The point A (2, 7) lies on the perpendicular
Question: The point A (2, 7) lies on the perpendicular bisector of the line segment joining the points P (5, 3) and Q (0, 4). Solution: False If $A(2,7)$ lies on perpendicular bisector of $P(6,5)$ and $Q(0,-4)$, then $A P=A Q$ $\therefore$ $A P=\sqrt{(6-2)^{2}+(5-7)^{2}}$ $=\sqrt{(4)^{2}+(-2)^{2}}$ $=\sqrt{16+4}=\sqrt{20}$ and $A=\sqrt{(0-2)^{2}+(-4-7)^{2}}$ $=\sqrt{(-2)^{2}+(-11)^{2}}$ $=\sqrt{4+121}=\sqrt{125}$ So, A does not lies on the perpendicular bisector of PQ. Alternate Method If the po...
Read More →The function
Question: The function $f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$ is not defined for $x=2$. In order to make $f(x)$ continuous at $x=2$, Here $f(2)$ should be defined as (a) 0 (b) 1 (C) 2 (d) 3 Solution: Here, $x^{3}+x^{2}-16 x+20$ $=x^{3}-2 x^{2}+3 x^{2}-6 x-10 x+20$ $=x^{2}(x-2)+3 x(x-2)-10(x-2)$ $=(x-2)\left(x^{2}+3 x-10\right)$ $=(x-2)(x-2)(x+5)$ $=(x-2)^{2}(x+5)$ So, the given function can be rewritten as $f(x)=\frac{(x-2)^{2}(x+5)}{x-2}$ $\Rightarrow f(x)=(x-2)(x+5)$ If $f(x)$ is continuous at...
Read More →A circle has its centre at the origin and a point
Question: A circle has its centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle. Solution: True First,we draw a circle and a point from the given information Now, distance between origin i.e., $O(0,0)$ and $P(5,0), O P=\sqrt{(5-0)^{2}+(0-0)^{2}}$ $\left[\because\right.$ Distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$ $=\sqrt{5^{2}+0^...
Read More →The function
Question: The function $f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$ is not defined for $x=2$. In order to make $f(x)$ continuous at $x=2$, Here $f(2)$ should be defined as (a) 0 (b) 1 (C) 2 (d) 3 Solution: Here, $x^{3}+x^{2}-16 x+20$ $=x^{3}-2 x^{2}+3 x^{2}-6 x-10 x+20$ $=x^{2}(x-2)+3 x(x-2)-10(x-2)$ $=(x-2)\left(x^{2}+3 x-10\right)$ $=(x-2)(x-2)(x+5)$ $=(x-2)^{2}(x+5)$ So, the given function can be rewritten as $f(x)=\frac{(x-2)^{2}(x+5)}{x-2}$ $\Rightarrow f(x)=(x-2)(x+5)$ If $f(x)$ is continuous at...
Read More →The points A(4, 3), B(6, 4), C(5, – 6)
Question: The points A(4, 3), B(6, 4), C(5, 6) and D(- 3, 5) are vertices of a parallelogram. Solution: False Now, distance between $A(4,3)$ and $B(6,4), A B=\sqrt{(6-4)^{2}+(4-3)^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$ $\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$ Distance between $B(6,4)$ and $C(5,-6), B C=\sqrt{(5-6)^{2}+(-6-4)^{2}}$ $=\sqrt{(-1)^{2}+(-1...
Read More →Which of the two rational numbers is greater in the given pair?
Question: Which of the two rational numbers is greater in the given pair? (i) $\frac{-4}{3}$ or $\frac{-8}{7}$ (ii) $\frac{7}{-9}$ or $\frac{-5}{8}$ (iii) $\frac{-1}{3}$ or $\frac{4}{-5}$ (iv) $\frac{9}{-13}$ or $\frac{7}{-12}$ (v) $\frac{4}{-5}$ or $\frac{-7}{10}$ (vi) $\frac{-12}{5}$ or $-3$ Solution: 1. The two rational numbers are $\frac{-4}{3}$ and $\frac{-8}{7}$. The LCM of the denominators 3 and 7 is 21. Now, $\frac{-4}{3}=\frac{-4 \times 7}{3 \times 7}=\frac{-28}{21}$ Also, $\frac{-8}{7}...
Read More →Solve this
Question: Let $f(x)=\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}$. The value which should be assigned to $f(x)$ at $x=\frac{\pi}{4}$, so that it is continuous everywhere is (a) 1 (b) $1 / 2$ (c) 2 (d) none of these Solution: (b) $\frac{1}{2}$ If $f(x)$ is continuous at $x=\frac{\pi}{4}$, then $\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$ $\Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}=f\left(...
Read More →In a survey of 100 students, the number of students studying the various
Question: In a survey of 100 students, the number of students studying the various languages is found as English only 18; English but not Hindi 23; English and Sanskrit 8; Sanskrit and Hindi 8; English 26; Sanskrit 48 and no language 24. Find (i) how many students are studying Hindi? (ii) how many students are studying English and Hindi both? Solution: Given: - Total number of students $=100$ - Number of students studying English(E) only $=18$ - Number of students learning English but not Hindi(...
Read More →The points A(3,1), B(12, – 2)
Question: The points A(3,1), B(12, 2) and C(0, 2) cannot be vertices of a triangle. Solution: True Let $\quad A \equiv\left(x_{1}, y_{1}\right) \equiv(3,1), B \equiv\left(x_{2}, y_{2}\right) \equiv(12,-2)$ and $\quad C=\left(x_{3}, y_{3}\right)=(0,2)$ $\therefore \quad$ Area of $\triangle A B C \Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[3-(2-2)+12(2-1)+0\{1-(-2)\}]$ $=\frac{1}{3}[3(-4)+12(1)+0]$ $=\frac{...
Read More →Solve this
Question: If $f(x)=\frac{1}{1-x}$, then the set of points discontinuity of the function $f(f(f(x)))$ is (a) $\{1\}$ (b) $\{0,1\}$ (c) $\{-1,1\}$ (d) none of these Solution: (b) $\{0,1\}$ Given: $f(x)=\frac{1}{1-x}$ Clearly, $f: R-\{1\} \rightarrow R$ Now, $f(f(x))=f\left(\frac{1}{1-x}\right)=\left(\frac{1}{1-\left(\frac{1}{1-x}\right)}\right)=\left(\frac{1-x}{-x}\right)=\left(\frac{x-1}{x}\right)$ $\therefore$ fof $: R-\{0,1\} \rightarrow R$ Now, $f(f(f(x)))=f\left(\frac{x-1}{x}\right)=\left(\fr...
Read More →Point P(0, 2) is the point of intersection
Question: Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(-l, 1) and B(3, 3). Solution: False We know that, the points lies on perpendicular bisector of the line segment joining the two points is equidistant from these two points. $\therefore$ $P A=\sqrt{[-4-(4)]^{2}+(6-2)^{2}}$ $=\sqrt{(0)^{2}+(4)^{2}}=4$ $P B=\sqrt{[-4-4]^{2}+(-6-2)^{2}} \sqrt{(0)^{2}+(-8)^{2}}=8$ $\because$ $P A \neq P B$ So, the point $P$ does not lie on th...
Read More →The value of b for which the function
Question: The value ofbfor which the function $f(x)=\left\{\begin{array}{ll}5 x-4, 0x \leq 1 \\ 4 x^{2}+3 b x, 1x2\end{array}\right.$ is continuous at every point of its domain, is (a) $-1$ (b) 0 (c) $\frac{13}{3}$ (d) 1 Solution: (a) $-1$ Given: $f(x)$ is continuous at every point of its domain. So, it is continuous at $x=1$. $\Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=f(1)$ $\Rightarrow \lim _{h \rightarrow 0} f(1+h)=f(1)$ $\Rightarrow \lim _{h \rightarrow 0}\left(4(1+h)^{2}+3 b(1+h)\right)=...
Read More →The value of b for which the function
Question: The value ofbfor which the function $f(x)=\left\{\begin{array}{ll}5 x-4, 0x \leq 1 \\ 4 x^{2}+3 b x, 1x2\end{array}\right.$ is continuous at every point of its domain, is (a) $-1$ (b) 0 (c) $\frac{13}{3}$ (d) 1 Solution: (a) $-1$ Given: $f(x)$ is continuous at every point of its domain. So, it is continuous at $x=1$. $\Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=f(1)$ $\Rightarrow \lim _{h \rightarrow 0} f(1+h)=f(1)$ $\Rightarrow \lim _{h \rightarrow 0}\left(4(1+h)^{2}+3 b(1+h)\right)=...
Read More →The points (0, 5), (0, -9)
Question: The points (0, 5), (0, -9) and (3, 6) are collinear. Solution: False Here, $x_{1}=0, x_{2}=0, x_{3}=3$ and $y_{1}=5, y_{2}=-9, y_{3}=6$ $\because \quad$ Area of triangle $\Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $\therefore \quad \Delta=\frac{1}{2}[0(-9-6)+0(6-5)+3(5+9)]$ $=\frac{1}{2}(0+0+3 \times 14)=21 \neq 0$ If the area of triangle formed by the points (0, 5), (0 9) and (3, 6) is zero, then the points...
Read More →If the function
Question: If the function $f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$ is continuous at each point of its domain, then the value off(0) is (a) 2 (b) $\frac{1}{3}$ (c) $-\frac{1}{3}$ (d) $\frac{2}{3}$ Solution: (b) $\frac{1}{3}$ Given: $f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$ If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{x\left(2-...
Read More →If the function
Question: If the function $f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$ is continuous at each point of its domain, then the value off(0) is (a) 2 (b) $\frac{1}{3}$ (c) $-\frac{1}{3}$ (d) $\frac{2}{3}$ Solution: (b) $\frac{1}{3}$ Given: $f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$ If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{x\left(2-...
Read More →Which of the two rational numbers is greater in the given pair?
Question: Which of the two rational numbers is greater in the given pair? (i) $\frac{3}{8}$ or 0 (ii) $\frac{-2}{9}$ or 0 (iii) $\frac{-3}{4}$ or $\frac{1}{4}$ (iv) $\frac{-5}{7}$ or $\frac{-4}{7}$ (v) $\frac{2}{3}$ or $\frac{3}{4}$ (vi) $\frac{-1}{2}$ or $-1$ Solution: We know: (i) Every positive rational number is greater than 0. (ii) Every negative rational number is less than 0. Thus, we have: (i) $\frac{3}{8}$ is a positive rational number. $\therefore \frac{3}{8}0$ (ii) $\frac{-2}{9}$ is a...
Read More →The point P(- 4, 2) lies on the line
Question: The point P(- 4, 2) lies on the line segment joining the points A(- 4, 6) and B(- 4, 6). Solution: True We plot all the points P(-4,2), A(-4, 6) and B(-4, 6) on the graph paper, From the figure, point P(- 4,2) lies on the line segment joining the points A(- 4,6) and B(- 4, 6),...
Read More →In a survey of 60 people, it was found that 25 people read newspaper
Question: In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read the newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, and 3 read all the three newspapers. Find (i) The number of people who read at least one of the newspapers, (ii) The number of people who read exactly one newspaper. Solution: Given: - Total number of people $=60$ - Number of people who read newspaper $\mathrm{H}=25$ - Number of people who read newspaper $\...
Read More →Δ ABC with vertices A(0 ,- 2,0), B(2, 0)
Question: Δ ABC with vertices A(0 ,- 2,0), B(2, 0) and C(0,2) is similar to ΔDEF with vertices D (- 4, 0), E(4, 0) and F(0, 4). Solution: True $\therefore$ Distance between $A(2,0)$ and $B(2,0), A B=\sqrt{[2-(2)]^{2}+(0-0)^{2}}=4$ $\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$ Similarly, distance between $B(2,0)$ and $C(0,2), B C=\sqrt{(0-2)^{2}+(2-0)^...
Read More →The function
Question: The function $f(x)=\left\{\begin{array}{ll}\frac{\sin 3 x}{x}, x \neq 0 \\ \frac{k}{2} , x=0\end{array}\right.$ is continuous at $x=0$, then $k=$ (a) 3 (b) 6 (c) 9 (d) 12 Solution: (b) 6 Given: $f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{x} \\ \frac{k}{2}, x=0\end{array}, x \neq 0\right.$ If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=f(0)$ $\Rightarrow 3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=\f...
Read More →The function f (x) = tan x is discontinuous on the set
Question: The functionf(x) = tanxis discontinuous on the set (a) $\{n \pi: n \in Z\}$ (b) $\{2 n \pi: n \in Z\}$ (c) $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$ (d) $\left\{\frac{n \pi}{2}: n \in Z\right\}$ Solution: (c) $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$ When $\tan (2 n+1) \frac{\pi}{2}=\tan \left(n \pi+\frac{\pi}{2}\right)=-\cot (n \pi)$, it is not defined at the integral points. $[n \in Z]$ Hence, $f(x)$ is discontinuous on the set $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right...
Read More →Express each of the following rational numbers in standard form:
Question: Express each of the following rational numbers in standard form: (i) $\frac{-12}{30}$ (ii) $\frac{-14}{49}$ (iii) $\frac{24}{-64}$ (iv) $\frac{-36}{-63}$ Solution: A rational number $\frac{a}{b}$ is said to be in the standard form if $a$ and $b$ have no common divisor other than unity and $b0$. Thus, (i) The greatest common divisor of 12 and 30 is 6 . $\therefore \frac{-12}{30}=\frac{-12 \div 6}{30 \div 6}=\frac{-2}{5}$ (In the standard form) (ii)The greatest common divisor of 14 and 4...
Read More →The function f (x) = tan x is discontinuous on the set
Question: The functionf(x) = tanxis discontinuous on the set (a) $\{n \pi: n \in Z\}$ (b) $\{2 n \pi: n \in Z\}$ (c) $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$ (d) $\left\{\frac{n \pi}{2}: n \in Z\right\}$ Solution: (c) $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$ When $\tan (2 n+1) \frac{\pi}{2}=\tan \left(n \pi+\frac{\pi}{2}\right)=-\cot (n \pi)$, it is not defined at the integral points. $[n \in Z]$ Hence, $f(x)$ is discontinuous on the set $\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right...
Read More →The value of a for which the function
Question: The value of a for which the function $f(x)=\left\{\begin{array}{ll}\frac{\left(4^{x}-1\right)^{3}}{\sin (x / a) \log \left\{\left(1+x^{2} / 3\right)\right\}}, x \neq 0 \\ 12(\log 4)^{3} , x=0\end{array}\right.$ may be continuous at $x=0$ is (a) 1 (b) 2 (c) 3 (d) none of these Solution: (d) none of these For $f(x)$ to be continuous at $x=0$, we must have $\lim _{x \rightarrow 0} f(x)=f(0)$ $\lim _{x \rightarrow 0}\left[\frac{\left(4^{x}-1\right)^{3}}{\sin \frac{x}{a} \log \left(1+\frac...
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