Which of the following statements is false:

Question: Which of the following statements is false: (a) $A-B=A \cap B^{\prime}$ (b) $A-B=A-(A \cap B)$ (c) $A-B=A-B^{\prime}$ (d) $A-B=(A \cup B)-B$. Solution: (c) $A-B=A-B$, It includes all those elements ofAwhich do not belongs to complement ofBwhich is equal toABbut not equal to $A-B$ Therefore, (c) is false ....

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Which of the following statements is false:

Question: Which of the following statements is false: (a) $A-B=A \cap B^{\prime}$ (b) $A-B=A-(A \cap B)$ (c) $A-B=A-B^{\prime}$ (d) $A-B=(A \cup B)-B$. Solution: (c) $A-B=A-B$, It includes all those elements ofAwhich do not belongs to complement ofBwhich is equal toABbut not equal to $A-B$ Therefore, (c) is false ....

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Prove

Question: $x \sqrt{1+2 x^{2}}$ Solution: Let $1+2 x^{2}=t$ $\therefore 4 x d x=d t$ $\Rightarrow \int x \sqrt{1+2 x^{2}} d x=\int \frac{\sqrt{t} d t}{4}$ $=\frac{1}{4} \int t^{\frac{1}{2}} d t$ $=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$ $=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$...

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For any two sets A and B,

Question: For any two sets $A$ and $B,(A-B) \cup(B-A)=$ (a) $(A-B) \cup A$ (b) $(B-A) \cup B$ (c) $(A \cup B)-(A \cap B)$ (d) $(A \cup B) \cap(A \cap B)$. Solution: (c) $(A \cup B)-(A \cap B)$ $(A-B) \cup(B-A)=\left(A \cap B^{\prime}\right) \cup\left(B \cap A^{\prime}\right)$ $=\left[A \cup\left(B \cap A^{\prime}\right)\right] \cap\left[B^{\prime} \cup\left(B \cap A^{\prime}\right)\right]$ [Using distribution law] $=\left[(A \cup B) \cap\left(A \cup A^{\prime}\right)\right] \cap\left[\left(B^{\p...

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Prove

Question: $x \sqrt{x+2}$ Solution: Let $(x+2)=t$ dx = dt $\Rightarrow \int x \sqrt{x+2} d x=\int(t-2) \sqrt{t} d t$ $=\int\left(t^{\frac{3}{2}}-2 t^{\frac{1}{2}}\right) d t$ $=\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t$ $=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$ $=\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+\mathrm{C}$ $=\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+\mathrm{C}$...

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What must be added to x3

Question: What must be added to $x^{3}-6 x^{2}-15 x+80$ so that the result is exactly divisible by $x^{2}+x-12$ Solution: Let, $p(x)=x^{3}-6 x^{2}-15 x+80$ $q(x)=x^{2}+x-12$ by division algorithm, when p(x) is divided by q(x) the remainder is a linear expression in x. so, let r(x) = ax + b is subtracted from p(x), so that p(x) - q(x) is divisible by q(x) let f(x) = p(x) - q(x) $q(x)=x^{2}+x-12$ $=x^{2}+4 x-3 x-12$ $=x(x+4)(-3)(x+4)$ $=(x+4),(x-3)$ clearly, (x - 3) and (x + 4) are factors of q(x)...

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The symmetric difference of A

Question: The symmetric difference ofA= {1, 2, 3} andB= {3, 4, 5} is (a) {1, 2} (b) {1, 2, 4, 5} (c) {4, 3} (d) {2, 5, 1, 4, 3} Solution: (b) {1, 2, 4, 5} Here, A= {1, 2, 3} andB= {3, 4, 5} The symmetric difference ofA andBis given by :- $(A-B) \cup(B-A)$ Now, we have: $(A-B)=\{1,2\}$ $(B-A)=\{4,5\}$ $(A-B) \cup(B-A)=\{1,2,4,5\}$...

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Solve the following system of equations graphically.

Question: Solve the following system of equations graphically. $2 x-3 y+6=0$ $2 x+3 y-18=0$ Also, find the area of the region bounded by these two lines andy-axis. Solution: The given equations are: $2 x-3 y+6=0$$\ldots \ldots . .(i)$ $2 x+3 y-18=0$$\ldots \ldots . .(i i)$ Putting $x=0$ in equation (i) we get: $\Rightarrow 2 \times 0-3 y=-6$ $\Rightarrow y=2$ $x=0, y=2$ Putting $y=0$ in equation (i) we get: $\Rightarrow 2 x-3 \times 0=-6$ $\Rightarrow x=-3$ $x=-3, \quad y=0$ Use the following ta...

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The symmetric difference of A

Question: The symmetric difference ofA= {1, 2, 3} andB= {3, 4, 5} is (a) {1, 2} (b) {1, 2, 4, 5} (c) {4, 3} (d) {2, 5, 1, 4, 3} Solution: (b) {1, 2, 4, 5} Here, A= {1, 2, 3} andB= {3, 4, 5} The symmetric difference ofA andBis given by :- $(A-B) \cup(B-A)$ Now, we have: $(A-B)=\{1,2\}$ $(B-A)=\{4,5\}$ $(A-B) \cup(B-A)=\{1,2,4,5\}$...

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Prove

Question: $\sqrt{a x+b}$ Solution: Let $a x+b=t$ $\Rightarrow a d x=d t$ $\therefore d x=\frac{1}{a} d t$ $\Rightarrow \int(a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t$ $=\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$ $=\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+\mathrm{C}$...

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The symmetric difference of A and B is

Question: The symmetric difference ofAandBis (a) $(A-B) \cap(B-A)$ (b) $(A-B) \cup(B-A)$ (c) $(A \cup B)-(A \cap B)$ (d) $\{(A \cup B)-A\} \cup\{(A \cup B)-B\}$ Solution: (b) $(A-B) \cup(B-A)$ The symmetric difference ofA andBis given by :- $(A-B) \cup(B-A)$...

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The symmetric difference of A and B is

Question: The symmetric difference ofAandBis (a) $(A-B) \cap(B-A)$ (b) $(A-B) \cup(B-A)$ (c) $(A \cup B)-(A \cap B)$ (d) $\{(A \cup B)-A\} \cup\{(A \cup B)-B\}$ Solution: (b) $(A-B) \cup(B-A)$ The symmetric difference ofA andBis given by :- $(A-B) \cup(B-A)$...

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What must be added to x3 − 3x2 − 12x + 19 so

Question: What must be added to $x^{3}-3 x^{2}-12 x+19$ so that the result is exactly divisible by $x^{2}+x-6$ Solution: Here, $p(x)=x^{3}-3 x^{2}-12 x+19$ $g(x)=x^{2}+x-6$ by division algorithm, when p(x) is divided by g(x) , the remainder will be a linear expression in x Let, r(x) = ax + b is added to p(x) ⟹ f(x) = p(x) + r(x) $=x^{3}-3 x^{2}-12 x+19+a x+b$ $f(x)=x^{3}-3 x^{2}+x(a-12)+19+b$ We know that, $g(x)=x^{2}+x-6$ First, find the factors for g(x) $g(x)=x^{2}+3 x-2 x-6$ = x(x + 3) -2(x +...

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Prove

Question: $\sin (a x+b) \cos (a x+b)$ Solution: $\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$ Let $2(a x+b)=t$ $\therefore 2 a d x=d t$ $\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t d t}{2 a}$ $=\frac{1}{4 a}[-\cos t]+\mathrm{C}$ $=\frac{-1}{4 a} \cos 2(a x+b)+\mathrm{C}$...

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If A = {1, 3, 5, B} and B = {2, 4},

Question: IfA= {1, 3, 5,B} andB= {2, 4}, then (a) $4 \in A$ (b) $\{4\} \subset A$ (c) $B \subset A$ (d) none of these. Solution: (d) none of these $4 \notin A$ $\{4\} \not \subset A$ $B \not \subset A$ Thus, we can say that none of these options satisfy the given relation....

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If A = {1, 3, 5, B} and B = {2, 4},

Question: IfA= {1, 3, 5,B} andB= {2, 4}, then (a) $4 \in A$ (b) $\{4\} \subset A$ (c) $B \subset A$ (d) none of these. Solution: (d) none of these $4 \notin A$ $\{4\} \not \subset A$ $B \not \subset A$ Thus, we can say that none of these options satisfy the given relation....

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For any two sets A and B,

Question: For any two sets $A$ and $B, A \cap(A \cup B)=$ (a)A (b)B (c) ϕ (d) none of these. Solution: (a)A $A \cap(A \cup B)=(A \cap A) \cup(A \cap B)=A \cup(A \cap B)=A$...

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Prove

Question: $\sin x \cdot \sin (\cos x)$ Solution: $\sin x \cdot \sin (\cos x)$ Let $\cos x=t$ $\therefore-\sin x d x=d t$ $\Rightarrow \int \sin x \cdot \sin (\cos x) d x=-\int \sin t d t$ $=-[-\cos t]+C$ $=\cos t+C$ $=\cos (\cos x)+C$...

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The number of subsets of a set containing n elements is

Question: The number of subsets of a set containingnelements is (a)n (b)2n 1 (c)n2 (d) 2n Solution: (d) 2nThe total number of subsets of a finite set consisting ofnelements is 2n....

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Draw the graphs of the following equations:

Question: Draw the graphs of the following equations: $2 x-3 y+6=0$ $2 x+3 y-18=0$ $y-2=0$ Find the vertices of the triangle so obtained. Also, find the area of the triangle. Solution: The given equations are $2 x-3 y+6=0$..(i) $2 x+3 y-18=0$..(ii) $y-2=0$..(iii) The two points satisfying (i) can be listed in a table as, The two points satisfying (ii) can be listed in a table as, It is seen that the coordinates of the vertices of the obtained triangle are $A(3,4), B(0,2), C(6,2)$ Area of $\trian...

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Prove

Question: $\frac{1}{x+x \log x}$ Solution: $\frac{1}{x+x \log x}=\frac{1}{x(1+\log x)}$ Let $1+\log x=t$ $\therefore \frac{1}{x} d x=d t$ $\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int_{t}^{1} d t$ $=\log |t|+\mathrm{C}$ $=\log |1+\log x|+\mathrm{C}$...

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Let A and B be two sets in the same universal set.

Question: Let $A$ and $B$ be two sets in the same universal set. Then, $A-B=$ (a) $A \cap B$ (b) $A^{\prime} \cap B$ (c) $A \cap B^{\prime}$ (d) none of these. Solution: (c) $A \cap B^{\prime}$ $A-B$ belongs to those elements of $A$ that do not belong to $B$. $\therefore A-B=A \cap B^{\prime}$...

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Let A and B be two sets in the same universal set.

Question: Let $A$ and $B$ be two sets in the same universal set. Then, $A-B=$ (a) $A \cap B$ (b) $A^{\prime} \cap B$ (c) $A \cap B^{\prime}$ (d) none of these. Solution: (c) $A \cap B^{\prime}$ $A-B$ belongs to those elements of $A$ that do not belong to $B$. $\therefore A-B=A \cap B^{\prime}$...

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Prove

Question: $\frac{(\log x)^{2}}{x}$ Solution: Let $\log |x|=t$ $\therefore \frac{1}{x} d x=d t$ $\Rightarrow \int \frac{(\log |x|)^{2}}{x} d x=\int t^{2} d t$ $=\frac{t^{3}}{3}+\mathrm{C}$ $=\frac{(\log |x|)^{3}}{3}+\mathrm{C}$...

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If both (x + 1) and (x - 1) are the factors of ax3

Question: If both $(x+1)$ and $(x-1)$ are the factors of $a x^{3}+x^{2}-2 x+b$, Find the values of $a$ and $b$ Solution: Here, $f(x)=a x^{3}+x^{2}-2 x+b$ (x + 1) and (x - 1) are the factors From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0 Let, x - 1= 0 ⟹ x = -1 Substitute x value in f(x) $f(1)=a(1)^{3}+(1)^{2}-2(1)+b$ = a + 1 - 2 + b = a + b - 1 ..... 1 Let, x + 1 = 0 ⟹ x = -1 Substitute x value in f(x) $f(-1)=a(-1)^{3}+(-1)^{2}-2(-1)+b$ = -a + 1 + 2 + b = -a...

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