The points A(x1, y1), B(x2 y2) and C(x3, y3)
Question: The points A(x1, y1), B(x2y2) and C(x3, y3) are the vertices of ΔABC. (i) The median from A meets BC at Find the coordinates of the point D. (ii) Find the coordinates of the point P on AD such that AP : PD = 2:1 (iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ:QE = 2:1 and CR:RF = 2:1 (iv ) What are the coordinates of the centroid of the ΔABC? Solution: Given that, the points A (x1,y1), B(x2, y2)andC (x3, y3)are the vertices of ΔABC. (i)We kn...
Read More →If A = {2, 3, 5} and B = {5, 7}, find:
Question: If A = {2, 3, 5} and B = {5, 7}, find: (i) $A \times B$ (ii) $\mathbf{B} \times \mathbf{A}$ (iii) $\mathbf{A} \times \mathbf{A}$ (iv) $\mathrm{B} \times \mathrm{B}$ Solution: (i) Given: A = {2, 3, 5} and B = {5, 7} To find: A B By the definition of the Cartesian product, Given two non empty sets P and Q. The Cartesian product P Q is the set of all ordered pairs of elements from P and Q, .i.e. $P \times Q=\{(p, q): p \in P, q \in Q\}$ Here, $A=\{2,3,5\}$ and $B=\{5,7\}$. So, $A \times B...
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Question: If $f(x)=x \sin \left(\frac{\pi}{4}\right)$ is everywhere continuous, then $f(0)=$______ Solution: $f(x)=x \sin \left(\frac{\pi}{4}\right)=x \times \frac{1}{\sqrt{2}}=\frac{x}{\sqrt{2}}$ Thus,f(x) is a polynomial function which is continuous everywhere. So, $f(x)$ is continuous at $x=0$. $\therefore f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{x}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times 0=0$ If $f(x)=x \sin \left(\frac{\pi}{4}\right)$ is everywhere continuous, then $f(0)=...
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) The product of a rational number and its reciprocal is ....... (ii) Zero has ....... reciprocal. (iii) The numbers ....... and ....... are their own reciprocals. (iv) zero is ....... the reciprocal of any number. (v) The reciprocal ofa, wherea 0, is ....... (vi) The reciprocal of $\frac{1}{a}$, where $a \neq 0$, is (vii) The reciprocal of a positive rational rational number is ....... (viii) The reciprocal of a negative rational number is ....... Solution: (i)1 ...
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Question: Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)=$ Solution: The given function is $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. It is given that, the function $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$. So, the function is continuous at $x=\frac{\pi}{4}$ $\therefore f\left(\frac{\pi}{4}\rig...
Read More →A(6,1), B (8, 2) and C(9, 4) are three vertices
Question: A(6,1), B (8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, then find the area of ΔADE. Solution: Given that, A (6,1), B (8,2) and C (9, 4) are three vertices of a parallelogram ABCD. Let the fourth vertex of parallelogram be (x, y). We know that, the diagonals of a parallelogram bisect each other. $\therefore$ Mid-point of $B D=$ Mid-point of $A C$ $\Rightarrow$$\left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{6+9}{2}, \frac{1+4}{2}\right...
Read More →Name the property of multiplication illustrated by each of the following statements:
Question: Name the property of multiplication illustrated by each of the following statements: (i) $\frac{-15}{8} \times \frac{-12}{7}=\frac{-12}{7} \times \frac{-15}{8}$ (ii) $\left(\frac{-2}{3} \times \frac{7}{9}\right) \times \frac{-9}{5}=\frac{-2}{3} \times\left(\frac{7}{9} \times \frac{-9}{5}\right)$ (iii) $\frac{-3}{4} \times\left(\frac{-5}{6}+\frac{7}{8}\right)=\left(\frac{-3}{4} \times \frac{-5}{6}\right)+\left(\frac{-3}{4} \times \frac{7}{8}\right)$ (iv) $\frac{-16}{9} \times 1=1 \times...
Read More →If P = {a, b} and Q = {x, y, z}, show that P × Q ≠ Q × P.
Question: If $P=\{a, b\}$ and $Q=\{x, y, z\}$, show that $P \times Q \neq Q \times P$. Solution: Given: P = {a, b} and Q = {x, y, z} To show: $P \times Q \neq Q \times P$ Now, firstly we find the $P \times Q$ and $Q \times P$ By the definition of the Cartesian product, Given two non - empty sets $P$ and $Q$. The Cartesian product $P \times Q$ is the set of all ordered pairs of elements from $\mathrm{P}$ and $\mathrm{Q}$, . i.e. $P \times Q=\{(p, q): p \in P, q \in Q\}$ Here, $P=(a, b)$ and $Q=(x...
Read More →Verify the following:
Question: Verify the following: (i) $\frac{3}{7} \times\left(\frac{5}{6}+\frac{12}{13}\right)=\left(\frac{3}{7} \times \frac{5}{6}\right)+\left(\frac{3}{7} \times \frac{12}{13}\right)$ (ii) $\frac{-15}{4} \times\left(\frac{3}{7}+\frac{-12}{5}\right)=\left(\frac{-15}{4} \times \frac{3}{7}\right)+\left(\frac{-15}{4} \times \frac{-12}{5}\right)$ (iii) $\left(\frac{-8}{3}+\frac{-13}{12}\right) \times \frac{5}{6}=\left(\frac{-8}{3} \times \frac{5}{6}\right)+\left(\frac{-13}{12} \times \frac{5}{6}\rig...
Read More →The set of points where f(x)
Question: The set of points wheref(x) =x [x] is discontinuous is _____________. Solution: The graph off(x) =x [x] is shown below. It can be seen that, thefunctionf(x) =x [x] is discontinuous at all integral values ofx. So, the set of point where wheref(x) =x [x] is discontinuous isZi.e. the set of integers. The set of points wheref(x) =x [x] is discontinuous is_____ the set of integers i.e.Z______....
Read More →If (- 4, 3) and (4, 3) are two vertices of an equilateral triangle,
Question: If (- 4, 3) and (4, 3) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle. Solution: Let the third vertex of an equilateral triangle be $(x, y)$. Let $A(-4,3), B(43)$ and $C(x, y)$. We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal. $\therefore$ $A B=B C=C \cdot A$ $\Rightarrow$ $A B^{2}=B C^{2}=C A^{2}$ ...(i) Now, taking first...
Read More →The set of points where f(x)
Question: The set of points wheref(x) =x [x] is discontinuous is _____________. Solution: The graph off(x) =x [x] is shown below. It can be seen that, thefunctionf(x) =x [x] is discontinuous at all integral values ofx. So, the set of point where wheref(x) =x [x] is discontinuous isZi.e. the set of integers. The set of points wheref(x) =x [x] is discontinuous is_____the set of integers i.e.Z______....
Read More →If A = {9, 1} and B = {1, 2, 3}, show that
Question: If A = {9, 1} and B = {1, 2, 3}, show that A B B A. Solution: Given: A = {9, 1} and B = {1, 2, 3} To show: A B B A Now, firstly we find the A B and B A By the definition of the Cartesian product, Given two non empty sets P and Q. The Cartesian product P Q is the set of all ordered pairs of elements from P and Q, .i.e. $P \times Q=\{(p, q): p \in P, q \in Q\}$ Here, $A=(9,1)$ and $B=(1,2,3) .$ So, $A \times B=(9,1) \times(1,2,3)$ $=\{(9,1),(9,2),(9,3),(1,1),(1,2),(1,3)\}$ $B \times A=(1...
Read More →If the function
Question: If the function $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1 \\ k, x=1\end{array}\right.$ is given to be continuous at $x=1$, then the value of $k$ is____________ Solution: The function $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1 \\ k, x=1\end{array}\right.$ is continuous at $x=1$. $\therefore f(1)=\lim _{x \rightarrow 1} f(x)$ $\Rightarrow k=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$ $\Rightarrow k=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}$ $\Righ...
Read More →If the function
Question: If the function $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1 \\ k, x=1\end{array}\right.$ is given to be continuous at $x=1$, then the value of $k$ is____________ Solution: The function $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1 \\ k, x=1\end{array}\right.$ is continuous at $x=1$. $\therefore f(1)=\lim _{x \rightarrow 1} f(x)$ $\Rightarrow k=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$ $\Rightarrow k=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}$ $\Righ...
Read More →Find the value of:
Question: Find the value of: (i) $\left(\frac{5}{8}\right)^{-1}$ (ii) $\left(\frac{-4}{9}\right)^{-1}$ (iii) $(-7)^{-1}$ (iv) $\left(\frac{1}{-3}\right)^{-1}$ Solution: We know that $a^{-1}=\frac{1}{a}$ or $a^{-1} \times a=1$ (i) $\left(\frac{5}{8}\right)^{-1}=\frac{8}{5}$ $\because \frac{5}{8} \times\left(\frac{5}{8}\right)^{-1}=1$ (ii) $\left(\frac{-4}{9}\right)^{-1}=\frac{9}{-4}=\frac{-9}{4}$ $\because \frac{-4}{9} \times\left(\frac{-4}{9}\right)^{-1}=1$ (iii) $(-7)^{-1}=\frac{1}{-7}=\frac{-1...
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Question: If $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-9}{x-3}, x \neq 3 \\ 2 x+k, x=3\end{array}\right.$ is continuous at $x=3$, then $k=$_________ Solution: The function $f(x)=\left\{\begin{array}{cc}\frac{x^{2}-9}{x-3}, x \neq 3 \\ 2 x+k, x=3\end{array}\right.$ is continuous at $x=3$. $\therefore f(3)=\lim _{x \rightarrow 3} f(x)$ $\Rightarrow 2 \times 3+k=\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}$ $\Rightarrow 6+k=\lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}$ $\Rightarrow 6+k=\lim _{x \r...
Read More →Find the values of a and b, when:
Question: Find the values of a and b, when: (i) $(a+3, b-2)=(5,1)$ (ii) $(a+b, 2 b-3)=(4,-5)$ (iii) $\left(\frac{\mathrm{a}}{3}+1, \mathrm{~b}-\frac{1}{3}\right)=\left(\frac{5}{3}, \frac{2}{3}\right)$ (iv) $(a-2,2 b+1=(b-1, a+2)$ Solution: Since, the ordered pairs are equal, the corresponding elements are equal. $\therefore, \mathrm{a}+3=5 \ldots$ (i) and $\mathrm{b}-2=1 \ldots$ (ii) Solving eq. (i), we get $a+3=5$ $\Rightarrow a=5-3$ $\Rightarrow a=2$ Solving eq. (ii), we get $b-2=1$ $\Rightarr...
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Question: If $f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$, is continuous at $x=0$, then $f(0)=$________ Solution: The function $f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$, is continuous at $x=0$. $\therefore f(0)$ $=\lim _{x \rightarrow 0} f(x)$ $=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x}$ $=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x} \times \frac{2+\sqrt{x+4}}{2+\sqrt{x+4}}$ $=\lim _{x \rightarrow 0} \frac{4-(x+4)}{\sin 2 x(2+\sqrt{x+4})}$ $=\lim _{x \rightarrow 0} \f...
Read More →Find the multiplicative inverse (i.e., reciprocal) of:
Question: Find the multiplicative inverse (i.e., reciprocal) of: (i) $\frac{13}{25}$ (ii) $\frac{-17}{12}$ (iii) $\frac{-7}{24}$ (iv) 18 (v) $-16$ (vi) $\frac{-3}{-5}$ (vii) $-1$ (viii) $\frac{0}{2}$ (ix) $\frac{2}{-5}$ (x) $\frac{-1}{8}$ Solution: (i)Reciprocal of $\frac{13}{25}$ is $\frac{25}{13}$. (ii) Reciprocal of $\frac{-17}{12}$ is $\frac{12}{-17}$, that is, $\frac{-12}{17}$. (iii) Reciprocal of $\frac{-7}{24}$ is $\frac{24}{-7}$, that is, $\frac{-24}{7}$. (iv) Reciprocal of 18 is $\frac{...
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Question: If $f(x)=\left\{\begin{array}{cl}\frac{1-\sin x}{\pi-2 x}, x \neq \frac{\pi}{2} \\ k, x=\frac{\pi}{0}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then $k=$___________ Solution: The function $f(x)=\left\{\begin{array}{cl}\frac{1-\sin x}{\pi-2 x}, x \neq \frac{\pi}{2} \\ k, x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$. $\therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)$ $\Rightarrow k=\lim _{x \rightarrow \frac{\pi}{2}} \f...
Read More →Find the ratio in which the line 2x+ 3y – 5 = 0
Question: Find the ratio in which the line 2x+ 3y 5 = 0 divides the line segment joining the points (8, 9) and (2,1). Also, find the coordinates of the point of division. Solution: Let the line 2x + 3y 5 = 0 divides the line segment joining the points A (8, 9) and B (2,1) in the ratio : 1 at point P. $\therefore$Coordinates of $P=\left\{\frac{2 \lambda+8}{\lambda+1}, \frac{\lambda-9}{\lambda+1}\right\}$ $\left[\because\right.$ internal division $\left.=\left\{\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}...
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) $\frac{-23}{17} \times \frac{18}{35}=\frac{18}{35} \times(\ldots . .)$ (ii) $-38 \times \frac{-7}{19}=\frac{-7}{19} \times(\ldots . .)$ (iii) $\left(\frac{15}{7} \times \frac{-21}{10}\right) \times \frac{-5}{6}=(\ldots . .) \times\left(\frac{-21}{10} \times \frac{-5}{6}\right)$ (iv) $\frac{-12}{5} \times\left(\frac{4}{15} \times \frac{25}{-16}\right)=\left(\frac{-12}{5} \times \frac{4}{15}\right) \times(\ldots . .)$ Solution: (i) $\frac{-23}{17} \times \frac{18}...
Read More →Find the values of k, if the points A(k + 1, 2k),
Question: Find the values of k, if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k 1, 5k) are colli near. Solution: We know that, if three points are collinear, then the area of triangle formed by these points is zero. Since, the points A(k + 1,2k), B(3k, 2k + 3) and C(5k -1, 5k) are collinear. Then, area of ΔABC = 0 $\Rightarrow$$\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]=0\right.$ Here, $\quad x_{1}=k+1, x_{2}=3 k, x_{3}=5 k-1$ a...
Read More →Verify each of the following:
Question: Verify each of the following: (i) $\left(\frac{5}{7} \times \frac{12}{13}\right) \times \frac{7}{18}=\frac{5}{7} \times\left(\frac{12}{13} \times \frac{7}{8}\right)$ (ii) $\frac{-13}{24} \times\left(\frac{-12}{5} \times \frac{35}{36}\right)=\left(\frac{-13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36}$ (iii) $\left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4}=\frac{-9}{5} \times\left(\frac{-10}{3} \times \frac{21}{-4}\right)$ Solution: (i) $\left(\frac{5}{7} \t...
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