Using factor theorem, factorize of the polynomials:

Question: Using factor theorem, factorize of the polynomials: $3 x^{3}-x^{2}-3 x+1$ Solution: Given, $f(x)=3 x^{3}-x^{2}-3 x+1$ The factors of constant term 1 is 1 The factors of the coefficient of $x^{2}=3$ The possible rational roots are 1, 1/3 Let, x 1 = 0 = x = 1 $f(1)=3(1)^{3}-(1)^{2}-3(1)+1$ = 3 1 3 + 1 = 0 So, x 1 is the factor of f(x) Now, divide f(x) with (x 1) to get other factors By long division method, $3 x^{2}+2 x-1$ $x-13 x^{3}-x^{2}-3 x+1$ $3 x^{3}-x^{2}$ (-) $(+)$ $2 x^{2}-3 x$ ...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $0.4 x+0.3 y=1.7$ $0.7 x-0.2 y=0.8$ Solution: The given equations are: $0.4 x+0.3 y=1.7 \ldots(i)$ $0.7 x-0.2 y=0.8 \ldots(i i)$ Multiply equation (i) by 2 and equation (ii) by 3 , and add both equations we get Put the value of $x$ in equation $(i)$ we get $0.4 \times 2+0.3 y=1.7$ $\Rightarrow 0.3 y=0.9$ $\Rightarrow y=3$ Hence the value of $x=2$ and $y=3$...

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If sets A and B are defined as A

Question: If sets $A$ and $B$ are defined as $A=\left\{(x, y): y=\frac{1}{x}, 0 \neq x \in R\right\}, B=\{(x, y): y=-x, x \in R\}$, then (a) $A \cap B=A$ (b) $A \cap B=B$ (c) $A \cap B=\phi$ (d) $A \cup B=A$ Solution: $A=\left\{(x, y): y=\frac{1}{x} ; 0 \neq x \in R\right\}$ $B=\{(x, y): y=-x ; x \in R\}$ then $(x, y) \in A \cap B$ i. e $(x, y) \in A$ and $(x, y) \in B$ i. e $y=\frac{1}{x}$ and $y=-x$ i. $\mathrm{e} \frac{1}{x}=-x ; \quad x \neq 0 \in R$ i. e $1=-x^{2} ; \quad x \neq 0 \in R$ i....

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Solve the following systems of equations:

Question: Solve the following systems of equations: $3 x-7 y+10=0$ $y-2 x-3=0$ Solution: The given equations are: $3 x-7 y+10=0$...(i) $y-2 x-3=0$...(ii) Multiply equation (i) by 2 and equation (ii) by 3 , and add both equations we get Put the value of $y$ in equation $(i)$ we get $3 x-7 \times 1+10=0$ $\Rightarrow 3 x=-3$ $\Rightarrow x=-1$ Hence the value of $x=-1$ and $y=1$...

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Prove

Question: $\sin ^{3}(2 x+1)$ Solution: Let $I=\int \sin ^{3}(2 x+1)$ $\begin{aligned} \Rightarrow \int \sin ^{3}(2 x+1) d x =\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ =\int\left(1-\cos ^{2}(2 x+1)\right) \sin (2 x+1) d x \end{aligned}$ Let $\cos (2 x+1)=t$ $\Rightarrow-2 \sin (2 x+1) d x=d t$ $\Rightarrow \sin (2 x+1) d x=\frac{-d t}{2}$ $\Rightarrow I=\frac{-1}{2} \int\left(1-t^{2}\right) d t$ $=\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}$ $=\frac{-1}{2}\left\{\cos (2 x+1)-\frac{\cos ^{3...

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Using factor theorem, factorize of the polynomials:

Question: Using factor theorem, factorize of the polynomials: $2 x^{4}-7 x^{3}-13 x^{2}+63 x-45$ Solution: Given, $f(x)=2 x^{4}-7 x^{3}-13 x^{2}+63 x-45$ The factors of constant term - 45 are 1, 3, 5, 9, 15, 45 The factors of the coefficient of $x^{4}$ is 2 . Hence possible rational roots of $f(x)$ are 1, 3, 5, 9, 15, 45, 1/2, 3/2, 5/2, 9/2, 15/2, 45/2 Let, x 1 = 0 = x = 1 f(1) = 2(1)4 7(1)3 13(1)2 + 63(1) 45 = 2 7 13 + 63 45 = 0 Let, x 3 = 0 = x = 3 $f(3)=2(3)^{4}-7(3)^{3}-13(3)^{2}+63(3)-45$ =...

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A survey shows that 63% of the people watch a News channel whereas 76% watch another channel.

Question: A survey shows that 63% of the people watch a News channel whereas 76% watch another channel. Ifx% of the people watch both channel, then (a)x= 35 (b)x= 63 (c) 39 x 63 (d)x= 39 Solution: Let A denote the percentage of people watching a news channel Let B denote the percentage of people watching other channel i.e n(A) = 63 n(B) = 76 Letn(AB) =x thenn(A⋃B) =n(A) +n(B) n(AB) = 63 + 76 x n(A⋃B) = 139 x i.ex= 139 n(A⋃B) Sincen(A⋃B) 100 i.e. 139 n(A⋃B) 139 100 = 39 i.e.x 39. also,n(AB) n(A) ...

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Solve the following systems of equations:

Question: Solve the following systems of equations: $11 x+15 y+23=0$ $7 x-2 y-20=0$ Solution: The given equations are: $11 x+15 y+23=0 \quad \ldots(i)$ $7 x-2 y-20=0 \quad \ldots(i i)$ Multiply equation $(i)$ by 2 and equation $(i i)$ by 15 , and add both equations we get Put the value of $x$ in equation $(i)$ we get $11 \times 2+15 y+23=0$ $\Rightarrow 15 y=-45$ $\Rightarrow y=-3$ Hence the value of $x=2$ and $y=-3$...

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If X

Question: If $X=\left\{8^{n}-7 n-1: n \in N\right\}$ and $Y=\{49 n-49: n \in N\}$. Then, (a) $X \subset Y$ (b) $Y \subset X$ (c) $X=Y$ (d) $X \cap Y=\phi$ Solution: $X=\left\{8^{n}-7 n-1, n \in N\right\}$ $Y=\{49 n-49 ; n \in N\}$ Since $8^{n}-7 n-1=(1+7)^{n}-7 n-1$ $=\left(1+7 n+{ }^{n} C_{2} 7^{2}+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}\right)-7 n-1$ $=1+7 n+{ }^{n} C_{2} 49+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}-7 n-1$ $=7^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+7^{n-2}\right)$ $=49\left({ }^{n} C_{...

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Given the linear equation 2x + 3y − 8 = 0,

Question: Given the linear equation 2x+ 3y 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is(i) intersecting lines(ii) Parallel lines(iii) coincident lines Solution: (i) For intersecting lines, Equation of another intersecting line to the given line is $2 x+5 y-3=0$ Since, condition for intersecting lines and unique solution is $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ (ii) For parallel lines, Equation of another parallel...

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Let

Question: LetF1be the set of all parallelograms,F2the set of all rectangles,F3the set of all rhombuses,F4the set of all squares andF5the set of trapeziums in a plane. ThenF1may be equal to (a) $F_{2} \cap F_{3}$ (b) $F_{3} \cap F_{4}$ (c) $F_{2} \cup F_{3}$ (d) $F_{2} \cup F_{3} \cup F_{4} \cup F_{1}$ Solution: We know that every rectangle, rhombus and square in a plane is a parallelogram but every trapezium is not a parallelogram. So, $F_{1}$ is either of $F_{1}$ or $F_{2}$ or $F_{3}$ or $F_{4}...

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Using factor theorem, factorize of the polynomials:

Question: Using factor theorem, factorize of the polynomials: $x^{4}+10 x^{3}+35 x^{2}+50 x+24$ Solution: Given, $f(x)=x^{4}+10 x^{3}+35 x^{2}+50 x+24$ The constant term in f(x) is equal to 24 The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 Let, x + 1 = 0 = x = -1 Substitute the value of x in f(x) $f(-1)=(-1)^{4}+10(-1)^{3}+35(-1)^{2}+50(-1)+24$ = 1-10 + 35 - 50 + 24 = 0 = (x + 1) is the factor of f(x) Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x) Since, f(x) is a polynomial o...

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Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis.

Question: Represent the following pair of equations graphically and write the coordinates of points where the lines intersectsy-axis. $x+3 y=6$ $2 x-3 y=12$ Solution: The given equations are $x+3 y=6 \quad \ldots \ldots .(i)$ $2 x-3 y=12 \quad \ldots \ldots . .(i i)$ Putting $x=0$ in equation (i) we get: $\Rightarrow 0+3 y=6$ $\Rightarrow y=2$ $x=0, \quad y=2$ Putting $y=0$ in equation $(i)$ we get: $\Rightarrow x+3 \times 0=6$ $\Rightarrow x=6$ $x=6, \quad y=0$ Use the following table to draw t...

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For any two sets A and B,

Question: For any two sets $A$ and $B, A \cap(A \cup B)^{\prime}$ is equal to (a)A (b)B (c) $\phi$ (d) $A \cap B$ ϕ Solution: $A \cap(A \cup B)^{\prime}$ $=A \cap\left(A^{\prime} \cap B^{\prime}\right)$ (De Morgan law) $=\left(A \cap A^{\prime}\right) \cap B^{\prime}$ $=\phi \cap B^{\prime}$ $=\phi$ Hence, the correct answer is option (c)....

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Prove

Question: $\sin ^{3}(2 x+1)$ Solution: Let $I=\int \sin ^{3}(2 x+1)$ $\begin{aligned} \Rightarrow \int \sin ^{3}(2 x+1) d x =\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ =\int\left(1-\cos ^{2}(2 x+1)\right) \sin (2 x+1) d x \end{aligned}$ Let $\cos (2 x+1)=t$ $\Rightarrow-2 \sin (2 x+1) d x=d t$ $\Rightarrow \sin (2 x+1) d x=\frac{-d t}{2}$ $\Rightarrow I=\frac{-1}{2} \int\left(1-t^{2}\right) d t$ $=\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}$ $=\frac{-1}{2}\left\{\cos (2 x+1)-\frac{\cos ^{3...

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For any two sets A and B,

Question: For any two sets $A$ and $B, A \cap(A \cup B)^{\prime}$ is equal to (a)A (b)B (c) $\phi$ (d) $A \cap B$ ϕ Solution: $A \cap(A \cup B)^{\prime}$ $=A \cap\left(A^{\prime} \cap B^{\prime}\right)$ (De Morgan law) $=\left(A \cap A^{\prime}\right) \cap B^{\prime}$ $=\phi \cap B^{\prime}$ $=\phi$ Hence, the correct answer is option (c)....

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Using factor theorem, factorize of the polynomials:

Question: Using factor theorem, factorize of the polynomials: $x^{4}-2 x^{3}-7 x^{2}+8 x+12$ Solution: Given, $f(x)=x^{4}-2 x^{3}-7 x^{2}+8 x+12$ The constant term f(x) is equal is 12 The factors of 12 are 1, 2, 3, 4, 6, 12 Let, x + 1 = 0 = x = -1 Substitute the value of x in f(x) $f(-1)=(-1)^{4}-2(-1)^{3}-7(-1)^{2}+8(-1)+12$ = 1 + 2 7 8 + 12 = 0 So, x + 1 is factor of f(x) Similarly, (x + 2), (x 2), (x 3) are also the factors of f(x) Since, f(x) is a polynomial of degree 4, it cannot have more ...

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Prove

Question: $\cos 2 x \cos 4 x \cos 6 x$ Solution: It is known that, $\cos A \cos B=\frac{1}{2}\{\cos (A+B)+\cos (A-B)\}$ $\therefore \int \cos 2 x(\cos 4 x \cos 6 x) d x=\int \cos 2 x\left[\frac{1}{2}\{\cos (4 x+6 x)+\cos (4 x-6 x)\}\right] d x$ $=\frac{1}{2} \int\{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)\} d x$ $=\frac{1}{2} \int\left\{\cos 2 x \cos 10 x+\cos ^{2} 2 x\right\} d x$ $=\frac{1}{2} \int\left[\left\{\frac{1}{2} \cos (2 x+10 x)+\cos (2 x-10 x)\right\}+\left(\frac{1+\cos 4 x}{2}\right)\...

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Two finite sets have m and n elements.

Question: Two finite sets havemandnelements. The number of subsets of the first set is 112 more than that of the second. The values ofmandnare respectively(a) 4, 7 (b) 7, 4 (c) 4, 4 (d) 7, 7 Solution: We know that if a setXcontainskelements, then the number of subsets ofXare 2k.It is given that the number of subsets of a set containingmelements is 112 more than the number of subsets of set containingnelements. $\therefore 2^{m}-2^{n}=112$ $\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2 \times 2 \time...

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Two finite sets have m and n elements.

Question: Two finite sets havemandnelements. The number of subsets of the first set is 112 more than that of the second. The values ofmandnare respectively(a) 4, 7 (b) 7, 4 (c) 4, 4 (d) 7, 7 Solution: We know that if a setXcontainskelements, then the number of subsets ofXare 2k.It is given that the number of subsets of a set containingmelements is 112 more than the number of subsets of set containingnelements. $\therefore 2^{m}-2^{n}=112$ $\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2 \times 2 \time...

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Prove

Question: $\sin 3 x \cos 4 x$ Solution: It is known that, $\sin A \cos B=\frac{1}{2}\{\sin (A+B)+\sin (A-B)\}$ $\therefore \int \sin 3 x \cos 4 x d x=\frac{1}{2} \int\{\sin (3 x+4 x)+\sin (3 x-4 x)\} d x$ $=\frac{1}{2} \int\{\sin 7 x+\sin (-x)\} d x$ $=\frac{1}{2} \int\{\sin 7 x-\sin x\} d x$ $=\frac{1}{2} \int \sin 7 x d x-\frac{1}{2} \int \sin x d x$ $=\frac{1}{2}\left(\frac{-\cos 7 x}{7}\right)-\frac{1}{2}(-\cos x)+\mathrm{C}$ $=\frac{-\cos 7 x}{14}+\frac{\cos x}{2}+\mathrm{C}$...

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Suppose

Question: Suppose $A_{1}, A_{2}, \ldots, A_{30}$ are thirty sets each having 5 elements and $B_{1}, B_{2}, \ldots, B_{n}$ are $n$ sets each with 3 elements. Let $\bigcup_{i=1}^{30} A_{i}=\bigcup_{j=1}^{n} B_{j}=S$ and each element of $S$ belong to exactly 10 of the $A_{i}$ 's and exactly 9 of theBj'sBj's, thennis equal to (a) 15 (b) 3 (c) 45 (d) 35 Solution: It is given that each set $A_{i}(1 \leq i \leq 30)$ contains 5 elements and $\bigcup_{i=1}^{30} A_{i}=S$. $\therefore n(S)=30 \times 5=150$...

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Using factor theorem, factorize of the polynomials:

Question: Using factor theorem, factorize of the polynomials: $x^{4}-7 x^{3}+9 x^{2}+7 x-10$ Solution: Given, $f(x)=x^{4}-7 x^{3}+9 x^{2}+7 x-10$ The constant term in f(x) is 10 The factors of 10 are 1, 2, 5, 10 Let, x 1 = 0 = x = 1 Substitute the value of x in f(x) $f(x)=14-7(1)^{3}+9(1)^{2}+7(1)-10$ = 1 7 + 9 + 7 10 = 10 10 = 0 (x 1)is the factor of f(x) Similarly, the other factors are (x + 1), (x 2), (x 5) Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear fact...

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Prove

Question: $\sin ^{2}(2 x+5)$ Solution: $\sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}$ $\Rightarrow \int \sin ^{2}(2 x+5) d x=\int \frac{1-\cos (4 x+10)}{2} d x$ $=\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x$ $=\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+\mathrm{C}$ $=\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+\mathrm{C}$...

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Suppose

Question: SupposeA1,A2,...,A30A1,A2,...,A30are thirty sets each having 5 elements andB1,B2,...,BnB1,B2,...,Bnarensets each with 3 elements.Let $\bigcup_{i=1}^{30} A_{i}=\bigcup_{j=1}^{n} B_{j}=S$ and each element of $S$ belong to exactly 10 of the $A_{i}$ 's and exactly 9 of theBj'sBj's, thennis equal to (a) 15 (b) 3 (c) 45 (d) 35 Solution: It is given that each set $A_{i}(1 \leq i \leq 30)$ contains 5 elements and $\bigcup_{i=1}^{30} A_{i}=S$. $\therefore n(S)=30 \times 5=150$ But, it is given ...

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