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Question: $\int_{0}^{1} \frac{d x}{1+x^{2}}$ Solution: Let $I=\int_{0}^{1} \frac{d x}{1+x^{2}}$ $\int \frac{d x}{1+x^{2}}=\tan ^{-1} x=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $\begin{aligned} I =\mathrm{F}(1)-\mathrm{F}(0) \\ =\tan ^{-1}(1)-\tan ^{-1}(0) \\ =\frac{\pi}{4} \end{aligned}$...
Read More →Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.
Question: Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m. Solution: Given, Perimeter of a rhombus = 80 m As we know, Perimeter of a rhombus =4 side=4 a 4 a= 80 m a = 20 m Let AC = 24 m Therefore OA =1/2 AC OA = 12 m In triangle AOB $\mathrm{OB}^{2}=\mathrm{AB}^{2}-\mathrm{OA}^{2}$ $\mathrm{OB}^{2}=20^{2}-12^{2}$ OB = 16 m Also, OB = OD because diagonal of rhombus bisect each other at90 Therefore, BD = 2 OB = 2 16 = 32 m Area of rhombus =1/2 BD AC Area of rho...
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Question: $\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$ Solution: Let $I=\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$ $\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $\begin{aligned} I =\mathrm{F}(1)-\mathrm{F}(0) \\ =\sin ^{-1}(1)-\sin ^{-1}(0) \\ =\frac{\pi}{2}-0 \\ =\frac{\pi}{2} \end{aligned}$...
Read More →If A = {1, 2, 3} and B = {x, y},
Question: If A = {1, 2, 3} and B = {x,y}, then the number of functions that can be defined from A into B is (a) 12 (b) 8 (c) 6 (d) 3 Solution: (b) 8 Given: Number of elements in set A = 3 Number of elements in set B = 2 Therefore, the number of functions that can be defined from A into B is = 23= 8....
Read More →One says, "Give me a hundred, friend! I shall then become twice as rich as you.
Question: One says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their respective capital? Solution: To find: (1) Total amount of A. (2) Total amount of B. Suppose A has Rsxand B has Rsy According to the given conditions, x+ 100 = 2(y100)x+ 100 = 2y 200x2y=300 ....(1)andy+ 10 = 6(x 10)y+ 10 = 6x 606xy= 70 ....(2) Multiplying equation (2) by 2 we get 12x 2y= 140 .....
Read More →If f(x)=
Question: If $f(x)=\log \left(\frac{1+x}{1-x}\right)$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$, then $f(g(x))$ is equal to (a) $f(3 x)$ (b) $\{f(x)\}^{3}$ (c) $3 f(x)$ (d) $-f(x)$ Solution: (c) $3 f(x)$ $f(x)=\log \left(\frac{1+x}{1-x}\right)$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$ Now, $\frac{1+g(x)}{1-g(x)}=\frac{1+\frac{3 x+x^{3}}{1+3 x^{2}}}{1-\frac{3 x+x^{3}}{1+3 x^{2}}}$ $=\frac{1+3 x^{2}+3 x+x^{3}}{1+3 x^{2}-3 x-x^{3}}$ $=\frac{(1+x)^{3}}{(1-x)^{3}}$ Then, $f(g(x))=\log \left(\frac{1+g(x)}{...
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Question: $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$ Solution: Let $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$ $\int \operatorname{cosec} x d x=\log |\operatorname{cosec} x-\cot x|=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}\left(\frac{\pi}{6}\right)$ $=\log \left|\operatorname{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|\operatorname{cosec} \frac{\pi}{6}-\cot \...
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Question: $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$ Solution: Let $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$ $\int \operatorname{cosec} x d x=\log |\operatorname{cosec} x-\cot x|=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}\left(\frac{\pi}{6}\right)$ $=\log \left|\operatorname{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|\operatorname{cosec} \frac{\pi}{6}-\cot \...
Read More →Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium?
Question: Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium? Solution: Given, Two parallel sides of trapezium are AB = 77 m and CD = 60 m The other two parallel sides of trapezium are BC = 26 m, AD = 25m Join AE and CF DE is perpendicular to AB and also, CF is perpendicular to AB Therefore, DC = EF = 60 m Let AE = x So, BF = 77 - 60 - x BF = 17 - x In triangle ADE, By using Pythagoras theorem, $\mathrm{DE}^{2}=\mathrm{AD}^{2...
Read More →Which of the following are functions?
Question: Which of the following are functions? (a) {(x,y) :y2=x,x,y R} (b) {(x,y) :y= |x|,x,y R} (c) {(x,y) :x2+y2= 1,x,y R} (d) {(x,y) :x2y2= 1,x,y R} Solution: (b) {(x,y) :y= |x|,x,y R} For every value ofx R, there is a unique value y R. i.e. there is a unique image for all values ofx R. Also, values of x occur only once in the ordered pairs. Thus, it is a function....
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Question: $\int_{0}^{\frac{\pi}{4}} \tan x d x$ Solution: Let$I=\int_{0}^{\frac{\pi}{4}} \tan x d x$ $\int \tan x d x=-\log |\cos x|=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}(0)$ $=-\log \left|\cos \frac{\pi}{4}\right|+\log |\cos 0|$ $=-\log \left|\frac{1}{\sqrt{2}}\right|+\log |1|$ $=-\log (2)^{-\frac{1}{2}}$ $=\frac{1}{2} \log 2$...
Read More →The range of f(x) = cos [x],
Question: The range off(x) = cos [x],for /2 x /2 is(a) {1, 1, 0} (b) {cos 1, cos 2, 1} (c) {cos 1, cos 1, 1} (d) [1, 1] Solution: (b) {cos 1, cos 2, 1} Since, $f(x)=\cos [x]$, where $\frac{-\pi}{2}x\frac{\pi}{2}$, $-\frac{\pi}{2}x\frac{\pi}{2}$ $\Rightarrow-1.57x1.57$ $\Rightarrow[x] \in\{-1,0,1,2\}$ Thus, $\cos [x]=\{\cos (-1), \cos 0, \cos 1, \cos 2\}$ Range of $f(x)=\{\cos 1,1, \cos 2\}$...
Read More →The range of f(x) = cos [x],
Question: The range off(x) = cos [x],for /2 x /2 is(a) {1, 1, 0} (b) {cos 1, cos 2, 1} (c) {cos 1, cos 1, 1} (d) [1, 1] Solution: (b) {cos 1, cos 2, 1} Since, $f(x)=\cos [x]$, where $\frac{-\pi}{2}x\frac{\pi}{2}$, $-\frac{\pi}{2}x\frac{\pi}{2}$ $\Rightarrow-1.57x1.57$ $\Rightarrow[x] \in\{-1,0,1,2\}$ Thus, $\cos [x]=\{\cos (-1), \cos 0, \cos 1, \cos 2\}$ Range of $f(x)=\{\cos 1,1, \cos 2\}$...
Read More →Let f(x) =
Question: Letf(x) = |x 1|. Then,(a)f(x2) = [f(x)]2 (b)f(x+y) =f(x)f(y) (c)f(|x| = |f(x)| (d) None of these Solution: (d) None of these $f(x)=|x-1|$ Since, $\left|x^{2}-1\right| \neq|x-1|^{2}$ $f\left(x^{2}\right) \neq(f(x))^{2}$ Thus, (i) is wrong. Since, $|x+y-1| \neq|x-1||y-1|$ $f(x+y) \neq f(x) f(y)$ Thus, (ii) is wrong. Since ||$x|-1| \neq|| x-1||=|x-1|$ $f(|x|) \neq|f(x)|$ Thus, (iii) is wrong. Hence, none of the given options is the answer....
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Question: $\int_{4}^{5} e^{x} d x$ Solution: Let $I=\int_{4}^{5} e^{x} d x$ $\int e^{x} d x=e^{x}=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $\begin{aligned} I =\mathrm{F}(5)-\mathrm{F}(4) \\ =e^{5}-e^{4} \\ =e^{4}(e-1) \end{aligned}$...
Read More →Find the values of a and b for which the following system of equations has infinitely many solutions:
Question: Find the values of a and b for which the following system of equations has infinitely many solutions: (i) $(2 a-1) x-3 y=5$ $3 x+(b-2) y=3$ (ii) $2 x-(2 a+5) y=5$ $(2 b+1) x-9 y=15$ (iii) $(a-1) x+3 y=2$ $6 x+(1-2 b) y=6$ $(i v) 3 x+4 y=12$ $(a+b) x+2(a-b) y=5 a-1$ $(v) 2 x+3 y=7$ $(a-b) x+(a+b) y=3 a+b-2$ $(v i) 2 x+3 y-7=0$ $\quad(a-1) x+(a+1) y=(3 a-1)$ (vii) $2 x+3 y=7$ $\quad(a-1) x+(a+2) y=3 a$ $($ viii $) x+2 y=1$ $(a-b) x+(a+b) y=a+b-2$ $(i x) 2 x+3 y=7$ $2 a x+a y=28-b y$ Solu...
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Question: $\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$ $\int \cos 2 x d x=\left(\frac{\sin 2 x}{2}\right)=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $\begin{aligned} I =\mathrm{F}\left(\frac{\pi}{2}\right)-\mathrm{F}(0) \\ =\frac{1}{2}\left[\sin 2\left(\frac{\pi}{2}\right)-\sin 0\right] \\ =\frac{1}{2}[\sin \pi-\sin 0] \\ =\frac{1}{2}[0-0]=0 \end{aligned}$...
Read More →If f(x) = cos (log x), then the value of
Question: If $f(x)=\cos (\log x)$, then the value of $f\left(x^{2}\right) f\left(y^{2}\right)-\frac{1}{2}\left\{f\left(\frac{x^{2}}{y^{2}}\right)+f\left(x^{2} y^{2}\right)\right\}$ is (a) 2 (b) 1 (c) 1/2 (d) None of these Solution: (d) None of these Given: $f(x)=\cos (\log x)$ $\Rightarrow f\left(x^{2}\right)=\cos \left(\log \left(x^{2}\right)\right)$ $\Rightarrow f\left(x^{2}\right)=\cos (2 \log (x))$ Similarly, $f\left(y^{2}\right)=\cos (2 \log (y))$ Now, $f\left(\frac{x^{2}}{y^{2}}\right)=\co...
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Question: $\int_{0}^{\frac{\pi}{4}} \sin 2 x d x$ Solution: Let $I=$ $\int_{0}^{\frac{\pi}{4}} \sin 2 x d x$ $\int \sin 2 x d x=\left(\frac{-\cos 2 x}{2}\right)=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}(0)$ $=-\frac{1 \pi}{2}\left[\cos 2\left(\frac{}{4}\right)-\cos 0\right]$ $=-\frac{1 \pi}{2}\left[\cos \left(-\frac{1}{2}\right)-\cos 0\right]$ $=-\frac{1}{2}[0-1]$ $=\frac{1}{2}$...
Read More →A park in the shape of a quadrilateral ABCD, has angle
Question: A park in the shape of a quadrilateral ABCD, has angle C = 90, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy? Solution: Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m. Let us join BD In triangle BCD, apply Pythagoras theorem $B D^{2}=B C^{2}+C D^{2}$ $B D^{2}=12^{2}+5^{2}$ BD = 13 m Area of triangle BCD =1/2 BC CD = 1/2 12 5 $=30 \mathrm{~m}^{2}$ Now, in triangle ABD Perimeter = 2s = 9 m + 8m + 13m s = 15 m By using Heron's Formula, A...
Read More →If f(x) = cos (log x), then the value of
Question: If $f(x)=\cos (\log x)$, then the value of $f\left(x^{2}\right) f\left(y^{2}\right)-\frac{1}{2}\left\{f\left(\frac{x^{2}}{y^{2}}\right)+f\left(x^{2} y^{2}\right)\right\}$ is (a) 2 (b) 1 (c) 1/2 (d) None of these Solution: (d) None of theseGiven: $f(x)=\cos (\log x)$ $\Rightarrow f\left(x^{2}\right)=\cos \left(\log \left(x^{2}\right)\right)$ $\Rightarrow f\left(x^{2}\right)=\cos (2 \log (x))$ Similarly, $f\left(y^{2}\right)=\cos (2 \log (y))$ Now, $f\left(\frac{x^{2}}{y^{2}}\right)=\cos...
Read More →Which one of the following is not a function?
Question: Which one of the following is not a function?(a) {(x,y) :x,y R,x2=y} (b) {(x,y) :x,y, R,y2=x} (c) {(x,y) :x,y R,x2=y3} (d) {(x,y) :x,y, R,y=x3} Solution: (b) {(x,y) :x,y, R,y2=x} $y^{2}=x$ gives two values of $y$ for a value of $x$. i.e. there are two images for a value of $x$. For example: $(2)^{2}=4$ and $(-2)^{2}=4$ Thus, it is not a function....
Read More →Which one of the following is not a function?
Question: Which one of the following is not a function?(a) {(x,y) :x,y R,x2=y} (b) {(x,y) :x,y, R,y2=x} (c) {(x,y) :x,y R,x2=y3} (d) {(x,y) :x,y, R,y=x3} Solution: (b) {(x,y) :x,y, R,y2=x} $y^{2}=x$ gives two values of $y$ for a value of $x$. i.e. there are two images for a value of $x$. For example: $(2)^{2}=4$ and $(-2)^{2}=4$ Thus, it is not a function....
Read More →If f :
Question: If $t: \mathrm{Q} \rightarrow \mathrm{Q}$ is defined as $f(x)=x^{2}$, then $f^{-1}(9)$ is equal to (a) 3 (b) 3 (c) {3, 3} (d) ϕ Solution: (c) {3, 3} If $f: A \rightarrow B$, 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 $f^{-1}\{9\}=x$ Then, $f(x)=9$ $\Rightarrow x^{2}=9$ $\Rightarrow x=\pm 3$ $\therefore f^{-1}\{9\}=\{-3,3\}$...
Read More →If f :
Question: If $t: \mathrm{Q} \rightarrow \mathrm{Q}$ is defined as $f(x)=x^{2}$, then $f^{-1}(9)$ is equal to (a) 3 (b) 3 (c) {3, 3} (d) ϕ Solution: (c) {3, 3} If $f: A \rightarrow B$, 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 $f^{-1}\{9\}=x$ Then, $f(x)=9$ $\Rightarrow x^{2}=9$ $\Rightarrow x=\pm 3$ $\therefore f^{-1}\{9\}=\{-3,3\}$...
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