Решения NCERT для класса 12. Математика. Часть I. Применение производных. Упражнение 6.2 | Набор 1
Вопрос 1. Покажите, что функция f (x) = 3x + 17 возрастает на R.
Решение:
If for a function f(x), f"(x) > 0 for all x, then the function is a strictly increasing function. (vice-versa is not true)
Given: f(x) = 3x + 17
f"(x) = 3 > 0 -(Always greater than zero)
Hence, 3x + 17 is strictly increasing on R.
Вопрос 2. Покажите, что функция задается формулой f (x) = e 2x и возрастает на R.
Решение:
If for a function f(x), f"(x) > 0 for all x, then the function is a strictly increasing function. (vice-versa is not true)
Given: f(x) = e2x
f’(x) = 2e2x > 0
Hence, f(x) = e2x is strictly increasing on ∞
Вопрос 3. Покажите, что функция, заданная формулой f (x) = sin x, является
(i) возрастание по (0, π/2)
(ii) убывание по (π/2, π)
(iii) не возрастает и не убывает по (0, π)
Решение:
Given: f(x) = sin x
So, f’(x) = d/dx(sin x) = cos x
(i) Now in (0, π/2), f’(x) = cos x > 0 (positive in first quadrant)
Hence, f(x) = sin x is strictly increasing in (0, π/2).
(ii) In (π/2, π), f’(x) = cos x < 0 -(negative in second quadrant)
Hence, f(x) = sin x is strictly decreasing in (π/2,π)
(iii) As we know that f’(x) = cos x is positive in interval(0, π/2)
and f’(x) = cos x is negative in interval (π/2, π)
So, it is neither increasing nor decreasing.
Вопрос 4. Найдите интервалы, в которых функция f, заданная формулой f(x) = 2x 2 – 3x,
(i) увеличение
(ii) уменьшение
Решение:
Given: f(x) = 2x2 – 3x
f"(x) =
= 4x – 3 -(1)
= x = 3/4
So the intervals are (-∞, 3/4) and (3/4, ∞)
(i) Interval (3/4, ∞) let take x = 1
So, from eq(1) f"(x) > 0
Hence, f is strictly increasing in interval (3/4, ∞)
(ii) Interval (-∞, 3/4) let take x = 0.5
So, from eq(1) f"(x) < 0
Hence, f is strictly decreasing in interval (-∞, 3/4)
Вопрос 5. Найдите интервалы, в которых функция f, заданная формулой f(x) = 2x 3 – 3x 2 – 36x + 7, является
(i) увеличение
(ii) уменьшение
Решение:
Given: f(x) = 2x3 – 3x2 – 36x + 7
f"(x) =
= 6x2 – 6x – 36 -(1)
f"(x) = 6(x2 – x – 6)
On putting f"(x) = 0, we get
6(x2 – x – 6) = 0
(x2 – x – 6) = 0
x = -2, x = 3
So, the intervals are (-∞, -2), (-2, 3), and (3, ∞)
For (-∞, -2) interval, take x = -3
From eq(1), we get
f"(x) = (+)(-)(-) = (+) > 0
So, f is strictly increasing in interval (-∞, -2)
For (-2, 3) interval, take x = 2
From eq(1), we get
f"(x) = (+)(+)(-) = (-) < 0
So, f is strictly decreasing in interval (-2, 3)
For (3, ∞)interval, take x = 4
From eq(1), we get
f"(x) = (+)(+)(+) = (+) > 0
So, f is strictly increasing in interval (3, ∞)
(i) f is strictly increasing in interval (-∞, -2) and (3, ∞)
(ii) f is strictly decreasing in interval (-2, 3)
Вопрос 6. Найдите промежутки, в которых строго возрастают или убывают следующие функции:
(и) х 2 + 2 х – 5
(ii) 10 – 6x – 2x 2
(iii) -2x 3 - 9x 2 - 12x + 1
(iv) 6 – 9х – х 2
(v) (х + 1) 3 (х – 3) 3
Решение:
(i) f(x) = x2 + 2x – 5
f"(x) = 2x + 2 -(1)
On putting f"(x) = 0, we get
2x + 2 = 0
x = -1
So, the intervals are (-∞, -1) and (-1, ∞)
For (-∞, -1) interval take x = -2
From eq(1), f"(x) = (-) < 0
So, f is strictly decreasing
For (-1, ∞) interval take x = 0
From eq(1), f"(x) = (+) > 0
So, f is strictly increasing
(ii) f(x) = 10 – 6x – 2x2
f"(x) = -6 – 4x
On putting f"(x) = 0, we get
-6 – 4x = 0
x = -3/2
So, the intervals are (-∞, -3/2) and (-3/2, ∞)
For (-∞, -3/2) interval take x = -2
From eq(1), f"(x) = (-)(-) = (+) > 0
So, f is strictly increasing
For (-3/2, ∞) interval take x = -1
From eq(1), f"(x) = (-)(+) = (-) < 0
So, f is strictly decreasing
(iii) f(x) = -2x3 – 9x2 – 12x + 1
f"(x) = -6x2 – 8x – 12
On putting f"(x) = 0, we get
-6x2 – 8x – 12 = 0
-6(x + 1)(x + 2) = 0
x = -1, x = -2
So, the intervals are (-∞, -2), (-2, -1), and (-1, ∞)
For (-∞, -2) interval take x = -3
From eq(1), f"(x) = (-)(-)(-) = (-) < 0
So, f is strictly decreasing
For (-2, -1) interval take x = -1.5
From eq(1), f"(x) = (-)(-)(+) = (+) > 0
So, f is strictly increasing
For (-1, ∞) interval take x = 0
From eq(1), f"(x) = (-)(+)(+) = (-) < 0
So, f is strictly decreasing
(iv) f(x) = 6 – 9x – x2
f"(x) = -9 – 2x
On putting f"(x) = 0, we get
-9 – 2x = 0
x = -9/2
So, the intervals are (-∞, -9/2) and (-9/2, ∞)
For f to be strictly increasing, f"(x) > 0
– 9 – 2x > 0
x > -9/2
So f is strictly increasing in interval (-∞, -9/2)
For f to be strictly decreasing, f"(x) < 0
-9 – 2x < 0
x < -9/2
So f is strictly decreasing in interval (-9/2, ∞)
(v) f(x) = (x + 1)3 (x – 3)3
f"(x) = (x + 3)3.3(x – 3)3 + (x – 3)3.3(x + 1)2
f"(x) = 6(x – 3)2(x + 1)2(x – 1)
Now, the factor of (x – 3)2 and (x + 1)2 are non-negative for all x
For f to be strictly increasing, f"(x) > 0
(x – 1) > 0
x > 1
So, f is strictly increasing in interval (1, ∞)
For f to be strictly decreasing, f"(x) < 0
(x – 1) < 0
x < 1
So, f is strictly decreasing in interval (-∞, 1)
Вопрос 7. Покажите, что y = log(1 + x) – 
, является возрастающей функцией x во всей области определения.
Решение:
f(x) = log(1+x)
f"(x)=
So, the domain of the given function is x > -1
Now, x2 > 0, (x + 2)2 ≥ 0, x + 1 > 0
From the above equation f"(x) ≥ 0 ∀ x in the domain(x > -1) and f is an increasing function.
Вопрос 8. Найдите значения x, при которых y = [x(x – 2)] 2 является возрастающей функцией.
Решение:
Given: y = f(x) = [x(x – 2)]2 = x2(x – 2x)2
= x4 – 4x3 + 4x2
f"(x) = 4x3 – 12x2 + 8x
f"(x) = 4x(x – 2)(x – 1)
x = 0, x = 1, x = 2
So, (∞, 0], [0, 1], [1, 2], [2,∞)
For (∞, 0], let x = -1
So, f"(x) = (-)(-)(-) = (-) ≤ 0
f(x) is decreasing
For [0, 1], let x = 1/2
So, f"(x) = (+)(-)(-) = (+) ≥ 0
f(x) is increasing
Similarly, for [1, 2], f(x) is decreasing
For [2,∞), f(x) is increasing
So, f(x) is increasing in interval [0, 1] and [2,∞)
Вопрос 9. Докажите, что y = 
является возрастающей функцией θ в [0, π/2].
Решение:
y = f(θ) =
Now 0 ≤ θ ≤ π/2, and we have 0 ≤ cosθ ≤ 1,
So, 4 – cosθ > 0
Therefore f"(θ) ≥ 0 for 0 ≤ θ ≤ π/2
Hence, f"(x) =
is a strictly increasing in the interval (θ, π/2).
Вопрос 10. Докажите, что логарифмическая функция возрастает на (0, ∞).
Решение:
Given: f(x) = log(x) -(logarithmic function)
f"(x) = 1/x ∀ x in (0, ∞)
Therefore, x > 0, so, 1/x > 0
Hence, the logarithmic function is strictly increasing in interval (0, ∞)