Class 11th Maths – RD Sharma Chapter 3 : Functions Multiple Choice Questions (MCQs) Solutions
Mark the correct alternative in each of the following:
-
- Let A = {1, 2, 3}, B = {2, 3, 4}, then which of the following is a function from A to B?
(a) {(1, 2), (1, 3), (2, 3), (3, 3)}
(b) {(1, 3), (2, 4)}
(c) {(1, 3), (2, 2), (3, 3)}
(d) {(1, 2), (2, 3), (3, 2), (3, 4)} Watch Solution - If f : Q → Q is defined as f(x) = x², then f⁻¹(9) is equal to
(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ Watch Solution - Which one of the following is not a function?
(a) {(x, y) : x, y ∈ R, x² = y}
(b) {(x, y) : x, y ∈ R, y² = x}
(c) {(x, y) : x, y ∈ R, x = y³}
(d) {(x, y) : x, y ∈ R, y = x³} Watch Solution - If f(x) = cos (log x), then f(x²) f(y²) − 1/2 { f(x²/y²) + f(x² y²) } has the value
(a) −2
(b) −1
(c) 1/2
(d) none of these Watch Solution - If f(x) = cos (log x), then f(x) f(y) − 1/2 { f(x/y) + f(xy) } has the value
(a) −1
(b) 1/2
(c) −2
(d) none of these Watch Solution - Let f(x) = |x − 1|. Then,
(a) f(x²) = [f(x)]²
(b) f(x + y) = f(x) f(y)
(c) f(|x|) = |f(x)|
(d) none of these Watch Solution - The range of f(x) = cos [x], for − π/2 < x < π/2 is
(a) {−1, 1, 0}
(b) {cos1, cos2, 1}
(c) {cos1, − cos1, 1}
(d) [−1, 1] Watch Solution - Which of the following are functions?
(a) {(x, y) : y² = x, x, y ∈ R}
(b) {(x, y) : y = |x|, x, y ∈ R}
(c) {(x, y) : x² + y² = 1, x, y ∈ R}
(d) {(x, y) : x² − y² = 1, x, y ∈ R} Watch Solution - If f(x) = log ((1 + x)/(1 − x)) and g(x) = (3x + x³)/(1 + 3x²), then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}³
(c) 3f(x)
(d) − f(x) Watch Solution - If A = {1, 2, 3}, B = {x, y}, then the number of functions that can be defined from A into B is
(a) 12
(b) 8
(c) 6
(d) 3 Watch Solution - If f(x) = log ((1 + x)/(1 − x)) , then f (2x/(1 + x²)) is equal to
(a) {f(x)}²
(b) {f(x)}³
(c) 2f(x)
(d) 3f(x) Watch Solution - If f(x) = cos (log x), then value of f(x) f(4) − 1/2 { f(x/4) + f(4x) } is
(a) 1
(b) −1
(c) 0
(d) ±1 Watch Solution - If f(x) = (2^x + 2^−x)/2 , then f(x + y) f(x − y) is equals to
(a) 1/2 {f(2x) + f(2y)}
(b) 1/2 {f(2x) − f(2y)}
(c) 1/4 {f(2x) + f(2y)}
(d) 1/4 {f(2x) − f(2y)} Watch Solution - If 2f(x) − 3f(1/x) = x² (x ≠ 0), then f(2) is equal to
(a) −7/4
(b) 5/2
(c) −1
(d) none of these Watch Solution - Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(−x) − f(x) =
(a) 2x
(b) 2|x|
(c) −2x
(d) −2|x| Watch Solution - The range of the function f(x) = (x² − x)/(x² + 2x) is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) none of these Watch Solution - If x ≠ 1 and f(x) = (x + 1)/(x − 1) is a real function, then f(f(f(2))) is
(a) 1
(b) 2
(c) 3
(d) 4 Watch Solution - If f(x) = cos (logₑ x), then f(1/x) f(1/y) − 1/2 { f(xy) + f(x/y) } is equal to
(a) cos (x − y)
(b) log (cos (x − y))
(c) 1
(d) cos (x + y) Watch Solution - Let f(x) = x, g(x) = 1/x and h(x) = f(x) g(x). Then, h(x) = 1 for
(a) x ∈ R
(b) x ∈ Q
(c) x ∈ R − Q
(d) x ∈ R, x ≠ 0 Watch Solution - If f(x) = (sin⁴x + cos²x)/(sin²x + cos⁴x) for x ∈ R, then f(2002) =
(a) 1
(b) 2
(c) 3
(d) 4 Watch Solution - The function f : R → R is defined by f(x) = cos²x + sin⁴x. Then, f(R) =
(a) [3/4, 1)
(b) (3/4, 1] (c) [3/4, 1] (d) (3/4, 1) Watch Solution - Let A = {x ∈ R : x ≠ 0, −4 ≤ x ≤ 4} and f : A → R be defined by f(x) = |x|/x for x ∈ A. Then A is
(a) {1, −1}
(b) {x : 0 ≤ x ≤ 4}
(c) {1}
(d) {x : −4 ≤ x ≤ 0} Watch Solution - If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x² + 7, then the values of x such that g(f(x)) = 8 are
(a) 1, 2
(b) −1, 2
(c) −1, −2
(d) 1, −2 Watch Solution - If f : [−2, 2] → R is defined by
f(x) = { −1, for −2 ≤ x ≤ 0
x − 1, for 0 ≤ x ≤ 2 }
then {x ∈ [−2, 2] : x ≤ 0 and f(|x|) = x} =
(a) {−1}
(b) {0}
(c) {−1/2}
(d) ϕ Watch Solution - If e^f(x) = (10 + x)/(10 − x), x ∈ (−10, 10) and f(x) = k f(200x/(100 + x²)), then k =
(a) 0.5
(b) 0.6
(c) 0.7
(d) 0.8 Watch Solution - If f is a real valued function given by f(x) = 27x³ + 1/x³ and α, β are roots of 3x + 1/x = 12. Then,
(a) f(α) ≠ f(β)
(b) f(α) = 10
(c) f(β) = −10
(d) none of these Watch Solution - If f(x) = 64x³ + 1/x³ and α, β are the roots of 4x + 1/x = 3. Then,
(a) f(α) = f(β) = −9
(b) f(α) = f(β) = 63
(c) f(α) ≠ f(β)
(d) none of these Watch Solution - If 3f(x) + 5f(1/x) = 1/x − 3 for all non-zero x, then f(x) =
(a) 1/14 (3/x + 5x − 6)
(b) 1/14 (−3/x + 5x − 6)
(c) 1/14 (−3/x + 5x + 6)
(d) none of these Watch Solution - If f : R → R be given by f(x) = 4^x /(4^x + 2) for all x ∈ R. Then,
(a) f(x) = f(1 − x)
(b) f(x) + f(1 − x) = 0
(c) f(x) + f(1 − x) = 1
(d) f(x) + f(x − 1) = 1 Watch Solution - If f(x) = sin [π²] x + sin [− π²] x, where [x] denotes the greatest integer less than or equal to x, then
(a) f(π/2) = 1
(b) f(π) = 2
(c) f(π/4) = −1
(d) none of these Watch Solution - The domain of the function f(x) = √(2 − 2x − x²) is
(a) [−√3, √3] (b) [−1 − √3, −1 + √3] (c) [−2, 2] (d) [−2 − √3, −2 + √3] - The domain of definition of f(x) = √((x + 3)/((2 − x)(x − 5))) is
(a) (−∞, −3] ∪ (2, 5)
(b) (−∞, −3) ∪ (2, 5)
(c) (−∞, −3] ∪ [2, 5] (d) none of these - The domain of the function f(x) = √((x + 1)(x − 3)/(x − 2)) is
(a) [−1, 2) ∪ [3, ∞)
(b) (−1, 2) ∪ [3, ∞)
(c) [−1, 2] ∪ [3, ∞)
(d) none of these - The domain of definition of the function f(x) = √(x − 1) + √(3 − x) is
(a) [1, ∞)
(b) (−∞, 3)
(c) (1, 3)
(d) [1, 3] - The domain of definition of the function f(x) = √((x − 2)/(x + 2)) + √((1 − x)/(1 + x)) is
(a) (−∞, −2] ∪ [2, ∞)
(b) [−1, 1] (c) ϕ
(d) none of these - The domain of definition of the function f(x) = log |x| is
(a) R
(b) (−∞, 0)
(c) (0, ∞)
(d) R − {0} - The domain of definition of f(x) = √(4x − x²) is
(a) R − [0, 4] (b) R − (0, 4)
(c) (0, 4)
(d) [0, 4] - The domain of definition of f(x) = √(x − 3 − 2√(x − 4)) − √(x − 3 + 2√(x − 4)) is
(a) [4, ∞)
(b) (−∞, 4] (c) (4, ∞)
(d) (−∞, 4) - The domain of the function f(x) = √(5|x| − x² − 6) is
(a) (−3, −2) ∪ (2, 3)
(b) [−3, −2] ∪ [2, 3)
(c) [−3, −2] ∪ [2, 3] (d) none of these - The range of the function f(x) = x/|x| is
(a) R − {0}
(b) R − {−1, 1}
(c) {−1, 1}
(d) none of these - The range of the function f(x) = (x + 2)/|x + 2| , x ≠ −2 is
(a) {−1, 1}
(b) {−1, 0, 1}
(c) {1}
(d) (0, ∞) - The range of the function f(x) = |x − 1| is
(a) (−∞, 0)
(b) [0, ∞)
(c) (0, ∞)
(d) R - Let f(x) = √(x² + 1). Then, which of the following is correct?
(a) f(xy) = f(x)f(y)
(b) f(xy) ≥ f(x)f(y)
(c) f(xy) ≤ f(x)f(y)
(d) none of these - If [x]² − 5[x] + 6 = 0, where [.] denotes the greatest integer function, then
(a) x ∈ [3, 4] (b) x ∈ (2, 3] (c) x ∈ [2, 3] (d) x ∈ [2, 4) - The range of f(x) = 1/(1 − 2 cos x) is
(a) [1/3, 1] (b) [−1, 1/3] (c) (−∞, −1] ∪ [1/3, ∞)
(d) [−1/3, 1] - The domain of the function f(x) √(4 − x) + 1/√(x² − 1) is equal to
(a) (−∞, −1) ∪ (1, 4)
(b) (−∞, −1] ∪ (1, 4] (c) (−∞, −1) ∪ [1, 4] (d) (−∞, −1) ∪ [1, 4) - Domain of f(x) = √(a² − x²), a > 0 is
(a) (−a, a)
(b) [−a, a] (c) [0, a] (d) (−a, 0] - If f(x) = ax + b, where a and b are integers, f(−1) = −5 and f(x) = 3, then a and b are equal
(a) a = −3, b = −1
(b) a = 2, b = −3
(c) a = 0, b = 2
(d) a = 2, b = 3 - The domain and range of the real function defined by f(x) = (4 − x)/(x − 4) is given by
(a) Domain = R, Range = {−1, 1}
(b) Domain = R − {1}, Range = R
(c) Domain = R − {4}, Range = {−1}
(d) Domain = R − {−4}, Range = {−1, 1} - The domain and range of real function f defined by f(x) = √(x − 1) is given by
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range = (0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = [1, ∞), Range = [0, ∞) - The domain of the function f given by f(x) = (x² + 2x + 1)/(x² − x − 6)
(a) R − {−2, 3}
(b) R − {−3, 2}
(c) R − {−2, 3] (d) R − (−2, 3) - The domain and range of the function f given by f(x) = 2 − |x − 5|, is
(a) Domain = R⁺, Range = (−∞, 1] (b) Domain = R, Range = (−∞, 2] (c) Domain = R, Range = (−∞, 2)
(d) Domain = R⁺, Range = (−∞, 2] - If f(x) = x³ − 1/x³ , then f(x) + f(1/x) is equal to
(a) 2x³
(b) 2/x³
(c) 0
(d) 1 - The domain of the function defined by f(x) = 1/√(x − |x|) is
(a) R₀
(b) R⁺
(c) R⁻
(d) none of these
- Let A = {1, 2, 3}, B = {2, 3, 4}, then which of the following is a function from A to B?