第12类NCERT解决方案–数学第I部分–第6章导数的应用–练习6.5 |套装1

问题1.找到以下给出的以下函数的最大值和最小值：

(i)f(x)=(2x – 1) ² + 3               

解决方案：

Given that, f(x) = (2x – 1)² + 3

From the given function we observe that

(2x – 1)² ≥ 0 ∀ x∈ R,

So, 

(2x – 1)² + 3 ≥ 3 ∀ x∈ R,

Now we find the minimum value of function f when 2x-1 = 0

So, x = 1/2

f = f(1/2) = (2(1/2) – 1)² + 3 = 3

Hence, the minimum value of the function is 3 and this function does not contain maximum value.

  (ii)f(x)= 9x ² + 12x + 2

解决方案：

Given that, f(x) = 9x² + 12x + 2

we can also write as f(x) = (3x + 2)² – 2

From the given function we observe that

(3x + 2)² ≥ 0 ∀ x∈ R,

So, 

(3x + 2)² – 2 ≥ 2 ∀ x∈ R,

Now we find the minimum value of function f when 3x + 2 = 0

So, x = -2/3

f = f(-2/3) = (3(-2/3) + 2)² – 2 = -2

Hence, the minimum value of the function is -2 and this function does not contain maximum value.

(iii)f(x)=-(x – 1) ² + 10

解决方案：

Given that, f(x) = -(x – 1)² + 10  

From the given function we observe that

(x – 1)² ≥ 0 ∀ x∈ R,

So, 

-(x – 1)² + 10 ≤ 10 ∀ x∈ R,

Now we find the maximum value of function f when x – 1 = 0

So, x = 1

f = f(1) = -(1 – 1)² +10 = 10

Hence, the maximum value of the function is 10 and this function does not contain minimum value.

(iv)g(x)= x ³ +1

解决方案：

Given that, g(x) = x³ + 1

When x —> ∞, then g(x) —> ∞

When x —> -∞, then g(x) —> -∞

So, this function has neither minimum nor maximum value

问题2。找到以下给出的以下函数的最大值和最小值：

(i)f(x)= | x + 2 | – 1

解决方案：

Given that, f(x) = |x + 2| – 1

$f(x)= \begin{cases} x+2-1; x\ge -2 \\ -x-2-1; x<2 \end{cases} \\ f(x)= \begin{cases} x+1;x\ge -2\\ -x-3;x<-2 \end{cases}\\ \text{Now}\space f'(x)=\begin{cases} 1;x\ge -2\\ -1;x<-2 \end{cases}$

At x = -2 f'(x) change sign from negative to positive, hence by first derivative test, x = -2

is a point of local minima. 

So, the minimum value f = f(-1)= |(-1)+ 2| – 1 = -1

So this function doesn’t contain maximum value

(ii)g(x)=-| x + 1 | + 3

解决方案：

Given that, g(x) = -|x + 1| + 3

$f(x)= \begin{cases} -x-1+3; x\ge -1 \\ x+1+3; x<-1 \end{cases} \\ f(x)= \begin{cases} 2-x;x\ge -1\\ x+4;x<-1 \end{cases}\\ f'(x)=\begin{cases} -1;x\ge -1\\ 1;x<-1 \end{cases}$

At x = -1 f'(x) change sign from positive to negative, hence by first derivative test, x = -1

is a point of local minima. 

So, the maximum value of f = f(-1) = -|(-1) + 1| + 3 = 3

So, this function doesn’t contain minimum value

(iii)h(x)=罪2x + 5

解决方案：

Given that, h(x) = sin(2x) + 5

On differentiate both side w.r.t x, we get

h'(x) = 2cos2x

Now put h'(x) = 0

2cos 2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

$x=±π/4,±\frac{3π}{4},±\frac{5π}{4}...$

Let’s perform second derivative test,

h”(x) = -4sin2x

h”(π/4) < 0 

$h''(\frac{3π}{4} )>0;h''(\frac{5π}{4})<0$

& so on.

So, $x=\frac{π}{4},\frac{-3π}{4},\frac{5π}{4}\frac{-7π}{4}\frac{9π}{4}.....\text{are \space points}$

of local maxima.

& $x=\frac{-π}{4},\frac{3π}{4},\frac{-5π}{4}....$are points of local minima.

So, the minimum value of the given function is 4 and the maximum value of the given function is 6

(iv)f(x)= | sin(4x + 3)|

解决方案：

Given that, f(x) = |sin(4x + 3)|

Now for any value of x, sin4x has the least value as -1. i.e., sin 4x + 3 ≥ 2

So f(x) = |sin(4x + 3)| = sin 4x + 3

On differentiate both side w.r.t x, we get

f'(x) = 4cos4x

Now put f'(x) = 0

4cos 4x = 0

4x = (2x – 1)π/2

x = (2x – 1)π/8

$x=±\frac{π}{8},±\frac{3π}{8},±\frac{7π}{8}....$  & son on

Let’s perform second derivative test,

f”(x) = -16 sin4x

f”(π/8) < 0; f”(3π/8) > 0;  f”(5π/8) < 0

So,$x=\frac{π}{8},\frac{-3π}{8},\frac{5π}{8},\frac{-7π}{8}$  …. are points of local maxima. Minimum value = 4.

& $x=\frac{-π}{8},\frac{3π}{8},\frac{-5π}{8},\frac{7π}{8}$are points of local minima. Minimum value = 2.

(v)h(x)= x + 1，x∈(-1，1)

解决方案：

Given that, h(x) = x + 1, x ∈ (-1, 1)

As we can clearly see from the function that h(x) is a strictly increasing function.

So, the minimum value of x will give minimum value of h(x).

Now, x ∈ (-1, 1)       

So, thi function has no minimum nor maximum value.

问题3.找到以下函数的局部最大值和局部最小值(如果有)。还要找到局部最大值和局部最小值(视情况而定)：

(i)f(x)= x ²

解决方案：

Given that f(x) = x²

On differentiate both side w.r.t x, we get

f'(x) = 2x

Now put f'(x) = 0

2x = 0

x = 0

Let’s do second derivative test,

f”(x) = 2 > 0

At, x = 0, f'(x) = 0 and f”(x) > 0,

So x = 0 is a point of local minima. Local minimum value.

(ii)g(x)= x ³ – 3x

解决方案：

Given that g(x) = x³ – 3x

On differentiate both side w.r.t x, we get

g'(x) = 3x² – 3

Now put g'(x) = 0

3x² – 3 = 0

x² = 1

x = ±1

Let’s do the second derivative test,

g”(x) = 6x ….(i)

g”(1) = 6 > 0

g”(-1) = -6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is 

g(1) = (1)³ – 3(1) = -2

So by second derivatives test, x = -1 is a point of local minima and the minimum value is 

g(-1) = (-1)³ – 3(-1) = 2

Hence, the local minimum value is -2 and the local maximum value is 2

(iii)h(x)= sin x + cos x，0

解决方案：

h(x) = sin x + cos x, x∈(0,π/2)

On differentiate both side w.r.t x, we get

h'(x) = cos x – sin x

Now put h'(x) = 0

cos x – sin x = 0

cos x = sin x, x ∈ (0, π/2)

Clearly x = π/4 [both cos x and sin x attain 1/√2 at π/4]

Let’s do second derivative test,

h”(x) = -sin x – cos x

h”(π/4) = $-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$

At $x=\frac{π}{4}$ is a point of local maxima and the maximum value is

h(π/4) = sin π/4 + cos π/4

= 1/√2 + 1/√2 = √2

(iv)f(x)= sin x – cos x，0

解决方案：

Given that, f(x) = sin x – cos x, x ∈ (0, 2π)

On differentiate both side w.r.t x, we get

f'(x) = cos x + sin x

Now put f'(x) = 0

cos x + sin x = 0

x = $\frac{3π}{4},\frac{7π}{4}\space$in (0, 2π)

Now let’s do the second derivative test 

f”(x) = -sin x + cos x

f”(3π​/4) = – √2 > 0

f”(7π​/4) = √2 > 0

So by second derivatives test, x = $\frac{3π}{4}$ is a point of local maxima and the maximum value is 

f($\frac{3π}{4}$) = -sin 3π​/4 + cos 3π​
4 = 1/√2 + 1/√2 = √2 > 0

So by second derivatives test, x = $\frac{7π}{4}$ is a point of local minima and the minimum value is 

f($\frac{7π}{4}$) = -sin 7π​/4 + cos 7π​
4 = -1/√2 – 1/√2 = -√2 > 0

Hence, the local minimum value is -√2 and the local maximum value is √2.

(v)f(x)= x ³ – 6x ² + 9x + 15

解决方案：

Given that, f(x) = x³ – 6x² + 9x + 15

On differentiate both side w.r.t x, we get

f'(x) = 3x² – 12x + 9

Now put f'(x) = 0

3x² – 12x + 9 = 0

3(x² – 4x + 3) = 0

x = 1, 3

Let’s do the second derivative test,

f”(x) = 6x – 12

f”(1) = -6 < 0

f”(3) = 6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is 

f'(1) = 3(1)² – 12(1) + 9 = 19

So by second derivatives test, x = 3 is a point of local minima and the minimum value is 

f'(3) = 3(3)² – 12(3) + 9 = 15

Hence, the local minimum value is 15 and the local maximum value is 19.

(六) $g(x)=\frac{x}{2}+\frac{2}{x}$ ，x> 0

解决方案：

Given that, $g(x)=\frac{x}{2}+\frac{2}{x}$, x > 0

On differentiate both side w.r.t x, we get

g'(x)=$\frac{1}{2}-\frac{2}{x^2}$

Now put g'(x) = 0

$\frac{1}{2}-\frac{2}{x^2}=0 \\ \frac{1}{2}=\frac{2}{x^2} \\ x^2=4 x=±2$but ‘x > 0’

So, x = 2

Now we will do the second derivative test,

g”(x)=$\frac{4}{x^3} \\ g''(2)=\frac{4}{2^3}=\frac{1}{2}>0$

Hence, x = 2 is a point of local minima.

Local maximum value = g(2) = 2

(vii) $g(x)=\frac{1}{x^2+2}$

解决方案：

Given that, $g(x)=\frac{1}{x^2+2} \\ g'(x)=\frac{-1}{(x^2+2)^2}=0 \\ x=0$

On differentiate both side w.r.t x, we get

$g'(x)=\frac{-2x}{(x^2+2)^2}$

Now put g'(x) = 0

$g'(x)=\frac{-2x}{(x^2+2)^2} = 0$

Now, let’s perform the second derivative test,

$g''(x)=\frac{(x^2+2)^2(-2)+2x(x^2+2)}{(x^2+2)^4}\\ g''(x)=\frac{2^2(-2)+2(0)(2)}{(0+4)^2}$

= -8/16 = -1/2 < 0

At x = 0, g'(x) = 0 and g”(x) < 0

Hence, ‘x = 0’ is a point of local maxima.

Now the domain of g(x) is (-∞, ∞).

$\lim_{x\to\infty} g(x)=0\space \& \space \lim_{x\to\infty} g(x)=0$

Value of g(x) at the extreme values of x is 0

So the global maxima of g(x)=$\frac{1}{x^2}$is at x = 0.

The maximum value is g(0) = 1/2

(viii) $f(x)=x\sqrt{1-x}$ ，x> 0

解决方案：

Given that, $f(x)=x\sqrt{1-x},x>0\\ f'(x)=\frac{x}{2\sqrt{1-x}}x-1+\sqrt{1-x}=\frac{-x+2(1-x)}{\sqrt{1-x}}=\frac{2-3x}{\sqrt{1-x}}$

Now put f'(x) = 0

$\sqrt{1-x}=\frac{x}{2\sqrt{1-x}}$

2(1 – x) = x

2 – 2x = x

3x = 2

x = 2/3

Now let’s do the second derivative test,

$f''(x)=\frac{\sqrt{1-x}(-3)-(2-3x).\frac{1}{2\sqrt{1-x}.(-1)}}{(1-x)}\\ f''(x)=\frac{-6(1-x)+(2-3x)}{2(1-x)^\frac{3}{2}}=\frac{3x-4}{(1-x)^{\frac{3}{2}}\\} \\f''(\frac{2}{3})=\frac{3.\frac{2}{3}-4}{(1-\frac{2}{3})^{\frac{3}{2}}}=\frac{-2}{(1/3)^{\frac{3}{2}}}<0$

x = 2/3 is a point of local maxima f(2/3) = $\frac{2}{3\sqrt{3}}$

Now, f (x) = x$x\sqrt{1-x},x>0$

For domain, 1 – x ≥ 0 or x ≤ 1

So x ∈ [0, 1]

Local maxima is at x = 2/3 and the local maximum value is $\frac{2}{3\sqrt{3}}$

问题4.证明以下函数没有最大值或最小值：

(i)f(x)= e ^x

解决方案：

Given that, f(x) = e^x

f'(x) = e^x

Now e^x > 0, f'(x) > 0

Hence, f(x) is a strictly increasing function with no maxima or minima.  

(ii)g(x)=对数x

解决方案：

Given that, g(x) = log x

g'(x) = 1/x

Now the domain of log x is x > 0

So, 1/x > 0, i.e., g'(x) > 0

Hence, g(x) is a strictly increasing function with no maxima or minima.  

(iii)h(x)= x ³ + x ² + x + 1

解决方案：

Given that, h(x) = x³ + x² + x + 1

h'(x) = 3x² + 2x + 1

Now for this quadratic expression 3x² + 2x + 1,

Its discriminant 0 = 2² – 4(3)(1) = -8 < 0

So, 3x² + 2x + 1 > 0

Hence, h(x) is a strictly increasing function with no maxima or minima.    

问题5.在给定的时间间隔中找到以下函数的绝对最大值和绝对最小值：

(i)f(x)= x ³ ，x∈[-2，2]

解决方案：

Given that, f(x) = x³, x ∈ [-2, 2]  

f'(x) = 3x²

f'(x) = 0 at x = 0 

f”(x) = 6x

f”(0) = 0, second derivative failure

Now f'(3⁺) > 0 and f'(3^–) > 0

f'(x) does not change sign at x = 0.

x = 0 is neither maxima nor minima

f(x) = x³ is a strictly increasing function.

(ii)f(x)= sin x + cos x，x∈[0，π]

解决方案：

Given that, f(x) = sin x + cos x, x ∈ [0, π]

First derivative

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

x = π/4

On applying second derivative test, 

f”(x) = -sin x – cos x

f”(π/4) = $\frac{-1}{\sqrt2}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$

Hence, x = π/4 is pof local maxima . f(π/4)=$\sqrt2$

Now, for global maxima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

For global maxima is at x = π/4 and the global maximum value is √2.

Now, for global minima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

Global minima is at x = π and the global minimum value is -1.

(iii) $f(x)=4x-\frac{1}{2}x^2,x∈[-2,\frac{9}{2}]$

解决方案：

Given that, $f(x)=4x-\frac{1}{2}x^2,x∈[-2,\frac{9}{2}]$

f'(x) = 4 – x

Now put f'(x) = 0

4 – x = 0

x = 4

Now applying second derivative test f”(x) = -1 < 0

Hence, x = 4 is a pt. of local maxima.

f(4) = $4.4-\frac{4^2}{2}=8$

Global maxima = max{f(-2), f(4), f(9/2)}

= max{-10, 8, 7.8}

= 8

Global maxima occur at x = 9/2 and global maximum value is f(9/2) = 8

Global minima = min{f(-2), f(4), f(9/2)}

= max{-10, 8, 16.9}

= -10

Global minima occur at x = -2 and the global minimum value is f(-2) = -10.

(iv)f(x)=(x – 1) ² + 3，x∈[-3，1]

解决方案：

Given that, f(x) = (x – 1)² + 3, x ∈ [-3, 1]

f'(x) = 2(x – 1)

Now put f'(x) = 0

2(x – 1) = 0

x = 1

Now applying second order derivative test,

f”(x) = 2 > 0

Hence, x = 1 is a point of local minima. f(1) = 3

Global maxima = max{f(-3), f(1)} 

= max{19, 3}

= 19

The global or absolute maxima occurs at x = -3 and the absolute maximum value is f(-3) = 19

Global minima = min{f(-3), f(1)}

= min{19, 3]

= 3

The global or absolute minima occurs at x = 1 and the absolute value is f(1) = 3

问题6.如果利润函数由p(x)= 41 – 24x – 18x ²给出，求出公司可以赚取的最大利润

解决方案：

Given that p(x) = 41 – 24x – 18x²

p'(x) = -24 – 36x

Now put p'(x) = 0

-24 – 36x = 0

x = -24/36

x = -2/3

Now, doing the second order derivative test,

p”(x) = -36 < 0

Hence, x = -2/3 is point of local maxima.

Now in quadratic function with domain R, if there is a local maxima, it is the global maxima also. BC₃ p(-∞)⇢ -∞ and p(+∞)⇢ -∞

The maximum profit is p(-2/3) = 49

If negative units (x) do not exist, then maximum profit is p(0) = 41.

问题7.在区间[0，3]上找到最大值和最小值3x ⁴ – 8x ³ + 12x ² – 48x + 25。

解决方案：

Given that f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25, x ∈ [0, 3]

f'(x) = 12x² – 24x² + 24x – 48

Now put f'(x) = 0

12x³ – 24x² + 24x – 48 = 0

12(x² – 2x² + 2x – 4) = 0

12(x²(x – 2) + 2(x – 2)) = 0

12(x² + 2)(x – 2) = 0

x = 2 because x² + 2 ≠ 0

Now applying second derivative test,

f”(x) = 12(3x² – 4x + 2)

f”(2) = 12(3.2² – 4.2 + 2)

f”(2) = 12.6 = 72 > 0

Hence, x = 2 is point of local minima.

f(2) = -39

Global maxima = max{f(0), f(2), f(3)}

= max{25, -39, 16}

= 25

Global maxima occur at x = 0 and the global maximum is 25.

Global minima = min{f(0), f(2), f(3)}

= min{25, -39, 16}

= -39

Global minima occur at x = 2 andthe global minimum value is -39.

问题8.在区间[0，2π]的哪些点，函数sin 2x是否达到最大值?

解决方案：

Given that, f(x) = sin 2x                     ,  x ∈ [0, 2π]

f'(x) = 2 cos 2x

Now put f'(x) = 0

2cos2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

x = π/4, 3π/4, 5π/4, 7π/4

Now let’s do second order derivative test.

f”(x) = -4 sin2x

f”(π/4) = $-4\sin\frac{π}{2}=-4<0\\ f''(\frac{3π}{4})=-4\sin\frac{3π}{2}=+4>0\\ f''(\frac{5π}{4})=-4\sin\frac{5π}{2}=-4<0\\ f''(\frac{7π}{4})=-4\sin\frac{7π}{2}=+4>0$

x = π/4 and x = 5π/4 are point of local maxima.

x = 3π/4 and x = 7π/4 are point of local minima.

f(π/4) = f(5π/4) = 1 and f(3π/4) = f(7π/4) = -1

Now, 

Global maxima = max{f(0), f(π/4), f(3π/4), f{5π/4}, f(7π/4), f(2π)}

= max{0, 1, -1, 1, -1, 0}

= 1

Global maxima occur at the points x = π/4 and x = 5π/4 and the absolute maximum value is 1.

问题9.函数sin x + cos x的最大值是多少?

解决方案：

Given that, f(x) = sin x + cos x

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

$x=\frac{π}{4},\frac{-3π}{4},\frac{5π}{4},\frac{-7π}{4},\frac{8π}{4}........$

= {-√2, √2, +√2, -√2, -√2}

Now, second order derivative test,

f”(x) = -sin x – cos x

f”(π/4) = f”(9π/4) = f”(17π/4)……….. = -√2 < 0

A liter ⇢ f(x) = sin x + cos x = $\sqrt{2}(\frac{1}{\sqrt{2}}\sin+\frac{1}{\sqrt{2}}\cos x)$ 

= $\sqrt{2}.\cos(\frac{π}{4}-x),f(x)_{max}=\sqrt{2}$

问题10.在间隔[1，3]中找到2x ³ – 24x + 107的最大值。在[-3，-1]中找到相同函数的最大值。

解决方案：

Given that f(x) = 2x³ – 24x + 107 

On differentiating w.r.t. x we get

f'(x) = 6x² – 24

Now, put f'(x) = 0

6x² = 24

x² = 4

x = ±2                    

Now second order test

f”(x) = 12x

f”(2) = 12.2 = 24 > 0

x = 2 is a pt. of local minima 

f(2) = 75

f”(-2) = 12(-2) = -24 < 0

x = -2 is point of local maxima. f(-2) = 139

Now, in the interval [1, 3]

Global maxima = max{f(1), f(2), f(3)}

= max{85, 75, 89}

= 89

Now, in the interval [-3,-1]

Global maxima = max{f(-3), f(-2), f(-1)}

= max{125, 139, 129}

= 139

问题11假定在x = 1时，函数x ⁴ – 62x ² + ax + 9在间隔[0，2]上达到最大值。查找a的值。

解决方案：

Give that, f(x) = x⁴ – 62x² + ax + 9

On differentiating w.r.t. x we get

f'(x) = 4x³ – 124x + a

The maximum value is attained at x = 1, and 1 lies between 0 and 2. 

So, at x = 1, there must be a local maxima

That means, f'(1) = 0

f'(1) = 4(1)³ – 124(1) + a = 0

-120 + a = 0

a = 120

问题12.在[0，2π]上找到x + sin2x的最大值和最小值。

解决方案：

Give that f(x) = x + sin2x, x ∈ [0, 2π]

On differentiating w.r.t. x we get

f'(x) = 1 + 2cos2x

Now put f'(x) = 0, we get

1 + 2cos2x = 0

cos2x = -1/2

$x=\frac{π}{3},\frac{2π}{3},\frac{4π}{3},\frac{5π}{3}$ ∈ [0, 2π]

Now, 

For global maxima = max{f(0), f(π/3), f(4π/3), f(2π)}

= max{0, π/3, $\frac{\sqrt{3}}{2},\frac{4π}{3}+\frac{\sqrt{3}}{2},2π$}

= 2π

Global maxima occur at x = 2π and the maximum value is f(2π) = 2π.

For global minima = min{f(0), f(2π/3), f(5π/3), f(2π)}

= min{0, $\frac{2π}{3}-\frac{\sqrt{3}}{2},\frac{5π}{3}-\frac{\sqrt{3}}{2},2π$}

= 0

Global minima occur at x = 0 and the minimum value is 0.