Prove
Question: $\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$ Solution: $\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$ $=\int\left(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}\right) d x$ $=3 \int x^{\frac{5}{2}} d x+2 \int x^{\frac{3}{2}} d x+3 \int x^{\frac{1}{2}} d x$ $=3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3 \frac{\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\mathrm{C}$ $=\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x...
Read More →In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali.
Question: In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Hindi only? How many can speak Bengali? How many can speak both Hindi and Bengali? Solution: LetABdenote the sets of the persons who can speak Hindi Bengali, respectively. Given: $n(A \cup B)=1000$ $n(A)=750$ $n(B)=400$ $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\Rightarrow 1000=750+400-n(A \cap B)$ $\Rightarrow n(A \cap B)=150$ Number of persons who can speak both Hindi and Bengali...
Read More →In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali.
Question: In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Hindi only? How many can speak Bengali? How many can speak both Hindi and Bengali? Solution: LetABdenote the sets of the persons who can speak Hindi Bengali, respectively. Given: $n(A \cup B)=1000$ $n(A)=750$ $n(B)=400$ $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\Rightarrow 1000=750+400-n(A \cap B)$ $\Rightarrow n(A \cap B)=150$ Number of persons who can speak both Hindi and Bengali...
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Question: $\int(1-x) \sqrt{x} d x$ Solution: $\int(1-x) \sqrt{x} d x$ $=\int\left(\sqrt{x}-x^{\frac{3}{2}}\right) d x$ $=\int x^{\frac{1}{2}} d x-\int x^{\frac{3}{2}} d x$ $=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+\mathrm{C}$ $=\frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+\mathrm{C}$...
Read More →Find the value of k if x - 3 is a factor of
Question: Find the value of $k$ if $x-3$ is a factor of $k^{2} x^{3}-k x^{2}+3 k x-k$ Solution: Let $f(x)=k^{2} x^{3}-k x^{2}+3 k x-k$ From factor theorem if x - 3 is the factor of f(x) then f(3) = 0 ⟹ x - 3 = 0 ⟹ x = 3 Substitute the value of x in f(x) $f(3)=k^{2}(3)^{3}-k(3)^{2}+3 k(3)-k$ $=27 k^{2}-9 k+9 k-k$ $=27 k^{2}-k$ = k(27k - 1) Equate f(3) to zero, to find k ⟹ f(3) = 0 ⟹ k(27k - 1) = 0 ⟹ k = 0 and 27k - 1 = 0 ⟹ k = 0 and 27k = 1 ⟹ k = 0 and k =1/27 When k = 0 and 1/27, (x - 3) will be...
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Question: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$ Solution: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$ On dividing, we obtain $=\int\left(x^{2}+1\right) d x$ $=\int x^{2} d x+\int 1 d x$ $=\frac{x^{3}}{x}+x+\mathrm{C}$...
Read More →Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0.
Question: Draw the graphs ofxy+ 1 = 0 and 3x+ 2y 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines andx-axis. Solution: The given equations are $x-y+1=0$...$.(i )$ $3 x+2 y-12=0$$...(i i)$ Putting $x=0$ in equation $(i)$, we get: $\Rightarrow 0-y=-1$ $\Rightarrow y=1$ $x=0, \quad y=1$ Putting $y=0$ in equation (i) we get: $\Rightarrow x-0=-1$ $\Rightarrow x=-1$ $x=-1, \quad y...
Read More →Find the value of a, if (x + 2) is a factor of
Question: Find the value of $a$, if $(x+2)$ is a factor of $4 x^{4}+2 x^{3}-3 x^{2}+8 x+5 a$ Solution: Here, $f(x)=4 x^{4}+2 x^{3}-3 x^{2}+8 x+5 a$ By factor theorem If (x + 2) is the factor of f(x) then, f(-2) = 0 ⟹ x + 2 = 0 ⟹ x = -2 Substitute the value of x in f(x) $f(-2)=4(-2)^{4}+2(-2)^{3}-3(-2)^{2}+8(-2)+5 a$ = 4(16) + 2(-8) - 3(4) - 16 + 5a = 64 - 16 - 12 - 16 + 5a = 5a + 20 equate f(-2) to zero f(-2) = 0 ⟹ 5a + 20 = 0 ⟹ 5a = - 20 ⟹ a =-20/5 ⟹ a = - 4 When a = - 4, (x + 2) will be factor...
Read More →Of the members of three athletic teams in a certain school,
Question: Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basket ball 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all? Solution: LetA,B Cbe the sets of members in basketball team, hockey team football team, respectively. Given: $n(A)=21$ $n(B)=26$ $n(C)=29$ $n(A \cap B)=14$ $n(B \cap C)=15$ $n(A \cap C)=12$ $n(...
Read More →Of the members of three athletic teams in a certain school,
Question: Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basket ball 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all? Solution: LetA,B Cbe the sets of members in basketball team, hockey team football team, respectively. Given: $n(A)=21$ $n(B)=26$ $n(C)=29$ $n(A \cap B)=14$ $n(B \cap C)=15$ $n(A \cap C)=12$ $n(...
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Question: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$ Solution: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$ $=\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x$ $=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{3}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+\mathrm{C}$ $=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+\mathrm{C}$ $=\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+\mathrm{C}$...
Read More →Find the value of a such that (x - 4) is a factor of
Question: Find the value of a such that $(x-4)$ is a factor of $5 x^{3}-7 x^{2}-a x-28$ Solution: Here, $f(x)=5 x^{3}-7 x^{2}-a x-28$ By factor theorem If (x - 4) is the factor of f(x) then, f(4) = 0 ⟹ x 4 = 0 ⟹ x = 4 Substitute the value of x in f(x) $f(4)=5(4)^{3}-7(4)^{2}-a(4)-28$ = 5(64) 7(16) 4a 28 = 320 112 4a 28 = 180 4 Equate f(4) to zero, to find a f(4) = 0 ⟹ 180 4a = 0 ⟹ -4a = -180 ⟹ 4a = 180 ⟹ a =180/4 ⟹ a = 45 When a = 45, (x - 4) will be factor of f(x)...
Read More →In a survey of 60 people, it was found that 25 people read newspaper H,
Question: In a survey of 60 people, it was found that 25 people read newspaperH, 26 read newspaperT, 26 read newspaperI, 9 read bothHandI, 11 read bothHandT, 8 read bothTandI, 3 read all three newspapers. Find: (i) the numbers of people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper. Solution: Given : $n(H)=25$ $n(T)=26$ $n(I)=26$ $n(H \cap I)=9$ $n(H \cap T)=11$ $n(T \cap I)=8$ $n(H \cap T \cap I)=3$ (i) We know: $n(H \cup T \cup I)=n(H)+n(T)+n...
Read More →In a survey of 60 people, it was found that 25 people read newspaper H,
Question: In a survey of 60 people, it was found that 25 people read newspaperH, 26 read newspaperT, 26 read newspaperI, 9 read bothHandI, 11 read bothHandT, 8 read bothTandI, 3 read all three newspapers. Find: (i) the numbers of people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper. Solution: Given : $n(H)=25$ $n(T)=26$ $n(I)=26$ $n(H \cap I)=9$ $n(H \cap T)=11$ $n(T \cap I)=8$ $n(H \cap T \cap I)=3$ (i) We know: $n(H \cup T \cup I)=n(H)+n(T)+n...
Read More →For what value of a is (x – 5) a factor of
Question: For what value of $a$ is $(x-5)$ a factor of $x^{3}-3 x^{2}+a x-10$ Solution: Here, $f(x)=x^{3}-3 x^{2}+a x-10$ By factor theorem If (x - 5) is the factor of f(x) then, f(5) = 0 ⟹ x - 5 = 0 ⟹ x = 5 Substitute the value of x in f(x) $f(5)=5^{3}-3(5)^{2}+a(5)-10$ = 125 - (3 * 25) + 5a - 10 = 125 - 75 + 5a - 10 = 5a + 40 Equate f(5) to zero f(5) = 0 ⟹ 5a + 40 = 0 ⟹ 5a = - 40 ⟹ a = 40/5 = - 8 When a = - 8, (x - 5) will be factor of f(x)...
Read More →Prove
Question: $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$ Solution: $\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$ $=\int\left(x+5-4 x^{-2}\right) d x$ $=\int x d x+5 \int 1 \cdot d x-4 \int x^{-2} d x$ $=\frac{x^{2}}{2}+5 x-4\left(\frac{x^{-1}}{-1}\right)+\mathrm{C}$ $=\frac{x^{2}}{2}+5 x+\frac{4}{x}+\mathrm{C}$...
Read More →Draw the graphs of the following equations on the same graph paper.
Question: Draw the graphs of the following equations on the same graph paper. $2 x+3 y=12$ $x-y=1$ Find the coordinates of the vertices of the triangle formed by the two straight lines and they-axis. Solution: The given equations are $2 x+3 y=12$$\ldots(i)$ $x-y=1$...(ii) Putting $x=0$ in equation $(i)$, we get: $\Rightarrow 2 \times 0+3 y=12$ $\Rightarrow y=4$ $x=0, \quad y=4$ Putting $y=0$ in equation $(i,$, we get: $\Rightarrow x+3 \times 0=6$ $\Rightarrow x=6$ $x=6, \quad y=0$ Use the follow...
Read More →Show that (x + 4), (x - 3) and (x - 7) are the factors of
Question: Show that $(x+4),(x-3)$ and $(x-7)$ are the factors of $x^{3}-6 x^{2}-19 x+84$ Solution: Here, $f(x)=x^{3}-6 x^{2}-19 x+84$ The factors given are (x + 4), (x - 3) and (x - 7) To prove that g(x) is the factor of f(x), The results of f(- 4) , f(3) and f(7) should be zero Let, x + 4 = 0 ⟹ x = - 4 Substitute the value of x in f(x) $f(-4)=(-4)^{3}-6(-4)^{2}-19(-4)+84$ $=-64-\left(6^{*} 16\right)-\left(19^{*}(-4)\right)+84$ = - 64 - 96 + 76 + 84 = 160 - 160 = 0 Let, x - 3 = 0 ⟹ x = 3 Substit...
Read More →In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea.
Question: In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: (i) how may drink tea and coffee both; (ii) how many drink coffee but not tea. Solution: LetABdenote the sets of the persons who drink tea coffee, respectively . Given: $n(A \cup B)=50$ $n(A)=30$ $n(A-B)=14$ (i) $n(A-B)=n(A)-n(A \cap B)$ $\Rightarrow 14=30-n(A \cap B)$ $\Rightarrow n(A \cap B)=16$ Thus, 16 persons drink tea and coffee both. (ii) $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\Rightarrow 50=30+n(B)-16$ ...
Read More →In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea.
Question: In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: (i) how may drink tea and coffee both; (ii) how many drink coffee but not tea. Solution: LetABdenote the sets of the persons who drink tea coffee, respectively . Given: $n(A \cup B)=50$ $n(A)=30$ $n(A-B)=14$ (i) $n(A-B)=n(A)-n(A \cap B)$ $\Rightarrow 14=30-n(A \cap B)$ $\Rightarrow n(A \cap B)=16$ Thus, 16 persons drink tea and coffee both. (ii) $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\Rightarrow 50=30+n(B)-16$ ...
Read More →Prove
Question: $\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$ Solution: $\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$ $=\int\left(x+\frac{1}{x}-2\right) d x$ $=\int x d x+\int \frac{1}{x} d x-2 \int 1 \cdot d x$ $=\frac{x^{2}}{2}+\log |x|-2 x+\mathrm{C}$...
Read More →In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea.
Question: In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find: (i) how may drink tea and coffee both; (ii) how many drink coffee but not tea. Solution: LetABdenote the sets of the persons who drink tea coffee, respectively . Given: $n(A \cup B)=50$ $n(A)=30$ $n(A-B)=14$ (i) $n(A-B)=n(A)-n(A \cap B)$ $\Rightarrow 14=30-n(A \cap B)$ $\Rightarrow n(A \cap B)=16$ Thus, 16 persons drink tea and coffee both. (ii) $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\Rightarrow 50=30+n(B)-16$ ...
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Question: $\int\left(2 x^{2}+e^{x}\right) d x$ Solution: $\int\left(2 x^{2}+e^{x}\right) d x$ $=2 \int x^{2} d x+\int e^{x} d x$ $=2\left(\frac{x^{3}}{3}\right)+e^{x}+\mathrm{C}$ $=\frac{2}{3} x^{3}+e^{x}+\mathrm{C}$...
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Question: $\int\left(a x^{2}+b x+c\right) d x$ Solution: $\int\left(a x^{2}+b x+c\right) d x$ $=a \int x^{2} d x+b \int x d x+c \int 1 . d x$ $=a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)+c x+\mathrm{C}$ $=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+\mathrm{C}$...
Read More →Show that (x – 2), (x + 3) and (x – 4) are the factors of
Question: Show that $(x-2),(x+3)$ and $(x-4)$ are the factors of $x^{3}-3 x^{2}-10 x+24$ Solution: Here, $f(x)=x^{3}-3 x^{2}-10 x+24$ The factors given are (x - 2), (x + 3) and (x - 4) To prove that g(x) is the factor of f(x), The results of f(2), f(-3) and f(4) should be zero Let, x - 2 = 0 ⟹ x = 2 Substitute the value of x in f(x) $f(2)=2^{3}-3(2)^{2}-10(2)+24$ = 8 - (3 * 4) - 20 + 24 = 8 - 12 - 20 + 24 = 32 - 32 = 0 Let, x + 3 = 0 ⟹ x = -3 Substitute the value of x in f(x) $f(-3)=(-3)^{3}-3(-...
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