Prove
Question: $\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals A. $\tan x+\cot x+\mathrm{C}$ B. $\tan x-\cot x+\mathrm{C}$ C. $\tan x \cot x+\mathrm{C}$ D. $\tan x-\cot 2 x+\mathrm{C}$ Solution: Let $\begin{aligned} I =\int \frac{d x}{\sin ^{2} x \cos ^{2} x} \\ =\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x \\ =\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x \\ =\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x \\ =\int \sec ^{2...
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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 the $B_{j}$ 's, then $n$ is 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...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}-6 x^{2}+3 x+10$ Solution: Let, $f(x)=x^{3}-6 x^{2}+3 x+10$ 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) $f(-1)=(-1)^{3}-6(-1)^{2}+3(-1)+10$ = -1 6 3 + 10 = 0 Similarly, the other factors (x 2) and (x 5) of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors. f(x) = k(x + 1)(x 2)(x 5) Substitute x = 0 on b...
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Question: $\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x$ equals (A) $10^{x}-x^{10}+\mathrm{C}$ (B) $10^{x}+x^{10}+\mathrm{C}$ (C) $\left(10^{x}-x^{10}\right)^{-1}+\mathrm{C}$ (D) $\log \left(10^{x}+x^{10}\right)+\mathrm{C}$ Solution: Let $x^{10}+10^{x}=t$ $\therefore\left(10 x^{9}+10^{x} \log _{e} 10\right) d x=d t$ $\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$ $=\log t+\mathrm{C}$ $=\log \left(10^{x}+x^{10}\right)+\mathrm{C}$ Hence, ...
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: Given, $f(x)=x^{3}+2 x^{2}-x-2$ The constant term in f(x) is -2 The factors of (-2) are 1, 2 Let, x 1 = 0 = x = 1 Substitute the value of x in f(x) $f(1)=(1)^{3}+2(1)^{2}-1-2$ = 1 + 2 1 2 = 0 Similarly, the other factors (x + 1) and (x + 2) of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors. f(x) = k(x 1)(x + 2)(x + 1 ) $x^{3}+2 x^{2}-x-2=k(x-1)(x+2)(x+1)$...
Read More →In a class of 175 students the following data shows the number of students opting one or more subjects.
Question: In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone? (a) 35 (b) 48 (c) 60 (d) 22 (e) 30 Solution: (c) 60 Let M, P and C denote the sets of students who have opted for mathematics, physics, and chemistry, respectively. Here,...
Read More →In a class of 175 students the following data shows the number of students opting one or more subjects.
Question: In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone? (a) 35 (b) 48 (c) 60 (d) 22 (e) 30 Solution: (c) 60 Let M, P and C denote the sets of students who have opted for mathematics, physics, and chemistry, respectively. Here,...
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Question: $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$ Solution: Let $x^{4}=t$ $\therefore 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} d x=\frac{1}{4} \int \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} d t$ ...(1) Let $\tan ^{-1} t=u$ $\therefore \frac{1}{1+t^{2}} d t=d u$ From (1), we obtain $\begin{aligned} \int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right) d x}{1+x^{8}} =\frac{1}{4} \int \sin u d u \\ =\frac{1}{4}(-\cos u)+\mathr...
Read More →Solve the following system of equations graphically:
Question: Solve the following system of equations graphically: Shade the region between the lines and the y-axis (i) $3 x-4 y=7$ $5 x+2 y=3$ (ii) $4 x-y=4$ $3 x+2 y=14$ Solution: The given equations are: $3 x-4 y=7$$. .(i)$ $5 x+2 y=3$$. .(i i)$ Putting $x=0$ in equation (i) we get: $\Rightarrow 3 \times 0-4 y=7$ $\Rightarrow y=-7 / 4$ $x=0, \quad y=-7 / 4$ Putting $y=0$ in equation (i) we get: $\Rightarrow 3 x-4 \times 0=7$ $\Rightarrow x=7 / 3$ $x=7 / 3, \quad y=0$ Use the following table to d...
Read More →Two finite sets have m and n elements.
Question: Two finite sets havemandnelements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second set. Then, the values ofmandnare: (a) 7, 6 (b) 6, 3 (c) 7, 4 (d) 3, 7 Solution: (c) 6, 4ATQ : $2^{m}-1=48+2^{n}-1$ $\Rightarrow 2^{m}-2^{n}=48$ $\Rightarrow 2^{m}-2^{n}=2^{6}-2^{4}$ By comparing we get: $m=6$ and $n=4$...
Read More →Two finite sets have m and n elements.
Question: Two finite sets havemandnelements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second set. Then, the values ofmandnare: (a) 7, 6 (b) 6, 3 (c) 7, 4 (d) 3, 7 Solution: (c) 6, 4ATQ : $2^{m}-1=48+2^{n}-1$ $\Rightarrow 2^{m}-2^{n}=48$ $\Rightarrow 2^{m}-2^{n}=2^{6}-2^{4}$ By comparing we get: $m=6$ and $n=4$...
Read More →Two finite sets have m and n elements.
Question: Two finite sets havemandnelements. The number of elements in the power set of first set is 48 more than the total number of elements in power set of the second set. Then, the values ofmandnare: (a) 7, 6 (b) 6, 3 (c) 7, 4 (d) 3, 7 Solution: (c) 6, 4ATQ : $2^{m}-1=48+2^{n}-1$ $\Rightarrow 2^{m}-2^{n}=48$ $\Rightarrow 2^{m}-2^{n}=2^{6}-2^{4}$ By comparing we get: $m=6$ and $n=4$...
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Question: $\frac{(x+1)(x+\log x)^{2}}{x}$ Solution: $\frac{(x+1)(x+\log x)^{2}}{x}=\left(\frac{x+1}{x}\right)(x+\log x)^{2}=\left(1+\frac{1}{x}\right)(x+\log x)^{2}$ Let $(x+\log x)=t$ $\therefore\left(1+\frac{1}{x}\right) d x=d t$ $\Rightarrow \int\left(1+\frac{1}{x}\right)(x+\log x)^{2} d x=\int t^{2} d t$ $=\frac{t^{3}}{3}+\mathrm{C}$ $=\frac{1}{3}(x+\log x)^{3}+\mathrm{C}$...
Read More →An investigator interviewed 100 students to determine the performance of three drinks:
Question: An investigator interviewed 100 students to determine the performance of three drinks: milk, coffee and tea. The investigator reported that 10 students take all three drinks milk, coffee and tea; 20 students take milk and coffee; 25 students take milk and tea; 12 students take milk only; 5 students take coffee only and 8 students take tea only. Then the number of students who did not take any of three drinks is (a) 10 (b) 20 (c) 25 (d) 30 Solution: Disclaimer: The question in the book ...
Read More →Using factor theorem, factorize of the polynomials:
Question: Using factor theorem, factorize of the polynomials: $x^{3}+6 x^{2}+11 x+6$ Solution: Given polynomial, $f(x)=x^{3}+6 x^{2}+11 x+6$ The constant term in f(x) is 6 The factors of 6 are 1, 2, 3, 6 Let, x + 1 = 0 = x = -1 Substitute the value of x in f(x) $f(-1)=(-1)^{3}+6(-1)^{2}+11(-1)+6$ = - 1 + 6 - 11 + 6 = 12 12 = 0 So, (x + 1) is the factor of f(x) Similarly, (x + 2) and (x + 3) are also the factors of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three...
Read More →If
Question: If $A \cap B-B$, then (a) $A \subset B$ (b) $B \subset A$ (c) $A=\Phi$ (d) $B=\Phi$ Solution: (b) $B \subset A$ Only this case is possible....
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Question: $\frac{(1+\log x)^{2}}{x}$ Solution: Let $1+\log x=t$ $\therefore \frac{1}{x} d x=d t$ $\Rightarrow \int \frac{(1+\log x)^{2}}{x} d x=\int t^{2} d t$ $=\frac{t^{3}}{3}+\mathrm{C}$ $=\frac{(1+\log x)^{3}}{3}+\mathrm{C}$...
Read More →In a city 20% of the population travels by car,
Question: In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus is (a) 80% (b) 40% (c) 60% (d) 70% Solution: (c) 60% Suppose C and B represents the population travel by car and Bus respectively. $n(C \cup B)=n(C)+n(B)-n(B \cap C)$ $=0.20+0.50-0.10$ $=0.6$ or $60 \%$...
Read More →If A = {x : x is a multiple of 3} and ,
Question: IfA= {x:xis a multiple of 3} and ,B= {x:xis a multiple of 5}, thenABis (a) $A \cap B$ (b) $A \cap \bar{B}$ (c) $\bar{A} \cap \bar{B}$ (d) $\overline{A \cap B}$ Solution: (b) $A \cap \bar{B}$ A= {x:xis a multiple of 3} $A=\{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48, \ldots\}$ B= {x:xis a multiple of 5.} $B=\{5,10,15,20,25,30,35,40,45,50, \ldots\}$ Now, we have: $A-B=\{3,6,9,12,18,21,24,27,33,36,39,42,48, \ldots\}$ $=A \cap \bar{B}$...
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Question: $\frac{\sqrt{\tan x}}{\sin x \cos x}$ Solution: Let $\begin{aligned} I =\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x \\ =\int \frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x} d x \\ =\int \frac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x \\ =\int \frac{\sec ^{2} x d x}{\sqrt{\tan x}} \end{aligned}$ Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ $\begin{aligned} \therefore I =\int \frac{d t}{\sqrt{t}} \\ =2 \sqrt{t}+\mathrm{C} \\ =2 \sqrt{\tan x}+\mathrm{C} \end{aligned}$...
Read More →If A = {x : x is a multiple of 3} and ,
Question: IfA= {x:xis a multiple of 3} and ,B= {x:xis a multiple of 5}, thenABis (a) $A \cap B$ (b) $A \cap \bar{B}$ (c) $\bar{A} \cap \bar{B}$ (d) $\overline{A \cap B}$ Solution: (b) $A \cap \bar{B}$ A= {x:xis a multiple of 3} $A=\{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48, \ldots\}$ B= {x:xis a multiple of 5.} $B=\{5,10,15,20,25,30,35,40,45,50, \ldots\}$ Now, we have: $A-B=\{3,6,9,12,18,21,24,27,33,36,39,42,48, \ldots\}$ $=A \cap \bar{B}$...
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Question: $\frac{1}{1-\tan x}$ Solution: Let $I=\int \frac{1}{1-\tan x} d x$ $=\int \frac{1}{1-\frac{\sin x}{\cos x}} d x$ $=\int \frac{\cos x}{\cos x-\sin x} d x$ $=\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} d x$ $=\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} d x$ $=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x$ $=\frac{x}{2}+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x$ Put $\cos x-\sin x=t \Rightarrow(-\sin x-\cos ...
Read More →If A and B are two given sets,
Question: If $A$ and $B$ are two given sets, then $A \cap(A \cap B)^{c}$ is equal to (a)A (b)B (c) Φ (d) $A \cap B^{c}$ Solution: (d) $A \cap B^{c}$ A and B are two sets. $A \cap B$ is the common region in both the sets. $(\mathrm{A} \cap \mathrm{B})^{c}$ is all the region in the universal set except $\mathrm{A} \cap \mathrm{B}$. Now, $\mathrm{A} \cap(\mathrm{A} \cap \mathrm{B})^{c}=\mathrm{A} \cap \mathrm{B}^{c}$...
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Question: $\frac{1}{1+\cot x}$ Solution: Let $I=\int \frac{1}{1+\cot x} d x$ $=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x$ $=\int \frac{\sin x}{\sin x+\cos x} d x$ $=\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x$ $=\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)} d x$ $=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x$ $=\frac{1}{2}(x)+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x$ Let $\sin x+\cos x=t \Rightarrow(\cos x-\si...
Read More →If A and B are two sets such that n
Question: If $A$ and $B$ are two sets such that $n(A)=70, n(B)=60, n(A \cup B)=110$, then $n(A \cap B)$ is equal to (a) 240 (b) 50 (c) 40 (d) 20 Solution: (d) 20 We have: $n(\mathrm{~A} \cap \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cup \mathrm{B})$ $=70+60-110$ $=20$...
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