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Question: Solution: $\int\left(4 x^{3}-5 x^{2}+6 x+9\right) d x=4\left(\frac{x^{4}}{4}\right)-5\left(\frac{x^{3}}{3}\right)+6\left(\frac{x^{2}}{2}\right)+9(x)$ $=x^{4}-\frac{5 x^{3}}{3}+3 x^{2}+9 x=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(2)-\mathrm{F}(1)$ $I=\left\{2^{4}-\frac{5 \cdot(2)^{3}}{3}+3(2)^{2}+9(2)\right\}-\left\{(1)^{4}-\frac{5(1)^{3}}{3}+3(1)^{2}+9(1)\right\}$ $=\left(16-\frac{40}{3}+12+18\right)-\left(1-\frac{5}{3}+3+9\right)$ $=16-\frac{40...
Read More →Let A = {1, 2, 3} and B = {2, 3, 4}.
Question: Let A = {1, 2, 3} and B = {2, 3, 4}. Then which of the following is a function from A to B?(a) {(1, 2), (1, 3), (2, 3), (3, 3)} (b) [(1, 3), (2, 4)] (c) {(1, 3), (2, 2), (3, 3)} (d) {(1, 2), (2, 3), (3, 2), (3, 4)} Solution: (c) {(1, 3), (2, 2), (3, 3)} We have R= {(1, 3), (2, 2), (3, 3)} We observe that each element of the given set has appeared as first component in one and only one ordered pair of R. So,R= {(1, 3), (2, 2), (3, 3)} is a function....
Read More →The sides of a quadrilateral, taken in order as 5 m, 12 m, 14 m, 15 m respectively,
Question: The sides of a quadrilateral, taken in order as 5 m, 12 m, 14 m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area. Solution: Given that the sides of the quadrilateral are AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m Join AC Now area of triangle ABC = AB BC = 1/2 5 12 $=30 \mathrm{~m}^{2}$ In triangle ABC, By applying Pythagoras theorem $A C^{2}=A B^{2}+B C^{2}$ $A C=\sqrt{5^{2}+12^{2}}$ AC = 13 m Now in triangle ADC, Perimeter = 2s = AD + DC ...
Read More →Let f(x) =
Question: Let $f(x)=x^{2}$ and $g(x)=2 x+1$ be two real functions. Find $(f+g)(x),(f-g)(x),(f g)(x)$ and $\left(\frac{f}{g}\right)(x)$. Solution: Given f(x) =x2andg(x) = 2x+ 1 Clearly,D(f) =RandD(g) =R $\therefore D(f \pm g)=D(f) \cap D(g)=R \cap R=R$ $D(f g)=D(f) \cap D(g)=R \cap R=R$ $D\left(\frac{f}{g}\right)=D(f) \cap D(g)-\{x: g(x)=0\}=R \cap R-\left\{-\frac{1}{2}\right\}=R-\left\{-\frac{1}{2}\right\}$ Thus, $(f+g)(x): R \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=x^{2}+2 x+1=(x+1)^{2} ....
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Question: $\int_{2}^{3} \frac{1}{x} d x$ Solution: Let $I=\int_{2}^{3} \frac{1}{x} d x$ $\int \frac{1}{x} d x=\log |x|=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $\begin{aligned} I =\mathrm{F}(3)-\mathrm{F}(2) \\ =\log |3|-\log |2|=\log \frac{3}{2} \end{aligned}$...
Read More →Let f :
Question: Let $t:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=\sqrt{x}$ and $g(x)=x$. Find $f+g, f-g, f g$ and $\frac{f}{g}$. Solution: It is given that $f:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ such that $f(x)=\sqrt{x}$ and $g(x)=x$. $D(f+g)=[0, \infty) \cap R=[0, \infty)$ So, $f+g:[0, \infty) \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$ $D(f-g)=D(f) \cap D(g)=[0, \infty) \cap R=[0, \infty)$ So,f-g: [0, ) Ris given by $(f-g)(x)=f(x)-g(x)=\sq...
Read More →The sides of a quadrilateral field, taken in order are 26 m,
Question: The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area. Solution: Here the length of the sides of the quadrilateral is given as AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m Diagonal AC is joined. Now, in triangle ADC By applying Pythagoras theorem $A C^{2}=A D^{2}+C D^{2}$ $A C^{2}=14^{2}+7^{2}$ AC = 25 m Now area of triangle ABC Perimeter = 2s = AB + BC + CA 2s = 26 m + 27 m + 25...
Read More →Let f :
Question: Let $t:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=\sqrt{x}$ and $g(x)=x$. Find $f+g, f-g, f g$ and $\frac{f}{g}$. Solution: It is given that $f:[0, \infty) \rightarrow R$ and $g: R \rightarrow R$ such that $f(x)=\sqrt{x}$ and $g(x)=x$. $D(f+g)=[0, \infty) \cap R=[0, \infty)$ So, $f+g:[0, \infty) \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$ $D(f-g)=D(f) \cap D(g)=[0, \infty) \cap R=[0, \infty)$ So,f-g: [0, ) Ris given by $(f-g)(x)=f(x)-g(x)=\sq...
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Question: $\int_{-1}^{1}(x+1) d x$ Solution: Let $I=\int_{-1}^{1}(x+1) d x$ $\int(x+1) d x=\frac{x^{2}}{2}+x=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(1)-\mathrm{F}(-1)$ $=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)$ $=\frac{1}{2}+1-\frac{1}{2}+1$ $=2$...
Read More →Let f, g :
Question: Let $t, g: \mathrm{R} \rightarrow \mathrm{R}$ be defined, respectively by $f(x)=x+1$ and $g(x)=2 x-3 .$ Find $f+g, f-g$ and $\frac{f}{g}$. Solution: f,g: R R is defined, respectively, byf(x) =x+ 1 andg(x) = 2x 3. (f+g) (x) =f(x) +g(x) $=(x+1)+(2 x-3)$ $=3 x-2$ $\therefore(f+g)(x)=3 x-2$ $(f-g)(x)=f(x)-g(x)$ $=(x+1)-(2 x-3)$ $=x+1-2 x+3$ $=-x+4$ $\therefore(f-g)(x)=-x+4$ $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, g(x) \neq 0, x \in \boldsymbol{R}$ $\Rightarrow\left(\frac{f}{g}\righ...
Read More →Let f, g :
Question: Let $t, g: \mathrm{R} \rightarrow \mathrm{R}$ be defined, respectively by $f(x)=x+1$ and $g(x)=2 x-3 .$ Find $f+g, f-g$ and $\frac{f}{g}$. Solution: f,g: R R is defined, respectively, byf(x) =x+ 1 andg(x) = 2x 3. (f+g) (x) =f(x) +g(x) $=(x+1)+(2 x-3)$ $=3 x-2$ $\therefore(f+g)(x)=3 x-2$ $(f-g)(x)=f(x)-g(x)$ $=(x+1)-(2 x-3)$ $=x+1-2 x+3$ $=-x+4$ $\therefore(f-g)(x)=-x+4$ $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, g(x) \neq 0, x \in \boldsymbol{R}$ $\Rightarrow\left(\frac{f}{g}\righ...
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Question: $\int_{0}^{4}\left(x+e^{2 x}\right) d x$ Solution: It is known that, $\int_{a}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h)+\ldots+f(a+(n-1) h)]$, where $h=\frac{b-a}{n}$ Here, $a=0, b=4$, and $f(x)=x+e^{2 x}$ $\therefore h=\frac{4-0}{n}=\frac{4}{n}$ $\Rightarrow \int_{0}^{4}\left(x+e^{2 x}\right) d x=(4-0) \lim _{n \rightarrow \infty} \frac{1}{n}[f(0)+f(h)+f(2 h)+\ldots+f((n-1) h)]$ $=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(0+e^{0}\right)+\left...
Read More →Find the area of the quadrilateral ABCD in which AB = 3 cm,
Question: Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Solution: For triangle ABC $A C^{2}=B C^{2}+A B^{2}$ 25 = 9 + 16 So, triangle ABC is a right angle triangle right angled at point R Area of triangle ABC =12 AB BC = 1/2 3 4 $=6 \mathrm{~cm}^{2}$ From triangle CAD Perimeter = 2s = AC + CD + DA 2s = 5 cm+ 4 cm+ 5 cm 2s= 14 cm s = 7 cm By using Heron's Formula Area of the triangle $\mathrm{CAD}=\sqrt{\mathrm{s} \times(\mathrm{s}-\mat...
Read More →The function f is defined by f(x)=
Question: The function $t$ is defined by $f(x)=\left\{\begin{aligned} 1-x, x0 \\ 1 \quad, x=0 . \text { Draw the graph of } f(x) . \\ x+1, x0 \end{aligned}\right.$ Solution: Here, f(x) = 1 xforx 0. So, $f(-4)=1-(-4)=5$ $f(-3)=1-(-3)=4$ $f(-2)=1-(-2)=3$ $f(-1)=1-(-1)=2$ etc. Also, $f(x)=1$ for $x=0$ Lastly, $f(x)=x+1$ for, $x0$. and $f(1)=2, f(2)=3, f(3)=4, f(4)=5$ and so on. Thus, the graph offis as shown below:...
Read More →The function f is defined by f(x)=
Question: The function $t$ is defined by $f(x)=\left\{\begin{aligned} 1-x, x0 \\ 1 \quad, x=0 . \text { Draw the graph of } f(x) . \\ x+1, x0 \end{aligned}\right.$ Solution: Here, f(x) = 1 xforx 0. So, $f(-4)=1-(-4)=5$ $f(-3)=1-(-3)=4$ $f(-2)=1-(-2)=3$ $f(-1)=1-(-1)=2$ etc. Also, $f(x)=1$ for $x=0$ Lastly, $f(x)=x+1$ for, $x0$. and $f(1)=2, f(2)=3, f(3)=4, f(4)=5$ and so on. Thus, the graph offis as shown below:...
Read More →If f, g and h are real functions defined by f(x)
Question: If $t, g$ and $h$ are real functions defined by $f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(x)=2 x^{2}-3$, find the values of $(2 f+g-h)(1)$ and $(2 f+g-h)(0)$. Solution: Given: $f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(\mathrm{x})=2 \mathrm{x}^{3}-3$ Clearly,f(x) is defined forx+ 1 0 . $\Rightarrow x \geq-1$ $\Rightarrow x \in[-1, \infty]$ Thus, domain (f) = [-1, ] . Clearly,g(x) is defined forx 0 . ⇒x R { 0} andh(x) is defined for allxsuch thatx R . Thus, domain (f) domain (g) domain...
Read More →If f, g and h are real functions defined by f(x)
Question: If $t, g$ and $h$ are real functions defined by $f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(x)=2 x^{2}-3$, find the values of $(2 f+g-h)(1)$ and $(2 f+g-h)(0)$. Solution: Given: $f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(\mathrm{x})=2 \mathrm{x}^{3}-3$ Clearly,f(x) is defined forx+ 1 0 . $\Rightarrow x \geq-1$ $\Rightarrow x \in[-1, \infty]$ Thus, domain (f) = [-1, ] . Clearly,g(x) is defined forx 0 . ⇒x R { 0} andh(x) is defined for allxsuch thatx R . Thus, domain (f) domain (g) domain...
Read More →Find the area of the shaded region in fig. below
Question: Find the area of the shaded region in fig. below Solution: Area of the shaded region =Area of ΔABC Area of ΔADB Now in triangle ADB $\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2} \ldots .$ (i) Given, AD = 12 cm, BD =16 cm Substituting the value of AD and BD in eq (i), we get $A B^{2}=12^{2}+16^{2}$ $=400 \mathrm{~cm}^{2}$ AB = 20 cm Now, area of a triangle = 1/2 AD BD $=96 \mathrm{~cm}^{2}$ Now in triangle ABC, S =1/2 (AB + BC + CA) =1/2 (52 + 48 + 20) = 60 cm By using Heron's Formul...
Read More →If f(x) = loge (1 − x) and g(x) = [x],
Question: If $f(x)=\log _{e}(1-x)$ and $g(x)=[x]$, then determine each of the following functions: (i)f+g (ii)fg (iii) $\frac{f}{g}$ (iv) $\frac{g}{f}$ Also, find $(f+g)(-1),(f g)(0),\left(\frac{f}{g}\right)\left(\frac{1}{2}\right),\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)$ Solution: Given: f(x) = loge(1 x) andg(x) = [x] Clearly,f(x) = loge(1 x) is defined for all ( 1-x) 0. $\Rightarrow 1x$ $\Rightarrow x1$ $\Rightarrow x \in(-\infty, 1)$ Thus, domain (f) = (-, 1) Again, g(x) = [x] is def...
Read More →If f(x) = loge (1 − x) and g(x) = [x],
Question: If $f(x)=\log _{e}(1-x)$ and $g(x)=[x]$, then determine each of the following functions: (i)f+g (ii)fg (iii) $\frac{f}{g}$ (iv) $\frac{g}{f}$ Also, find $(f+g)(-1),(f g)(0),\left(\frac{f}{g}\right)\left(\frac{1}{2}\right),\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)$ Solution: Given: f(x) = loge(1 x) andg(x) = [x] Clearly,f(x) = loge(1 x) is defined for all ( 1-x) 0. $\Rightarrow 1x$ $\Rightarrow x1$ $\Rightarrow x \in(-\infty, 1)$ Thus, domain (f) = (-, 1) Again, g(x) = [x] is def...
Read More →If f(x) = loge (1 − x) and g(x) = [x],
Question: If $f(x)=\log _{e}(1-x)$ and $g(x)=[x]$, then determine each of the following functions: (i)f+g (ii)fg (iii) $\frac{f}{g}$ (iv) $\frac{g}{f}$ Also, find $(f+g)(-1),(f g)(0),\left(\frac{f}{g}\right)\left(\frac{1}{2}\right),\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)$ Solution: Given: f(x) = loge(1 x) andg(x) = [x] Clearly,f(x) = loge(1 x) is defined for all ( 1-x) 0. $\Rightarrow 1x$ $\Rightarrow x1$ $\Rightarrow x \in(-\infty, 1)$ Thus, domain (f) = (-, 1) Again, g(x) = [x] is def...
Read More →Find the values of p and q for which the following system of linear equations has infinite number of solutions:
Question: Find the values ofpandqfor which the following system of linear equations has infinite number of solutions: $2 x+3 y=9$ $(p+q) x+(2 p-q) y=3(p+q+1)$ Solution: GIVEN: $2 x+3 y=9$ $(p+q) x+(2 p-q) y=3(p+q+1)$ To find: To determine for what value ofkthe system of equation has infinitely many solutions We know that the system of equations $a_{1} x+b_{1} y=c_{1}$ $a_{2} x+b_{2} y=c_{2}$ For infinitely many solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here $\frac{2}...
Read More →Show that
Question: Solution: ...(1) It is known that, $\int_{a}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h) \ldots f(a+(n-1) h)]$, where $h=\frac{b-a}{n}$ Here, $a=-1, b=1$, and $f(x)=e^{x}$ $\therefore h=\frac{1+1}{n}=\frac{2}{n}$ $\therefore I=(1+1) \lim _{n \rightarrow \infty} \frac{1}{n}\left[f(-1)+f\left(-1+\frac{2}{n}\right)+f\left(-1+2 \cdot \frac{2}{n}\right)+\ldots+f\left(-1+\frac{(n-1) 2}{n}\right)\right]$ $=2 \lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{-1}+e^{\...
Read More →Let f and g be two real functions defined by f(x)
Question: Let $f$ and $g$ be two real functions defined by $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^{2}}$. Then, describe each of the following functions: (i)f+g (ii)gf (iii)f g (iv) $\frac{f}{q}$ (v) $\frac{g}{f}$ (vi) $2 f-\sqrt{5} g$ (vii) $f^{2}+7 f$ (viii) $\frac{5}{8}$ Solution: Given: $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^{2}}$ Clearly, $f(x)=\sqrt{x+1}$ is defined for all $x \geq-1$. Thus, domain $(f)=[-1, \infty]$ Again, $g(x)=\sqrt{9-x^{2}}$ is defined for $9-x^{2} \geq 0 \Rightarrow x^{2...
Read More →Let f and g be two real functions defined by f(x)
Question: Let $f$ and $g$ be two real functions defined by $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^{2}}$. Then, describe each of the following functions: (i)f+g (ii)gf (iii)f g (iv) $\frac{f}{q}$ (v) $\frac{g}{f}$ (vi) $2 f-\sqrt{5} g$ (vii) $f^{2}+7 f$ (viii) $\frac{5}{8}$ Solution: Given: $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^{2}}$ Clearly, $f(x)=\sqrt{x+1}$ is defined for all $x \geq-1$. Thus, domain $(f)=[-1, \infty]$ Again, $g(x)=\sqrt{9-x^{2}}$ is defined for $9-x^{2} \geq 0 \Rightarrow x^{2...
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