WBCHSE Solutions For Class 12 Maths Relations And Functions

Relations And Functions

Relations And Functions Exercise 1A Review Of Facts And Formulae

1. A relation R in a set A is a subset of A x A

If (a, b) ∈ R, then a R b.

If (a, b) ∉ R, then a is not related to b.

Domain (R) = {a: (a, b) ∈ R] and Range (R) = {b: (a, b) ∈ R}.

2. Let R be a relation on a non-empty set A. Then, R is said to be:

1. Reflexive, if (a, a) ∈ R for each a ∈ A

i.e., if a R a for each a ∈ R.

2. Symmetric, if (a, b) ∈ R ⇒ (b, a) ∈ R

i.e., a R b ⇒ b Ra

3. Transitive, if (a, b) e R and (b, c)e R ⇒ (a, c) ∈ R

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i.e., a R b, b R c ⇒  a R c.

3. A relation R on a set A is said to be an equivalence relation if it is reflexive, symmetric, and transitive.

4. Equivalence class, determined by an element a is defined as a

⇒ [a] = [b ∈ A : (a, b) ∈ R}

Relations And Functions Exercise 1A Multiple Choice Questions

Question 1. Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 2. Reflexive and transitive but not symmetric

R is reflexive and transitive but not symmetric

 

Question 2. Let A = {a, b, c} and let R= {(a, a), (a, b), (b, a)}. Then, Ris

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 3. Symmetric and transitive but not reflexive

R is symmetric and transitive but not reflexive

Question 3. Let A = {1,2, 3} and let R= {(1,1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}. Then, R is

1. Reflexive and symmetric but not transitive
2. Symmetric and transitive but not reflexive
3. Reflexive and transitive but not symmetric
4. An equivalence relation

Answer: 1. Reflexive and symmetric but not transitive

(1,2)∈ R and (2,3) ∈ R, But, (1,3) ∉ R So, R is not transitive.

Question 4. Let S be the set of all straight lines in a plane. Let R be a relation on S Defined by a R b ⇔ a ⊥ b. Then, R is

1. Reflexive but neither symmetric nor transitive
2. Symmetric but neither reflexive nor transitive
3. Transitive but neither reflexive nor symmetric
4. An equivalence relation

Answer: 2. Symmetric but neither reflexive nor transitive

a ⊥ b is not true. So R is not reflexive

a ⊥ b and b ⇒ b ⊥ a is always true.

Question 5. Let S be the set of all straight lines in a plane. Let R be a relation on S Defined by a R b ⇔ a || b. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 6. Let Z be the set of all integers and let R be a relation on Z defined by a R b o (a-b) is divisible by 3. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 7. Let R be a relation on the set N of all natural numbers, defined by a R b ⇔  a is a factor of b. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 2. Reflexive and transitive but not symmetric

Question 8. Let Z be the set of all integers and let R be a relation on Z defined by a R b ⇔  a>b. Then, R is

1. Symmetric and transitive but not reflexive
2. Reflexive and symmetric but not transitive
3. Reflexive and transitive but not symmetric
4. An equivalence relation

Answer: 3. Reflexive and transitive but not symmetric

Question 9. Let S be the set of all real numbers and let R be a relation on S defined by a R b ⇔ IaI ≤  b. Then, R is

1. Reflexive but neither symmetric nor transitive
2. Symmetric but neither reflexive nor transitive
3. Transitive but neither reflexive nor symmetric
4. None of these

Answer: 3. Transitive but neither reflexive nor symmetric

⇒  I-3I ≤ -3 is not true. So, R is not reflexive.

⇒ I-1I ≤ 1 ⇒ (-1) R1. But I-1I ≤ 1 is not true.

∴ R is not symmetric

3. a R b, b R c ⇒ IaI ≤  b and IbI ≤  c ⇒ IaI ≤  c.

Question 10. Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ Ia-bI ≤ 1. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 1. Reflexive and symmetric but not transitive

⇒  Ia- aI = 0 ≤ 1 is always true.

⇒  a R b ⇒ Ia- b I≤1 ⇒ I-(a-b)I<1 ⇒ Ib-aI <1 ⇒  b R a.

⇒  2 R 1 and 1 R \(\frac{1}{2}\).

But, 2 is not related to \(\frac{1}{2}\).  So, R is not transitive.

Question 11. Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔(1 + ab) > 0. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. None of these

Answer: 1. Reflexive and symmetric but not transitive

a R a, since (1 + a²) > 0.

a R b ⇒ (1 + ab) > 0 ⇒ (1 + ba) > 0 ⇒b R a.

Let a = \(\frac{-1}{2}\), b = \(\frac{-1}{2}\) and c = R.

Then, a R b and b R c. But, a is not related to c.

Question 12. Let S be the set of all triangles in a plane and let R be a relation on S defined by Δ₁S Δ₂ ⇔ Δ₁ ≡ Δ₂. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 13. Let S be the set of all real numbers and let R be a relation on S defined by a R b ⇔ a² + b² = 1. Then, R is

1. Symmetric but neither reflexive nor transitive
2. Reflexive but neither symmetric nor transitive
3. Transitive but neither reflexive nor symmetric
4. None of these

Answer: 1. Symmetric but neither reflexive nor transitive

(l² + 1²) ≠ 1. So,1 R 1 is not true.

a R b ⇒ a²+ b² =1 ⇒ b² + a² =1 ⇒  b R a.

1 R 0 and 0 R 1. But,1 is not related to 1.

Question 14. Let JR be a relation on N x N, defined by (a, b) R (c, d)⇔ a+d= b+ c. Then, R is 

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 15. Let A be the set of all points in a plane and let O be the origin. Let R = {(P, Q): OP= OQ}. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 16. Let Q be the set of all rational numbers and * be the binary operation, defined by a * b = a + 2b, then

1. * is commutative but not associative
2. * is associative but not commutative
3. * is neither commutative nor associative
4. * is both commutative and associative

Answer: 3. * is neither commutative nor associative

Question 17. Let a * b = a + ab for all a, b ∈ Q. Then,

1. * Is not a binary composition
2. * Is not commutative
3. * Is commutative but not associative
4. * Is both commutative and associative

Answer: 2. * Is not commutative

Question 18. Let Q⁺ be the set of all positive rationals. Then, the operation * on Q⁺ defined by a * b \(=\frac{a b}{2}\) for all a, b ∈ Q⁺ is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Question 19. Let Z be the set of all integers and let a*b = a – b + ab. Then, * is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 3. Neither commutative nor associative

Question 20. Let Z be the set of all integers. Then, the operation * on Z defined by a*b= a + b – ab is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Question 21. Let Z+be the set of all positive integers. Then, the operation * on Z+ defined by a * b = a^b is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 3. Neither commutative nor associative

Question 22. Define * on Q- {-1} by a * a+ b+ ab. Then, * on Q- {-1} is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Relations And Functions Exercise 1B Review Of Facts And Formulae

1. Function: Let A and B be two non-empty sets. Then, a rule / which associates to each x e A, a unique element fix) of B, is called a function from A to B, and we write, \(f\): A → B.

Dom (f) = {x ∈ A :f(x) ∈ B}, range  (f)= {f(x): x ⇒ A}.

2. Let f: A→ B. Then

⇒ f is one-one ⇔ {f(x₁) =f(x₂) ⇒ x₁ = x₂}

⇒  f is onto ⇔ range (f) = B.

3. Two functions/ and g are said to be equal if they have the same domain and f(x)=g(x) ∀ x.

4. Invertible functions: A function is said to be invertible if/is one-one and

onto.f{x) = y ⇒  f^-1(y) = x.

5. Composite of two functions:

⇒ Let/: A → B and g: B→ C, then g o f: A →  C is called the composite of f and g.

⇒ Dom (g o f) = dom (f)

g o f is defined only when range (f) ⊆ dom (g)

⇒ Dom (f o g) = dom(g).

f o g is defined only when range (g) ⊆ dom (f)

Relations And Functions Exercise 1B Multiple Choice Questions

Question 1. f : N → N : f(x) = 2x is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 2. One-one and into

⇒ 2x =  3

⇒ x =  \(\frac{3}{2}\)

∴ f is into.

Question 2. f : N → N : f(x) = x²+ x+1 is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 2. One-one and into

⇒  f(x₁) = f(x₂)

= x²₁+x₁ +1

= x²₂+x₂ +1

= (x²₁– x²₂) + (x₁-x₂) = 0

⇒ (x₁-x₂)(x₁+x₂+1) = 0

= x₁– x₂= 0

=  x₁ = x₂.

∴ f is one-one.

f(x) =1 ⇒  x²+ x+1 =1 ⇒  x(x+ 1) = 0 ⇒ x= 0 or x=-1.

And, none of 0 and -1 is in N. So, f is into.

Question 3. f: R→ R : f(x) = x² is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 4. Many-one and into

Question 4. f: R→ R : f(x) = x³ is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 1. Many-one and into

Question 5. f : R⁺→+R⁺: f(x) = e^x is

1. Many-one and into
2. Many-one and onto
3. One-one and into
4. One-one and onto

Answer: 4. One-one and onto

⇒ f(x₁) = f(x₂) => e^x₁ = e^x₂

⇒ x₁ = x₂

∴ f is one-one.

For each x ∈ R⁺ ∃ log x ∈ R⁺ s.t. f(log x)= x.

So,/is onto.

Question 6. \(f:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \) f(x) = sin x is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 2. One-one and onto

Question 7. f : R → R :f(x) = cos x is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 3. Many-one and into

Cos (2π- θ) = cos θ ⇒ f is Many-one.

Range (f) = [-1, 1] ⊂ R =>  f is into.

Question 8. f : C→ R : f(z) = IzI is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 3. Many-one and into

i ≠-i. But f(i) = f(-i)=1. So, fis many-one.

-1 ∈ R having no pre-image in C. So, fis into.

Question 9. Let A = R- {3} and B =R- {1}. Then f: A → B : f(x) = \(\frac{(x-2)}{(x-3)}\) is 

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 2. One-one and onto

⇒ f(x₁)= f(x₂)

⇒ \(\frac{\left(x_1-2\right)}{\left(x_1-3\right)}=\frac{\left(x_2-2\right)}{\left(x_3-3\right)}\)

⇒ x₁= x₂, So f is one one.

Let \(\frac{x-2}{x-3}\). Then, x\(=\frac{3 y-2}{y-1}\). Clearly y ≠1 and x≠3

Question 10.  Let f: N → N: f(n)=

\(\frac{1}[2}\)(n+1) when n is odd

\(\frac{n}{2}\),when n is even

1. One-one and into
2. One-one and onto
3. Many-one and onto
4. Many-one and into

Answer: 4. Many-one and into

f (1) = f(2) shows that/is many-one.

If n is odd, then (2n- 1) is odd, and f(2n- 1) =n.

If n is even, then 2n is even, and f (2n) = n.

∴ f is onto.

Question 11. Let A and B be two non-empty sets and let f: (A x B) → (B x A): f(a, b) = (b, a). Then, f is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 1. One-one and onto

Question 12. Let f: Q→ Q: f(x) = (2x + 3). Then,f^-1(y) = ?

1. (2y- 3)
2. \(\\)
3. \(\frac{1}{2}\)(y-3)
4. None of these

Answer: 3. \(\frac{1}{2}\)(y-3)

y = 2x+3

x = \(\frac{1}{2}(y-3)\)

⇒ \(f^{-1}(y)=\frac{4 y}{(4-3 y)}\)

Question 13.\(\text { Let } f: R-\left\{\frac{-4}{3}\right\} \rightarrow R-\left\{\frac{4}{3}\right\}: f(x)=\frac{4 x}{(3 x+4)} \text {. Then, } f^{-1}(y)=\text { ? }\)

1. \(\frac{4 y}{(4-3 y)}\)
2. \(\frac{4 y}{(4+3 y)}\)
3. \(\frac{4 y}{(3 y-4)}\)
4. None of these

Answer: 1. \(\frac{4 y}{(4-3 y)}\)

⇒ \(y\frac{4 x}{3 x+4}\)

⇒ \(x\frac{4 y}{(4-3 y)}\)

⇒ \(f^{-1}(y)=\frac{4 y}{(4-3 y)}\)

Question 14. Let f : N→X : f(x) = 4x² + 12x+ 15. Then f^-1(y) =?

1. \(\frac{1}{2}(\sqrt{y-4}+3)\)
2. \(\frac{1}{2}(\sqrt{y-6}-3)\)
3. \(\frac{1}{2}(\sqrt{y-4}+5)\)
4. None of these

Answer: 2. \(\frac{1}{2}(\sqrt{y-6}-3)\)

y = 4x² +12x+15

= (2x+3)² +6

⇒ \(x\frac{1}{2}(\sqrt{y-6}-3)\)

⇒ \(f^{-1}(y)=\frac{1}{2}(\sqrt{y-6}-3)\)

Question 15. \(\text { If } f(x)=\frac{(4 x+3)}{(6 x-4)}, x \neq \frac{2}{3} ; \text { then }(f \circ f)(x)=\text { ? }\)

1. x
2. (2x-3)
3. \(\frac{4 x-6}{3 x+4}\)
4. None of these

Answer: 1. x

f(x)= \(\frac{4 x+3}{6 x-4}\) = y(say)

Then f(y)= \(\frac{4 y+3}{6 y-4}\)

= \(=\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}\)

= \(=\frac{34 x}{34}\)

= x

⇒ f[f(x)] = x ⇒ (f o f) (x) = x

Question 16. If f(x) = (x² — 1) and g(x) = (2x + 3), then (g o f)(x) = ?

1. (2X²+ 3)
2. (3X² + 2)
3. (2x² + 1)
4. None of these

Answer: 3. (2x²+ 1)

(g o f )(x) = g [f(x)] = g(x²-1)

= 2(x²-1)+3 = (2x²+1)

Question 17. \(\text { If } f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}, \text { then } f(x)=?\)

1. x²
2. x-1
3. x-2
4. None of these

Answer: 3. x-2

Let  \(x+\frac{1}{x}\)= z Then,

f(z) = \(f\left(x+\frac{1}{x}\right)=\left(x^2+\frac{1}{x^2}\right)=\left(x+\frac{1}{x}\right)^2-2\)= (z-2)

f(z)= (x²-2)

Question 18. \(\text { If } f(x)=\frac{1}{(1-x)^{\prime}} \text {, then }(f \circ f \circ f)(x)=\text { ? }\)

1. \(\frac{1}{(1-3 x)}\)
2. \(\frac{x}{(1+3 x)}\)
3. x
4. None of these

Answer: 3. x

(f o f)(x) = f{f(x)}

= \(f\left(\frac{1}{1-x}\right)=\frac{1}{\left(1-\frac{1}{1-x}\right)}\)

= \(\frac{1-x}{-x}=\frac{x-1}{x}\)

{f o(f o f)}(x) ={f o(f o f)(x)}]= \(=f\left(\frac{x-1}{x}\right)=\frac{1}{1-\frac{x-1}{x}}=x^1/3\)

Question 19. \(\text { If } f(x)=\sqrt[3]{3-x^3}, \text { then }(f \circ f)(x)=\text { ? }\)

1. x1/3
2. x
3. (1-x1/3)
4. None of these

Answer: 2. x

(fof)(x) =f{f(x)}- {(3- x3) ^1/3} = f/(y), where y= (3- x³) ^1/3

= (3- y³)^1/3= [3- {3- x³}]^1/3= (x³)^1/3= x

Question 20. If f(x) = x²– 3x + 2, then (f o f)(x) = ?

1. x
2. x⁴-6x³
3. x⁴– 6x³+10x²
4. None of these

Answer: 4. None of these

(fof)(x) =f{f(x)} =f{x²– 3x + 2) =f(y), where y= (x²– 3x+ 2)

= y²– 3y+ 2= (x²– 3x+ 2)²– 3(x²– 3x+ 2) + 2= (x⁴– 6x³ + 10x²– 3x).

Question 21. If f(x) = 8x³ and g(x) = x^1/3, then (g o f)(x) = ?

1. x
2. 2x
3. \(\frac{x}{2}\)
4. 3×2

Answer: 2. 2x

(g of)(x) = g{f(x)] = g(8x³) = (8x³)^1/3= 2x.

Question 22. If f (x) = x², g(x) = tan x and h(x) = log x, then {ho(gof}}\(\left(\sqrt{\frac{\pi}{4}}\right)\)

1. 0
2. 1
3. \(\frac{1}{x}\)
4. \(\frac{1}{2} \log \frac{\pi}{4}\)

Answer: 1. 0

{ho(g o f)}(x) = (ho g){f(x)} = (h o g)(x²)

= h{g(x²)} = h(tan x²) = log(tan x²)

∴ \(\{h \circ(g \circ f)\} \sqrt{\frac{\pi}{4}}=\log \left(\tan \frac{\pi}{4}\right)\)

= log 1

= 0

Question 23. If/= {(1, 2), (3, 5), (4, 1)} andg= {(2, 3), (5, 1), (1, 3)}, then (g o f) = ?

1. {(3,1), (1,3), (3, 4)}
2. {(1,3), (3, 1), (4, 3)}
3. {(3,1), (1,3), (3, 4)}
4. {(2, 5), (5, 2), (1,5)}

Answer: 2. {(1,3), (3, 1), (4, 3)}

⇒ Dom (gof) = Dom (f) = {1, 3, 4}

⇒ (g o f )(l) =g{f(l)} = g(2) = 3,(g of)(3) =g{f(3)} =g(5) =1

⇒ (g o f)(4) = g[f(4)) =g(1) = 3

⇒ g o f= {(1/3), (3, 1), (4, 3)}.

Question 24. Let/(x) = \(\sqrt{9-x^2}\). Then, dom (f) =?

1. [-3, 3]
2. (-∞, -3]
3. [3, ∞)
4. (-∞, -3] u (4, ∞)

Answer: 1. [-3, 3]

f(x) is defined only when 9- x² > 0 => x² <9 =» -3 < x < 3.

∴ Dom (f) =  [-3, 3].

Question 25. Let f(x) \(\sqrt{\frac{x-1}{x-4}}\). Then, dom (f) = ?

1. [1,4)
2. [1,4]
3. (-∞,4]
4. (-∞,1]∪ (4,∞)

Answer: 4. (-∞,1]∪ (4,∞)

f(x) is defined when-4≠ 0 and \(\frac{x-1}{x-4}\)

⇒ x ≠ 4 and (x ≥ 4 or x<l) => (x>4 or x<l)

⇒  Dom (f) = (-∞, 1]u (4, ∞).

Question 26. Let f(x) \(f(x)=e^{\sqrt{x^2-1}}\). Then, dom (f) =?

1. (∞-,1]
2. [-1,∞)
3. (1,∞)
4. (-∞,-1](1,∞)

Answer: 3. (1,∞)

f(x) is defined only when (x²-1) > 0 and (x- 1) > 0

⇒ (x – l)(x+ 1) ≥ 0 and (x- 1) > 0 => x+1 > 0 and x-1 > 0⇒ x >1

∴ dom (f) = (1, ∞).

Question 27. Let f(x) = \(\frac{x}{\left(x^2-1\right)}\) . Then, dom (f) =?

1. R
2. R-1
3. R- {-1}
4. R- {-1,1}

Answer: 4. R- {-1,1}

f(x) is not defined when (x- 1) = 0, i.e., when (x- l)(x+ 1) = 0,

i.e., when x=1 or x= -1.

dom (f) = R—{1, -1}.

Question 28. Let f(x)= \(\). Then ,dom(f) = ?

1. (-1,1)
2. [-1/ 1]
3. [-l, 1]- {0}
4. None of these

Answer: 3. [-l, 1]- {0}

⇒ \(\frac{\sin ^{-1} x}{x}\) is defined only when x≠ 0 and x ∈ [-1,1].

∴ dom (f)= [-1,1]- {0}.

Question 29. Let f(x) = cos^-12x. then, dom (f) =?

1. [-1,1]
2. \(\left[\frac{-1}{2}, \frac{1}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. \(\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]\)

Answer: 2. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

sin^-12x is defined only when 2x ∈ [-1, 1] => x \(\left[\frac{-1}{2}, \frac{1}{2}\right]\)

Question 30. Let f(x) = Cos^-1(3x-1) . Then, dom (f) =?

1. \(\left(0, \frac{2}{3}\right)\)
2. \(\left[0, \frac{2}{3}\right]\)
3. \(\left[\frac{-2}{3}, \frac{2}{3}\right]\)
4. None of these

Answer: 2. \(\left[0, \frac{2}{3}\right]\)

cos^-1 (3x- 1) is defined only when (3x-1) ∈ [-1, 1]

⇒  3x ∈ [0, 2] => x ∈ \(\left[0, \frac{2}{3}\right]\)

⇒ dom(f)= \(\left[0, \frac{2}{3}\right]\)

Question 31. Let f(x)= \(\sqrt{\cos x}\). Then, dom (f) = ?

1. \(\left[0, \frac{\pi}{2}\right]\)
2. \(\left[\frac{3 \pi}{2}, 2 \pi\right]\)
3. \(\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)
4. None of These

Answer: 3. \(\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)

f(x) is defined only when cos x > 0.

⇒ x liesin 1st or 4th quadrant.

⇒  \(=\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)

Question 32. Let f(x) = \(\sqrt{\log \left(2 x-x^2\right)}\). then, dom (f) = ?

1. (0,2)
2. [1,2]
3. (-,1)
4. None of these

Answer: 4. None of these

f (x) is defined only when log (2x- x²) > 0

⇒  (2x- x²) >1 => (1 + x²– 2x) < 0

⇒  (1- x)² < 0

⇒  (1- x) = 0

⇒   x=1.

⇒ dom (f) = {1}.

Question 33. Let f(x) = x². Then, dom (/) and range (/) are respectively

1. R and R
2. R⁺ and R⁺
3. R and R⁺
4. R and R- {0}

Answer: 3. R and R⁺

f(x) = x² is clearly defined for each x ∈ R.

So, dom (f)=R. y= x² x=± \(x= \pm \sqrt{ } y .\)

When y < 0, there is no real value of x. So, y ≥ 0.

∴Range (f) = R.

Question 34. Let f(x) = x³. Then, dom (f) and range (f) are respectively.

1. R and R
2. R⁺ and R⁺
3. R and R⁺
4. R⁺ and R

Answer: 1. R and R

f (x) = x³ is defined for each x ∈ R.

So, dom (f)=R.

For each y∈ R,y^1/3 ∈ R and so x= is real.

∴ range (f) =R.

Question 35. Let f(x)= \(\log (1-x)+\sqrt{x^2-1}\).Then ,dom(f) = ?

1. (1,∞)
2. (-∞,-1)
3. [-1,1]
4. (0,1)

Answer: 2. (-∞,-1)

Let f (x) = g{x) + h(x), where g(x) = log (1- x) and h(x) = \(\sqrt{x^2-1}\)

g(x) is defined only whenl-x>0 => x<l.So, dom (g) = (∞, 1)

h(x) is defined only when x2-1 > 0 => x >1 or x <-1

∴ dom (h) = (-∞, -1]∪[1, ∞)

∴ dom (f) = dom (g)∩ dom(h) = (-∞, -1].

Question 36. Let \(=\frac{1}{\left(1-x^2\right)}\). Then ,range(f)= ?

1. [-∞,1)
2. [1,∞)
3. [-1,1]
4. None of these

Answer: 2. [0,1)

y =  \(\frac{1}{\left(1-x^2\right)}\)

x = \(\sqrt{1-\frac{1}{y}}\)

Clearly, x is not defined when y = 0 or 1\(-\frac{1}{y}\) < 0, i.e., y = 0 or y < 1.

∴ range (f) = [1, ∞).

Question 37. Let f(x)\(=\frac{x^2}{\left(1+x^2\right)}\)

1. [1,∞]
2. [0,1)
3. [-1,1]
4. (0,1)

Answer: 2. [0,1)

y = \(\frac{x^2}{\left(1+x^2\right)} \Rightarrow x=\sqrt{\frac{y}{1-y}}\)

Clearly, x is defined only when \(\frac{y}{(1-y)}\) and(1-y)≠ 0, i.e., when 0 ≤ y ≤ 1.

So, range (f) = [0, 1).

Question 38. The range of f(x) = \(x+\frac{1}{x}\) is

1. [-2,2]
2. [2,∞)
3. (-∞,-2)
4. None of these

Answer: 4. None of these

y\(=\frac{x^2+1}{x}\)

⇒ x²– xy + 1= 0

x= \(\frac{y \pm \sqrt{y^2-4}}{2}\)

Question 39. The range of f(x) = a^x, Where a> 0 is

1. ]-∞,0]
2. ]∞-,0)
3. [0,∞)
4. (0,∞)

Answer: 4. (0,∞)

Clearly, a^x> 0 whatever may be the value of x.

∴ Range (f) = (0, ∞)

Relations And Functions Exercise 1C Results At A Glance

1. sin^-1x = θ => x = sin θ

cos^-1x = θ => x = cos θ

tan^-1x = θ =» x = tan θ

2. Domain and range of inverse functions

 

3. sin (sin^-1x) = x, if-1 ≤ x ≤ 1

cos (cos^-1x) = x, if-1 ≤ x ≤ 1

tan (tan^-1x) = x, if ∞ < x < ∞

4.

sin^-1(sin x) = x, if x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

cos ^-1(cos x) = x, if x ∈ [0, π]

tan^-1(tan x) = x, if x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

5.

sin^-1x = cosec^-1\(\left(\frac{1}{x}\right)\) , x≠0

cos^-1x = sec^-1\(\left(\frac{1}{x}\right)\) , x≠0

tan^-1x= cot^-1\(\left(\frac{1}{x}\right)\) , x≠0

cosec^-1x= sin^-1 \(\left(\frac{1}{x}\right)\),(x ≤ -1 or x ≥ 1)

sec^-1x= cos^-1\(\left(\frac{1}{x}\right)\), when (x ≤1 or x ≥ 1)

cot^-1x = tan^-1\(\left(\frac{1}{x}\right)\), when x > 0

6. sin^-1(-x) = -sin^-1x

cos^-1(-x) = 7i- cos^-1x

tan^-1(-x) = -tan^-1x

cot^-1(-x) =n- cot^-1x

sec^-1(-x) =n- sec^-1x

cosec^-1(-x) = -cosec^-1x

7. sin^-1-x + cos^-1-x =\(\frac{\pi}{2}\) -1 ≤ x ≤1

tan^-1-x + cot^-1-x = \(\frac{\pi}{2}\) , x ≠ 0

sec ^-1x + cosec ^-1x = \(\frac{\pi}{2}\),  (x ≤ -1 or x ≥ 1)

8.

1. For 0 <x< 1, we have

sin^-1-x=  \(\cos ^{-1} \sqrt{1-x^2}\)

= \(\tan ^{-1} \frac{x}{\sqrt{1-x^2}}=\cot ^{-1} \frac{\sqrt{1-x^2}}{x}\)

= \(\sec ^{-1} \frac{1}{\sqrt{1-x^2}}\)

= cosec^-1 \(\frac{1}{x}\)

2. Cos^-1-x= \(\sin ^{-1} \sqrt{1-x^2}\)

= \(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\)

= \(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\)

= \(\sec ^{-1} \frac{1}{x}\)

= cosec^-1 \(\frac{1}{\sqrt{1-x^2}}\)

3. For x>0, We Have

tan^-1x= \(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\)

= \(\cos ^{-1} \frac{1}{\sqrt{1+x^2}}=\cot ^{-1} \frac{1}{x}\)

= \(\sec ^{-1} \sqrt{1+x^2}\)

= cosec^-1 \(\frac{\sqrt{1+x^2}}{x}\)

9. 

2sin^-1x = sin^-1(\(2 x \sqrt{1-x^2}\)) = cos^-1 (1- 2x²)

2cos^-1x = cos^-1(2x²– 1) = sin ^-1\(\left(2 x \sqrt{1-x^2}\right.\))

2tan^-1x= tan^-1\(\left(\frac{2 x}{1-x^2}\right)\)  = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\)

10.

3sin^-1x = sin^-1(3x- 4x³)

3cos^-1x = cos^-1(4x³– 3x)

3tan^-1x = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\)

11.

When x > 0, y > 0 and xy < 1, we have

tan^-1x + tan y^-1 =  tan^-1\(\left(\frac{x+y}{1-x y}\right)\)

When x > 0, y > 0 and xy > 1, we have

tan^-1x + tan y^-1 = π + tan^-1\(\left(\frac{x+y}{1-x y}\right)\)

12.

sin^–1x + sin^–1y = sin^–1(\(x \sqrt{1-y^2}+y \sqrt{1-x^2}\))

sin^–1x – sin^–1y = sin^–1(\(x \sqrt{1-y^2}-y \sqrt{1-x^2}\))

cos^–1x + cos^–1y= cos^–1(\(x y-\sqrt{1-x^2} \cdot \sqrt{1-y^2}\))

cos^–1x – cos^–1y = cos^–1(\(x y+\sqrt{1-x^2} \cdot \sqrt{1-y^2}\))

cot ^–1x+ cot ^–1x= cot^–1\(\left(\frac{x y-1}{x+y}\right)\)

cot ^–1x- cot ^–1y = cot^–1\(\left(\frac{x y+1}{x-y}\right)\)

Relations And Functions Exercise 1C Multiple Choice Questions

Question 1. The principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

1. \(\frac{\pi}{6}\frac{\pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. None of these

Answer: 1. \(\frac{\pi}{6}\frac{\pi}{6}\)

1. Let cos^-1\(\left(\frac{\sqrt{3}}{2}\right)\), where x ∈ [0, π].

Then, cos x=\(\frac{\sqrt{3}}{2}\)= cos\(\frac{\pi}{6}\)

⇒  x= \(\frac{\pi}{6}\)

Question 2. The principal value of cosec^-1(2) is

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{6}\)
3. \(\frac{2 \pi}{3}\)
4. \(\frac{5 \pi}{6}\)

Answer: 2. \(\frac{\pi}{6}\)

Let cosec^-1(2) = x, where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)-{0}

Then, cosec x= 2cosec \(\frac{\pi}{6}\)

⇒ x= \(\frac{\pi}{6}\)

Question 3. The principal value of \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{5 \pi}{4}\)

Answer: 3. \(\frac{3 \pi}{4}\)

Let cos^-1 \(\left(\frac{-1}{\sqrt{2}}\right)\) = x, where x∈ [0,π]

Then,cos x = \(\frac{-1}{\sqrt{2}}\)

=  \(-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)\)

= \(\cos \frac{3 \pi}{4} \Rightarrow x=\frac{3 \pi}{4}\)

Question 4. The principal value of \(\sin ^{-1}\left(\frac{-1}{2}\right)\) is

1. \(\frac{-\pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. None of these

Answer: 1. \(\frac{-\pi}{6}\)

Let sin^-1 \(\left(\frac{-1}{2}\right)\) = x where x∈\(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Then, sin x  = \(\frac1{-1}{2}\)

= \(-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)\)

⇒ x= \(frac{-\pi}{6}\)

Question 5. The principal value of \(\cos ^{-1}\left(\frac{-1}{2}\right)\) is

1. \(\frac{-\pi}{3}\)
2. \(\frac{2 \pi}{3}\)
3. \(\frac{4 \pi}{3}\)
4. \(\frac{\pi}{3}\)

Answer: 2. \(\frac{2 \pi}{3}\)

Let cos^-1 \(\frac{-1}{2}\) = x where x∈[0,π]

Then, cos x  = \(\frac{-1}{2}\)

= \(-\cos \frac{\pi}{3}\)

= \(\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}\)

x = \(\frac{2 \pi}{3}\)

Question 6. The principal value of tan \(\tan ^{-1}(-\sqrt{3})\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{4 \pi}{3}\)
3. \(\frac{-\pi}{3}\)
4. None of these

Answer: 3. \(\frac{-\pi}{3}\)

Let tan^-1\(-\sqrt{3}\)= x, where x ∈\(\left(\frac{-\pi}{2},\frac{\pi}{2}\right)\)

Then, tan x= \(-\sqrt{3}\)

= \(-\tan \frac{\pi}{3}=\tan \left(\frac{-\pi}{3}\right)\)

⇒ \(x=\frac{-\pi}{3}\)

Question 7. The principal value of cot^-1(-1) is

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{5 \pi}{4}\)
4. \(\frac{3 \pi}{4}\)

Answer: 4. \(\frac{3 \pi}{4}\)

Let co^-1(-1) = x, where x ∈ [0, π].

Then, cot x=-1

= \(-\cot \frac{\pi}{4}=\cot \left(\pi-\frac{\pi}{4}\right)\)

= \(\cot \frac{3 \pi}{4}\)

⇒ x= \(\frac{-\pi}{3}\)

Question 9. The principal value of cosec^-1 \((-\sqrt{2})\) is

1. \(\frac{-\pi}{4}\)
2. \(\frac{3 \pi}{4}\)
3. \(\frac{5 \pi}{4}\)
4. None of these

Answer: 1. \(\frac{-\pi}{4}\)

Let cosec\((-\sqrt{2})\)= x, where x ∈\(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)– {0}

Then , cosec x \(-\sqrt{2}\)

= cosec \(\frac{-\pi}{4}\)

= cosec \(\frac{-\pi}{4}\)

x = \(\frac{-\pi}{4}\)

Question 10. The principal value of \(\cot ^{-1}(-\sqrt{3})\)

1. \(\frac{-\pi}{6}\)
2. \(\frac{-\pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. \(\frac{5 \pi}{6}\)

Answer: 4. \(\frac{5 \pi}{6}\)

Let cot\((-\sqrt{3})\)= x, where x [0,π]

Then , cot x \(-\sqrt{3}\)

= -cot \(\frac{\pi}{6}\)

\(=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6}\)

x= \(\frac{5 \pi}{6}\)

Question 11. The value of \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{5 \pi}{3}\)
3. \(\frac{\pi}{3}\)
4. None of these

Answer: 3. \(\frac{\pi}{3}\)

Let sin^-1\(\left(\sin \frac{2 \pi}{3}\right)\) = x , where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Then,sin x = \(\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}\)

x= \(\frac{\pi}{3}\)

Question 12. The value of \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)

1. \(\frac{13 \pi}{6}\)
2. \(\frac{7 \pi}{6}\)
3. \(\frac{5 \pi}{6}\)
4. \(\frac{\pi}{6}\)

Answer: 4. \(\frac{\pi}{6}\)

Let cos \(\left(\cos \frac{13 \pi}{6}\right)\)= x, where x [0,]

Then,cos x = \(\cos \frac{13 \pi}{6}=\cos \left(2 \pi+\frac{\pi}{6}\right)\)

x= \(\frac{\pi}{6}\)

Question 13. The value of \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)

1. \(\frac{7 \pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{\pi}{6}\)
4. None of these

Answer: 3. \(\frac{\pi}{6}\)

Let tan^-1\(\left(\tan \frac{7 \pi}{6}\right)\) = x,where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Then, tan x = tan \(\left(\tan \frac{7 \pi}{6}\right)\)

= tan \(\left(\pi+\frac{\pi}{6}\right)=\tan \frac{\pi}{6}\)

x=  \(\frac{\pi}{6}\)

Question 14. The value of \(\cot ^{-1}\left(\cot \frac{5 \pi}{4}\right)\)

1. \(\frac{\pi}{4}\)
2. \(\frac{-\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. None of these

Answer: 1. \(\frac{\pi}{4}\)

Let cos^-1\(\left(\cot \frac{5 \pi}{4}\right)\) = x, where x [0,π]

Then ,cot x = cot \(\frac{5 \pi}{4}\)

= \(\cot \frac{5 \pi}{4}=\cot \left(\pi+\frac{\pi}{4}\right)\)

x\(=\frac{\pi}{4}\)

Question 15. The value of \(\sec ^{-1}\left(\sec \frac{8 \pi}{5}\right)\)

1. \(\frac{2 \pi}{5}\)
2. \(\frac{3 \pi}{5}\)
3. \(\frac{8 \pi}{5}\)
4. None of these

Answer: 1. \(\frac{2 \pi}{5}\)

Let sec^-1\(\left(\sec \frac{8 \pi}{5}\right)\)= x where x∈[0,π]- \(\left\{\frac{\pi}{2}\right\}\)

Then, sec x \(\frac{8 \pi}{5}\)

= sec \(\sec \left(2 \pi-\frac{2 \pi}{5}\right)=\sec \frac{2 \pi}{5}\)

x = \(\frac{2 \pi}{5}\)

Question 16. The value of cosec^-1 (cosec\(\frac{4 \pi}{3}\)) is

1. \(\frac{\pi}{3}\)
2. \(\frac{-\pi}{3}\)
3. \(\frac{2 \pi}{3}\)
4. None of these

Answer: 2. \(\frac{-\pi}{3}\)

Let cosec^-1 \(\frac{4 \pi}{3}\)= x where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Then,cosec x= cosec \(\frac{4 \pi}{3}\)

= cosec\(\left(\pi+\frac{\pi}{3}\right)\)

=- cosec \(\frac{\pi}{3}\)

=cosec \(\frac{-\pi}{3}\)

x =\(\frac{-\pi}{3}\)

Question 17. The value of \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)

1. \(\frac{3 \pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{-\pi}{4}\)
4. None of these

Answer: 3. \(\frac{-\pi}{4}\)

Let tan^-1\(\left(\tan \frac{3 \pi}{4}\right)\)= x, where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Then,tan x= \(\tan \frac{3 \pi}{4}\)

= tan\(\left(\pi-\frac{\pi}{4}\right)=\tan \left(\frac{-\pi}{4}\right)\)

x = \(\frac{-\pi}{4}\)

Question 18. \(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\)

1. 0
2. \(\frac{2 \pi}{3}\)
3. \(\frac{\pi}{2}\)
4. π

Answer: 3. \(\frac{\pi}{2}\)

Let ,sin^-1 x= \(\left(\frac{-1}{2}\right)\)= x, where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Then , sin x = \(-\frac{1}{2}\)

-sin = \(\frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)\)

x= \(\frac{-\pi}{6}\)

Given example = \(\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{6}\right)Question \)

= \(\frac{3 \pi}{6}=\frac{\pi}{2}\)

Question 19. The value of sin \(\left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)\)

1. 0
2. 1
3. -1
4. None of these

Answer: 2. 1

⇒ \(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}=\frac{\pi}{2}\)

⇒ \(\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)=\sin \frac{\pi}{2}\)

= 1

20. If x ≠ 0, then cos (tan^-1x+ cot^-1x)

1. -1
2. 1
3. 0
4. None of these

Answer: 3. 0

⇒ \(\cos \left(\tan ^{-1}+\cot ^{-1} x\right)\)

= \(\cos \frac{\pi}{2}\)

= 0

Question 21. The value of sin \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)

1. \(\frac{2}{5}\)
2. \(\frac{4}{5}\)
3. \(\frac{-2}{5}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

cos^-1 x \(=\sin ^{-1} \sqrt{1-x^2}\)

= \(\cos ^{-1} \frac{3}{5}=\sin ^{-1} \sqrt{1-\frac{9}{25}}\)

= \(\sin ^{-1} \frac{4}{5}\)

∴  \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)

Question 22. \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

1. \(\frac{4 \pi}{3}\)
2. \(\frac{\pi}{2}\)
3. \(\frac{3 \pi}{4}\)
4. π

Answer: 4. \(\frac{3 \pi}{4}\)

cos^-1 \(\left(\cos \frac{2 \pi}{3}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

= \(=\cos ^{-1}\left\{\cos \left(\pi-\frac{\pi}{3}\right)\right\}+\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}\)

= \(\cos ^{-1}\left(-\cos \frac{\pi}{3}\right)+\sin ^{-1}\left\{\sin \frac{\pi}{3}\right\}=\cos ^{-1}\left(\frac{-1}{2}\right)+\frac{\pi}{3}\)

= \(\left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)=\pi\)

Question 23. \(\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)\)

1. \(\frac{\pi}{3}\)
2. Question \(\frac{-\pi}{3}\)
3. \(\frac{5 \pi}{3}\)
4. None of these

Answer: 2. \(\frac{-\pi}{3}\)

⇒ \(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)\)

= \(\tan ^{-1} \sqrt{3}-\left(\pi-\sec ^{-1} 2\right)\)

= \(\frac{\pi}{3}-\pi+\frac{\pi}{3}=\frac{-\pi}{3}\)

24. \(\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{3 \pi}{2}\)
3. 2π
4. None of these

Answer: 1. \(\frac{2 \pi}{3}\)

= \(\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}\)

= \(\frac{\pi}{3}+\left(2 \times \frac{\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\)

= \(\frac{2 \pi}{3}\)

Question 25. \(\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)\)

1. π
2. \(\frac{2 \pi}{3}\frac{2 \pi}{3}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{\pi}{2}\)

Answer: 3. \(\frac{3 \pi}{4}\)

⇒ \(\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)\)

= \(\frac{\pi}{4}+\left(\pi-\cos ^{-1} \frac{1}{2}\right)-\sin ^{-1} \frac{1}{2}\)

= \(\left(\frac{\pi}{4}+\pi-\frac{\pi}{3}-\frac{\pi}{6}\right)\)

= \(\frac{3 \pi}{4}\)

Question 26. tan[2 tan^-1\(\frac{1}{5}-\frac{\pi}{4}\)]= ?

1. \(\frac{7}{17}\)
2. \(\frac{-7}{17}\)
3. \(\frac{7}{12}\)
4. \(\frac{-7}{12}\)

Answer: 2. \(\frac{-7}{17}\)

Let 2\(\tan ^{-1} \frac{1}{5}=\theta\) .

Then \(\tan ^{-1} \frac{1}{5}=\frac{1}{2} \theta \Rightarrow \tan \frac{1}{2} \theta=\frac{1}{5}\)

∴ \(\tan \theta=\frac{2 \tan \frac{1}{2} \theta}{1-\tan ^2 \frac{1}{2} \theta}=\frac{\left(2 \times \frac{1}{5}\right)}{\left(1-\frac{1}{25}\right)}\)

= \(\left(\frac{2}{5} \times \frac{25}{24}\right)=\frac{5}{12}\)

∴ \(\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]=\tan \left(\theta-\frac{\pi}{4}\right)=\frac{\tan \theta-\tan \frac{\pi}{4}}{1+\tan \theta \cdot \tan \frac{\pi}{4}}\)

= \(\frac{\left(\frac{5}{12}-1\right)}{\left(1+\frac{5}{12} \times 1\right)}=\frac{\left(\frac{-7}{12}\right)}{\left(\frac{17}{12}\right)}=\frac{-7}{17}\)

Question 27. \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)\)= ?

1. \(\frac{(3-\sqrt{5})}{2}\)
2. \(\frac{(3+\sqrt{5})}{2}\)
3. \(\frac{(5-\sqrt{3})}{2}\)
4. \(\frac{(5+\sqrt{3})}{2}\)

Answer: 1. \(\frac{(3+\sqrt{5})}{2}\)

Let \(\cos ^{-1} \frac{\sqrt{5}}{3}=\theta\).Then,cos \(\cos \theta=\frac{\sqrt{5}}{3}\)

⇒ \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan \frac{1}{2} \theta=\frac{\sin (\theta / 2)}{\cos (\theta / 2)}\)

⇒ \(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\left\{\frac{\left(1-\frac{\sqrt{5}}{3}\right)}{\left(1+\frac{\sqrt{5}}{3}\right)}\right\}^{1 / 2}\)

= \(\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)^{1 / 2}=\left\{3-\frac{\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3+\sqrt{5}}\right\}^{1 / 2}\)

= \(\frac{(3-\sqrt{5})}{2}\)

Question 28. \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)

1. \(\frac{3}{4}\)
2. \(\frac{4}{5}\)
3. \(\frac{3}{5}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

Let cos^-1 \(\frac{3}{5}\)= x ∈ where [0,π] .Then ,cos x= \(\frac{3}{5}\)

∴ Since x[0,π],sinx > 0

∴ Since x\(\sqrt{1-\frac{9}{25}}\)

= \(\sqrt{\frac{16}{25}}=\frac{4}{5}\)

⇒ \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\frac{4}{5}\)

Question 29. \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)

1. \(\frac{3}{5}\)
2. \(\frac{4}{5}\)
3. \(\frac{4}{9}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

Let \(\tan ^{-1} \frac{3}{4}\) = x where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

∴tan x= \(\frac{3}{4}\) and since x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\),we have cos x>0

∴cos x = \(\frac{1}{\sec x}=\frac{1}{\sqrt{1+\tan ^2 x}}\)

= \(\frac{1}{\sqrt{1+\frac{9}{16}}}=\frac{4}{5}\)

= \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)

cos x=\(\frac{4}{5}\)

Question 30. \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\)

1. 1
2. 0
3. \(\frac{-1}{2}\)
4. None of these

Answer: 1. 1

⇒ \(\sin \left\{\frac{\pi}{3}-\sin ^{-3}\left(\frac{-1}{2}\right)\right\}=\sin \left\{\frac{\pi}{3}+\sin ^{-1} \frac{1}{2}\right\}\)

sin = \( \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \frac{\pi}{2}\)

= 1

Question 31. \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)\)

1. \(\frac{1}{\sqrt{5}}\)
2. \(\frac{2}{\sqrt{5}}\)
3. \(\frac{1}{\sqrt{10}}\)
4. \(\frac{2}{\sqrt{10}}\)

Answer: 3. \(\frac{1}{\sqrt{10}}\)

Let \(\cos ^{-1} \frac{4}{5}=\) = x where x∈ [0,π].

Then cos x= \(\frac{4}{5}\)

Since x∈ [0,π] ⇒ \(\frac{1}{2} x \in\left[0, \frac{\pi}{2}\right] \Rightarrow \sin \frac{1}{2} x>0\)

sin = \( \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \frac{1}{2} x\)

= \(\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{\left(1-\frac{4}{5}\right)}{2}}=\)

= \(\frac{1}{\sqrt{10}}\)

Question 32. \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\)

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\)

= \(\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}\)

=  \(\tan ^{-1}\left\{2 \times \frac{1}{2}\right\}\)

tan^-1 1= \(\frac{\pi}{4}\)

Question 33. \(\text { If } \cot ^{-1}\left(\frac{-1}{5}\right)=x, \text { then } \sin x=?\)

1. \(\frac{1}{\sqrt{26}}\)
2. \(\frac{5}{\sqrt{26}}\)
3. \(\frac{1}{\sqrt{24}}\)
4. None of these

Answer: 2. \(\frac{5}{\sqrt{26}}\)

cot^-1\(\left(\frac{-1}{5}\right)\)

cot x= \(\left(\frac{-1}{5}\right)\),where x∈ (0,π)

sin x>0 in (0,π)

sin x= 1/ cosecx

= \(\frac{1}{\sqrt{1+\cot ^2 x}}\)

⇒\(\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}}\)

Question 34. \(\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=?\)

1. \(\frac{\pi}{2}\)
2. π
3. \(\frac{3 \pi}{2}\)
4. None of these

Answer: 3. \(\frac{3 \pi}{2}\)

Range of \(\cos ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

cos^-1= \(\left(\frac{-\sqrt{3}}{2}\right)\)

Question x = \(\frac{-\sqrt{3}}{2}\)

– cos = \(\frac{\pi}{6}=\cos \left(\pi-\frac{\pi}{6}\right)=\cos \frac{5 \pi}{6}\)

x= \(\frac{5 \pi}{6}\)

sin^-1= \(\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)=

= \(-\sin ^{-1} \frac{1}{2}+2 \times \frac{5 \pi}{6}\)

= \(-\frac{\pi}{6}+\frac{5 \pi}{3}=\frac{4 \pi}{6}\)

=\(\frac{3 \pi}{2}\)

35. \(\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{ } 2}\right)=?\)

1. \(\frac{\pi}{2}\)
2. π
3. \(\frac{3 \pi}{2}\)
4. \(\frac{2 \pi}{2}\)

Answer: 1. \(\frac{\pi}{2}\)

Range of \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

⇒ \(\tan ^{-1}(-1)=x \Rightarrow \tan x=-1\)

= – \(\tan \frac{\pi}{4}=\tan \left(\frac{-\pi}{4}\right) \Rightarrow\)

= \(x\frac{-\pi}{4}\)

The range of cos^-1 is [0,π]

= \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) = y

cos y = \(\frac{-1}{\sqrt{2}}\)

= \(-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}\)

y = \(\frac{3 \pi}{4}\)

Given example = \(-\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{2 \pi}{4}=\frac{\pi}{2}\)

Question 36. \(\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=?\)

1. 1
2. \(\frac{1}{2}\)
3. 0
4. None of these

Answer: 3. 0

Given example = \(\cot \frac{\pi}{2}\)= 0

Question 37. \(\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=?\)

1. \(\tan ^{-1} \frac{4}{3}\)
2. \(\tan ^{-1} \frac{2}{3}\)
3. \(\tan ^{-1} 2\)
4. \(\tan ^{-1} 3\)

Answer: 3. \(\tan ^{-1} 2\)

⇒ \(\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}\)

\(=\tan ^{-1}\left\{\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\right\}\)

= \(\tan ^{-1} 2\)

Question 38. \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=?\)

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{2}\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{4}\)

⇒ \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}\)

tan^-1\(\left\{\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right\}\)

tan^-1\(=\frac{\pi}{4}\)

Question 39. \(2 \tan ^{-1} \frac{1}{3}=?\)

1. \(\tan ^{-1} \frac{3}{2}\)
2. \(\tan ^{-1} \frac{3}{4}\)
3. \(\tan ^{-1} \frac{4}{3}\)
4. None of these

Answer: 2. \(\tan ^{-1} \frac{3}{4}\)

Use \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)

Question 40. \(\cos \left(2 \tan ^{-1} \frac{1}{2}\right)=?\)

1. \(\frac{3}{5}\)
2. \(\frac{4\frac{3}{5}[}{5}\)
3. \(\frac{7}{8}\)
4. [None of these

Answer: 1. \(\frac{3}{5}\)

Use \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)

Question 41. \(\sin \left[2 \sin ^{-1} \frac{4}{5}\right]\)

1. \(\frac{12}{25}\)
2. \(\frac{80}{89}\)
3. \(\frac{75}{128}\)
4. None of these

Answer: 2. \(\frac{80}{89}\)

Use \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)

Question 42. \(\sin \left[2 \sin ^{-1} \frac{4}{5}\right]\)

1. \(\frac{2}{25}\)
2. \(\frac{16}{25}\)
3. \(\frac{24}{25}\)
4. None of these

Answer: 3. \(\frac{16}{25}\)

Use \(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right]\)

Question 43. \(\text { If } \tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3} \text {, then } x=?\)

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{6}\)
4. None of these

Answer: 1. \(\frac{1}{2}\)

\(\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}=\tan ^{-1} 1-\tan ^1 \frac{1}{3}\) \(\tan ^{-1}\left\{\frac{\left(1-\frac{1}{3}\right)}{\left(1+\frac{1}{3}\right)}\right\}=\tan ^{-1} \frac{1}{2}\)

= \(\frac{1}{2}\)

Question 44. \(\text { If } \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \text {, then } x=\text { ? }\)

1. 1
2. -1
3. 0
4. \(\frac{1}{2}\)

Answer: 3. 0

We know that \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

= \(\tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\frac{\pi}{2}\)

= \(\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}\)

(1-x)= \(\frac{1}{(1+x)}\)(1-x) = 1

x= 0

Question 45. \(\text { If } \sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}, \text { then }\left(\cos ^{-1} x+\cos ^{-1} y\right)=?\)

1. \(\frac{\pi}{6}\)
2. \(\frac{\pi}{3}\)
3. \(\pi\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{3}\)

⇒ \(\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}\)

∴ \(\left(\frac{\pi}{2}-\cos ^{-1} x\right)+\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\frac{2 \pi}{3}\)

⇒ \(\cos ^{-1} x+\cos ^{-1} y=\left(\pi-\frac{2 \pi}{3}\right)=\frac{\pi}{3}\)

Question 46. \(\left(\tan ^{-1} 2+\tan ^{-1} 3\right)=?\)

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\pi\)

Answer: 3. \(\frac{3 \pi}{4}\)

( x = 2,y = 3) = xy>1

∴  \(\pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)=\pi+\tan ^{-1}(-1)\)

= \(\pi-\tan (1)=\left(\pi-\frac{\pi}{4}\right)\)

= \(\frac{3 \pi}{4}\)

Question 47. \(\text { If } \tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8, \text { then } x=\text { ? }\)

1. \(\frac{1}{3}\)
2. \(\frac{1}{5}\)
3. 3
4. 5

Answer: 2. \(\frac{1}{5}\)

⇒ \(\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8\)

= \(\frac{3+x}{1-3 x}\)= 8

3+x = 8-24x

x= \(\frac{1}{5}\)

Question 48. \(\text { If } \tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4} \text {, then } x=\text { ? }\)

1. \(\frac{1}{2} \text { or }-2\)
2. \(\frac{1}{3} \text { or }-3\)
3. \(\frac{1}{4} \text { or }-2\)
4. \(\frac{1}{6} \text { or }-1\)

Answer: 4. \(\frac{1}{6} \text { or }-1\)

⇒ \(\tan ^{-1}\left(\frac{3 x+2 x}{1-6 x^2}\right)=\frac{\pi}{4}\)

∴ \(\frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}=1\)

= \(\frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}\)= 1

⇒ 6x²+5x-1 = 0

(x+1)(6x-1)= 0

x= -1 or x= \(\frac{1}{6}\)

Question 49. \(\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?\)

1. \(\frac{13}{6}\)
2. \(\frac{17}{6}\)
3. \(\frac{19}{6}\)
4. \(\frac{23}{6}\)

Answer: 2. \(\frac{17}{6}\)

cos^-1 x= \(\frac{\sqrt{1-x^2}}{x}\)

cos^-1\(\frac{4}{5}\) = tan^-1\(\frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}\)

tan^-1 = \(\frac{3}{4}\)

∴ \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\)

tan^-1 = \(\frac{\left(\frac{3}{4}+\frac{2}{3}\right)}{\left(1-\frac{3}{4} \times \frac{2}{3}\right)}\)

tan^-1 = \(\frac{17}{6}\)

\(\cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}=\tan ^{-1} \frac{3}{4 \)

∴ Given example = \(\tan \left\{\tan ^{-1} \frac{17}{6}\right\}=\frac{17}{6}\)

Question 50. cot^-1 + cosec^-1\(\frac{\sqrt{41}}{4}\)

1. \(\frac{\pi}{6}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{3}\)
4. \(\frac{3 \pi}{4}\)

Answer: 2. \(\frac{\pi}{4}\)

cosec x^-1 = \(\cot ^{-1} \sqrt{x^2-1}\)

cosec x^-1 = \(\frac{\sqrt{41}}{4}\)

cot^-1 = \(\frac{\sqrt{41}}{4}\) -1

cot^-1 = \(\frac{5}{4}\)

= \(\cot ^{-1} 9+\cot ^{-1} \frac{5}{4}\)

= \(\tan ^{-1} \frac{1}{9}+\tan ^{-1} \frac{4}{5}\)

Question 51. Range of sin^-1x is

1. \(\left[0, \frac{\pi}{2}\right]\)
2. [0, π]
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

Answer: 3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Question 52. The range of cos^-1x is

1. [0, π]
2. \(\left[0, \frac{\pi}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

 Answer: 1. [0, π]

Question 53. The range of tan^-1x is

1. \(\left(0, \frac{\pi}{2}\right)\)
2. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

Answer: 2. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Question 54. The range of sec^-1x is

1. \(\left[0, \frac{\pi}{2}\right]\)
2. [0, π]
3. \([0, \pi]-\left\{\frac{\pi}{2}\right\}\)
4. None of these

Answer: 3. \([0, \pi]-\left\{\frac{\pi}{2}\right\}\)

Question 55. Range of cosec^-1x is

1. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
2. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
4. None of these

Answer: 3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Question 56. Domain of cos^-1x is

1. [0,1]
2. [-1,1]
3. [-1,0]
4. None of these

Answer: 2. [-1,1]

Question 57. Domain of sec^-1 x is

1. [-1,1]
2. R- {0}
3. R- [-1,1]
4. R- {-1,1}

Answer: 3. R- [-1,1]