The distance between the points (0, 5)
Question: The distance between the points (0, 5) and (- 5, 0) is (a) 5 (b) 52 (c)25 (d) 10 Solution: $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ Here, $x_{1}=0, y_{1}=5$ and $x_{2}=-5, y_{2}=0$ $\therefore$ Distance between the points $(0,5)$ and $(-5,0)$ $=\sqrt{(-5-0)^{2}+(0-5)^{2}}$ $=\sqrt{25+25}=\sqrt{50}=5 \sqrt{2}$...
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Question: If $f(x)=\left\{\begin{array}{cc}\frac{1-\sin x}{(\pi-2 x)^{2}} \cdot \frac{\log \sin x}{\log \left(1+\pi^{2}-4 \pi x+4 x^{2}\right)}, x \neq \frac{\pi}{2} \\ k , x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\pi / 2$, then $\mathrm{k}=$ (a) $-\frac{1}{16}$ (b) $-\frac{1}{32}$ (c) $-\frac{1}{64}$ (d) $-\frac{1}{28}$ Solution: (c) $\frac{-1}{64}$ If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then $\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$ If $\frac{\...
Read More →If you have a collection of 6 pairs of white socks and 3 pairs of black socks.
Question: If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white? (ii) black? Solution: Number of pairs of white socks $=6$ Number of pairs of black socks $=3$ Total number of pairs of socks $=6+3=9$ $\therefore$ Number of possible outcomes $=9$ (i) Let $A$ be the event of getting a pair of white socks. $\therefore \mathrm{P}(\mathrm{A})=\frac{6}{9}=\frac{2}{3}$ (ii) Let $\mathrm{B}$ be the event of...
Read More →The distance of the point P(- 6, 8)
Question: The distance of the point P(- 6, 8) from the origin is (a) 8 (b) 27 (c) 10 (d) 6 Solution: (c)Distance between the points (x1,y2)and (x2, y2) $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ Here, $\quad x_{1}=-6, y_{1}=8$ and $x_{2}=0, y_{2}=0$ $\therefore$ Distance between $P(-6,8)$ and origin i.e., $O(0,0)$, $P O=\sqrt{[0-(-6)]^{2}+(0-8)^{2}}$ $=\sqrt{(6)^{2}+(-8)^{2}}$ $=\sqrt{36+64}=\sqrt{100}=10$...
Read More →Let A ⊂ B ⊂ U. Exhibit it in a Venn diagram.
Question: Let $A \subset B \subset U$. Exhibit it in a Venn diagram. Solution: Given: A\subsetB\subsetU. Corresponding Venn diagram...
Read More →If we have 15 boys and 5 girls in a class which carries a higher probability?
Question: If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value? Solution: Number of boys in the class $=15$ Number of girls in the class $=5$ Total number of students in the class $=15+5=20$ $\therefore$ Number of possible outcomes $=20$ Since the number of boys is more than the number of girls, boys will have a higher probability. Hence, there is the higher probability of getting a copy belonging to a ...
Read More →Let A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}.
Question: Let A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}. Using Venn diagrams, verify the following (i) $A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$ (ii) $A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$. Solution: (i) Given: A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}. $B^{\cap} C=\{e, o\}$ and $A^{U}\left(B^{\cap} C\right)=\{a, e, I, o, u\}$ RHS: $A^{U} B=\{a, d, e, I, o, u, v\}$ and $A^{U} C=\{a, e, I, o, u, t, m\}$ $\left(A^{U} B\right) \cap(A \cup C)=\{a, e, ...
Read More →The distance between the points A(0, 6)
Question: The distance between the points A(0, 6) and 5(0,- 2) is (a) 6 (b) 8 (c) 4 (d) 2 Solution: (b)v Distance between the points (x1, y1) and (x2, y2), $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ Here,$x_{1}=0, y_{1}=6$ and $x_{2}=0, y_{2}=-2$ $\therefore$ Distance between $A(0,6)$ and $B(0,-2)$, $A B=\sqrt{(0-0)^{2}+(-2-6)^{2}}$ $=\sqrt{0+(-8)^{2}}=\sqrt{8^{2}}=8$...
Read More →If you put 21 consonants and 5 vowels in a bag. What would carry greater probability?
Question: If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability. Solution: Number of consonants $=21$ Number of vowels $=5$ Total number of possible outcomes $=21+5=26$ Let $\mathrm{C}$ be the event of getting a consonant and $\mathrm{V}$ be the event of getting a vowel. $\therefore \mathrm{P}(\mathrm{C})=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{21}{26}$ And, $\ma...
Read More →The distance of the point P(2, 3)
Question: The distance of the point P(2, 3) from the X-axis is (a) 2 (b) 3 (c) 1 (d) 5 Solution: (b)We know that, if (x, y) is any point on the cartesian plane in first quadrant. Then, x = Perpendicular distance from Y-axis and y = Perpendicular distance from X-axis Distance of the point P(2, 3) from the X-axis = Ordinate of a point P(2, 3)= 3....
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Question: If $f(x)$ defined by $f(x)=\left\{\begin{array}{ll}\frac{\left|x^{2}-x\right|}{x^{2}-x}, x \neq 0,1 \\ 1 , \quad x=0 \\ -1 , \quad x=1\end{array}\right.$ then $f(x)$ is continuous for all (a) $x$ (b) $x$ except at $x=0$ (c) $x$ except at $x=1$ (d) $x$ except at $x=0$ and $x=1$. Solution: (d) $x$ except at $x=0$ and $x=1$. Given: $f(x)=\left\{\begin{array}{cc}\frac{\left|x^{2}-x\right|}{x^{2}-x} , x \neq 0,1 \\ 1 , x=0 \\ -1 , x=1\end{array}\right.$ $\Rightarrow f(x)=\left\{\begin{array...
Read More →A bag contains 5 red marbles, 8 white marbles, 4 green marbles.
Question: A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be (i) red (ii) white (iii) not green Solution: Number of red marbles $=5$ Number of white marbles $=8$ Number of green marbles $=4$ Total number of marbles in the bag $=5+8+4=17$ $\therefore$ Total number outcomes $=17$ (i) Let $A$ be the event of drawing a red ball. $\therefore \mathrm{P}(\mathrm{A})=\frac{\text { Number of favourable o...
Read More →Prove that the area of the equilateral triangle
Question: Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangle drawn on the other two sides of the triangle. Solution: Let a right triangle BAC in which A is right angle and AC = y, AB = x. Three equilateral triangles ΔAEC, Δ AFB and ΔCBD are drawn on the three sides of ΔABC. Again let area of triangles made on AC, AS and BC are A1, A2and A3, respectively. To prove A3= A1+ A2 Proof $\ln \De...
Read More →A bag contains 3 red balls and 5 black balls.
Question: A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) red (ii) black Solution: Number of red balls $=3$ Number of black balls $=5$ Total number of balls $=3+5=8$ (i) Let $A$ be the event of drawing a red ball. $\therefore P(A)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{8}$ (ii) Let $\mathrm{B}$ be the event of drawing a black ball. $\therefore P(B)=\frac{...
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Question: If $f(x)=\left\{\begin{array}{cl}\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}, x \neq 0 \\ k , x=0\end{array}\right.$ is continuous at $x=0$, then $k$ equals (a) $16 \sqrt{2} \log 2 \log 3$ (b) $16 \sqrt{2} \ln 6$ (c) $16 \sqrt{2} \ln 2 \ln 3$ (d) none of these Solution: (c) $16 \sqrt{2} \ln 2 \ln 3$ Given: $f(x)=\left\{\begin{array}{l}\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}, x \neq 0 \\ k, x=0\end{array}\right.$ If $f(x)$ is continuous at $x=0$, then $\lim _{x \...
Read More →A bag contains 4 red, 5 black and 6 white balls.
Question: A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: (i) white (ii) red (iii) not black (iv) red or white Solution: Number of red balls $=4$ Number of black balls $=5$ Number of white balls $=6$ Total number of balls in the bag $=4+5+6=15$ Therefore, the total number of cases is 15 . (ii) Let A denote the event of getting a white ball. Number of favourable outcomes, i.e. number of white balls $=6$ $\mathrm...
Read More →Prove that the area of the semi-circle drawn
Question: Prove that the area of the semi-circle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semi-circles drawn on the other two sides of the triangle. Solution: Let ABC be a right triangle, right angled at B and AB = y, BC = x. Three semi-circles are drawn on the sides AB, BC and AC, respectively with diameters AB, BC and AC, respectively. Again, let area of circles with diameters AB, BC and AC are respectively A1, A2and A3. To prove $A_{3}=A_{1}+A...
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Question: If $f(x)=\left|\log _{10} x\right|$, then at $x=1$ (a) $f(x)$ is continuous and $f\left(1^{+}\right)=\log _{10} e$ (b) $f(x)$ is continuous and $f\left(1^{+}\right)=\log _{10} e$ (c) $f(x)$ is continuous and $f\left(1^{-}\right)=\log _{10} e$ (d) $f(x)$ is continuous and $f\left(1^{-}\right)=-\log _{10} e$ Solution: (a) $f(x)$ is continuous and $f^{\prime}\left(1^{+}\right)=\log _{10} e$ (d) $f(x)$ is continuous and $f^{\prime}\left(1^{-}\right)=-\log _{10} e$ Given: $f(x)=\left|\log _...
Read More →Let A = {2, 4, 6, 8, 10}, B = {4, 8, 12, 16} and C = {6, 12, 18, 24}
Question: Let A = {2, 4, 6, 8, 10}, B = {4, 8, 12, 16} and C = {6, 12, 18, 24}. Using Venn diagrams, verify that: (i) $(A \cup B) \cup C=A \cup(B \cup C)$ (ii) $(A \cap B) \cap C=A \cap(B \cap C)$. Solution: (i) LHS: $A^{U} B=\{2,4,6,8,10,12,16\}\left(A^{U} B\right)^{U} C=\{2,4,6,8,10,12,16,18,24\}$ A B RHS: $\mathrm{B}^{U} \mathrm{C}=\{4,6,8,10,12,16,18,24\}$ $A^{U}\left(B^{U} C\right)=\{2,4,6,8,10,12,16,18,24\}$ LHS = RHS. [Verified] (ii) LHS: $A^{\cap} B=\{4,8\}\left(A^{\cap} B\right)^{\cap} ...
Read More →A bag contains 5 white and 7 red balls.
Question: A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white? Solution: Number of white balls $=5$ Number of red balls $=7$ Total number of balls $=5+7=12$ $\therefore$ The total number of cases $=12$ Again, there are 5 white balls. Therefore, the number of favourable outcomes is 5 . $\therefore \mathrm{P}($ the drawn ball is white $)=\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{5}{12}$...
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Question: If $f(x)=|x-a| \phi(x)$, where $\phi(x)$ is continuous function, then (a) $f\left(a^{+}\right)=\phi(a)$ (b) $f\left(a^{-}\right)=-\phi(a)$ (c) $f\left(a^{+}\right)=f\left(a^{-}\right)$ (d) none of these Solution: (a) $f^{\prime}\left(a^{+}\right)=\phi(a)$ (b) $f^{\prime}\left(a^{-}\right)=-\phi(a)$ Here, $f(x)=|x-a| \phi(x)$ $f^{\prime}\left(a^{+}\right)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _{h \rightarrow 0} \frac{|h+a-a| \phi(a+h)-|a-a| \phi(a)}{h}=\lim _{h \rightarrow ...
Read More →A bag contains 6 red, 8 black and 4 white balls.
Question: A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black? Solution: Number of red balls $=6$ Number of black balls $=8$ Number of white balls $=4$ Total number of balls $=6+8+4=18$ $\therefore$ Total number of cases $=18$ Again, number of balls that are not black $=18-8=10$ Thus, the number of favourable cases is 10 . $\therefore P($ the drawn ball is not black $)=\frac{\text { Number of favourable cases }}{\text {...
Read More →In figure, line segment DF intersects the side AC
Question: In figure, line segment DF intersects the side AC of a ΔABC at the point E such that E is the mid-point of CA and $\angle \mathrm{AEF}=\angle \mathrm{AFE}$. Prove that $\frac{B D}{C D}=\frac{B F}{C F}$. Solution: Given $\triangle A B C, E$ is the mid-point of $C A$ and $\angle A E F=\angle A F E$ To prove $\frac{B D}{C D}=\frac{B F}{C F}$ Construction Take a point $G$ on $A B$ such that $C G \| E F$. Proof Since, $E$ is the mid-point of $C A$. $\therefore$ $C E=A E$ ...(i) In $\triangl...
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Question: The function $f(x)=\frac{4-x^{2}}{4 x-x^{3}}$ (a) discontinuous at only one point (b) discontinuous exactly at two points (c) discontinuous exactly at three points (d) none of these Solution: (C) discontinuous exactly at three points Given: $f(x)=\frac{4-x^{2}}{4 x-x^{3}}$ $\Rightarrow f(x)=\frac{4-x^{2}}{x\left(4-x^{2}\right)}$ $\Rightarrow f(x)=\frac{1}{x}, x \neq 0$ and $4-x^{2} \neq 0$ or $x \neq 0, \pm 2$ Clearly, $f(x)$ is defined and continuous for all real numbers except $\{0, ...
Read More →What is the probability that a number selected from the numbers 1, 2, 3, ...,
Question: What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4? Solution: There are 15 numbers from $1,2, \cdots, 15$. Hence, the total number of cases is 15 . Again, the multiples of 4 are 4,8 and 12 . Therefore, the total number of favourable cases is 3 . $\therefore \mathrm{P}($ the number is a multiple of 4$)=\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{3}{15}=\frac{1}{5}$...
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