Prove that:
Question: Prove that: $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}$ Solution: To Prove: $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}$ Formula Used: 1) $\sin A=2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}$ 2) $1+\cos A=2 \cos ^{2} \frac{A}{2}$ Proof: $\mathrm{LHS}=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ $=\tan ^{-1}\left(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)$ $=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}...
Read More →Let n be a fixed positive integer.
Question: Let n be a fixed positive integer. Define a relation R in Z as follows: a, b Z, aRb if and only if a b is divisible by n. Show that R is an equivalance relation. Solution: Given a, b Z, aRb if and only if a b is divisible by n. Now, for aRa ⇒ (a a) is divisible by n, which is true for any integer a as 0 is divisible by n. Thus, R is reflective. Now, aRb So, (a b) is divisible by n. ⇒ (b a) is divisible by n. ⇒ (b a) is divisible by n ⇒ bRa Thus, R is symmetric. Let aRb and bRc Then, (a...
Read More →Let f: R → R be the function defined
Question: Let f: R R be the function defined by f(x) = 1/(2 cos x) x R. Then, find the range of f. Solution: Given, f(x) = 1/(2 cos x) x R Let y = 1/(2 cos x) 2y ycos x = 1 cos x = (2y 1)/ y cos x = 2 1/y Now, we know that -1 cos x 1 So, -1 2 1/y 1 -3 1/y -1 1 1/y 3 1/3 y 1 Thus, the range of the given function is [1/3, 1]....
Read More →If functions f: A → B and g: B → A satisfy g o f = IA,
Question: If functions f: A B and g: B A satisfy g o f = IA, then show that f is one-one and g is onto. Solution: Given, f: A B and g: B A satisfy g o f = IA Its clearly seen that function g is inverse of f. So, f has to be one-one and onto. Hence, g is also one-one and onto....
Read More →Let X = {1, 2, 3} and Y = {4, 5}.
Question: Let X = {1, 2, 3} and Y = {4, 5}. Find whether the following subsets of X Y are functions from X to Y or not. (i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} (ii) g = {(1, 4), (2, 4), (3, 4)} (iii) h = {(1,4), (2, 5), (3, 5)} (iv) k = {(1,4), (2, 5)}. Solution: Given, X = {1, 2, 3} and Y = {4, 5} So, X Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} (i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} f is not a function as f(1) = 4 and f(1) = 5 Hence, pre-image 1 has not unique image. (ii) g = {(1,...
Read More →Let the function f: R → R be defined
Question: Let the function f: R R be defined by f (x) = cos x, x R. Show that f is neither one-one nor onto. Solution: We have, f: R R, f(x) = cos x Now, f (x1) = f (x2) cos x1= cos x2 x1= 2n x2, n Z Its seen that the above equation has infinite solutions for x1and x2 Hence, f(x) is many one function. Also the range of cos x is [-1, 1], which is subset of given co-domain R. Therefore, the given function is not onto....
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x$ Solution: $I=\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x$ $\frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{2 x+1}$ $2 x^{2}+7 x-3=A x(2 x+1)+B(2 x+1)+C x^{2}$ Equating constants $-3=B$ Equating coefficients of $x$ $7=A+2 B$ $7=A-6$ $A=13$ Equating coefficients of $x^{2}$ $2=2 A+C$ $2=26+C$ $C=-24$ Thus, $I=\int \frac{13 d x}{x}-\int \frac{3 d x}{x^{2}}-24 \int \frac{d x}{2 x+1}$ $I=13...
Read More →Let C be the set of complex numbers.
Question: Let C be the set of complex numbers. Prove that the mapping f: C R given by f (z) = |z|, z C, is neither one-one nor onto. Solution: Given, f: C R such that f (z) = |z|, z C Now, let take z = 6 + 8i Then, f (6 + 8i) = |6 + 8i| = (62+ 82) = 100 = 10 And, for z = 6 8i f (6 8i) = |6 8i| = (62+ 82) = 100 = 10 Hence, f (z) is many-one. Also, |z| 0, z C But the co-domain given is R Therefore, f(z) is not onto....
Read More →If the mappings f and g are given by
Question: If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write f o g. Solution: Given, f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} Now, fog (2) = f(g(2)) = f(3) = 5 fog (5) = f(g(5)) = f(1) = 2 fog (1) = f(g(1)) = f(3) = 5 Thus, fog = {(2, 5), (5, 2), (1, 5)}...
Read More →Are the following set of ordered pairs functions?
Question: Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective. (i) {(x, y): x is a person, y is the mother of x}. (ii){(a, b): a is a person, b is an ancestor of a}. Solution: (i) Given, {(x, y): x is a person, y is the mother of x} Its clearly seen that each person x has only one biological mother. Hence, the above set of ordered pairs make a function. Now more than one person may have same mother. Thus, the function is many-many one a...
Read More →Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function?
Question: Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g (x) = x + , then what value should be assigned to and . Solution: Given, g = {(1, 1), (2, 3), (3, 5), (4, 7)} Its seen that every element of domain has a unique image. So, g is function. Now, also given that g (x) = x + So, we have g (1) = (1) + = 1 + = 1 .. (i) And, g (2) = (2) + = 3 2 + = 3 .. (ii) Solving (i) and (ii), we have = 2 and = -1 Therefore, g (x) = 2x 1...
Read More →Prove that
Question: Prove that $\tan ^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3 / 2}}\right)=\tan ^{-1} x+\tan ^{-1} \sqrt{x}$ Solution: We know that, $\tan ^{-1}\left(\frac{A+B}{1-A B}\right)=\tan ^{-1} A+\tan ^{-1} B$ Now, taking $A=x$ and $B=\sqrt{x}$ We get, $\tan ^{-1} x+\tan ^{-1} \sqrt{x}=\tan ^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3 / 2}}\right)$ As, $x \cdot x^{1 / 2}=x^{3 / 2}$ Hence, Proved....
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+x-1}{(x+1)^{2}(x+2)} d x$ Solution: $I=\int \frac{x^{2}+x-1}{(x+1)^{2}(x+2)} d x$ $\frac{\mathrm{x}^{2}+\mathrm{x}-1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{(\mathrm{x}+1)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+2}$ $x^{2}+x-1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}$ Put $x=-2$ $1=C$ $C=1$ Put $x=-1$ $-1=B$ $B=-1$ Equating coefficients of constants $-1=2 A+2 B+C$ $-1=2 A-2+1$ $A=0$ Thus, $I=0 \times \int \f...
Read More →If f: R → R is defined by
Question: If f: R R is defined by f (x) = x2 3x + 2, write f (f (x)). Solution: Given, f (x) = x2 3x + 2 Then, f (f (x)) = f (x2 3x + 2) = (x2 3x + 2)2 3(x2 3x + 2) + 2, = x4+ 9x2+ 4 6x3+ 4x2 12x 3x2+ 9x 6 + 2 = x4 6x3+ 10x2 3x Thus, f (f (x)) = x4 6x3+ 10x2 3x...
Read More →Prove that:
Question: Prove that: $\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x y}}\right)=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{y}$ Solution: To Prove: $\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x y}}\right)=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{y}$ We know that, $\tan A+\tan B=\frac{\tan A+\tan B}{1-\tan A \tan B}$ Also, $\tan ^{-1}\left(\frac{\mathrm{A}+\mathrm{B}}{1-\mathrm{AB}}\right)=\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}$ Taking $A=\sqrt{x}$ and $B=\sqrt{y}$ We get, $\tan ^{-1}\left...
Read More →If A = {a, b, c, d} and the function f = {(a, b),
Question: If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f1. Solution: Given, A = {a, b, c, d} and f = {(a, b), (b, d), (c, a), (d, c)} So, f-1= {(b, a), (d, b), (a, c), (c, d)}...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x$ Solution: $I=\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x$ $\frac{\mathrm{x}^{2}}{(\mathrm{x}-1)(\mathrm{x}+1)^{2}}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}+1}+\frac{\mathrm{C}}{(\mathrm{x}+1)^{2}}$ $\mathrm{x}^{2}=\mathrm{A}(\mathrm{x}+1)^{2}+\mathrm{B}(\mathrm{x}-1)(\mathrm{x}+1)+\mathrm{C}(\mathrm{x}-1)$ Put $x=1$ $1=4 \mathrm{~A}$ $A=\frac{1}{4}$ Put $x=-1$ $1=-2 C$ $C=-\frac{1}{2}$ Equating coeff...
Read More →Let f: R → R be the function defined
Question: Let f: R R be the function defined by f (x) = 2x 3, x R. write f1. Solution: Given function, f (x) = 2x 3, x R Let y = 2x 3 x = (y + 3)/ 2 Thus, f-1(x) = (x + 3)/ 2...
Read More →Prove that:
Question: Prove that: $\cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$ Solution: To Prove: $\cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$ Formula Used: 1) $\tan \left(\frac{\pi}{4}+\mathrm{A}\right)=\frac{1+\tan \mathrm{A}}{1-\tan \mathrm{A}}$ 2) $\operatorname{cosec}^{2} A=1+\cot ^{2} A$ 3) $1-\cos A=2 \sin ^{2}\left(\frac{A}{2}\right)$ 4) $\sin A=2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)$ Proof: $\mathrm{LHS}...
Read More →Let f, g: R → R be defined by f(x) = 2x + 1
Question: Let f, g: R R be defined by f(x) = 2x + 1 and g (x) = x2 2, x R, respectively. Then, find g o f. Solution: Given, f(x) = 2x + 1 and g (x) = x2 2, x R Thus, g o f = g (f (x)) = g (2x + 1) = (2x + 1)2 2 = 4x2+ 4x + 1 2 = 4x2+ 4x 1...
Read More →Let D be the domain of the real valued
Question: Let D be the domain of the real valued function f defined by f(x) = (25 x2). Then, write D. Solution: Given, f(x)=(25 x2) The function is defined if 25 x2 0 So, x2 25 -5 x 5 Therefore, the domain of the given function is [-5, 5]....
Read More →Let A = {a, b, c} and the relation R be defined
Question: Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make R reflexive and transitive. Solution: Given relation, R = {(a, a), (b, c), (a, b)} To make R as reflexive we should add (b, b) and (c, c) to R. Also, to make R as transitive we should add (a, c) to R. Hence, the minimum number of ordered pairs to be added are (b, b), (c, c) and (a, c) i.e. 3....
Read More →Prove that:
Question: Prove that: $\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=2 \cos ^{-1} x$ Solution: To Prove: $\sec ^{-1}\left(\frac{1}{2 \mathrm{x}^{2}-1}\right)=2 \cos ^{-1} \mathrm{x}$ Formula Used: 1) $\cos 2 A=2 \cos ^{2} A-1$ 2) $\cos ^{-1} \mathrm{~A}=\sec ^{-1}\left(\frac{1}{\mathrm{~A}}\right)$ Proof: $\mathrm{LHS}=\sec ^{-1}\left(\frac{1}{2 \mathrm{x}^{2}-1}\right)$ $=\cos ^{-1}\left(2 x^{2}-1\right) \ldots$(1) Let $x=\cos A \ldots(2)$ Substituting (2) in (1), $\mathrm{LHS}=\cos ^{-1}\left(2 \...
Read More →Prove that:
Question: Prove that: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 \mathrm{~A}=2 \cos ^{2} \mathrm{~A}-1$ Proof: $\mathrm{LHS}=\cos ^{-1}\left(2 \mathrm{x}^{2}-1\right) \ldots(1)$ $\operatorname{Let} x=\cos A \ldots(2)$ Substituting (2) in (1), $\mathrm{LHS}=\cos ^{-1}\left(2 \cos ^{2} \mathrm{~A}-1\right)$ $=\cos ^{-1}(\cos 2 \mathrm{~A})$ $=2 \mathrm{~A}$ From $(2), A=\cos ^{-1} x$ $=\mathrm{RHS}$ ...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x}{(x-1)^{2}(x+2)} d x$ Solution: $I=\int \frac{x}{(x-1)^{2}(x+2)} d x$ $\frac{\mathrm{x}}{(\mathrm{x}-1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{(\mathrm{x}-1)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+2}$ $x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}$ Put $x=-2$ $-2=9 c$ $C=-\frac{2}{9}$ Put $x=1$ $1=3 B$ $B=\frac{1}{3}$ Equating coefficients of constants $0=-2 A+2 B+C$ $0=-2 A+2 * \frac{1}{3}-\frac{2}{9}$ $A=\frac{2}{9}$ Thus, $I=...
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