - \beta$
and,
$2^{20} + 2^{19}\alpha + 2\beta = 4$
$2^{20} + 2^{19}(-\beta) + 2\beta = 4$
$\Rightarrow 2\beta(1 - 2)^{18} = 2^2 (1 - 2^{18})$
$\Rightarrow \beta = 2$
$\therefore \alpha = -2$
$\Rightarrow \beta - \alpha = 4$
Problem 27 of 30
Let $\alpha$ and $\beta$ be two real numbers such that $\alpha + \beta = 1$ and $\alpha \beta = -1$. Let $p_n = (\alpha)^n + (\beta)^n$, $p_{n-1} = 11$ and $p_{n + 1} = 29$ for some integer $n \geqslant 1$. Then the value of $p_n^2$ is ______.Solution:
$p_{n-1} = 11$$\Rightarrow \alpha ^ {\ n-1} + \beta ^ {\ n-1} = 11$
$\Rightarrow \dfrac{\alpha ^ n}{\alpha} + \dfrac{\beta ^ n}{\beta} = 11$
$\Rightarrow \dfrac{\beta (\alpha ^ n) + \alpha (\beta ^ n)}{\alpha \beta } = 11$
$\Rightarrow \dfrac{\beta (\alpha ^ n) + \alpha (\beta ^ n)}{-1} = 11$
$\Rightarrow \beta (\alpha ^ n) + \alpha (\beta ^ n) = -11$ $...(i)$
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$p_{n+1} = 29$
$\Rightarrow \alpha ^ {\ n+1} + \beta ^ {\ n+1} = 29$
$\Rightarrow \alpha (\alpha ^ n) + \beta (\beta ^ n) = 29$ $...(ii)$
$(i) + (ii)$
$\Rightarrow \beta (\alpha ^ n) + \alpha (\beta ^ n) + \alpha (\alpha ^ n) + \beta (\beta ^ n) = 29 + (-11)$
$\Rightarrow (\alpha + \beta) (\alpha ^ n + \beta ^ n)= 18 $
$\Rightarrow (1) (\alpha ^ n + \beta ^ n)= 18 $
$\therefore p_n = (\alpha ^ n + \beta ^ n) = 18$
$\Rightarrow p_n ^ 2 = 18^2 = 324$
Problem 28 of 30
The total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$, is ______.Solution:
Any given number $N$ and $18$ will have a gcd of $3$, only if:$N \equiv \pm 3\ (mod 18)$, that is, the number is of the form:
$18n + 3$ or $18n - 3$
Considering the $4$-digit numbers of the form $18n - 3$, we get:
$999 \lt 18n - 3 \lt 10000$
$\Rightarrow 1002 \lt 18n \lt 10003$
$\Rightarrow \dfrac{1002}{18} \lt n \lt \dfrac{10003}{18}$
$\Rightarrow 55\dfrac{12}{18} \lt n \lt 555\dfrac{13}{18}$
$\because n$ is an integer,
$56 \le n \le 555$
$\therefore$ we have $555 - 55 = 500$ integer values of $n$ satisfying the above inequality.
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Similarly, considering the $4$-digit numbers of the form $18n + 3$, we get:
$999 \lt 18n + 3 \lt 10000$
$\Rightarrow 996 \lt 18n \lt 9997$
$\Rightarrow 55 \dfrac{1}{3} \lt n \lt 555 \dfrac{7}{18}$
$\therefore \Rightarrow 56 \le n \le 555$
and there are $555 - 55 = 500$ integer values of $n$ satisfying this inequality.
Therefore, there are $500 \times 2 = 1000$ $4$-digit numbers which will have a gcd of $3$ with $18$.
Problem 29 of 30
If the arithmetic mean and geometric mean of the $p\xasuper{th}$ and $q\xasuper{th}$ terms of the sequence $-16, 8, -4, 2, ...$ satisfy the equation $4x^2 - 9x + 5 = 0$, the $p + q$ is equal to ______.Solution:
The given sequence is a series in geometric progression where the first term is $a = -16$ and the common ratio is $r = -\dfrac{1}{2}.$
Let the $p^{th}$ and $q^{th}$ term be $ar^{p-1}$ and $ar^{q-1}$ respectively.
From the $A.M - G.M$ inequality we know that $A.M \geq G.M$
Therefore, the $A.M$ is $\dfrac{5}{4}$ and the $G.M$ is $1.$
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$G.M = 1$
$\therefore \displaystyle {\sqrt{(ar^{\ p-1})(ar^{\ q-1})} = 1}$
$\Rightarrow \displaystyle {\sqrt{(a^{\ 2})(r^{\ p+q-2})} = 1}$
$\Rightarrow a \displaystyle {\sqrt{(r^{\ p+q-2})} = 1}$
$\Rightarrow \displaystyle {\sqrt{(r^{\ p+q-2})} = - \dfrac{1}{16}}$
$\Rightarrow \displaystyle {r^{\ p+q-2} = \left(- \dfrac{1}{16}\right)^2}$
$\Rightarrow r^{\ p+q-2} = \dfrac{1}{2^8}$
$\Rightarrow \dfrac{1}{(-2)^{p+q-2}} = \dfrac{1}{2^8}$
$\Rightarrow \dfrac{1}{(-2)^{p+q-2}} = \dfrac{1}{(-2)^8}$
$\Rightarrow p+q-2 = 8$
$\therefore p+q=10$
Problem 30 of 30
Let $L$ be a common tangent line to the curves $4x^2 + 9y^2 = 36$ and $(2x)^2 + (2y)^2 = 31$. Then the square of the slope of the line $L$ is ______.Solution:
The general equation of the tangent of an ellipse of the form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$, where $m$ is the slope of the tangent at the given point.The given ellipse $4x^2 + 9y^2 = 36$ can be written as:
$\dfrac{x^2}{3^2} + \dfrac{y^2}{2^2} = 1$, .
Therefore, the tangent $L_1$ is:
$y = mx \pm \sqrt{9m^2 + 4}$
The general equation of the tangent of a circle of the form $x^2 + y^2 = r^2$ is $y = mx \pm \sqrt{r^2m^2 + r^2}$, where $m$ is the slope of the tangent at the given point
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Therefore, the equation of the tangent to the circle $4x^2 + 4y^2 = 31$ is,
$L_2: y = mx \pm \sqrt{\dfrac{31}{4}m^2 + \dfrac{31}{4}}$
Since $L$ is the common tangent to both the curves,
$L_1 = L_2$
$\Rightarrow mx \pm \sqrt{9m^2 + 4} = mx \pm \sqrt{\dfrac{31}{4}m^2 + \dfrac{31}{4}}$
$\Rightarrow \pm \sqrt{9m^2 + 4} = \pm \sqrt{\dfrac{31}{4}m^2 + \dfrac{31}{4}}$
$\Rightarrow 9m^2 + 4 = \dfrac{31}{4}m^2 + \dfrac{31}{4}$
$\Rightarrow 36m^2 + 16 = 31m^2 + 31$
$\Rightarrow 5m^2 = 15$
$\Rightarrow m^2 = 3$
Therefore, the square of the slope of the tangent is $3$