Problem 1 of 30
If for the matrix, $A = \begin{bmatrix} 1 & - \alpha \\ \alpha & \beta \end{bmatrix}$, $AA^T=I_2,$ then the value of $\alpha^4 + \beta^4$ is :Solution:
$A = \begin{bmatrix} 1 & - \alpha \\ \alpha & \beta \end{bmatrix}$$\Rightarrow A^T = \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix}$
$\Rightarrow AA^T = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha \beta & \alpha^2 + \beta^2 \end{bmatrix}$
$\therefore AA^T = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha \beta & \alpha^2 + \beta^2 \end{bmatrix} = I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Therefore,
$ 1 + \alpha^2 = 1 \Rightarrow \alpha = 0$
and
$\alpha^2 + \beta^2 = 1$
$\Rightarrow 0 + \beta^2 = 1$
$\Rightarrow \beta = \pm 1$
These values satisfy the third condition, $\alpha - \alpha \beta = 0$
Therefore,
$\alpha^4 + \beta^4 = (0)^4 + (\pm 1)^4 = 1$
Problem 2 of 30
Let $A$ be a $3 \times 3$ matrix with det$(A)=4.$ Let $R_i$ denote the $i^{th}$ row of $A.$ If a matrix $B$ is obtained by performing the operation $R_2 \rightarrow 2R_2 + 5 R_3$ on $2A,$ then det$(B)$ is equal toSolution:
$det(A) = 4$Multiplying a square matrix of order $N$ by a scalar $\lambda$, multiplies the determinant by $\lambda^{N}$
$\therefore det(2A) = 2^3(4) = 32$
$\because R_2 \rightarrow 2R_2 + 5R_3$
Multiplying a single row/column by $2$ means the determinant is also multiplied by $2.$
$\therefore$ det$(B) = 2 \times$ det$(2A) = 2 \times 32 = 64$
Problem 3 of 30
The following system of linear equations$\phantom{0000} 2x + 3y + 2z = 9$
$\phantom{0000} 3x + 2y + 2z = 9$
$\phantom{0000} x - y + 4z = 8$
Solution:
$\begin{bmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 9 \\ 8 \end{bmatrix}$$A = \begin{bmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{bmatrix}$, $B = \begin{bmatrix} 9 \\ 9 \\ 8 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
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Computing the of $A$, we get:
$|A| = -20$
Since the determinant of the coefficient matrix is non-zero, the given system of equations has a unique solution, but we still need to validate the option giving the relationship between $\alpha, \beta, \gamma$. Therefore, we need to complete solving for these values.
$A^{-1} = \dfrac{adj(A)}{|A|} = \begin{bmatrix} - \dfrac{1}{2} & \dfrac{7}{10} & - \dfrac{1}{10}\\ \dfrac{1}{2} & - \dfrac{3}{10} & - \dfrac{1}{10} \\ \dfrac{1}{4} & - \dfrac{1}{4} & \dfrac{1}{4} \end{bmatrix}$
$X = A^{-1}B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$
The solution of the given system of linear equations is $x = 1, y = 1$ and $z = 2$
$x ^2 + y^2 + z^2 = 1^2 + 1^2 + 2^2 = 6 \neq 12$
$\therefore$ The given system has a unique solution but the sum of the squares of the solutions is not equal to $12.$
Problem 4 of 30
If $I_n = \displaystyle{\int\limits_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}} \cot^n x dx$, then :Solution:
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$I_n$ $= \displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^n x \ dx$
$ = \left[- \dfrac{cot^{n-1}x}{n-1}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} - \displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \ dx $
$= \left[- \dfrac{cot^{n-1}x}{n-1}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} - I_{n-2}$
Therefore,
$I_4= \left[- \dfrac{cot^{3}x}{3}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} - I_{2}$
$I_5= \left[- \dfrac{cot^{4}x}{4}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} - I_{3}$
$I_6= \left[- \dfrac{cot^{5}x}{5}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} - I_{4}$
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$I_2 + I_4 = \left[- \dfrac{cot^{3}x}{3}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} = \dfrac{1}{3}$
$I_3 + I_5 = \left[- \dfrac{cot^{4}x}{4}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} = \dfrac{1}{4}$
$I_4 + I_6 = \left[- \dfrac{cot^{5}x}{5}\right]_{\frac{\pi}{2}}^\frac{\pi}{4} = \dfrac{1}{5}$
$\dfrac{1}{I_2 + I_4}, \dfrac{1}{I_3 + I_5}, \dfrac{1}{I_4 + I_6}$ that is, $3,4,5$are in $A.P.$
Problem 5 of 30
A function $f(x)$ is given by $f(x) = \dfrac{5^x}{5^x + 5}$, then the sum of the series$f\left( \dfrac{1}{20} \right) + f\left( \dfrac{2}{20} \right) + f\left( \dfrac{3}{20} \right) + ... + f\left( \dfrac{39}{20} \right)$
is equal to:
Solution:
Let us observe the given series before we begin to solve it.
The series has $39$ terms, the middle term being $f\left(\dfrac{20}{20}\right) = f(1)$, and each term on the left is of the form $f(1 - a)$ with a corresponding term $f(1 + a)$ on the right.
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$f(1 + a) = \dfrac{5^{1 + a}}{5^{1 + a} + 5} = \dfrac{5 \cdot 5^a}{ 5\cdot 5^a + 5} = \dfrac{5^a}{5^a + 1}$
$f(1 - a) = \dfrac{5^{1 - a}}{5^{1 - a} + 5} = \dfrac{\dfrac{5}{5^a}}{\dfrac{5}{5^a} + 5} = \dfrac{5}{5 + 5 \cdot 5^a} = \dfrac{1}{1 + 5^a}$
Therefore,
$f(1 - a) + f(1 + a) = \dfrac{1}{1 + 5^a} + \dfrac{5^a}{5^a + 1} = 1$
Now we can rewrite the given series as:
$f\left( \dfrac{1}{20} \right) + f\left( \dfrac{2}{20} \right) + f\left( \dfrac{3}{20} \right) + ... + f\left( \dfrac{39}{20} \right)$
$= f\left( 1 - \dfrac{19}{20} \right) + f\left( 1 - \dfrac{18}{20} \right) + ... + f\left( \dfrac{20}{20} \right) + ... + f\left(1 + \dfrac{18}{20}\right)$
$+ f\left(1 + \dfrac{19}{20} \right)$
$= \left[\displaystyle{\sum\limits_{n = 1}^{19}} \left\{f\left(1 - \dfrac{n}{20} \right) + f\left(1 + \dfrac{n}{20} \right)\right\} \right] + f\left(\dfrac{20}{20} \right)$
$= \left[\displaystyle{\sum\limits_{n = 1}^{19}} 1 \right] + f\left(1 \right) = 19 + \dfrac{5^1}{5^1 + 5} = 19 \dfrac{1}{2} = \dfrac{39}{2}$
Problem 6 of 30
Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \geqslant 1$, then the value of $\dfrac{a_{10} - 2a_8}{3a_9}$Solution:
If $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0,$$\alpha^2 - 6\alpha - 2 = 0$
$\Rightarrow \alpha^2 - 2 = 6\alpha$ $...(i)$
$\beta^2 - 6\beta- 2 = 0$
$\Rightarrow \beta^2 - 2 = 6\beta$ $...(ii)$
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$\dfrac{a_{10} - 2a_8}{3a_9}$
$= \dfrac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{3(\alpha^9 - \beta^9)}$
$= \dfrac{\alpha^{10} - 2\alpha^8 - \beta^{10} + 2\beta^8}{3(\alpha^9 - \beta^9)}$
$= \dfrac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{3(\alpha^9 - \beta^9)}$
Substituting values of $\alpha^2 - 2 = 6 \alpha $ and $\beta^2 - 2 = 6 \beta$ from equations $(i)$ and $(ii)$, we get:
$\dfrac{\alpha^8(6\alpha) - \beta^8(6\beta)}{3(\alpha^9 - \beta^9)}$
$= \dfrac{6(\alpha^9 - \beta^9)}{3(\alpha^9 - \beta^9)}$
$= 2$
Problem 7 of 30
The minimum value of $f(x) = a^{a^x} + a^{1 - a^x}$, where $a, x \in \mathbb{R}$ and $a \gt 0$, is equal to:Solution:
Using the AM/GM inequality, $AM \geq GM$
$\Rightarrow \dfrac{a^{a^x} + \dfrac{a}{a^{a^x}}}{2} \geq \sqrt{a^{a^x} \cdot \dfrac{a}{a^{a^x}}}$
$\Rightarrow a^{a^x} + \dfrac{a}{a^{a^x}} \geq 2 \sqrt{a}$
$\therefore$ The minimum value of $a^{a^x} + \dfrac{a}{a^{a^x}}$ is $2\sqrt{a}$
Problem 8 of 30
The integral $\displaystyle{\int} \dfrac{e^{3\log_{e}2x} + 5e^{2\log_{e}2x}}{e^{4\log_{e}x} + 5e^{3\log_{e}x} - 7e^{2\log_{e}x}} dx$, $x \gt 0$, is equal to:(where $c$ is a constant of integration)
Solution:
We can write the given integral as:
$\displaystyle{\int} \dfrac{e^{3\log_{e}2x} + 5e^{2\log_{e}2x}}{e^{4\log_{e}x} + 5e^{3\log_{e}x} - 7e^{2\log_{e}x}} dx$
$= \displaystyle{\int} \dfrac{(e^{\log_{e}2x})^{3} + 5(e^{\log_{e}2x})^{2}}{(e^{\log_{e}x})^{4} + 5(e^{\log_{e}x})^{3} - 7(e^{\log_{e}x})^{2}} dx$
$= \displaystyle{\int} \dfrac{(2x)^{3} + 5(2x)^{2}}{(x)^{4} + 5(x)^{3} - 7(x)^{2}} dx$
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$= \displaystyle{\int} \dfrac{\cancel{x^2}(8x + 20)}{\cancel{x^2}(x^2 + 5x - 7)} dx$
$= \displaystyle{\int} \dfrac{4(x + 5)}{x^2 + 5x - 7} dx$
$= \displaystyle 4{\int} \dfrac{2x + 5}{x^2 + 5x - 7} dx$
We can observe that the numerator of the above function is the derivative of the denominator. Therefore, substituting
$x^2 + 5x - 7 = z$
$\Rightarrow \dfrac{d}{dx}z = 2x + 5$
$\Rightarrow dx = \dfrac{dz}{2x + 5}$, and we get our integral as:
$\displaystyle 4{\int} \dfrac{2x + 5}{x^2 + 5x - 7} dx$
$= \displaystyle 4{\int} \dfrac{2x + 5}{z} \dfrac{1}{2x + 5} dz$
$= \displaystyle 4{\int} \dfrac{dz}{z}$
$= 4 \ln |z| + c$
$= 4\ln |x^2 + 5x - 7| + c$
Problem 9 of 30
If $\alpha, \beta \in \mathbb{R}$ are such that $1 - 2i$ (here $i^2 = -1$) is a root of the equation $z^2 + \alpha z + \beta = 0$, then $(\alpha - \beta)$ is equal to:Solution:
Substituting $z = 1 - 2i$ in the given equation, we get:$(1 - 2i)^2 + \alpha (1 - 2i) + \beta = 0$
$\Rightarrow 1 -4i + 4i^2 + \alpha - 2\alpha i + \beta = 0$
$\Rightarrow 1 - 4i + 4(-1) + \alpha - 2\alpha i + \beta = 0$
$\Rightarrow (\alpha + \beta - 3) + (-4 -2\alpha)i = 0$
Equating both the real and imaginary parts of the above equation to $0$, we get:
$\alpha + \beta - 3 = 0 \Rightarrow \alpha + \beta = 3$
and
$-4 - 2\alpha = 0$
$\Rightarrow \alpha = -2$
$\therefore \beta - 2 = 3 \Rightarrow \beta = 5$
$\therefore \alpha - \beta = -2 - 5 = -7$
Problem 10 of 30
If the curve $x^2 + 2y^2 = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$, then the angle subtended by the line segment $PQ$ at the origin is:Solution:
$x + y = 1$
$\Rightarrow y = 1 - x$
$\Rightarrow x^2 + 2(1 - x)^2 = 2$
$\Rightarrow 3x^2 - 4x = 0$
$\Rightarrow x = \dfrac{4}{3}$ or $x=0$
$\therefore$ The two points $P$ and $Q$ are $\left( \dfrac{4}{3}, -\dfrac{1}{3} \right)$ and $(0, 1)$
Slope of $OP = \tan \theta _1 = - \dfrac{1}{4}$
$\theta _1 = \tan ^{-1} \left(- \dfrac{1}{4} \right)$
$Q$ lies on the $y-axis,$ $\therefore \theta_2 = \dfrac{\pi}{2}$
$\angle POQ = \theta_2 - \theta _1 = \dfrac{\pi}{2} - \tan ^{-1} \left(- \dfrac{1}{4} \right) = \dfrac{\pi}{2} + \tan ^{-1} \left(\dfrac{1}{4} \right)$
Problem 11 of 30
The shortest distance between the line $x - y = 1$ and the curve $x^2 = 2y$ is:Solution:
The shortest distance of the given line from the curve, will be at a point where the tangent to the curve is parallel to the given line.Finding the slope of the given equation of the curve using , we get:
$\dfrac{dx^2}{dx} = 2 \dfrac{dy}{dx}$
$\Rightarrow \dfrac{dy}{dx} = x$
The slope of the given line is $-\dfrac{a}{b} = -\dfrac{1}{-1} = 1$.
Therefore, if the point of tangency is $(x_1, y_1)$ then $\dfrac{dy}{dx} = x_1 = 1$ and $y_1 = \dfrac{1^2}{2} = \dfrac{1}{2}$
Therefore, the point $\left(1, \dfrac{1}{2}\right)$ on the curve is closest from the line $x - y - 1 = 0$, and its distance from the line is given by,
$\dfrac{\left\lvert 1 - \left(-\dfrac{1}{2}\right) - 1 \right\rvert}{1^2 + (-1)^2}$
$= \dfrac{1}{2\sqrt{2}}$
Problem 12 of 30
A hyperbola passes through the foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:Solution:
The eccentricity, $e_1$ of the ellipse, as explained , is given by:
$e_1 = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5}$Foci $= (\pm3, 0)$
Let equation of hyperbola be $\dfrac{x^2}{A^2} - \dfrac{y^2}{B^2} = 1$
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Since the hyperbola passes through the focii of the ellipse,
$\dfrac{(\pm 3)^2}{A^2} - \dfrac{0^2}{B^2} = 1$
$\Rightarrow A^2 = 9 \Rightarrow A = 3$
Given that $e_1e_2 = 1$, therefore
$e_2 = \dfrac{1}{e_1} = \dfrac{5}{3}$
Using formula for eccentricity of ,
$(e_2)^2 = 1 + \dfrac{B^2}{A^2}$
$\Rightarrow \dfrac{25}{9} = 1 + \dfrac{B^2}{9}$
$\Rightarrow B^2 = 16$
Therefore, the equation of the given hyperbola is:
$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$
Problem 13 of 30
A plane passes through the points $A(1, 2, 3), B(2, 3, 1)$ and $C(2, 4, 2)$. If $O$ is the origin and $P$ is $(2, -1, 1)$, then the projection of $\overrightarrow{OP}$ on this plane is of length:Solution:
Equation of the plane $X$ passing through $A, B, C$ will be: $\begin{vmatrix} x - 1 & y - 2 & z - 3 \\ 2 - 1 & 3 - 2 & 1 - 3 \\ 2 - 1 & 4 - 2 & 2 - 3 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x - 1 & y - 2 & z - 3 \\ 1 & 1 & -2 \\ 1 & 2 & -1 \end{vmatrix} = 0$
$\Rightarrow (x - 1) (-1 + 4) - (y - 2)(-1 + 2) + (z - 3)(2 - 1) = 0$
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$\Rightarrow (x - 1)(3) - (y - 2)(1) + (z - 3)(1) = 0$
$\Rightarrow 3x - 3 - y + 2 + z - 3 = 0$
$\Rightarrow 3x - y + z - 4 = 0$
$\therefore X : 3x - y + z - 4 = 0$
Let $L_1$ be the perpendicular from $O$ to the plane $X$, intersecting it at $O'$
$L_1: \dfrac{x - 0}{3} = \dfrac{y - 0}{-1} = \dfrac{z - 0}{1} = - \dfrac{3(0) - (0) + (0) - 4 }{9 + 1 + 1} = \dfrac{4}{11}$
$\Rightarrow O' = \left( \dfrac{12}{11}, - \dfrac{4}{11}, \dfrac{4}{11} \right)$
Let $L_2$ be the perpendicular from $P$ to $X$, intersecting it at $P'$
$L_2: \dfrac{x - 2}{3} = \dfrac{y - (-1)}{-1} = \dfrac{z - 1}{1} = - \dfrac{3(2) - (-1) + (1) - 4 }{9 + 1 + 1} = \dfrac{8}{11}$
$\Rightarrow P' = \left( \dfrac{10}{11}, -\dfrac{7}{11}, \dfrac{7}{11} \right)$
$\overrightarrow{O'P'}$ is the projection of $\overrightarrow{OP}$ on $X$.
$O'P' = \sqrt{ \left( \dfrac{10}{11} - \dfrac{12}{11} \right)^2 + \left( - \dfrac{7}{11} - \dfrac{4}{11} \right)^2 + \left( \dfrac{7}{11} - \dfrac{4}{11} \right)^2}$
$= \dfrac{\sqrt{22}}{11} = \sqrt{\dfrac{2}{11}}$
Problem 14 of 30
$\lim\limits_{n \rightarrow \infty} \left[ \dfrac{1}{n} + \dfrac{n}{(n + 1)^2} + \dfrac{n}{(n + 2)^2} + ... + \dfrac{n}{(2n - 1)^2}\right]$ is equal to:Solution:
$\lim\limits_{n \rightarrow \infty} \left[ \dfrac{1}{n} + \dfrac{n}{(n + 1)^2} + \dfrac{n}{(n + 2)^2} + ... + \dfrac{n}{(2n - 1)^2}\right]$
$= \lim\limits_{n \rightarrow \infty} \left[ \dfrac{n}{(n + 0)^2} + \dfrac{n}{(n + 1)^2} + \dfrac{n}{(n + 2)^2} + ... + \dfrac{n}{[n + (n - 1)]^2}\right]$
$= \lim \limits_{n \rightarrow \infty} \sum \limits_{r = 0}^{n-1} \dfrac{n}{(n+r)^2}$
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$= \lim \limits_{n \rightarrow \infty} \sum \limits_{r = 0}^{n-1} \dfrac{n}{n^2 \left(1+\dfrac{r}{n} \right)^2}$
$= \lim \limits_{n \rightarrow \infty} \sum \limits_{r = 0}^{n-1} \dfrac{1}{\left(1+\dfrac{r}{n} \right)^2} \cdot \dfrac{1}{n}$
Using the Riemann sum method, we can convert the above sum into an integral of the form:
$\displaystyle \int \limits_0^1 \dfrac{1}{(1+x)^2} dx$
$= - \left[ \dfrac{1}{1 + x}\right]_0^1$
$= - \left[ \dfrac{1}{2} - 1\right] $
$= \dfrac{1}{2}$
Problem 15 of 30
In a group of $400$ people, $160$ are smokers and non-vegetarian; $100$ are smokers and vegetarian and the remaining $140$ are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35\%$, $20\%$ and $10\%$ respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is:Solution:
No. of smokers and non-vegetarians $= 160$
No. of smokers and vegetarians $= 100$
No. of non-smokers and vegetarian $= 140$
No. of smokers and non-vegetarians and chest disorder $\dfrac{35}{100} \times 160 = 56$
No. of smokers and vegetarians with chest disorder $\dfrac{20}{100} \times 100 = 20$
No. of non-smokers and vegetarian with chest disorder $\dfrac{10}{100} \times 140 = 14$
Total no. of people with chest disorder $= 56 + 20 + 14 = 90$
No. of smokers and non-vegetarians and chest disorder $\dfrac{35}{100} \times 160 = 56$
$P($Smoker & non-vegetarian$) = \dfrac{56}{90} = \dfrac{28}{45}$
Problem 16 of 30
Let $A$ be a set of all $4$-digit natural numbers whose exactly one digit is $7$. Then the probability that a randomly chosen element of $A$ leaves a remainder of $2$ when divided by $5$ is:Solution:
For a number to leave a remainder of $2$ when divided by $5$ the number has to end with $2$ or $7$.For numbers ending with $2$, one of the other digits has to be $7$.
We can choose the place of $7$ in $3$ ways.
And the remaining two digits, can be $0$ to $9$ (excluding $7$), that is $9$ ways.
Therefore, there are $3 \times 9 \times 9 = 243$ such numbers.
Now, we need to eliminate the number starting with $0$ from the above count.
If we fix $0$ at the first position, then the $7$ can be placed anywhere in the remaining two digits and the other digit can be $0$ to $9$ excluding $7$.
Therefore, there are $2 \times 9 = 18$ such numbers.
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Therefore, there are $243 - 18 = 225$ four-digit numbers containing exactly one $7$ and ending with $2$.
Similarly, for numbers ending with $7$, the other digits can be filled up with any of the digits from $0$ to $9$ excl. $7$.
Therefore, there are $9 \times 9 \times 9 = 729$ such numbers.
To count the numbers starting with $0$ and ending with $7$, the remaining two digits can be filled up with any digit from $0$ to $9$ excluding $7$.
Therefore, there are $9 \times 9 = 81$ such numbers starting with $0$ and ending with $7$.
Therefore, there are $729 - 81 = 648$ four-digit numbers ending with $7$ and containing exactly one $7$.
Similarly, the total count of $4$-digit numbers containing exactly one $7$:
The $7$ can be placed in any one of the $4$ places.
The remaining three digits can be filled up in $9$ ways ($0$ to $9$ excl $7$).
Therefore, there are $4 \times 9 \times 9 \times 9 = 2916$ such numbers.
The count of numbers beginning with $0$ and containing one $7$ can be found out by fixing $0$ at the beginning, and $7$ can be placed in any one of the $3$ digits, and the remaining two digits can be filled up in $9$ ways each.
Therefore, there are $3 \times 9 \times 9 = 243$ such numbers.
Therefore, there are $2916 - 243 = 2673$ four-digit numbers containing exactly one $7$.
Therefore, the required probability is:
$\dfrac{225 + 648}{2673}$
$= \dfrac{25 + 72}{297} = \dfrac{97}{297}$
Problem 17 of 30
If $0 \lt x, y \lt \pi$ and $\cos x + \cos y - \cos{(x + y)} = \dfrac{3}{2}$, then $\sin x + \cos y$ is equal to:Solution:
$\cos x + \cos y - \cos{(x + y)} = \dfrac{3}{2}$
$\Rightarrow \cos x + \cos y - 1 - \cos{(x + y)} = \dfrac{1}{2}$
$\Rightarrow \cos x + \cos y - \left[1 + \cos \left( 2 \cdot \dfrac{(x + y)}{2} \right) \right] = \dfrac{1}{2}$
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Using the compound angle formulae $\cos (A + B) + \cos (A - B) = 2 \cos A \ \cos B$ and $\cos 2A = 2 \cos^2 \dfrac{A}{2} - 1$ from and , we get:
$2 \cos \left(\dfrac{x + y}{2} \right) \cos \left(\dfrac{x - y}{2} \right) - 2 \cos^2\left(\dfrac{x + y}{2}\right) = \dfrac{1}{2}$
$\Rightarrow 4 \cos \left(\dfrac{x + y}{2} \right) \cos \left(\dfrac{x - y}{2} \right) - 4 \cos^2\left(\dfrac{x + y}{2}\right) = 1$
$\Rightarrow 4 \cos^2\left(\dfrac{x + y}{2}\right) - 4 \cos \left(\dfrac{x + y}{2} \right) \cos \left(\dfrac{x - y}{2} \right) + 1 = 0$ $...eqn\ (i)$
Let $\cos \left(\dfrac{x + y}{2}\right) = a$
$\therefore 4a^2 - 4a \cos \left(\dfrac{x - y}{2} \right) + 1 = 0$
The above is a quadratic equation in $a$, and for $a$ to have a real solution,
$16 \cos^2 \left(\dfrac{x - y}{2} \right) - 16 \ge 0$
$\Rightarrow \cos^2 \left(\dfrac{x - y}{2} \right) - 1 \ge 0$
$\Rightarrow \cos^2 \left(\dfrac{x - y}{2} \right) \ge 1$
Therefore, the only possible value of $\cos \left(\dfrac{x - y}{2} \right)$ is $1$
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Therefore, $\dfrac{x - y}{2} = 0$
$\Rightarrow x = y$
Substituting $y = x$ and $\cos \left(\dfrac{x - y}{2} \right) = 1$ in $eqn\ (i)$,
$4 \cos^2\left(x\right) - 4 \cos \left(x \right) + 1 = 0$
$\Rightarrow (2 \cos x - 1)^2 = 0$
$\Rightarrow \cos x = \dfrac{1}{2}$
Therefore, $\sin x = \dfrac{\sqrt{3}}{2}$
Therefore,
$\sin x + \cos y = \sin x + \cos x = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}$
$= \dfrac{1 + \sqrt{3}}{2}$
Problem 18 of 30
Let $x$ denote the total number of one-one functions from a set $A$ with $3$ elements to a set $B$ with $5$ elements and $y$ denote the total number of one-one functions from set $A$ to the set $A \times B$. Then:Solution:
Number of one-one functions from $A$ to $B = x = \xaperm{5}{3} = 60$$A \times B = 3 \times 5 = 15$
Number of one-one functions from $A$ to $A \times B = y = \xaperm{15}{3} = 2730$
Therefore,
$\dfrac{y}{x} = \dfrac{2730}{60} = \dfrac{91}{2}$
$\Rightarrow2y = 91x$
Problem 19 of 30
$\text{cosec} \left[2\cot^{-1}(5) + \cos^{-1}\left( \dfrac{4}{5} \right)\right]$ is equal to:Solution:
Let $\cot \theta_1 = 5$ and
$\cos \theta_2 = \dfrac{4}{5}$
$\therefore \sin \theta_2 = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5}$
$\text{cosec} \left(2 \theta_1 + \theta_2 \right)$
$= \dfrac{1}{\sin \left(2 \theta_1 + \theta_2 \right)}$
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$= \dfrac{1}{\sin (2\theta_1) \cos (\theta_2) + \sin(\theta_2) \cos(2\theta_1)}$
$= \dfrac{5}{4 \times \sin (2\theta_1) + 3 \times \cos(2\theta_1)}$
$= \dfrac{5}{4 \times 2 \sin (\theta_1) \cos(\theta_1) + 3 \times (\cos^2 \theta_1 - sin^2 \theta_1)}$
$= \dfrac{5}{4 \times 2 \sin \theta_1 \cos\theta_1 + 3 \cos^2 \theta_1 - 3 sin^2 \theta_1}$
$= \dfrac{5}{4 \times 2 \times \dfrac{1}{\sqrt{26}} \times \dfrac{5}{\sqrt{26}} + 3 \times \dfrac{25}{26} - 3 \times \dfrac{1}{26}}$
$= \dfrac{5}{\dfrac{40}{26} + \dfrac{75}{26} - \dfrac{3}{26}}$
$= \dfrac{5 \times 26}{112}$
$= \dfrac{65}{56}$
Problem 20 of 30
The contrapositive of the statement $\unicode{0x201C} \text{If you will work, you will earn money} \unicode{0x201D}$ is:Solution:
$A =$ "You will work"
$B =$ "You will earn money"
Contrapositive of $A \longrightarrow B$ is $\neg B \longrightarrow \neg A$
The contrapositive of "If you will work, you will earn money" is "If you will not earn money, you will not work"
Problem 21 of 30
A function $f$ is defined on $[-3, 3]$ as$f(x) = \left\{\begin{array}{cc} \min\{|x|, 2-x^2\}, & -2 \le x \le 2 \\ \lfloor |x| \rfloor, & 2 \lt |x| \le 3 \end{array} \right.$
where $\lfloor x \rfloor$ denotes the greatest integer $\le x$. The number of points, where $f$ is not differentiable in $(-3, 3)$ is ______.
Solution:
We will solve this problem using simple visual inspection by plotting the approximate graphs of the functions.
Since we are required to find the non-differentiable points in the open interval $(-3, 3)$ we can redefine our function as:
$f(x) = \left\{\begin{array}{cc} \min\{|x|, 2-x^2\}, & -2 \le x \le 2 \\ \lfloor |x| \rfloor, & 2 \lt |x| \lt 3 \end{array} \right.$
Observe that we replaced that $|x| \le 3$ with $|x| \lt 3$
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The functions plotted independently looks approximately as shown in the following figure:
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After considering the intervals for which the functions are defined, they will look as shown below:
As we can see, there are two points of discontinuity and totally five points where the function is non-differentiable.
Problem 22 of 30
If the curve, $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y)dx + xdy = 0$, passes through the intersection of the lines, $2x - 3y = 1$ and $3x + 2y = 8$, then $|y(1)|$ is equal to ______.Solution:
$(2xy - y)dx + xdy = 0$$\Rightarrow (2xy - y) dx = -x dy$
$\Rightarrow \dfrac{dy}{dx} = \dfrac{2xy^2 - y}{-x}$
$\Rightarrow \dfrac{dy}{dx} - \dfrac{y}{x} = -2y^2$
$\Rightarrow y^{-2}\dfrac{dy}{dx} - \dfrac{y^{-1}}{x} = -2$
Let $z=y^{-1}$ and $1-n=-1$
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Integrating factor $= e^{(1-n)\int - \frac{1}{x}dx} = e^{log_e x} = x$
$\therefore xz = (1-n) \int -2x dx$
$\Rightarrow xz = x^2 + C$
$\Rightarrow z = x + \dfrac{C}{x}$
$\Rightarrow \dfrac{1}{y} = x + \dfrac{C}{x}$
The intersection point of $2x - 3y = 1$ and $3x +2y = 8$ is $(2,1).$ Therefore the curve passes through $(2,1)$
$\therefore \dfrac{1}{y} = x + \dfrac{C}{x}$
$\Rightarrow \dfrac{1}{1} = 2 + \dfrac{C}{2}$
$\Rightarrow C = -2$
$\therefore \dfrac{1}{y} = x - \dfrac{2}{x}$
$\Rightarrow y = \dfrac{1}{x - \dfrac{2}{x}}$
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$\Rightarrow y(1) = \dfrac{1}{1 - \dfrac{2}{1}}$
$\Rightarrow y(1) = -1$
$\Rightarrow |y(1)| = 1$
Problem 23 of 30
The total number of two digit numbers $'n'$, such that $3^n + 7^n$ is a multiple of $10$ is ______.Solution:
If $3^n + 7^n$ is divisible by $10$, then,$3^n + 7^n \equiv 0\ (mod\ 10)$
For $n = 0, 1, 2, 3, 4,...$
$3^n \equiv \{1, 3, 9, 7, 1,...\}\ (mod\ 10)$ and
$7^n \equiv \{1, 7, 9, 3, 1,...\}\ (mod\ 10)$
We can see that for $n = 1, 3, 5, 7,...$, that is, $n \equiv 1\ (mod\ 4)$ or $n \equiv 3\ (mod\ 4)$
$3^n + 7^n \equiv 0\ (mod\ 10)$
Therefore, $n$ has to be equal to either $4m + 1$ or $4m + 3$ for any integer $m$.
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Therefore, for $n = 4m + 1$
$9 \lt 4m + 1 \lt 100$
$\Rightarrow 8 \lt 4m \lt 99$
$\Rightarrow 2 \lt m \lt 24\dfrac{3}{4}$
$\Rightarrow 3 \le m \le 24$
Therefore, there are $22$ possible values of $m$ for this case.
Similarly, for $n = 4m + 3$,
$9 \lt 4m + 3 \lt 100$
$\Rightarrow 6 \lt 4m \lt 97$
$\Rightarrow 1\dfrac{2}{4} \lt 4m \lt 24{1}{4}$
$\Rightarrow 2 \le m \le 24$
Therefore, there are $23$ possible values of $m$ for this case.
Therefore, there are a total of $22 + 23 = 45$ possible $2$-digit values of $n$ satisfying the given condition.
Problem 24 of 30
If $\lim\limits_{x \rightarrow 0} \dfrac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$ exists and is equal to $b$, then the value of $a - 2b$ is ______.Solution:
$b = \lim\limits_{x \rightarrow 0} \dfrac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$Since the above limit is of the indeterminate form $\dfrac{0}{0}$, we can apply , and get:
$b = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}[ax - (e^{4x} - 1)]}{\dfrac{d}{dx}[ax(e^{4x} - 1)]}$
$b = \lim\limits_{x \rightarrow 0} \dfrac{a - 4e^{4x}}{ax(e^{4x}) + a(e^{4x}) - a}$
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$b = \dfrac{a-4}{0}$
Since it is given that the limit exists, the above must be of an indeterminate form, that is $\dfrac{0}{0}$
$\therefore a = 4 $
$\lim\limits_{x \rightarrow 0} \dfrac{a - 4e^{4x}}{ax(e^{4x}) + a(e^{4x}) - a}$
$= \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}(a - 4e^{4x})}{\dfrac{d}{dx}\left[ax(e^{4x}) + a(e^{4x}) - a \right]}$ (applying L'Hospitals rule again)
$=\lim\limits_{x \rightarrow 0} \dfrac{- 16e^{4x}}{ax(16e^{4x}) + a(4e^{4x}) + a(4e^{4x})}$
$= -\dfrac{2}{a}$
$b=-\dfrac{2}{a} = -\dfrac{1}{2}$
$a - 2b = 4 - 2(-\dfrac{1}{2}) = 5$
Problem 25 of 30
If the curves $x = y^4$ and $xy = k$ cut at right angles, then $(4k)^6$ is equal to ______.Solution:
Let $(x_1, y_1)$ be the intersection point of the two given curves, and $s_1$ and $s_2$ be the slopes of the two curves at $x_1, y_1$.Therefore,
$s_1 \times s_2 = -1$
Differentiating both sides of the equation of the first curve w.r.t $x$,
$\dfrac{d}{dx}(x) = \dfrac{d}{dx}(y^4)$
$\Rightarrow 1 = 4y^3 \dfrac{dy}{dx}$
$\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{4y^3}$
$\therefore s_1 = \dfrac{1}{4{y_1}^3}$
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Differentiating both sides of the equation of the second curve w.r.t $x$,
$\dfrac{d(xy)}{dx} = 0$
$\Rightarrow y\dfrac{dx}{dx} + x\dfrac{dy}{dx} = 0$
$\Rightarrow y + x\dfrac{dy}{dx} = 0$
$\Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}$
$\therefore s_2 = -\dfrac{y_1}{x_1}$
Since, $(x_1, y_1)$ satisfies both equations, putting them in the first equation gives us:
$x_1 = {y_1}^4$,
and because $s_1 \times s_2 = -1$
$\therefore -\dfrac{1}{4{y_1}^3} \times \dfrac{y_1}{x_1} = -1$
$\Rightarrow x_1 = \dfrac{1}{4{y_1}^2}$
$\therefore {y_1}^4 = \dfrac{1}{4{y_1}^2}$
$\Rightarrow y_1^6 = 2^{-2}$
$\Rightarrow y_1 = 2^{-\frac{1}{3}}$
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$\therefore x_1 = {y_1}^4 = 2^{-\frac{4}{3}}$
$\because x_1 y_1 = k$,
$k = 2^{-\frac{1}{3}} \times 2^{-\frac{4}{3}} = 2^{-\frac{5}{3}}$
$(4k)^6 = \left(2^2 \times 2^{-\frac{5}{3}}\right)^6$
$= \left(2^{\frac{1}{3}}\right)^6 = 2^2 = 4$
Problem 26 of 30
The value of $\displaystyle{\int\limits_{-2}^{2}} |3x^2 - 3x - 6| dx$ is ______.Solution:
We can rewrite the function inside the integral as:
$f(x) = 3|x^2 - x - 2|$
$= 3|(x - 2)(x + 1)|$
Therefore, $x^2 - x - 2$ is negative for $-1 \lt x \lt 2$ and positive for all other values of $x$.
We can write the function as:
$f(x) = \left\{\begin{array}{ll} -3(x^2 - x - 2) & \text{for } -1 \lt x \lt 2 \\ 3(x^2 - x - 2) & \text{otherwise} \end{array}\right.$
Therefore, we can break up the given integral into two parts:
$\displaystyle{\int\limits_{-2}^{2}} |3x^2 - 3x - 6| dx$
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$= \displaystyle \int\limits_{-2}^{-1} (3x^2 - 3x - 6) dx + \int\limits_{-1}^{2} -(3x^2 - 3x - 6) dx$
$= \left[ \dfrac{3x^3}{3} - \dfrac{3x^2}{2} - 6x \right]_{-2}^{-1} + \left[ -\dfrac{3x^3}{3} + \dfrac{3x^2}{2} + 6x \right]_{-1}^{2}$
$= \left[ \dfrac{3(-1)^3}{3} - \dfrac{3(-1)^2}{2} - 6(-1) \right] - \left[ \dfrac{3(-2)^3}{3} - \dfrac{3(-2)^2}{2} - 6(-2) \right] $
$+ \left[ -\dfrac{3(2)^3}{3} + \dfrac{3(2)^2}{2} + 6(2) \right] - \left[ -\dfrac{3(-1)^3}{3} + \dfrac{3(-1)^2}{2} + 6(-1) \right]$
$= \left[ -\dfrac{3(1)^3}{3} - \dfrac{3(1)^2}{2} + 6(1) \right] - \left[ -\dfrac{3(2)^3}{3} - \dfrac{3(2)^2}{2} + 6(2) \right] $
$+ \left[ -\dfrac{3(2)^3}{3} + \dfrac{3(2)^2}{2} + 6(2) \right] - \left[ \dfrac{3(1)^3}{3} + \dfrac{3(1)^2}{2} - 6(1) \right]$
$= -\dfrac{3(1)^3}{3} - \dfrac{3(1)^2}{2} + 6(1) +\cancel{\dfrac{3(2)^3}{3}} + \dfrac{3(2)^2}{2} - \cancel{6(2)}$
$- \cancel{\dfrac{3(2)^3}{3}} + \dfrac{3(2)^2}{2} + \cancel{6(2)} - \dfrac{3(1)^3}{3} - \dfrac{3(1)^2}{2} + 6(1)$
$= 3(2)^2 - 3(1)^2 - 2(1)^3 + 12(1)$
$= 19$
Problem 27 of 30
If the remainder when $x$ is divided by $4$ is $3$, then the remainder when $(2020 + x)^{2022}$ is divided by $8$ is ________.Solution:
We will solve this problem using the concept of .
$2020 \equiv 4 (mod \ 8).$
$\therefore 2020 + x \equiv 7$ or $11 (mod \ 8)$
$\Rightarrow 2020 + x \equiv -1$ or $3 (mod \ 8)$
$\Rightarrow (2020 + x)^{2022} \equiv (-1)^{2022}$ or $(3)^{2022} (mod \ 8)$
$Case \ I$
$(2020 + x)^{2022} \equiv (-1)^{2022} \equiv 1 (mod \ 8)$
$Case \ II$
$(2020 + x)^{2022} \equiv (3)^{2022} \equiv (3^2)^{1011} \equiv 1 (mod \ 8)$
Since both the cases lead to the value $(2020 + x)^{2022} \equiv 1\ (mod\ 8)$ the correct answer is $1$
Problem 28 of 30
A line $'l'$ passing through the origin is perpendicular to the lines$l_1 : \overrightarrow{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$
$l_2 : \overrightarrow{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$
If the co-ordinates of the point in the first octant on $'l_2'$ at a distance of $\sqrt{17}$ from the point of intersection of $'l'$ and $'l_1'$ are $(a, b, c)$ then $18(a + b + c)$ is equal to ______.
Solution:
$l_1 : \overrightarrow{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$
$\Rightarrow L_1: \dfrac{x - 3}{1} = \dfrac{y + 1}{2} = \dfrac{z - 4}{2} $
$\therefore$ direction ratios of $L_1 = (1, 2, 2)$
$l_2 : \overrightarrow{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$
$\Rightarrow L_2: \dfrac{x - 3}{2} = \dfrac{y - 3}{2} = \dfrac{z - 2}{1}$
$\therefore$ direction ratios of $L_2 = (2, 2, 1)$
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Let the direction ratios of required line, $L$, be $a, b, c.$
Since $L$ is perpendicular to $L_1$ and $L_2,$ we have,
$1a + 2b + 2c = 0$ $...(i)$
$2a + 2b + 1c = 0$ $...(ii)$
Upon solving for the direction ratios of $L$ from $(i)$ and $(ii)$ we get the ratios as $(-2, 3, -2)$
$L : \dfrac{x}{2} = \dfrac{y}{-3} = \dfrac{z}{2} $
Solving for the intersection point of $L$ and $L_1,$ we obtain,
$(2 \lambda, -3 \lambda, 2 \lambda) = (\mu+3, 2 \mu - 1, 2 \mu + 4)$
$\Rightarrow \mu = -1, \lambda = 1$
Therefore, the intersection point $P = (2, -3, 2)$
Let a point on $L_2$ be $Q = (2v + 3, 2v + 3, v + 2 )$
$PQ = \sqrt{(2v + 3 - 2 )^2 + (2v + 3 + 3)^2 + (v + 2 - 2)^2} = \sqrt{17} $
$\Rightarrow (2v + 1)^2 + (2v + 6)^2 + (v)^2 = 17$
$\Rightarrow 9v^2 + 28v + 20 = 0$
$\Rightarrow v = -2 $ or $v = - \dfrac{10}{9}$
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$\therefore Q = (-1, -1, 0)$ or $\left( \dfrac{7}{9}, \dfrac{7}{9}, \dfrac{8}{9} \right) $
Since $Q$ must lie in the first octant, $Q \neq (-1, -1, 0)$
$\therefore Q = \left( \dfrac{7}{9}, \dfrac{7}{9}, \dfrac{8}{9} \right)$
$18(a+b+c) = 18 \left( \dfrac{7}{9} + \dfrac{7}{9} + \dfrac{8}{9}\right) = 44$
Problem 29 of 30
A line is a common tangent to the circle $(x-3)^2 + y^2 = 9$ and the parabola $y^2 = 4x.$ If the two points of contact $(a,b)$ and $(c,d)$ are distinct and lie in the first quadrant, then $2(a+c)$ is equal to ________Solution:
We know from , that the tangent to a parabola $y^2 = 4ax$ is given by:
$y = mx + \dfrac{a}{m}$
In the given problem, the parabola is $y^2 = 4x$, therefore, the equation of the tangent will be:
$y = mx + \dfrac{1}{m}$
$\Rightarrow m^2x - my + 1 = 0$
The given circle is centred at $(3, 0)$ and has a radius of $3$.
Since the tangent $y = mx + \dfrac{1}{m}$ is also a tangent to the circle, its distance from the centre is $3$.
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Therefore,
$ \dfrac{m^2(3) - m(0) + 1}{\sqrt{m^4 + m^2}} = 3$
$\Rightarrow 3m^2 + 1 = 3\sqrt{m^4 + m^2}$
$\Rightarrow 9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$
$\Rightarrow 3m^2 = 1$
$\Rightarrow m = \pm \dfrac{1}{\sqrt{3}}$
Therefore, the equation of the tangent is:
$y = \pm \left( \dfrac{x}{\sqrt{3}} + \sqrt{3} \right)$
$y = \pm \left( \dfrac{x}{\sqrt{3}} + \sqrt{3} \right)$
$\Rightarrow y^2 = \dfrac{x^2}{3} + 2x + 3$
This line passes through $(a, b)$,
$\therefore b^2 = \dfrac{a^2}{3} + 2a + 3$
Since $(a, b)$ is on the circle,
$b^2 = 9 - (a - 3)^2$
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Therefore,
$9 - (a - 3)^2 = \dfrac{a^2}{3} + 2a + 3$
$\Rightarrow 9 - a^2 + 6a - 9 = \dfrac{a^2}{3} + 2a + 3$
$\Rightarrow 18a - 3a^2 = a^2 + 6a + 9$
$\Rightarrow 4a^2 - 12a + 9 = 0$
$\Rightarrow (2a - 3)^2 = 0$
$\Rightarrow a = \dfrac{3}{2}$
Similarly, since the point $(c, d)$ lies on the tangent and the parabola, we get:
$4c = \dfrac{c^2}{3} + 2c + 3$
$\Rightarrow 12c = c^2 + 6c + 9$
$\Rightarrow c^2 - 6c + 9 = 0$
$\Rightarrow (c - 3)^2 = 0$
$\Rightarrow c = 3$
Therefore,
$2(a + c) = 2 \left( \dfrac{3}{2} + 3 \right)= 9$
Problem 30 of 30
Let $\overrightarrow{a} = \hat{i} + \alpha \hat{j} + 3\hat{k}$ and $\overrightarrow{b} = 3\hat{i} - \alpha \hat{j} + \hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $8\sqrt{3}$ square units, then $\overrightarrow{a} \cdot \overrightarrow{b}$ is equal to ______.Solution:
Area, $A$, of parallelogram with adjacent sides $\overrightarrow{a}$ and $\overrightarrow{b}$, $= |\overrightarrow{a} \times \overrightarrow{b}|$$\Rightarrow A = |(\hat{i} + \alpha \hat{j} + 3 \hat{k}) \times (3 \hat{i} - \alpha \hat{j} + \hat{k})|$
$\Rightarrow A = |4\alpha \hat{i} + 8 \hat{j} - 4 \alpha \hat{k}|$
$\therefore 8 \sqrt{3} = \sqrt{(4 \alpha)^2 + (8)^2 + (-4 \alpha)^2 }$
$\Rightarrow 64 \times 3 = 16\alpha ^2 + 64 + 16\alpha^2$
$\Rightarrow \alpha^2 = 4 $
Therefore, $\overrightarrow{a}\cdot\overrightarrow{b} = (\hat{i} + \alpha \hat{j} + 3 \hat{k})\cdot (3 \hat{i} - \alpha \hat{j} + \hat{k}) = 3 - \alpha^2 + 3 = 2$