If f(x)
Question: If $t(x)=x^{2}$, find $\frac{f(1.1)-f(1)}{(1.1)-1} .$ Solution: Given: f(x) =x2 Therefore, $\frac{f(1.1)-f(1)}{(1.1)-1}=\frac{(1.1)^{2}-(1)^{2}}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$...
Read More →In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.
Question: In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 25 cm, find the area of the rectangle. Solution: Given OD = 10 cm and OE = 25cm By using Pythagoras theorem $\therefore \mathrm{OD}^{2}=\mathrm{OE}^{2}+\mathrm{DE}^{2}$ $\Rightarrow \mathrm{DE}=\sqrt{\mathrm{OD}^{2}-\mathrm{OE}^{2}}=\sqrt{10^{2}-(2 \sqrt{5})^{2}}=4 \sqrt{5} \mathrm{~cm}$ Area of rectangle OCDE = OE DE $=2 \sqrt{5} \times 4 \sqrt{5} \mathrm{~cm}^{2}=40 \mathrm{~cm}^{2}$...
Read More →The function f is defined by
Question: The function $f$ is defined by $f(x)=\left\{\begin{array}{lc}x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10\end{array}\right.$ The relation $g$ is defined by $g(x)= \begin{cases}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2x10\end{cases}$ Show thatfis a function andgis not a function. Solution: The function $f$ is defined by $f(x)= \begin{cases}x^{2} 0 \leqslant x \leqslant 3 \\ 3 x 3 \leqslant x \leqslant 10\end{cases}$ It is observed that for 0 x 3,f(x) =x2. 3 x 10,f(x) = 3x Also, atx= 3,f(x) = ...
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Question: $\frac{5 x}{(x+1)\left(x^{2}+9\right)}$ Solution: Let $\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}$ ...(1) $\Rightarrow 5 x=A\left(x^{2}+9\right)+(B x+C)(x+1)$ $\Rightarrow 5 x=A x^{2}+9 A+B x^{2}+B x+C x+C$ Equating the coefficients of $x^{2}, x$, and constant term, we obtain $A+B=0$ $B+C=5$ $9 A+C=0$ On solving these equations, we obtain $A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{9}{2}$ From equation (1), we obtain $\frac{5 x}{(x+1)\lef...
Read More →In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4
Question: In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD. Solution: Draw ALDC, BMDC then, AL = BM = 4 cm and LM = 7 cm. In Δ ADL, we have $A D^{2}=A L^{2}+D L^{2}$ $\Rightarrow 25=16+\mathrm{DL}^{2}$ ⇒ DL = 3 cm Similarly, $\mathrm{MC}=\sqrt{\mathrm{BC}^{2}-\mathrm{BM}^{2}}=\sqrt{25-16}=3 \mathrm{~cm}$ x = CD = CM + ML + LD = (3 + 7 + 3) cm = 13 cm $\operatorname{ar}(\operatornam...
Read More →Let A = [9, 10, 11, 12, 13]
Question: Let A = [9, 10, 11, 12, 13] and letf: A N be defined byf(n) = the highest prime factor ofn. Find the range off. Solution: Given: A={9, 10, 11, 12, 13} f:AN be defined byf(n) = the highest prime factor ofn. f(9) = the highest prime factor of 9 = 3 f(10) = the highest prime factor of 10 = 5 f(11) = the highest prime factor of 11 = 11 f(12) = the highest prime factor of 12 = 3 f(13) = the highest prime factor of 13 = 13 The range of $f$ is the set of all $f(n)$, where $n \in A$. Therefore...
Read More →Let A = [p, q, r, s] and B = [1, 2, 3].
Question: Let A = [p,q,r,s] and B = [1, 2, 3]. Which of the following relations from A to B is not a function? (a) R1= [(p, 1), (q, 2), (r, 1), (s, 2)] (b) R2= [(p, 1), (q, 1), (r, 1), (s, 1)] (c) R3= [(p, 1), (q, 2), (p, 2), (s, 3) (d) R4= [(p, 2), (q, 3), (r, 2), (s, 2)]. Solution: (c) R3= [(p, 1), (q, 2), (p, 2), (s, 3) All the relations in (a), (b) and (d) have a unique image in B for all the elements in A. R3is not a function from A to B becausep A has two images, 1 and 2, in B. Hence, opti...
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Question: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\left[\operatorname{Hint}: \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} \operatorname{Put} x=t^{6}\right]$ Solution: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$ Let $x=t^{6} \Rightarrow d x=6 t^{5} d t$ $\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=\int \frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)} d x$ $=\int ...
Read More →In figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm.
Question: In figure, AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB. Solution: Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices CA = CB = OC ⇒ CA = CB = 6.5 cm ⇒ AB = 13 cm In right angled triangle OAB, we have $A B^{2}=O B^{2}+O A^{2}$ $\Rightarrow 13^{2}=\mathrm{OB}^{2}+12^{2}$ $\Rightarrow \mathrm{OB}^{2}=13^{2}-12^{2}=169-144=25$ ⇒ OB = 5 $\therefore \operatorname{ar}(\triangle \mathrm{AOB})=(1 / 2)(12 \times 5)=30 \mathrm{~cm}^...
Read More →If f : R
Question: Iff: R R be defined byf(x) =x2+ 1, then findf1[17] andf1[3]. Solution: If $f: A \rightarrow B$ is such that $y \in B$, then $f^{-1}\{y\}=\{x \in A: f(x)=y\}$. In other words, $f^{-1}\{y\}$ is the set of pre - images of $y$. Let $f^{-1}\{17\}=x$. Then,f(x) =17 . $\Rightarrow x^{2}+1=17$ $\Rightarrow x^{2}=17-1=16$ $\Rightarrow x=\pm 4$ $\therefore f^{-1}\{17\}=\{-4,4\}$ Again, let $f^{-1}\{-3\}=x$. Then, $f(x)=-3$ $\Rightarrow x^{2}+1=-3$ $\Rightarrow x^{2}=-3-1=-4$ $\Rightarrow x=\sqrt...
Read More →If f : R
Question: Iff: R R be defined byf(x) =x2+ 1, then findf1[17] andf1[3]. Solution: If $f: A \rightarrow B$ is such that $y \in B$, then $f^{-1}\{y\}=\{x \in A: f(x)=y\}$. In other words, $f^{-1}\{y\}$ is the set of pre - images of $y$. Let $f^{-1}\{17\}=x$. Then,f(x) =17 . $\Rightarrow x^{2}+1=17$ $\Rightarrow x^{2}=17-1=16$ $\Rightarrow x=\pm 4$ $\therefore f^{-1}\{17\}=\{-4,4\}$ Again, let $f^{-1}\{-3\}=x$. Then, $f(x)=-3$ $\Rightarrow x^{2}+1=-3$ $\Rightarrow x^{2}=-3-1=-4$ $\Rightarrow x=\sqrt...
Read More →A man travels 600 km partly by train and partly by car.
Question: A man travels 600 km partly by train and partly by car. If the covers 400 km by train and the rest by car, it takes him 6 hours 30 minutes. But, if the travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car. Solution: Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases: Case I: When a man travels 600Km by train and the rest by car Time taken by a man to travel $400 \mathrm{Km}$...
Read More →Let A = (12, 13, 14, 15, 16, 17)
Question: Let A = (12, 13, 14, 15, 16, 17) andf: A Z be a function given by f(x) = highest prime factor ofx. Find range off. Solution: Given: A={12, 13, 14, 15, 16, 17} f:AZbe defined byf(x) = the highest prime factor ofx. f(12) = the highest prime factor of 12 = 3 f(13) = the highest prime factor of 13 = 13 f(14) = the highest prime factor of 14 = 7 f(15) = the highest prime factor of 15 = 5 f(16) = the highest prime factor of 16 = 2 f(17) = the highest prime factor of 17 = 17 The range of $f$ ...
Read More →Compute the area of trapezium PQRS in figure
Question: Compute the area of trapezium PQRS in figure Solution: We have, ar(trap. PQRS) = ar(rect. PSRT) + ar(ΔQRT) ⇒ ar(trap. PQRS) = PT RT + 1/2(QT RT) = 8 RT + 1/2(8 RT) = 12 RT In ΔQRT, we have $\mathrm{QR}^{2}=\mathrm{QT}^{2}+\mathrm{RT}^{2}$ $\Rightarrow \mathrm{RT}^{2}=\mathrm{QR}^{2}-\mathrm{QT}^{2}$ $\Rightarrow R T^{2}=17^{2}-8^{2}=225$ ⇒ RT = 15 Hence, Area of trapezium $=12 \times 15 \mathrm{~cm}^{2}=180 \mathrm{~cm}^{2}$...
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Question: $\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$ Solution: $\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$ Multiplying and dividing by $x^{-3}$, we obtain $\frac{x^{-3}}{x^{2} \cdot x^{-3}\left(x^{4}+1\right)^{\frac{3}{4}}}=\frac{x^{-3}\left(x^{4}+1\right)^{\frac{-3}{4}}}{x^{2} \cdot x^{-3}}$ $=\frac{\left(x^{4}+1\right)^{\frac{-3}{4}}}{x^{5} \cdot\left(x^{4}\right)^{-\frac{3}{4}}}$ $=\frac{1}{x^{5}}\left(\frac{x^{4}+1}{x^{4}}\right)^{-\frac{3}{4}}$ $=\frac{1}{x^{5}}\left(1+\fra...
Read More →Let X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
Question: Let X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16} Determine which of the following sets are functions from X to Y. (a)f1= {(1, 1), (2, 11), (3, 1), (4, 15)} (b)f2= {(1, 1), (2, 7), (3, 5)} (c)f3= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} Solution: (a) Given: f1= {(1, 1), (2, 11), (3, 1), (4, 15)} f1is a function from X to Y. (b) Given: f2= {(1, 1), (2, 7), (3, 5)} f2is not a function from X to Y because 2 X has no image in Y. (c) Given: f3= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} f3...
Read More →In figure, PQRS is a square and T and U are, respectively, the midpoints of PS and QR.
Question: In figure, PQRS is a square and T and U are, respectively, the midpoints of PS and QR. Find the area of ΔOTS if PQ = 8 cm. Solution: From the figure, T and U are mid points of PS and QR respectively TU ∥ PQ ⇒ TO ∥ PQ Thus, inΔPQS, T is the midpoint of PS and TO ∥ PQ TO = (1/2) PQ = 4 cm Also, TS = (1/2) PS = 4 cm $\therefore \operatorname{ar}(\Delta \mathrm{OTS})=(1 / 2)(\mathrm{TO} \times \mathrm{TS})=(1 / 2)(4 \times 4) \mathrm{cm}^{2}=8 \mathrm{~cm}^{2}$...
Read More →f, g, h are three function defined from R to R as follows:
Question: f,g,hare three function defined from R to R as follows: (i)f(x) =x2 (ii)g(x) = sinx (iii)h(x) =x2+ 1 Find the range of each function. Solution: (i) Given: f(x) =x2 Range off(x) = R+(set of all positive integers) = {y R| y 0} (ii) Given: g(x) = sinx Range ofg(x) = {y R : -1 y 1}(iii) Given: h(x) =x2+ 1 Range ofh(x) = {y R :y 1}...
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Question: $\frac{1}{x \sqrt{a x-x^{2}}}$ [Hint: Put $x=\frac{a}{t}$ ] Solution: $\frac{1}{x \sqrt{a x-x^{2}}}$ Let $x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^{2}} d t$ $\Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-\left(\frac{a}{t}\right)^{2}}}\left(-\frac{a}{t^{2}} d t\right)$ $=-\int \frac{1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^{2}}}} d t$ $=-\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^{2}}{t}-\frac{t^{2}}{t^{2}}}} d t$ $=-\frac...
Read More →Let f :
Question: Letf: R R andg: C C be two functions defined asf(x) =x2andg(x) =x2. Are they equal functions? Solution: It is given that f: R R andg: C C are two function defined asf(x) =x2andg(x) =x2. Thus, domain (f) = R and domain (g) = C . Since, domain (f) domain (g), f(x) andg(x) are not equal functions....
Read More →In figure, compute the area of quadrilateral ABCD.
Question: In figure, compute the area of quadrilateral ABCD. Solution: Given: DC = 17 cm, AD = 9 cm and BC = 8 cm In ΔBCD we have $C D^{2}=B D^{2}+B C^{2}$ $\Rightarrow 17^{2}=B D^{2}+8^{2}$ $\Rightarrow B D^{2}=289-64$ $\Rightarrow B D=15$ In ΔABD we have $A B^{2}+A D^{2}=B D^{2}$ $\Rightarrow 15^{2}=\mathrm{AB}^{2}+9^{2}$ $\Rightarrow \mathrm{AB}^{2}=225-81=144$ ⇒ AB = 12 ar(quad ABCD) = ar(ΔABD) + ar(ΔBCD) $\operatorname{ar}(q u a d ~ A B C D)=1 / 2(12 \times 9)+1 / 2(8 \times 17)=54+68=122 \...
Read More →Write the following relations as sets of ordered pairs and find which of them are functions:
Question: Write the following relations as sets of ordered pairs and find which of them are functions: (a) {(x,y) :y= 3x,x {1, 2, 3},y [3,6, 9, 12]} (b) {(x,y) :yx+ 1,x= 1, 2 andy= 2, 4, 6} (c) {(x,y) :x+y= 3,x,y, [0, 1, 2, 3]} Solution: (a) Given: {(x,y) :y= 3x,x {1, 2, 3},y [3,6, 9, 12]} On substitutingx= 1, 2, 3 inx, we get : y = 3, 6, 9, respectively. R = {(1, 3) , (2, 6), (3, 9)} Hence, we observe that each element of the given set has appeared as the first component in one and only one ord...
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Question: $\frac{1}{\sqrt{x+a}+\sqrt{(x+b)}}$ Solution: $\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$ $=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}$ $=\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b}$ $\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x$ $=\frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]$ $=\frac{2}{3(a-b)}\left...
Read More →Let f :
Question: Letf: R+ R, where R+is the set of all positive real numbers, such thatf(x) = logex. Determine (a) the image set of the domain off (b) {x:f(x) = 2 (c) whetherf(xy) =f(x) :f(y) holds Solution: Given: f: R+ R andf(x) = logex.............(i) (a)f: R+ R Thus, the image set of the domainf=R. (b) {x:f(x) =-2 ⇒f(x) =2 .....(ii) From equations (i) and (ii), we get : $\log _{e} x=-2$ $\Rightarrow x=e^{-2}$ Hence, $\{x: f(x)=-2\}=\left\{e^{-2}\right\} . \quad\left[\right.$ Since $\left.\log _{a} ...
Read More →Let f :
Question: Letf: R+ R, where R+is the set of all positive real numbers, such thatf(x) = logex. Determine (a) the image set of the domain off (b) {x:f(x) = 2 (c) whetherf(xy) =f(x) :f(y) holds Solution: Given: f: R+ R andf(x) = logex.............(i) (a)f: R+ R Thus, the image set of the domainf=R. (b) {x:f(x) =-2 ⇒f(x) =2 .....(ii) From equations (i) and (ii), we get : $\log _{e} x=-2$ $\Rightarrow x=e^{-2}$ Hence, $\{x: f(x)=-2\}=\left\{e^{-2}\right\} . \quad\left[\right.$ Since $\left.\log _{a} ...
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