Prove
Question: $\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$ Solution: $\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}=\frac{-2 \sin \frac{2 x+2 \alpha}{2} \sin \frac{2 x-2 \alpha}{2}}{-2 \sin \frac{x+\alpha}{2} \sin \frac{x-\alpha}{2}} \quad\left[\cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\right]$ $=\frac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}$ $=\frac{\left[2 \sin \left(\frac{x+\alpha}{2}\right) \cos \left(\fra...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $2 y^{3}-5 y^{2}-19 y+42$ Solution: Given, $f(x)=2 y^{3}-5 y^{2}-19 y+42$ The constant in f(x) is + 42 The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42 Let, y 2 = 0 = y = 2 $f(2)=2(2)^{3}-5(2)^{2}-19(2)+42$ = 16 20 38 + 42 = 0 So, (y 2) is the factor of f(y) Now, divide f(y) with (y 2) to get other factors By, long division method $2 y^{2}-y-21$ $y-22 y^{3}-5 y^{2}-19 y+42$ $2 y^{3}-4 y^{2}$ $(-) \quad(+)$ $-y^{2}-19 y$ $-y^{2}+2 y$ (+...
Read More →If A and B are two sets,
Question: IfAandBare two sets, thenA (AB)' is equal to ____________. Solution: For setsAandB A (A⋃B)' =A⋂ (A'B') = (AA') B'(using associative properly of sets) =ϕB' Hence,A (A⋃B)' =ϕ...
Read More →If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
Question: IfU= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},A= {1, 2, 3, 5),B= {2, 4, 6, 7) andC= {2, 3, 4, 8}. Then, ____________. (i) (BC)'=_____ (ii) (CA)'=_____ Solution: IfU= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A= {1, 2, 3, 5),B= {2, 4, 6, 7) andC= {2, 3, 4, 8} A' = {4, 6, 7, 8, 9, 10},B'= (1, 3, 5, 8, 9, 10) andC'={1, 5, 6, 7, 9, 10} Then (BC)' =B'⋂C' = {1, 3, 5, 8, 9, 10} ⋂ {1, 5, 6, 7, 9, 10} i.e (BC)' = {1, 5, 9, 10} and (CA)' = (C⋂A')' =C' A = {1, 5, 6, 7, 9, 10} {1, 2, 3, 5} (CA)' = {1, 2, 3, 5, 6, 7,...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{1}{7 x}+\frac{1}{6 y}=3$ $\frac{1}{2 x}-\frac{1}{3 y}=5$ Solution: The given equations are: $\frac{1}{7 x}+\frac{1}{6 y}=3$$\cdots(i)$ $\frac{1}{2 x}-\frac{1}{3 y}=5$...(ii) Multiply equation (ii) by $\frac{1}{2}$ and add both equations we get $\frac{1}{7 x}+\frac{1}{6 y}=3$ Put the value of $x$ in equation $(i)$, we get $\frac{1}{7 \times \frac{1}{14}}+\frac{1}{6 y}=3$ $\Rightarrow \frac{1}{6 y}=1$ $\Rightarrow y=\frac{1}{6}$ Hence the ...
Read More →Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}.
Question: Given the setsA= {1, 3, 5},B= {2, 4, 6} andC= {0, 2, 4, 6, 8}. Then the universal set of all the three setsA,BandCcan be____________. Solution: A= {1, 3, 5} B= {2, 4, 6} C= {0, 2, 4, 6, 8} Then universal let forA, BandCis such that $A \subseteq U$ $B \subseteq U$ and $C \subseteq U$ i.e $U=\{0,1,2,3,4,5,6,8\}$...
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Question: $\frac{\sin ^{2} x}{1+\cos x}$ Solution: $\frac{\sin ^{2} x}{1+\cos x}=\frac{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)^{2}}{2 \cos ^{2} \frac{x}{2}}\left[\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} ; \cos x=2 \cos ^{2} \frac{x}{2}-1\right]$ $=\frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$ $=2 \sin ^{2} \frac{x}{2}$ $=1-\cos x$ $\therefore \int \frac{\sin ^{2} x}{1+\cos x} d x=\int(1-\cos x) d x$ $=x-\sin x+\mathrm{C}$...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $y^{3}-2 y^{2}-29 y-42$ Solution: Given, $f(x)=y^{3}-2 y^{2}-29 y-42$ The constant in f(x) is - 42 The factors of -42 are 1, 2, 3, 6, 7, 14, 21, 42 Let, y + 2 = 0 = y = 2 $f(-2)=(-2)^{3}-2(-2)^{2}-29(-2)-42$ = -8 -8 + 58 42 = 0 So, (y + 2) is the factor of f(y) Now, divide f(y) with (y + 2) to get other factors By, long division $y^{2}-4 y-21$ $y+2 y^{3}-2 y^{2}-29 y-42$ $y^{3}+2 y^{2}$ $(-) \quad(-)$ $-4 y^{2}-29 y$ $-4 y^{2}-8 y$ (+...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $0.5 x+0.7 y=0.74$ $0.3 x+0.5 y=0.5$ Solution: The given equations are: $0.5 x+0.7 y=0.74 \ldots(i)$ $0.3 x+0.5=0.5 \ldots(i i)$ Multiply equation $(i)$ by $0.5$ and $(i i)$ by 2 and subtract equation (ii) from (i) we get Put the value of $x$ in equation $(i)$, we get $0.5 \times 0.5+0.7 y=0.74$ $\Rightarrow 0.7 y=0.49$ $\Rightarrow y=0.7$ Hence the value of $x=0.5$ and $y=0.7$...
Read More →Prove
Question: $\cos ^{4} 2 x$ Solution: $\cos ^{4} 2 x=\left(\cos ^{2} 2 x\right)^{2}$ $=\left(\frac{1+\cos 4 x}{2}\right)^{2}$ $=\frac{1}{4}\left[1+\cos ^{2} 4 x+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $=\frac{1}{4}\left[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $\therefore \int \cos ^{4} 2 x d x=\int\left(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}\right) d x$ $...
Read More →For all sets A and B,
Question: For all setsAandB,B (AB) is equal to ____________. Solution: For setAandB B (AB) =B (AB)' (By defination of negation) =B (A'⋃B') = (BA') ⋃ (BB') = (BA') ⋃ϕ =BA' =BA' =B A Hence,B (AB) =B A...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $2 x-\frac{3}{y}=9$ $3 x+\frac{7}{y}=2, y \neq 0$ Solution: The given equations are: $2 x-\frac{3}{y}=9 \ldots(i)$ $3 x+\frac{7}{y}=2 \ldots(i i)$ Multiply equation $(i)$ by 3 and $(i i)$ by 2 and subtract equation (ii) from (i) we get $6 x-\frac{9}{y}=27$ Put the value of $y$ in equation $(i)$, we get $2 x-\frac{3}{-1}=9$ $\Rightarrow 2 x=6$ $\Rightarrow x=3$ Hence the value of $x=3$ and $y=-1$...
Read More →Prove
Question: $\sin ^{4} x$ Solution: $\sin ^{4} x=\sin ^{2} x \sin ^{2} x$ $=\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$ $=\frac{1}{4}(1-\cos 2 x)^{2}$ $=\frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$ $=\frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$ $=\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ $=\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ $\therefore \int \sin ^{4} x d x=\frac{1}{4} \int\left[...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}-10 x^{2}-53 x-42$ Solution: Given, $f(x)=x^{3}-10 x^{2}-53 x-42$ The constant in f(x) is - 42 The factors of - 42 are 1, 2, 3, 6, 7, 14, 21, 42 Let, x + 1 = 0 = x = - 1 $f(-1)=(-1)^{3}-10(-1)^{2}-53(-1)-42$ = -1 10 + 53 42 = 0 So., (x + 1) is the factor of f(x) Now, divide f(x) with (x + 1) to get other factors By long division, $x^{2}-11 x-42$ $x+1 x^{3}-10 x^{2}-53 x-42$ $x^{3}+x^{2}$ (-) (-) $-11 x^{2}-53 x$ $-11 x^{2}-11 x$...
Read More →For all sets A and B,
Question: For all setsAandB,A (AB) is equal to ____________. Solution: For setAandB A (AB) =A (AB)' (By defination ofAB) =A (A'⋃B') (By De-Morgan's law) = (AA') ⋃ (AB') (By Distributive law) =ϕ⋃ (AB') = AB' i.eA (AB) =A B...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $3 x-\frac{y+7}{11}+2=10$ $2 y+\frac{x+11}{7}=10$ Solution: The given equations are: $3 x-\frac{y+7}{11}+2=10$ $\Rightarrow 3 x-\frac{(y+7)}{11}=8$ $\Rightarrow \frac{33 x-y-7}{11}=8$ $\Rightarrow 33 x-y-7=88$ $\Rightarrow 33 x-y-7=88$ $\Rightarrow 33 x-y=95 \quad \ldots \ldots \ldots(1)$ $2 y+\frac{x+11}{7}=10$ $\Rightarrow \frac{14 y+x+11}{7}=10$ $\Rightarrow 14 y+x+11=70$ $\Rightarrow 14 y+x=59$ $\Rightarrow x+14 y=59 \quad \ldots \ldots \ld...
Read More →Prove
Question: $\cos ^{4} 2 x$ Solution: $\cos ^{4} 2 x=\left(\cos ^{2} 2 x\right)^{2}$ $=\left(\frac{1+\cos 4 x}{2}\right)^{2}$ $=\frac{1}{4}\left[1+\cos ^{2} 4 x+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $=\frac{1}{4}\left[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $\therefore \int \cos ^{4} 2 x d x=\int\left(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}\right) d x$ $...
Read More →Prove
Question: $\cos ^{4} 2 x$ Solution: $\cos ^{4} 2 x=\left(\cos ^{2} 2 x\right)^{2}$ $=\left(\frac{1+\cos 4 x}{2}\right)^{2}$ $=\frac{1}{4}\left[1+\cos ^{2} 4 x+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$ $=\frac{1}{4}\left[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $=\frac{1}{4}\left[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x\right]$ $\therefore \int \cos ^{4} 2 x d x=\int\left(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}\right) d x$ $...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\sqrt{2} x-\sqrt{3} y=0$ $\sqrt{3} x-\sqrt{8} y=0$ Solution: The given equations are: $\sqrt{2} x-\sqrt{3} y=0 \ldots(i)$ $\sqrt{3} x-\sqrt{8} y=0 \ldots($ ii $)$ Multiply equation $(i)$ by $\sqrt{3}$ and equation $(i i)$ by $\sqrt{2}$ and subtract equation (ii) from (i), we get $\sqrt{6} x-3 y=0$ $\Rightarrow y=0$ Put the value of $y$ in equation $(i)$, we get $\sqrt{2} x+0 \times y=0$ $\Rightarrow x=0$ Hence the value of $x=0$ and $y=0$...
Read More →The power set of set A
Question: The power set of setA= {1, 2} is ____________. Solution: LetA= {1, 2} Then number of subject ofAare 22= 4 i.e.ϕ, {1}, {2}, {1, 2} P(A) = {{1}, {2},ϕ, {1, 2}} i.eP(A) = {ϕ, {1}, {2}, {1, 2}}...
Read More →Prove
Question: $\sin ^{4} x$ Solution: $\sin ^{4} x=\sin ^{2} x \sin ^{2} x$ $=\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$ $=\frac{1}{4}(1-\cos 2 x)^{2}$ $=\frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$ $=\frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$ $=\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ $=\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$ $\therefore \int \sin ^{4} x d x=\frac{1}{4} \int\left[...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $x+2 y=\frac{3}{2}$ $2 x+y=\frac{3}{2}$ Solution: The given equations are: $x+2 y=\frac{3}{2}$$\ldots(i)$ $2 x+y=\frac{3}{2} \ldots(i i)$ Multiply equation (ii) by 2 and subtract equation (ii) from (i) we get $x+2 y=\frac{3}{2}$ Put the value of $x$ in equation $(i)$, we get $\frac{1}{2}+2 y=\frac{3}{2}$ $\Rightarrow 2 y=1$ $\Rightarrow y=\frac{1}{2}$ Hence the value of $x=\frac{1}{2}$ and $y=\frac{1}{2}$...
Read More →When A =
Question: WhenA= ϕ, then the number of elements inP(P(A)) is ____________. Solution: A=ϕ Thenn(P(A)) = 1 $\Rightarrow n(P(P(A)))=2^{n(P(A))}$ $=2^{\prime}=2$ i.e.n(P(P(A))) = 2...
Read More →Prove
Question: $\frac{\cos x}{1+\cos x}$ Solution: $\frac{\cos x}{1+\cos x}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$ $\left[\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right.$ and $\left.\cos x=2 \cos ^{2} \frac{x}{2}-1\right]$ $=\frac{1}{2}\left[1-\tan ^{2} \frac{x}{2}\right]$ $\therefore \int \frac{\cos x}{1+\cos x} d x=\frac{1}{2} \int\left(1-\tan ^{2} \frac{x}{2}\right) d x$ $=\frac{1}{2} \int\left(1-\sec ^{2} \frac{x}{2}+1\right) d x$ $=\frac{1}{2} \int...
Read More →When A
Question: WhenA= ϕ, then the number of elements inP(A) is ____________. Solution: LetA= ϕ Then number of elements inP(A) = 1 = 20 i.e. $P(A)=\{\{\phi\}\}$...
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