Prove
Question: $\frac{x}{e^{x^{2}}}$ Solution: Let $x^{2}=t$ $\therefore 2 x d x=d t$ $\Rightarrow \int \frac{x}{e^{x^{2}}} d x=\frac{1}{2} \int \frac{1}{e^{t}} d t$ $=\frac{1}{2} \int e^{-t} d t$ $=\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+\mathrm{C}$ $=-\frac{1}{2} e^{-x^{2}}+\mathrm{C}$ $=\frac{-1}{2 e^{x^{2}}}+\mathrm{C}$...
Read More →Let A and B be two sets that
Question: Let $A$ and $B$ be two sets that $n(A)=16, n(B)=14, n(A \cup B)=25$. Then, $n(A \cap B)$ is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution: We know: $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ Now, $n(A \cap B)=n(A)+n(B)-n(A \cup B)$ $=16+14-25$ $=5$...
Read More →Let A and B be two sets that
Question: Let $A$ and $B$ be two sets that $n(A)=16, n(B)=14, n(A \cup B)=25$. Then, $n(A \cap B)$ is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution: We know: $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ Now, $n(A \cap B)=n(A)+n(B)-n(A \cup B)$ $=16+14-25$ $=5$...
Read More →Let A and B be two sets that
Question: Let $A$ and $B$ be two sets that $n(A)=16, n(B)=14, n(A \cup B)=25$. Then, $n(A \cap B)$ is equal to (a) 30 (b) 50 (c) 5 (d) none of these Solution: We know: $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ Now, $n(A \cap B)=n(A)+n(B)-n(A \cup B)$ $=16+14-25$ $=5$...
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Question: $e^{2 x+3}$ Solution: Let $2 x+3=t$ $\therefore 2 d x=d t$ $\Rightarrow \int e^{2 x+3} d x=\frac{1}{2} \int e^{t} d t$ $=\frac{1}{2}\left(e^{t}\right)+\mathrm{C}$ $=\frac{1}{2} e^{(2 x+3)}+\mathrm{C}$...
Read More →In each of the following two polynomials, find the value of a, if (x + a) is a factor:
Question: In each of the following two polynomials, find the value of a, if (x + a) is a factor: 1. $x^{3}+a x^{2}-2 x+a+4$ 2. $x^{4}-a^{2} x^{2}+3 x-a$ Solution: 1. $x^{3}+a x^{2}-2 x+a+4$ let, $f(x)=x^{3}+a x^{2}-2 x+a+4$ here, x + a = 0 ⟹ x = - a Substitute the value of x in f(x) $f(-a)=(-a)^{3}+a(-a)^{2}-2(-a)+a+4$ $=(-a)^{3}+a^{3}-2(-a)+a+4$ = 3a + 4 Equate to zero ⟹ 3a + 4 = 0 ⟹ 3a = -4 ⟹ a = 4/3 So, when (x + a) is a factor of f(x) then a = 4/3 2. $x^{4}-a^{2} x^{2}+3 x-a$ let, $f(x)=x^{4...
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Question: $\frac{x}{9-4 x^{2}}$ Solution: Let $9-4 x^{2}=t$ $\therefore-8 x d x=d t$ $\Rightarrow \int \frac{x}{9-4 x^{2}} d x=\frac{-1}{8} \int \frac{1}{t} d t$ $=\frac{-1}{8} \log |t|+C$ $=\frac{-1}{8} \log \left|9-4 x^{2}\right|+C$...
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Question: $\frac{1}{x(\log x)^{m}}, x0$ Solution: Let $\log x=t$ $\therefore \frac{1}{x} d x=d t$ $\Rightarrow \int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{(t)^{m}}$\ $=\left(\frac{t^{-m+1}}{1-m}\right)+\mathrm{C}$ $=\frac{(\log x)^{1-m}}{(1-m)}+\mathrm{C}$...
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Question: $\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}$ Solution: Let $2+3 x^{3}=t$ $\therefore 9 x^{2} d x=d t$$\Rightarrow \int \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}} d x=\frac{1}{9} \int \frac{d t}{(t)^{3}}$ $=\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+\mathrm{C}$ $=\frac{-1}{18}\left(\frac{1}{t^{2}}\right)+\mathrm{C}$ $=\frac{-1}{18\left(2+3 x^{3}\right)^{2}}+\mathrm{C}$...
Read More →Let U be the universal set containing 700 elements.
Question: Let $U$ be the universal set containing 700 elements. If $A, B$ are sub-sets of $U$ such that $n(A)=200, n(B)=300$ and $(A \cap B)=100$. Then $n\left(A^{\prime} \cap B^{\prime}\right)=$ (a) 400 (b) 600 (c) 300 (d) none of these. Solution: (c) 300 $n\left(A^{\prime} \cap B^{\prime}\right)=n(A \cup B)^{\prime}$ $=n(U)-n(A \cup B)$ $=700-\{200+300-100\}$ $=300$...
Read More →In each of the following two polynomials, find the value of a, if (x - a) is a factor:
Question: In each of the following two polynomials, find the value of a, if (x - a) is a factor: 1. $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$ 2. $x^{5}-a^{2} x^{3}+2 x+a+1$ Solution: (1) $x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$ let, $f(x)=x^{6}-a x^{5}+x^{4}-a x^{3}+3 x-a+2$ here, x - a = 0 ⟹ x = a Substitute the value of x in f(x) $f(a)=a^{6}-a(a)^{5}+(a)^{4}-a(a)^{3}+3(a)-a+2$ $=a^{6}-a^{6}+(a)^{4}-a^{4}+3(a)-a+2$ = 2a + 2 Equate to zero ⟹ 2a + 2 = 0 ⟹ 2(a + 1) = 0 ⟹ a = -1 So, when (x - a) is a fact...
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Question: $\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}$ Solution: Let $x^{3}-1=t$ $\therefore 3 x^{2} d x=d t$ $\Rightarrow \int\left(x^{3}-1\right)^{\frac{1}{3}} x^{5} d x=\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} d x$ $=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}$ $=\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t$ $=\frac{1}{3}\left[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]+\mathrm{C}$ $=\frac{1}{3}\left[\frac{3}{7} t^{\frac{7}{3}}+...
Read More →Let U be the universal set containing 700 elements.
Question: Let $U$ be the universal set containing 700 elements. If $A, B$ are sub-sets of $U$ such that $n(A)=200, n(B)=300$ and $(A \cap B)=100$. Then $n\left(A^{\prime} \cap B^{\prime}\right)=$ (a) 400 (b) 600 (c) 300 (d) none of these. Solution: (c) 300 $n\left(A^{\prime} \cap B^{\prime}\right)=n(A \cup B)^{\prime}$ $=n(U)-n(A \cup B)$ $=700-\{200+300-100\}$ $=300$...
Read More →Draw the graphs of the equations 5x − y = 5 and 3x − y = 3.
Question: Draw the graphs of the equations 5xy= 5 and 3xy= 3. Determine the co-ordinates of the vertices of the triangle formed by these lines andy-axis. Calculate the area of the triangle so formed. Solution: The given equations are: $5 x-y=5$$. .(i)$ $3 x-y=3$....(ii) Putting $x=0$ in equation (i) we get: $\Rightarrow 5 \times 0-y=5$ $\Rightarrow y=-5$ $x=0, \quad y=-5$ Putting $y=0$ in equation (i) we get: $\Rightarrow 5 x-0=5$ $\Rightarrow x=1$ $x=1, \quad y=0$ Use the following table to dra...
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Question: $\frac{x}{\sqrt{x+4}}, x0$ Solution: Let $\mathrm{I}=\int \frac{x}{(x+4)} d x$put $x+4=t$ $\Rightarrow d x=d t$ Now, $\mathrm{I}=\int \frac{(t-4)}{\sqrt{t}} d t$ $=\int\left(\sqrt{t}-4 t^{-1 / 2}\right) d t$ $=\frac{2}{3} t^{3 / 2}-4\left(2 t^{1 / 2}\right)+\mathrm{C}$ $=\frac{2}{3} \cdot t . t^{1 / 2}-8 t^{1 / 2}+\mathrm{C}$ $=\frac{2}{3}(x+4) \sqrt{x+4}-8 \sqrt{x+4}+\mathrm{C}$ $=\frac{2}{3} x \sqrt{x+4}+\frac{8}{3} \sqrt{x+4}-8 \sqrt{x+4}+\mathrm{C}$ $=\frac{2}{3} x \sqrt{x+4}-\frac...
Read More →Let A
Question: Let $A=\{x: x \in R, x \geq 4\}$ and $B=\{x \in R: x5\} .$ Then, $n\left(A^{\prime} \cap B^{\prime}\right)=$ (a) $(4,5]$ (b) $(4,5)$ (c) $[4,5)$ (d) $[4,5]$ Solution: (c) $[4,5)$ $A=\{x: x \in R, x \geq 4\}$ and $B=\{x \in R: x5\}$ $A \cap B=[4,5)$...
Read More →Let A
Question: Let $A=\{x: x \in R, x \geq 4\}$ and $B=\{x \in R: x5\} .$ Then, $n\left(A^{\prime} \cap B^{\prime}\right)=$ (a) $(4,5]$ (b) $(4,5)$ (c) $[4,5)$ (d) $[4,5]$ Solution: (c) $[4,5)$ $A=\{x: x \in R, x \geq 4\}$ and $B=\{x \in R: x5\}$ $A \cap B=[4,5)$...
Read More →If x - 2 is a factor of each of the following two polynomials, find the value of a in each case:
Question: If x - 2 is a factor of each of the following two polynomials, find the value of a in each case: 1. $x^{3}-2 a x^{2}+a x-1$ 2. $x^{5}-3 x^{4}-a x^{3}+3 a x^{2}+2 a x+4$ Solution: (1) Let $f(x)=x^{3}-2 a x^{2}+a x-1$ from factor theorem if (x - 2) is the factor of f(x) the f(2) = 0 let, x - 2 = 0 ⟹ x = 2 Substitute x value in f(x) $f(2)=2^{3}-2 a(2)^{2}+a(2)-1$ = 8 - 8a + 2a - 1 = - 6a + 7 Equate f(2) to zero ⟹ - 6a + 7 = 0 ⟹ - 6a = -7 ⟹ a=76 When, (x - 2) is the factor of f(x) then a =...
Read More →For any three sets A, B and C
Question: For any three sets A, B and C (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ (b) $A \cap(B-C)=(A \cap B)-C$ (c) $A \cup(B-C)=(A \cup B) \cap\left(A \cup C^{\prime}\right)$ (d) $A \cup(B-C)=(A \cup B)-(A \cup C)$. Solution: (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ Let $x$ be any arbitrary element of $A \cap(B-C)$. Thus, we have, $x \in[A \cap(B-C)] \Rightarrow x \in A$ and $x \in(B-C)$ $\Rightarrow x \in A$ and $(x \in B$ and $x \notin C)$ $\Rightarrow(x \in A$ and $x \in B)$ and $(x \in A$ and...
Read More →For any three sets A, B and C
Question: For any three sets A, B and C (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ (b) $A \cap(B-C)=(A \cap B)-C$ (c) $A \cup(B-C)=(A \cup B) \cap\left(A \cup C^{\prime}\right)$ (d) $A \cup(B-C)=(A \cup B)-(A \cup C)$. Solution: (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ Let $x$ be any arbitrary element of $A \cap(B-C)$. Thus, we have, $x \in[A \cap(B-C)] \Rightarrow x \in A$ and $x \in(B-C)$ $\Rightarrow x \in A$ and $(x \in B$ and $x \notin C)$ $\Rightarrow(x \in A$ and $x \in B)$ and $(x \in A$ and...
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Question: $\frac{1}{x-\sqrt{x}}$ Solution: $\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}$ Let $(\sqrt{x}-1)=t$ $\therefore \frac{1}{2 \sqrt{x}} d x=d t$ $\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x=\int \frac{2}{t} d t$ $=2 \log |t|+\mathrm{C}$ $=2 \log |\sqrt{x}-1|+\mathrm{C}$...
Read More →For any three sets A, B and C
Question: For any three sets A, B and C (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ (b) $A \cap(B-C)=(A \cap B)-C$ (c) $A \cup(B-C)=(A \cup B) \cap\left(A \cup C^{\prime}\right)$ (d) $A \cup(B-C)=(A \cup B)-(A \cup C)$. Solution: (a) $A \cap(B-C)=(A \cap B)-(A \cap C)$ Let $x$ be any arbitrary element of $A \cap(B-C)$. Thus, we have, $x \in[A \cap(B-C)] \Rightarrow x \in A$ and $x \in(B-C)$ $\Rightarrow x \in A$ and $(x \in B$ and $x \notin C)$ $\Rightarrow(x \in A$ and $x \in B)$ and $(x \in A$ and...
Read More →Solve the following system of linear equations graphically.
Question: Solve the following system of linear equations graphically. $4 x-5 y-20=0$ $3 x+5 y-15=0$ Determine the vertices of the triangle formed by the lines representing the above equation and they-axis. Solution: The given equations are: $4 x-5 y-20=0$$.(i)$ $3 x+5 y-15=0$(ii) Putting $x=0$ in equation $(i)$, we get: $\Rightarrow 4 \times 0-5 y=20$ $\Rightarrow y=-4$ $x=0, \quad y-4$ Putting $y=0$ in equation (i) we get: $\Rightarrow 4 x-5 \times 0=20$ $\Rightarrow x=5$ $x=5, \quad y=0$ Use t...
Read More →What must be added to 3x3
Question: What must be added to $3 x^{3}+x^{2}-22 x+9$ so that the result is exactly divisible by $3 x^{2}+7 x-6$ Solution: Let, $p(x)=3 x^{3}+x^{2}-22 x+9$ and $q(x)=3 x^{2}+7 x-6$ By division theorem, when p(x) is divided by q(x), the remainder is a linear equation in x. Let, r(x) = ax + b is added to p(x), so that p(x) + r(x) is divisible by q(x) f(x) = p(x) + r(x) $\Rightarrow f(x)=3 x^{3}+x^{2}-22 x+9(a x+b)$ $\Rightarrow=3 x^{3}+x^{2}+x(a-22)+b+9$ We know that, $q(x)=3 x^{2}+7 x-6$ $=3 x^{...
Read More →Prove
Question: $(4 x+2) \sqrt{x^{2}+x+1}$ Solution: Let $x^{2}+x+1=t$ $\therefore(2 x+1) d x=d t$ $\int(4 x+2) \sqrt{x^{2}+x+1} d x$ $=\int 2 \sqrt{t} d t$ $=2 \int \sqrt{t} d t$ $=2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$ $=\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\mathrm{C}$...
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