If f : R → R defined by
Question: If $f: R \rightarrow R$ defined by $f(x)=\left\{\begin{array}{cl}\frac{\cos 3 x-\cos x}{x^{2}}, x \neq 0 \\ \lambda, x=0\end{array}\right.$ is continuous at $x=0$, then $\lambda=$___________ Solution: The function $f(x)=\left\{\begin{array}{cl}\frac{\cos 3 x-\cos x}{x^{2}}, x \neq 0 \\ \lambda, x=0\end{array}\right.$ is continuous at $x=0$ $\therefore f(0)=\lim _{x \rightarrow 0} f(x)$ $\Rightarrow \lambda=\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^{2}}$ $\Rightarrow \lambda=\lim...
Read More →Find the coordinates of the point
Question: Find the coordinates of the point $R$ on the line segment joining the points $P(-1,3)$ and $Q(2,5)$ such that $P R=\frac{3}{5} P Q$. Solution: According to the question, Given that,$P R=\frac{3}{5} P Q$ $\Rightarrow$ $P R=\frac{3}{5} P Q$ $\Rightarrow$ $\frac{P R+R Q}{P R}=\frac{5}{3}$ $\Rightarrow$ $1+\frac{R Q}{P R}=\frac{5}{3}$ $\Rightarrow$ $\frac{P Q}{P R}=\frac{5}{3}-1=\frac{2}{3}$ $\therefore$ $R Q: P R=2: 3$ or $P R: R Q=3: 2$ Suppose, $R(x, y)$ be the point which divides the l...
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Question: If $f(x)=\left\{\begin{array}{cl}\frac{x-4}{|x-4|}+a, x4 \\ a+b, x=4 \\ \frac{x-4}{|x-4|}+b, x4\end{array}\right.$. Then $f(x)$ is continuous at $x=4$, then $a+b=$ Solution: The function $f(x)=\left\{\begin{array}{ll}\frac{x-4}{|x-4|}+a, x4 \\ a+b, x=4 \\ \frac{x-4}{|x-4|}+b, x4\end{array}\right.$ is continuous at $x=4$. $\therefore f(4)=\lim _{x \rightarrow 4} f(x)$ $\Rightarrow f(4)=\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)$ $|x-4|= \begin{cases}-(x-4), x4 \\ ...
Read More →Verify each of the following:
Question: Verify each of the following: (i) $\frac{3}{7} \times \frac{-5}{9}=\frac{-5}{9} \times \frac{3}{7}$ (ii) $\frac{-8}{7} \times \frac{13}{9}=\frac{13}{9} \times \frac{-8}{7}$ (iii) $\frac{-12}{5} \times \frac{7}{-36}=\frac{7}{-36} \times \frac{-12}{5}$ (iv) $-8 \times \frac{-13}{12}=\frac{-13}{12} \times(-8)$ Solution: (i) $\frac{3}{7} \times \frac{-5}{9}=\frac{-5}{9} \times \frac{3}{7}$ $\mathrm{LHS}=\frac{3 \times(-5)}{7 \times 9}$ $=-\frac{15}{63}$ Simplifying, we get: $-\frac{15}{63}...
Read More →If the points A (2, 9), B (a, 5) and C (5, 5)
Question: If the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ABC right angled at B, then find the values of a and hence the area of ΔABC. Solution: Given that, the points A (2, 9), B(a, 5) and C(5, 5) are the vertices of a ΔABC right angled at B. By Pythagoras theorem, AC2= AB2+ BC2 ,,.(i) $\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right)=\sqrt{\left(x_{2}-x_{1}^{2}\right)+\left(y_{2}-y_{1}\right)^{2}}\right]$ $=...
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Question: If $A=\{x: x \in R, x5\}$ and $B=\{x: x \in R, x4\}$, find $A \cap B$. Solution: $A=\{x: x \in R, x5\}$ As $x$ takes all real values upto 5 hence the set $A$ will contain all numbers from $-\infty$ to 5 $A=(-\infty, 5)$ $B=\{x: x \in R, x4\}$ As $x$ takes all real values greater than 4 hence the set $B$ will contain values from 4 to $\infty$ $B=(4,-\infty)$ Hence their intersection or the common part between sets $A$ and $B$ would be values from 4 to 5 Hence $A \cap B=(4,5)$ Representi...
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Question: If $f(x)=\left\{\begin{array}{cl}x+k, x3 \\ 4, x=3 \\ 3 x-5, x3\end{array}\right.$ is continuous at $x=3$, then $k=$___________ Solution: The function $f(x)=\left\{\begin{array}{cl}x+k, x3 \\ 4, x=3 \\ 3 x-5, x3\end{array}\right.$ is continuous at $x=3$. $\therefore f(3)=\lim _{x \rightarrow 3} f(x)$ $\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$ $\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$ ...(1) Now, $f(3)=...
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Question: If $f(x)=\left\{\begin{array}{cl}x+k, x3 \\ 4, x=3 \\ 3 x-5, x3\end{array}\right.$ is continuous at $x=3$, then $k=$___________ Solution: The function $f(x)=\left\{\begin{array}{cl}x+k, x3 \\ 4, x=3 \\ 3 x-5, x3\end{array}\right.$ is continuous at $x=3$. $\therefore f(3)=\lim _{x \rightarrow 3} f(x)$ $\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$ $\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$ ...(1) Now, $f(3)=...
Read More →Find each of the following products:
Question: Find each of the following products: (i) $\frac{3}{5} \times \frac{-7}{8}$ (ii) $\frac{-9}{2} \times \frac{5}{4}$ (iii) $\frac{-6}{11} \times \frac{-5}{3}$ (iv) $\frac{-2}{3} \times \frac{6}{7}$ (v) $\frac{-12}{5} \times \frac{10}{-3}$ (vi) $\frac{25}{-9} \times \frac{3}{-10}$ (vii) $\frac{5}{-18} \times \frac{-9}{20}$ (viii) $\frac{-13}{15} \times \frac{-25}{26}$ (ix) $\frac{16}{-21} \times \frac{14}{5}$ (x) $\frac{-7}{6} \times 24$ (xi) $\frac{7}{24} \times(-48)$ (xii) $\frac{-13}{5}...
Read More →Prove that
Question: Prove that $A-B=A \cap B .^{\prime}$ Solution: Let $x$ be some element in set $A-B$ that is $x \in(A-B)$ Now if we prove that $x \in\left(A \cap B^{\prime}\right)$ then $(A-B)=\left(A \cap B^{\prime}\right)$ $x \in(A-B)$ means $x \in A$ and $x \notin B$ Now $x \notin B$ means $x \in B$.' Hence we can say that $x \in A$ and $x \in B$.' Hence $x \in A \cap B$.' And as $x \in A \cap B^{\prime}$ and also $x \in A-B$ we can conclude that $A-B=A \cap B .^{\prime}$...
Read More →Prove the following
Question: If $D\left(\frac{-1}{2}, \frac{5}{2}\right) E(7,3)$ and $F\left(\frac{7}{2}, \frac{7}{2}\right)$ are the mid-points of sides of $\triangle A B C$, then find the area of the $\triangle A B C$. Solution: Let $A \equiv\left(x_{1}, y_{1}\right), B \equiv\left(x_{2}, y_{2}\right)$ and $C \equiv\left(x_{3}, y_{3}\right)$ are the vertices of the $\triangle A B C$. Gives, $D\left(-\frac{1}{2}, \frac{5}{2}\right), E(7,3)$ and $F\left(\frac{7}{2}, \frac{7}{2}\right)$ be the mid-points of the sid...
Read More →If the function
Question: If the function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, 0 \leq x1 \\ 2, x=1 \\ x+1, 1x \leq 2\end{array}\right.$ is continuous at $x=1$, then $a-b=$________________ Solution: The function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, 0 \leq x1 \\ 2, x=1 \\ x+1, 1x \leq 2\end{array}\right.$ is continuous at $x=1$. $\therefore f(1)=\lim _{x \rightarrow 1} f(x)$ $\Rightarrow f(1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$ .....(1) Now, $f(1)=2$ ....(2) $\lim _{x \right...
Read More →Find the symmetric difference A Δ B,
Question: Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}. Solution: A = {1, 2, 3} B = {3, 4, 5} The symmetric difference A Δ B is given by $A \Delta B=(A-B) \cup_{(B-A)}$ Venn diagram representation: Representing the given sets A and B through venn diagram Hence as seen the elements in A Δ B are 1, 2, 4 and 5 Hence the symmetric difference A Δ B = {1, 2, 4, 5}...
Read More →If the function
Question: If the function $f(x)=\left\{\begin{array}{cl}\frac{\sin ^{2} a x}{x^{2}}, x \neq 0 \\ 1, x=0\end{array}\right.$ is continuous at $x=0$, then $a=$_____________ Solution: The function $f(x)=\left\{\begin{array}{cl}\frac{\sin ^{2} a x}{x^{2}}, x \neq 0 \\ 1, x=0\end{array}\right.$ is continuous at $x=0$. $\therefore \lim _{x \rightarrow 0} f(x)=f(0)$ $\Rightarrow \lim _{x \rightarrow 0} \frac{\sin ^{2} a x}{x^{2}}=1$ $\Rightarrow a^{2}\left(\lim _{x \rightarrow 0} \frac{\sin a x}{a x}\ri...
Read More →Prove that
Question: Prove that $A \cap\left(A^{U} B\right)^{\prime}=\phi$ Solution: $\mathrm{LHS}=\mathrm{A} \cap\left(\mathrm{A}^{\cup} \mathrm{B}\right)^{\prime}$ Using De-Morgan's law $\left(A^{U} B\right)^{\prime}=\left(A^{\prime} \cap B^{\prime}\right)$ $\Rightarrow L H S=A \cap\left(A^{\prime} \cap B^{\prime}\right)$ $\Rightarrow L H S=\left(A \cap A^{\prime}\right) \cap\left(A \cap B^{\prime}\right)$ We know that $A \cap A^{\prime}=\phi$ $\Rightarrow L H S=\phi \cap\left(A \cap B^{\prime}\right)$ W...
Read More →(i) Which rational number is its own additive inverse?
Question: (i) Which rational number is its own additive inverse? (ii) Is the difference of two rational numbers a rational number? (iii) Is addition commutative on rational numbers? (iv) Is addition associative on rational numbers? (v) Is subtraction commutative on rational numbers? (vi) Is subtraction associative on rational numbers? (vii) What is the negative of a negative rational number? Solution: 1. Zero is a rational number that is its own additive inverse. 2. Yes Consider $\frac{a}{b}-\fr...
Read More →Solve this
Question: If $f(x)=\left\{\begin{array}{cl}\frac{x^{3}-a^{3}}{x-a}, x \neq a \\ b, x=a\end{array}\right.$ is continuous at $x=a$, then $b=$ Solution: It is given that, the function $f(x)=\left\{\begin{array}{cl}\frac{x^{3}-a^{3}}{x-a}, x \neq a \\ b, x=a\end{array}\right.$ is continuous at $x=a$. $\therefore f(a)=\lim _{x \rightarrow a} f(x)$ $\Rightarrow b=\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a}$ $\Rightarrow b=3 a^{2} \quad\left(\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}...
Read More →The line segment joining the points A(3, 2)
Question: The line segment joining the points A(3, 2) and B(5,1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x 18y + k = 0. Find the value of k. Solution: Given that, the line segment joining the points 4(3,2) and 6(5,1) is divided at the point P in the ratio 1 : 2. $\therefore \quad$ Coordinate of point $P=\left\{\frac{5(1)+3(2)}{1+2}, \frac{1(1)+2(2)}{1+2}\right\}$ $=\left(\frac{5+6}{3}, \frac{1+4}{3}\right)=\left(\frac{11}{3}, \frac{5}{3}\right)$ $\left[\because\righ...
Read More →What number should be subtracted from
Question: What number should be subtracted from $\frac{-2}{3}$ to get $\frac{-1}{6} ?$ Solution: Let the required number bex. Now, $\frac{-2}{3}-x=\frac{-1}{6}$ $\Rightarrow \frac{-2}{3}-x+x=\frac{-1}{6}+x \quad$ (Adding $x$ to both the sides) $\Rightarrow \frac{-2}{3}=\frac{-1}{6}+x$ $\Rightarrow \frac{-2}{3}+\frac{1}{6}=\frac{-1}{6}+x+\frac{1}{6} \quad$ (Adding $\frac{1}{6}$ to both the sides) $\Rightarrow\left(\frac{-4}{6}+\frac{1}{6}\right)=x$ $\Rightarrow\left(\frac{-4+1}{6}\right)=x$ $\Rig...
Read More →If A = {3, {2}}, find P(A).
Question: If A = {3, {2}}, find P(A). Solution: A = {3, {2}} We have to find $P(A)$ which is power set of $A$ The power set of set $A$ is collection of all possible subsets of $A$ The possible subsets of $A$ are $\{\phi\},\{3\},\{\{2\}\},\{3,\{2\}\}$ Hence the power set $P(A)$ will be $\mathrm{P}(\mathrm{A})=\{\{\phi\},\{3\},\{\{2\}\},\{3,\{2\}\}$...
Read More →The function
Question: The function $f(x)=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, \text { if } x \neq 0 \\ k, \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is (a) 3(b) 2(c) 1(d) 1.5 Solution: It is given that, the function $f(x)=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, \text { if } x \neq 0 \\ k, \text { if } x=0\end{array}\right.$ is continuous at $x=0$. $\therefore f(0)=\lim _{x \rightarrow 0} f(x)$ $\Rightarrow k=\lim _{x \rightarrow 0}\left(\frac{\sin ...
Read More →If the centre of a circle is (2a, a – 7),
Question: If the centre of a circle is (2a, a 7), then Find the values of a, if the circle passes through the point (11, 9) and has diameter 102 units. Solution: By given condition, Distance between the centre C(2a, a 7) and the point P( 11, 9), which lie on the circle = Radius of circle $\therefore \quad$ Radius of circle $=\sqrt{(11-2 a)^{2}+(-9-a+7)^{2}}$$\ldots($ i) $\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)=\sqrt{\left.\lef...
Read More →If A = {5, 6, 7}, find P(A).
Question: If A = {5, 6, 7}, find P(A). Solution: A = {5, 6, 7} We have to find $P(A)$ which is power set of $A$ The power set of set $A$ is collection of all possible subsets of $A$ The possible subsets of $A$ are $\{\phi\},\{5\},\{6\},\{7\},\{5,6\},\{5,7\},\{6,7\},\{5,6,7\}$ Hence the power $\operatorname{set} P(A)$ will be $P(A)=\{\{\phi\},\{5\},\{6\},\{7\},\{5,6\},\{5,7\},\{6,7\},\{5,6,7\}\}$...
Read More →What number should be added to -1 so as to get
Question: What number should be added to $-1$ so as to get $\frac{5}{7} ?$ Solution: Let the required number bex. Now, $-1+x=\frac{5}{7}$ $\Rightarrow-1+x+1=\frac{5}{7}+1 \quad$ (Adding 1 to both the sides) $\Rightarrow x=\left(\frac{5+7}{7}\right)$ $\Rightarrow x=\frac{12}{7}$ Hence, the required number is $\frac{12}{7}$....
Read More →The function f(x) = [x] is continuous at
Question: The functionf(x) = [x] is continuous at (a) 4 (b) $-2$ (c) 1 (d) $1.5$ Solution: The graph off(x) = [x] is shown below. It can be seen that, the functionf(x) = [x] is discontinuous at all integral values ofx. It is continuous at all points except the integer points.Thus, the functionf(x) = [x] is continuous atx= 1.5 and discontinuous atx= 4,x= 2 andx= 1.Hence, the correct answer is option (d)....
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