A and B are any two non-empty sets and A is proper subset of B.
Question: AandBare any two non-empty sets andAis proper subset ofB. Ifn(A) = 5, then the minimum possible value ofn(A∆B) is___________. Solution: GivenAB=ϕ $A \subseteq B$ and $n(A)=5$ Then minimum possible value ofn(A∆B) Since $A \subsetneq B \quad$ i.e $n(A) \subsetneq n(B)$ ⇒A⋃B=B AB= n(A∆B) =n(A⋃B) n(AB) =n(B) n(A) =n(B) 5 i.e n(A∆B) = n(B) 5 n(A) 5 = 0 i.e.n(A∆B) 0 Minimum possible value ofn(A∆B) = 1...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}-2 x^{2}-x+2$ Solution: Let, $f(x)=x^{3}-2 x^{2}-x+2$ Theconstant term is 2 The factors of 2 are 1, 1/2 Let, x 1= 0 = x = 1 $f(1)=(1)^{3}-2(1)^{2}-(1)+2$ = 1 2 1 + 2 = 0 So, (x 1) is the factor of f(x) Divide f(x) with (x 1) to get other factors By, long division $x^{2}-x-2$ $x-1 x^{3}-2 x^{2}-y+2$ $x^{3}-x^{2}$ (-) (+) $-x^{2}-x$ $-x^{2}+x$ (+) (-) 2x + 2 2x + 2 (+) (-) 0 $=x^{3}-2 x^{2}-y+2=(x-1)\left(x^{2}-x-2\right)$ Now, $x...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{2}{x}+\frac{5}{y}=1$ $\frac{60}{x}+\frac{40}{y}=19, x \neq 0, y \neq 0$ Solution: The given equations are: $\frac{2}{x}+\frac{5}{y}=1$$\ldots(i)$ $\frac{60}{x}+\frac{40}{y}=19 \ldots(i i)$ Multiply equation $(i)$ by 8 and subtract (ii) from equation (i), we get $-\frac{44}{x}=-11$ $\Rightarrow x=4$ Put the value of $x$ in equation $(i)$, we get $\Rightarrow \frac{2}{4}+\frac{5}{y}=1$ $\Rightarrow \frac{5}{y}=1-\frac{2}{4}$ $\Rightarrow \...
Read More →A and B are any two non-empty sets and A is proper subset of B.
Question: AandBare any two non-empty sets andAis proper subset ofB. Ifn(A) = 5, then the minimum possible value ofn(A∆B) is___________. Solution: GivenAB=ϕ $A \subseteq B$ and $n(A)=5$ Then minimum possible value ofn(A∆B) Since $A \subsetneq B \quad$ i.e $n(A) \subsetneq n(B)$ ⇒A⋃B=B AB= n(A∆B) =n(A⋃B) n(AB) =n(B) n(A) =n(B) 5 i.e n(A∆B) = n(B) 5 n(A) 5 = 0 i.e.n(A∆B) 0 Minimum possible value ofn(A∆B) = 1...
Read More →Prove
Question: $\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}$ Solution: $\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}=\frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}+\frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}$ $=\frac{\sin x}{\cos ^{2} x}+\frac{\cos x}{\sin ^{2} x}$ $=\tan x \sec x+\cot x \operatorname{cosec} x$ $\begin{aligned} \therefore \int \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x =\int(\tan x \sec x+\cot x \operatorname{cosec} x) d x \\ =\sec x-\operatorname...
Read More →If n(A ∩ B) = 5, n(A ∩ C) = 7 and n(A ∩ B ∩ C) = 3,
Question: Ifn(AB) = 5,n(AC) = 7 andn(ABC) = 3, then the minimum possible value ofn(BC) is ____________. Solution: Ifn(AB) = 5 n(AC) = 7 n(ABC) = 3 Then the minimum possible value ofn(BC) Sincen(A⋃B⋃C) =n(A) +n(B) +n(C) n(AB) n(BC) n(CA)+n(ABC) SinceABCBC ⇒n(ABC) n(BC) ⇒ 3 n(BC) minimum possible value ofn(BC) = 3...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{3}{x}-\frac{1}{y}=-9$ $\frac{2}{x}+\frac{3}{y}=5$ Solution: The given equations are: $\frac{3}{x}-\frac{1}{y}=-9$$\ldots(i)$ $\frac{2}{x}+\frac{3}{y}=5 \ldots(i i)$ Multiply equation $(i)$ by 3 and add both equations, we get $\frac{9}{x}-\frac{3}{y}=-27$ Put the value of $x$ in equation $(i)$, we get $\frac{3}{\frac{-1}{2}}-\frac{1}{y}=-9$ $\Rightarrow \frac{-1}{y}=-3$ $\Rightarrow y=\frac{1}{3}$ Hence the value of $x=-\frac{1}{2}$ and $...
Read More →If n(A ∩ B) = 5, n(A ∩ C) = 7 and n(A ∩ B ∩ C) = 3,
Question: Ifn(AB) = 5,n(AC) = 7 andn(ABC) = 3, then the minimum possible value ofn(BC) is ____________. Solution: Ifn(AB) = 5n(AC) = 7n(ABC) = 3Then the minimum possible value ofn(BC) Sincen(A⋃B⋃C) =n(A) +n(B) +n(C) n(AB) n(BC) n(CA)+n(ABC) SinceABCBC ⇒n(ABC) n(BC) ⇒ 3 n(BC) minimum possible value ofn(BC) = 3...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $2 y^{3}+y^{2}-2 y-1$ Solution: Given, $f(y)=2 y^{3}+y^{2}-2 y-1$ The constant term is 2 The factors of 2 are 1, 1/2 Let, y 1= 0 = y = 1 $f(1)=2(1)^{3}+(1)^{2}-2(1)-1$ = 2 + 1 2 1 = 0 So, (y 1) is the factor of f(y) Divide f(y) with (y 1) to get other factors By, long division $2 y^{2}+3 y+1$ $y-1,2 y^{3}+y^{2}-2 y-1$ $2 y^{3}-2 y^{2}$ (-) (+) $3 y^{2}-2 y$ $3 y^{2}-3 y$ (-) (+) y 1 y 1 (-) (+) 0 $=2 y^{3}+y^{2}-2 y-1=(y-1)\left(2 y^{...
Read More →If A, B and C are any three non-empty sets such that any two of them are disjoint,
Question: IfA,BandCare any three non-empty sets such that any two of them are disjoint, then (ABC) (ABC) = ____________. Solution: IfA, BandCare three non-empty sets such that any two of there are disjoint say AB =ϕ BC =ϕandA C =ϕ ThenABC=ϕ ⇒ (A⋃B⋃C) (ABC) = (A⋃B⋃C) ϕ = ϕ...
Read More →Prove
Question: $\tan ^{4} x$ Solution: $\tan ^{4} x$ $=\tan ^{2} x \cdot \tan ^{2} x$ $=\left(\sec ^{2} x-1\right) \tan ^{2} x$ $=\sec ^{2} x \tan ^{2} x-\tan ^{2} x$ $=\sec ^{2} x \tan ^{2} x-\left(\sec ^{2} x-1\right)$ $=\sec ^{2} x \tan ^{2} x-\sec ^{2} x+1$ $\begin{aligned} \therefore \int \tan ^{4} x d x =\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 \cdot d x \\ =\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+\mathrm{C} \end{aligned}$ ...(1) Consider $\int \sec ^{2} x \tan ^{2} x d x$...
Read More →If n(A ∩ B) = 10, n(B ∩ C) = 20 and n(A ∩ C) = 30,
Question: Ifn(AB) = 10,n(BC) = 20 andn(AC) = 30, then the greatest possible value ofn(ABC) is ____________. Solution: Ifn(AB) = 10 n(BC) = 20 n(AC) = 30 To find the greatest possible value ofn(ABC) SinceABCAB,ABCACandABCBC ⇒n(ABC) n(A B),n(ABC) n(AC) andn(ABC) n(BC) ⇒n(ABC) min{n(AB),n(AC),n(BC)} min {10, 20, 30} = 10 i.e maximum / greatest possible value ofn(ABC) is 10....
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{15}{u}+\frac{2}{\nu}=17$ $\frac{1}{u}+\frac{1}{\nu}=\frac{36}{5}$ Solution: The given equations are: $\frac{15}{u}+\frac{2}{v}=17$ $\frac{1}{u}+\frac{1}{v}=\frac{36}{5} \ldots(i i)$ Multiply equation (ii) by 2 and subtract (ii) from (i), we get $\frac{15}{u}+\frac{2}{v}=17$ Put the value of $u$ in equation $(i)$, we get $\frac{15}{5}+\frac{2}{v}=17$ $\Rightarrow \frac{2}{v}=14$ $\Rightarrow v=\frac{1}{7}$ Hence the value of $u=5$ and $v=...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}-3 x^{2}-9 x-5$ Solution: Given, $f(x)=x^{3}-3 x^{2}-9 x-5$ The constant in f(x) is -5 The factors of -5 are 1, 5 Let, x + 1 = 0 = x = -1 $f(-1)=(-1)^{3}-3(-1)^{2}-9(-1)-5$ = -1 3 + 9 5 = 0 So, (x + 1) is the factor of f(x) Divide f(x) with (x + 1) to get other factors By, long division $x^{2}-4 x-5$ $x+1 x^{3}-3 x^{2}-9 x-5$ $x^{3}+x^{2}$ (-) (-) $-4 x^{2}-9 x$ $-4 x^{2}-4 x$ (+) (+) - 5x 5 - 5x 5 (+) (+) 0 $=x^{3}-3 x^{2}-9 x-...
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 number less than 20} Then,n(S) +n(P) = ____________. Solution: LetS={x:xis a positive multiple of 3 less than 100} S= {3, 6, 9, ....... 99} P= {x:xis a prime number less than 20}= {2, 3, 5, 7, 11, 13, 17, 19} n(S) = 33 n(P) = 8 n(S) +n(P) = 41...
Read More →Prove
Question: $\tan ^{3} 2 x \sec 2 x$ Solution: $\tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 x \sec 2 x$ $=\left(\sec ^{2} 2 x-1\right) \tan 2 x \sec 2 x$ $=\sec ^{2} 2 x \cdot \tan 2 x \sec 2 x-\tan 2 x \sec 2 x$ $\therefore \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x$ $=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+\mathrm{C}$ Let $\sec 2 x=t$ $\therefore 2 \sec 2 x \tan 2 x d x=d t$ $\therefore \int \tan ^{3} 2 x \sec 2 x d x=...
Read More →For any three sets A, B and C,
Question: For any three setsA,BandC, (ABC) (A B' C') C' is equal to ____________. Solution: (ABC) (AB' C') C'...
Read More →If A and B are two sets,
Question: IfAandBare two sets, then (AB') (BC) is equal to ____________. Solution: For setA, BandC (AB') (BC)...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{x+y}{x y}=2$ $\frac{x-y}{x y}=6$ Solution: The given equations are: $\frac{x+y}{x y}=2$ $x+y=2 x y$ ..( $(i)$ $\frac{x-y}{x y}=6$ $x-y=6 x y$$\ldots($ ii $)$ Adding both equations, we get Put the value of $y$ in equation $(i)$, we get $x+\frac{1}{4}=2 x \times \frac{1}{4}$ $\Rightarrow \frac{-x}{2}=\frac{1}{4}$ $\Rightarrow x=-\frac{1}{2}$ Hence the value of $x=-\frac{1}{2}$ and $y=\frac{1}{4}$...
Read More →For any three sets A,
Question: For any three setsA,BandC, (AB) (CB) is equal to ____________. Solution: ForA,BandC (AB) (C B) = (ABC) (CBC) =ABCBCC (∵ intersection is associative) =A (BCBC) C =ABCC (∵BCBC=BC) =ACBC Hence, (A B) (C B) = (AC) B...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}+13 x^{2}+32 x+20$ Solution: Given, $f(x)=x^{3}+13 x^{2}+32 x+20$ The constant in f(x) is 20 The factors of 20 are 1, 2, 4, 5, 10, 20 Let, x + 1 = 0 = x = -1 $f(-1)=(-1)^{3}+13(-1)^{2}+32(-1)+20$ = -1 + 13 32 + 20 = 0 So, (x + 1) is the factor of f(x) Divide f(x) with (x + 1) to get other factors By, long division $x^{2}+12 x+20$ $x+1, x^{3}+13 x^{2}+32 x+20$ $x^{3}+x^{2}$ (-) (-) $12 x^{2}+32 x$ $12 x^{2}+12 x$ (-) (-) 20x 20 2...
Read More →Prove
Question: $\frac{\cos x-\sin x}{1+\sin 2 x}$ Solution: $\frac{\cos x-\sin x}{1+\sin 2 x}=\frac{\cos x-\sin x}{\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cos x}$ $\left[\sin ^{2} x+\cos ^{2} x=1 ; \sin 2 x=2 \sin x \cos x\right]$ $=\frac{\cos x-\sin x}{(\sin x+\cos x)^{2}}$ Let $\sin x+\cos x=t$ $\therefore(\cos x-\sin x) d x=d t$ $\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$ $=\int \frac{d t}{t^{2}}$ $=\int t^{-2} d t$ $=-t^{-1}+\m...
Read More →For any two sets A and B,
Question: For any two setsAandB, [B' (B' A)]' is equal to ____________. Solution: For setAandB,[B'⋃ (B' A')]' = (B')'⋂ (B'A')' [By De-Morgan's Law] = (B')'⋂ (B'⋂ (A')')' = (B')'⋂ (B'⋂A)' = (B')'⋂ (B⋃A') [By De-Morgan's Law] =B⋂ (B⋃A') =B⋂ (B⋃A') = (B⋂B) ⋃ (B⋂A') =B⋃ (B⋂A') [Since $B \cap A^{\prime} \subseteq B$ ] =B i.e [B'⋃ (B'A')]' =B...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{1}{2 x}+\frac{1}{3 y}=2$ $\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}$ Solution: The given equations are: $\frac{1}{2 x}+\frac{1}{3 y}=2...(i) $\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}$..$(i i)$ Multiply equation $(i)$ by $\frac{1}{2}$ and $(i i)$ by $\frac{1}{3}$ and subtract equation (ii) from (i) we get $\frac{1}{4 x}+\frac{1}{6 y}=1$ Put the value of $x$ in equation $(i)$, we get $\frac{1}{2 \times \frac{1}{2}}+\frac{1}{3 y}=2$ $\frac{1...
Read More →If A and B are two sets,
Question: IfAandBare two sets, then ((A' B') A)' is equal to ____________. Solution: For setsAandB, ((A' ⋃B') A)' = ((A' ⋃B')A')' = ((AB)' A') ' (using De-Morgan's Law) = (((AB)' )' ⋃ (A')') (using De-Morgan's Law) = ((AB) ⋃ A) =A i.e ((A' ⋃B') A)' =A...
Read More →