If A = {a, b, c, d, e}, B = {a, c, e, g}, and C = {b, e, f, g} verify that:
Question: If A = {a, b, c, d, e}, B = {a, c, e, g}, and C = {b, e, f, g} verify that: (i) $A \cap(B-C)=(A \cap B)-(A \cap C)$ (ii) $A-(B \cap C)=(A-B) \cup(A-C)$ Solution: (i) B - C represents all elements in B that are not in C $B-C=\{a, c\}$ $A^{\cap}(B-C)=\{a, c\}$ $A^{\cap} B=\{a, c, e\}$ $A^{\cap} C=\{b, e\}$ $\left(A^{\cap} B\right)-\left(A^{\cap} C\right)=\{a, c\}$ $\Rightarrow A \cap_{(B-C)}=\left(A^{\cap} B\right)-\left(A^{\cap} C\right)$ Hence proved (ii) $B^{\cap} C=\{e, g\}$ $A-\left...
Read More →The probability that it will rain tomorrow is 0.85.
Question: The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow? Solution: Let $A$ be the event of raining tomorrow. The probability that it will rain tomorrow, $\mathrm{P}(\mathrm{A})$, is $0.85$. Since the event of raining tomorrow and not raining tomorrow are complementary to each other, the probability of not raining tomorrow is: $\mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})=1-0.85=0.15$...
Read More →In ΔPQR, PD ⊥ QR such that D lies on QR,
Question: In ΔPQR, PD QR such that D lies on QR, if PQ = a, PR = b, QD = c and DR = d, then prove that (a + b)(a -b) = (c + d) (c -d). Solution: Given In A PQR, PD 1 QR, PQ = a, PR = b,QD = c and DR =d To prove (a + b) (a-b) = (c + d)(c-d) Proof In right angled ΔPDQ, $P Q^{2}=P D^{2}+Q D^{2}$ [by Pythagoras theorem] $\Rightarrow \quad a^{2}=P D^{2}+c^{2}$ $\Rightarrow \quad P D^{2}=a^{2}-c^{2}$ $\ldots$ (i) In right angled $\triangle P D R, \quad P R^{2}=P D^{2}+D R^{2} \quad$ [by Pythagoras the...
Read More →In given figure PQR is a right triangle,
Question: In given figure PQR is a right triangle, right angled at Q and QS PR. If PQ = 6 cm and PS = 4 cm, then find QS, RS and QR. Solution: Given, ΔPQR in which Q = 90, QS PR and PQ = 6 cm, PS = 4 cm In ΔSQP and ΔSRQ, $\angle P S Q=\angle R S Q$ [each equal to $90^{\circ}$ ] $\angle S P Q=\angle S Q R$ [each equal to $90^{\circ}-\angle R$ ] $\therefore \quad \Delta S Q P \sim \Delta S R Q$ Then, $\frac{S Q}{P S}=\frac{S R}{S Q}$ $\Rightarrow$ $S Q^{2}=P S \times S R$ $\ldots$ (i) $\begin{arra...
Read More →Show that the function g (x) = x − [x] is discontinuous at all integral points
Question: Show that the function $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer function. Solution: Given: $\mathrm{g}(x)=x-[x]$ It is evident that $g$ is defined at all integral points. Let $n \in Z$. Then, $\mathrm{g}(n)=n-[n]=n-n=0$ The left hand limit of $f$ at $x=n$ is, $\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{-}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]=n-(n-1)=1$ The right hand limit offatx=nis, $\...
Read More →The pie chart (as shown in the figure 25.23) represents the amount spent on different sports by a sports club in a year.
Question: The pie chart (as shown in the figure 25.23) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport. Solution: Amount spent on cricket $=\frac{\text { Central angle of the corresponding sector } \times \text { Total Money spent }}{360^{\circ}}$ $=\frac{150^{\circ} \times 108000}{360^{\circ}}=$ Rs 45,000 Amount spent on hockey $=\frac{\text { Central angle of the corre...
Read More →The following pie-chart shows the monthly expenditure of Shikha on various items.
Question: The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16000 per month, answer the following questions:(i) How much does she spend on rent? (ii) How much does she spend on education? (iii) What is the ratio of expenses on food and rent? Solution: (i) Money spent on rent $=\frac{\text { Central angle of the corresponding sector } \times \text { Total Money spent }}{360^{\circ}}$ $=\frac{81^{\circ} \times 16000}{360^{\circ}}=\operatorname{Rs} 3...
Read More →In Fig. 25.21, the pie-chart shows the marks obtained by a student in various subjects.
Question: In Fig. 25.21, the pie-chart shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects. Solution: Marks scored in mathematics $=\frac{\text { Central angle of corresponding sector } \times \text { Total Marks }}{360^{\circ}}$ $135=\frac{90^{\circ} \times \text { Total }}{360^{\circ}}$ Total Marks $=540$ Marks scored in Hindi = (Central angle of Hindi x ...
Read More →In given figure, ABC is a triangle right angled
Question: In given figure, ABC is a triangle right angled at B and BD AC. If AD = 4 cm and CD = 5 cm, then find BD and AB. Solution: Given, $\triangle A B C$ in which $\angle B=90^{\circ}$ and $B D \perp A C$ Also, $A D=4 \mathrm{~cm}$ and $C D=5 \mathrm{~cm}$ In $\triangle A D B$ and $\triangle C D B, \quad \angle A D B=\angle C D B$[each equal to $90^{\circ}$ ] and $\quad \angle B A D=\angle D B C \quad$ [each equal to $90^{\circ}-\angle C$ ] $\therefore \quad \Delta D B A \sim \triangle D C B...
Read More →In Fig. 25.20, the pie-chart shows the marks obtained by a student in an examination.
Question: In Fig. 25.20, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects. Solution: Marks secured in mathematics = (108 x 440)/360 marks = 132 marks Marks secured in science = (81 x 440)/360 marks = 99 marks Marks secured in English = (72 x 440)/360 marks = 88 marks Marks secured in Hindi = (54 x 440)/360 marks = 66 marks Marks secured in social science = (45 x 440)/360 marks = 55 ma...
Read More →Prove that
Question: Prove that $f(x)= \begin{cases}\frac{\sin x}{x}, x0 \\ x+1, x \geq 0\end{cases}$ is everywhere continuous. Solution: When $x0$, we have $f(x)=\frac{\sin x}{x}$ We know that $\sin x$ as well as the identity function $x$ are everywhere continuous. So, the quotient function $\frac{\sin x}{x}$ is continuous at each $x0$. When $x0$, we have $f(x)=x+1$, which is a polynomial function. Therefore, $f(x)$ is continuous at each $x0$. Now, Let us consider the point $x=0$. Given: $f(x)=\left\{\beg...
Read More →The following pie-chart shows the number of students admitted in different faculties of a college.
Question: The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:(i) What is the total number of students? (ii) What is the ratio of students in science and arts? Solution: (i) Students in science $=\frac{\text { Central angle of the corresponding sector } \times \text { Total students }}{360^{\circ}}$ $1000=\frac{100^{\circ} \times \text { Total students }}{360^{\circ}}$ $\therefore$ Total ...
Read More →A street light bulb is fixed on a pole 6 m
Question: A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, then find how far she is away from the base of the pole. Solution: Let A be the position of the street bulb fixed on a pole AB = 6 m and CD = 1.5 mbe the height of a woman and her shadow be ED = 3 m. Let distance between pole and woman be x m. Here, woman and pole both are standing vertically. So, $C D \| A B$ In $\triangle C D E$ and $\triangle A B E, \quad \angl...
Read More →The pie-chart given in Fig. 25.18 shows the annual agricultural production of an Indian state.
Question: The pie-chart given in Fig. 25.18 shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of (i) Wheat (ii) Sugar (iii) Rice (iv) Maize (v) Gram Solution: (i) Production of wheat $=\frac{\text { Central angle for wheat } \times \text { Total production }}{360^{\circ}}$ $=\frac{120^{\circ} \times 81000}{360^{\circ}}=27000$ tonnes (ii) Production of sugar $=\frac{\text { Central angle for...
Read More →Discuss the continuity of f(x) = sin | x |.
Question: Discuss the continuity of $f(x)=\sin |x|$. Solution: Let $f(x)=\sin |x|$ This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as, $f=hog$, where $g(x)=|x|$ and $h(x)=\sin x$ $[\because h o g(x)=h(g(x))=h(|x|)=\sin |x|]$ It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions. $g(x)=|x|$ can be written as $g(x)= \begin{cases}-x, \text { if } x0 \\ x, \text { if } x \geq 0\end{cases}$ Clearly, $g$ is de...
Read More →The pie-chart given in Fig. 25.17 represents the expenditure on different items in constructing a flat in Delhi.
Question: The pie-chart given in Fig. 25.17 represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 112500, find the following:(i) Total cost of the flat. (ii) Expenditure incurred on labour. Solution: (i) Expenditure incurred on cement $=\frac{\text { Central angle of the corresponding sector } \times \text { Total cost }}{360^{\circ}}$ Total cost of the flat $=\frac{360^{\circ} \times 112500}{75^{\circ}}=\operatorname{Rs} 54000...
Read More →A flag pole 18 m high casts a shadow 9.6 m long.
Question: A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow. Solution: Let BC = 18 m be the flag pole and its shadow be AB = 9.6 m. The distance of the top of the pole, C from the far end i.e., A of the shadow is AC. In right angled $\triangle A B C, \quad A C^{2}=A B^{2}+B C^{2} \quad$ [by Pythagoras theorem] $\Rightarrow \quad A C^{2}=(9.6)^{2}+(18)^{2}$ $A C^{2}=92.16+324$ $\Rightarrow \quad A C^{2}=41616$ $\therefore \qua...
Read More →Draw a pie-diagram for the following data of the investment pattern in a five year plan:
Question: Draw a pie-diagram for the following data of the investment pattern in a five year plan: Solution: We know: Central angle of a component $=\left(\right.$ component value $/$ sum of component values $\left.\times 360^{\circ}\right)$ Here the total percentage = 100% Thus, the central angle for each component can be calculated as follows: Item Amount(in %) Sector angle Agriculture 14 14/100 x 360 = 50.4 Irrigation and Power 16 16/100 x 360 = 57.6 Small Industries 29 29/100 x 360 =104.4 Tr...
Read More →For going to a city B from city A there is a route
Question: For going to a city B from city A there is a route via city C such that AC CB, AC = 2x km and CB = 2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway. Solution: Given, AC CB, km,CB = 2(x + 7) km and AB = 26 km On drawing the figure, we get the right angled Δ ACB right angled at C. Now, In ΔACB, by Pythagoras theorem, $A B^{2}=A C...
Read More →Following is the break up of the expenditure of a family on different items of consumption:
Question: Following is the break up of the expenditure of a family on different items of consumption: Draw a pie-diagram to represent the above data. Solution: We know: Central angle of a component $=\left(\right.$ component value $/$ sum of component values $\left.\times 360^{\circ}\right)$ Here, total expenditure = Rs 3000 Thus, central angle for each component can be calculated as follows: Item Expenditure (in Rs) Sector angle Food 1600 1600/3000 360 = 192 Clothing 200 200/3000360 = 24 Rent 6...
Read More →A 5m long ladder is placed leaning towards
Question: A 5m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall. Solution: Let AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed. If the foot of the ladder is moved 1.6 m towards the wall i.e, AD = 1.6 m, then the ladder is slide upward i.e. ,CE = x m. In...
Read More →Following is the break up of the expenditure of a family on different items of consumption:
Question: Following is the break up of the expenditure of a family on different items of consumption: Draw a pie-diagram to represent the above data. Solution: We know: Central angle of a component $=\left(\right.$ component value $/$ sum of component values $\left.\times 360^{\circ}\right)$ Here, total expenditure = Rs 3000Thus, central angle for each component can be calculated as follows: Item Expenditure (in Rs) Sector angle Food 1600 1600/3000 360 = 192 Clothing 200 200/3000360 = 24 Rent 60...
Read More →Discuss the continuity of the function
Question: Discuss the continuity of the function $f(x)= \begin{cases}2 x-1, \text { if } x2 \\ \frac{3 x}{2}, \text { if } x \geq 2\end{cases}$ Solution: When $x2$, we have $f(x)=2 x-1$ We know that a polynomial function is everywhere continuous. So, $f(x)$ is continuous for each $x2$. When $x2$, we have $f(x)=\frac{3 x}{2}$ The functions $3 x$ and 2 are continuous being the polynomial and constant function, respectively. Thus, the quotient function $\frac{3 x}{2}$ is continuous at each $x2$. No...
Read More →In the given figure, if PQRS is a parallelogram
Question: In the given figure, if PQRS is a parallelogram and AB || PS, then prove that 0C || SR. Solution: Given PQRS is a parallelogram, so PQ || SR and PS || QR. Also, AB || PS. To prove $O C \| S R$ Proof in $\triangle O P S$ and $\triangle O A B$, $P S \| A B$ $\angle P O S=\angle A O B$ [common angle] $\angle O S P=\angle O B A$ [corresponding angles] $\therefore \quad \Delta O P S \sim \Delta O A B \quad$ [by AAA similarity criterion] Then, $\frac{P S}{\cap AR}=\frac{O S}{\cap QR}$ $\ldot...
Read More →Solve this
Question: If $f(x)=\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}$ for $x \neq \pi / 4$, find the value which can be assigned to $f(x)$ at $x=\pi / 4$ so that the function $f(x)$ becomes continuous every where in $[0, \pi / 2]$. Solution: When $x \neq \frac{\pi}{4}, \tan \left(\frac{\pi}{4}-x\right)$ and $\cot 2 x$ are continuous in $\left[0, \frac{\pi}{2}\right]$. Thus, the quotient function $\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}$ is continuous in $\left[0, \frac{\pi}{2}\right]...
Read More →