Solve the following systems of equations:
Question: Solve the following systems of equations: $x+\frac{y}{2}=4$ $\frac{x}{3}+2 y=5$ Solution: The given equations are: $x+\frac{y}{2}=4 \ldots$$\ldots(i)$ $\frac{x}{3}+2 y=5$$. .($ ii $)$ Multiply equation $(i)$ by 4 and subtract equations $(i)-(i i)$, we get $4 x+2 y=16$ Put the value of $x$ in equation $(i)$, we get $3+\frac{y}{2}=4$ $\Rightarrow \frac{y}{2}=1$ $\Rightarrow y=2$ Hence the value of $x$ and $y$ are $x=3$ and $y=2$...
Read More →If A and B are any two sets,
Question: IfAandBare any two sets, thenABis equal to ____________. Solution: LetAandBbe any two sets thenA B=A⋂BC i.eA B= A⋂BC...
Read More →If A and B are finite sets such that
Question: IfAandBare finite sets such thatAB, thenn(AB) = ____________. Solution: Letn(A) =m n(B) =n, since both are finite set SinceABthenAB=B ⇒n(AB) =n(B) n(AB) =n=n(B)...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $y^{3}-7 y+6$ Solution: Given, $f(y)=y^{3}-7 y+6$ The constant term in f(y) is 6 The factors are 1, 2, 3, 6 Let, y 1 = 0 = y = 1 $f(1)=(1)^{3}-7(1)+6$ = 1 7 + 6 = 0 So, (y 1) is the factor of f(y) Similarly, (y 2) and (y + 3) are also the factors Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors = f(y) = k(y 1)( y 2)(y + 3) $\Rightarrow y^{3}-7 y+6=k(y-1)(y-2)(y+3)--1$ Substitute k = 0 in eq 1 =...
Read More →Prove
Question: $\frac{1-\cos x}{1+\cos x}$ Solution: $\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$ $\left[2 \sin ^{2} \frac{x}{2}=1-\cos x\right.$ and $\left.2 \cos ^{2} \frac{x}{2}=1+\cos x\right]$ $=\tan ^{2} \frac{x}{2}$ $=\left(\sec ^{2} \frac{x}{2}-1\right)$ $\therefore \int \frac{1-\cos x}{1+\cos x} d x=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x$ $=\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]+\mathrm{C}$ $=2 \tan \frac{x}{2}-x+\mathrm{C}$...
Read More →The set
Question: The set {xR: 1 x 2} can be written as ____________. Solution: The set {xR: 1 x 2} is in interval with real values from 1 to 2, including 1. i.e {xR: 1 x 2} = [1, 2]...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{4}{x}+3 y=8$ $\frac{6}{x}-4 y=-5$ Solution: The given equations are: $\frac{4}{x}+3 y=8$$\ldots(i)$ $\frac{6}{x}-4 y=-5$$\cdots(i i)$ Multiply equation (i) by 4 and equation (ii) by 3 and add both equations we get $\frac{16}{x}+12 y=32$ Put the value of $x$ in equation $(i)$ we get $\frac{4}{2}+3 y=8$ $\Rightarrow 3 y=6$ $\Rightarrow y=2$ Hence the value of $x=2$ and $y=2$....
Read More →If A is a finite set containing n elements,
Question: IfAis a finite set containingnelements, then the number of subsets ofAis ____________. Solution: Letn(A) =n; ieAhas elements Then number of subsets ofAis 2n....
Read More →If A and B are two finite sets,
Question: IfAandBare two finite sets, thenn(A) +n(B) is equal to ____________. Solution: LetAandBbe two finite sets Letn(A) =m n(B) =n Thenn(A) +n(B) =m+nwhich is also finite...
Read More →Let S = the set of points inside the square,
Question: LetS= the set of points inside the square,T= the set of points inside the triangle andC= the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then, (a)STC= ϕ (b)STC=C (c)STC=S (d)ST=SC Solution: LetS= the set of points inside the square T= the set of points inside the triangle C= the set of points inside circle Given triangle and circle intersect each other and are contained in a square i.eTandCare in square ⇒S⋃T⋃C = S Henc...
Read More →In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students play both the games.
Question: In a class of 60 students, 25 students play cricket and 20 students play tennis and 10 students play both the games. Then the number of students who play neither is (a) 0 (b) 25 (c) 35 (d) 45 Solution: Let ⋃ denote the universal set Let C denote the set of students playing circket Let T denote the set of students playing tennis n(⋃) = 60,n(C) = 25,n(T) = 20 n(C⋂T) = 10 Thenn(C⋃T)'=n(⋃) n(C⋃T) n(⋃) [n(C) +n(T) n(C⋂T)] = 60 [25 + 20 10] = 60 [45 10] = 60 35 n(C⋃T)'= 25 Hence, the number ...
Read More →Prove
Question: $\sin 4 x \sin 8 x$ Solution: It is known that, $\sin A \cdot \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$ $\therefore \int \sin 4 x \sin 8 x d x$ $=\int \frac{1}{2}\{\cos (4 x-8 x)-\cos (4 x+8 x)\} d x$ $=\frac{1}{2} \int\{\cos (-4 x)-\cos (12 x)\} d x$ $=\frac{1}{2} \int\{\cos 4 x-\cos 12 x\} d x$ $=\frac{1}{2}\left(\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right)+\mathrm{C}$...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{x}{3}+\frac{y}{4}=11$ $\frac{5 x}{6}-\frac{y}{3}=-7$ Solution: The given equations are: $\frac{x}{3}+\frac{y}{4}=11 \ldots$$(i)$ $\frac{5 x}{6}-\frac{y}{3}=-7$...$(i i)$ Multiply equation $(i)$ by $\frac{1}{3}$ and equation $(i i)$ by $\frac{1}{4}$ and add both equations we get $\frac{x}{9}+\frac{y}{12}=\frac{11}{3}$ Put the value of $x$ in equation $(i)$ we get $\frac{6}{3}+\frac{y}{4}=11$ $\Rightarrow \frac{y}{4}=9$ $\Rightarrow y=36$ ...
Read More →In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both.
Question: In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 both. Then the number of persons who read neither is (a) 210 (b) 290 (c) 180 (d) 260 Solution: Total persons in a town is 840 Let H denote set of persons who read Hindi Let E denote set of persons who read English Thenn(⋃) = 840,n(H) = 450,n(E) = 300 Thenn(H⋃E)'=n(⋃) n(H⋃E) =n(⋃) [n(H) +n(E) n(H⋂E)] = 840 [450 + 300 200] = 840 550 n(H⋃E)'= 290 Therefore, the number of persons who read neither is 290. Hence, the ...
Read More →Prove
Question: $\sin x \sin 2 x \sin 3 x$ Solution: It is known that, $\sin A \sin B=\frac{1}{2}\{\cos (A-B)-\cos (A+B)\}$ $\therefore \int \sin x \sin 2 x \sin 3 x d x=\int\left[\sin x \cdot \frac{1}{2}\{\cos (2 x-3 x)-\cos (2 x+3 x)\}\right] d x$ $=\frac{1}{2} \int(\sin x \cos (-x)-\sin x \cos 5 x) d x$ $=\frac{1}{2} \int(\sin x \cos x-\sin x \cos 5 x) d x$ $=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$ $=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin...
Read More →Let S = {x : x is a positive multiple of 3 less than 100},
Question: LetS= {x:xis a positive multiple of 3 less than 100},P= {x:xis a prime less than 20}. Then,n(S) +n(P) is (a) 34 (b) 31 (c) 33 (d) 30 Solution: LetS= {x:xis a positive multiple of 3 less than 100} P= {x:xis prime less than 20} HereS= {3, 6, 9, 12, ........99} n(S) = 33 andP= {2, 3, 5, 7, 11, 13, 17, 19} n(P) = 8 ⇒n(S) +n(P) = 33 + 8 = 41...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{x}{7}+\frac{y}{3}=5$ $\frac{x}{2}-\frac{y}{9}=6$ Solution: The given equations are: $\frac{x}{7}+\frac{y}{3}=5 \ldots$(i) $\frac{x}{2}-\frac{y}{9}=6 \ldots$(ii) Multiply equation (i) by $\frac{1}{3}$ and add both equations we get $\frac{x}{21}+\frac{y}{9}=\frac{5}{3}$ Put the value of $x$ in equation $(i)$ we get $\frac{14}{7}+\frac{y}{3}=5$ $\Rightarrow \frac{y}{3}=3$ $\Rightarrow y=9$ Hence the value of $x=14$ and $y=9$....
Read More →If A and B are two sets,
Question: IfAandBare two sets, thenA (AB) equals (a)A (b)B (c) ϕ (d)AB Solution: A (A⋃B) = (AA) ⋃ (AB) = A ⋃ (AB) A (A⋃B) =A (sinceABA) Hence, the correct answer is option A....
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}-23 x^{2}+142 x-120$ Solution: Let, $f(x)=x^{3}-23 x^{2}+142 x-120$ The constant term in f(x) is -120 The factors of -120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 Let, x - 1 = 0 = x = 1 $f(1)=(1)^{3}-23(1)^{2}+142(1)-120$ = 1 - 23+ 142 - 120 = 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}-22 x+120$ $x-1 x^{3}-23 x^{2}+142 x-120$ $x^{3}-x^{2}$ $(-)(...
Read More →The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ⋃ C′ is equal to
Question: The set (ABC) (ABC)⋃Cis equal to (a)BC (b)AC (c)BC (d)AC Solution: (A⋃B⋃C) (ABC)⋃CHence (ABC) (ABC) ⋃C' = A C'Hence , the correct answer is option D....
Read More →Two finite sets have m and n elements respectively.
Question: Two finite sets havemandnelements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The value ofmandnis respectively are: (a) 7, 6 (b) 5, 1 (c) 6, 3 (d) 8, 7 Solution: Let us suppose two finite sets are A and BLet A hasmelements Let B hasnelements Then total number of subjects of A is 2mand total number of subjects of B is 2n. According to given condition, 2m 2n= 56 i.e 2n(2m n 1) = 56 Since 56 = 8 7 $=2^{3} \times 7$ ...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $7(y+3)-2(x+2)=14$ $4(y-2)+3(x-3)=2$ Solution: The given equations are: $7(y+3)-2(x+2)=14$$\ldots(i)$ $7 y-2 x=-3$ $4(y-2)+3(x-3)=2$$\ldots(i i)$ $4 y+3 x=19$ Multiply equation $(i)$ by 3 and equation $(i i)$ by 2 and add both equations we get Put the value of $x$ in equation $(i)$ we get $7 \times 1-2 x=-3$ $\Rightarrow-2 x=-10$ $\Rightarrow x=5$ Hence the value of $x=5$ and $y=1$...
Read More →Prove
Question: $\sin ^{3} x \cos ^{3} x$ Solution: Let $\begin{aligned} I =\int \sin ^{3} x \cos ^{3} x \cdot d x \\ =\int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x \\ =\int \cos ^{3} x\left(1-\cos ^{2} x\right) \sin x \cdot d x \end{aligned}$ Let $\cos x=t$ $\Rightarrow-\sin x \cdot d x=d t$ $\Rightarrow I=-\int t^{3}\left(1-t^{2}\right) d t$ $=-\int\left(t^{3}-t^{5}\right) d t$ $=-\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}+\mathrm{C}$ $=-\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x...
Read More →Each set
Question: Each set $X_{r}$ contains 5 elements and each set $Y_{r}$ contains 2 elements and $\bigcup_{r=1}^{20} X_{r}=S=\bigcup_{r=1}^{n} Y_{r}$. If each element of $S$ belongs to exactly 10 of the $X_{r}^{\prime s}$ and to exactly 4 of the $Y_{r}^{\prime} s$, then $n$ is (a) 10 (b) 20 (c) 100 (d) 50 Solution: Let us suppose Each $x_{r}$ contains 5 elements and each $y_{r}$ contains 2 elements such that $\bigcup_{r=1}^{20} X_{r}=S=\bigcup_{r=1}^{n} Y_{r}$ $\therefore n(S)=20 \times 5 \quad\left(...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{x}{2}+y=0.8$ $\frac{7}{x+\frac{y}{2}}=10$ Solution: The given equations are: $\frac{x}{2}+y=0.8 \ldots$$\ldots(i)$ $\frac{7}{x+\frac{y}{2}}=10$ $\Rightarrow 2 x+y=1.4 \ldots(i i)$ Subtract (ii) from (i) we get $\frac{x}{2}+y=0.8$ Put the value of $x$ in equation $(i i)$ we get $2 \times 0.4+y=1.4$ $\Rightarrow y=0.6$ Hence the value of $x=0.4$ and $y=0.6$....
Read More →