类12 NCERT解决方案-数学第I部分-第6章导数的应用-第6章的其他练习|套装2

第6章衍生物的应用–第6章的其他练习|套装1

问题12.三角形的斜边上的一点与三角形的边相距a和b。证明斜边的最小长度为$(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$

解决方案：

Given, a triangle ABC

Let, PE = a & PD = b

In the △ABC, ∠B = 90

Let ∠C = θ, so, ∠ DPA = θ

DP|| BC.

Now in △ADDP,

cosθ = DP/AP = b/AP 

AP = b/cosθ

In △EPC,

sinθ = EP/CP = a/CP

CP = a/sin θ               

Now AC = h = PA + PC

h = $\frac{b}{\cos θ}+\frac{a}{\sin θ}$

h(θ) = b sec θ + a cosec θ

Put h'(θ) = $\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}$

$\frac{b}{\cos θ}.\frac{\sin θ}{\cos θ}=\frac{a}{\sin θ}.\frac{\cos θ}{\sin θ}$

b sin³θ = a cos ³θ

tan3θ = a/b

tanθ = (a/b)^1/3 

secθ = $\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$

cosecθ = $\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}$

h_max = $b.\frac{\sqrt{b^{2/3}+a^{2/3}}}{b^{1/3}}+a.\frac{\sqrt{b^{2/3}+a^{2/3}}}{a^{1/3}}$

h_max = (b^2/3+a^2/3)^3/2

问题13。找出由f(x)=(x – 2) ⁴ (x + 1) ³给出的函数f具有的点

(i)局部最大值

(ii)局部最小值

(iii)拐点

解决方案：

f(x) = (x – 2)⁴(x + 1)³

On differentiating w.r.t x, we get

f'(x) = 4(x – 2)³(x + 1)³ + 3(x + 1)²(x – 2)⁴

Put f'(x) = 0

(x – 2)³(x + 1)² [4(x + 1) + 3(x – 2)] = 0

(x – 2)³(x + 1)²(7x – 2) = 0

Now,

Around x = -1, sign does not change, i.e

x = -1 is a point of inflation

Around x = 2/7, sign changes from +ve to -ve i.e.,

x = 2/7 is a point of local maxima.

Around x = 2, sign changes from -ve to +ve i.e.,

x = 2 is a point of local minima

问题14。找出由f(x)= cos ² x + sin x，x∈[0，π]给出的函数f的绝对最大值和最小值

解决方案：

f(x) = cos²x + sin x; x ϵ [0, π]

On differentiating w.r.t x, we get

f'(x) = 2cos x(-sin x) + cos x = cos x – sin2x

Put f'(x) = 0

cos x(1 – 2sin x) = 0

cos x = 0; sin x = 1/2

In x ϵ[0, π] if cos x = 0, then x = π/2

and if sin x = 1/2, then x = π/6 & 5π/6

Now, f”(x) = -sin x – 2 cos2x

f”(π/2) = -1 + 2 = 1 > 0

x = π/2 is a point of local minima f(π/2) = 1

f”(π/6) = $\frac{-1}{2}-2.\frac{1}{2}=\frac{-3}{2}<0$

x = π/6 is a point of local maxima f(π/6) = 5/4

$f''(\frac{5π}{6})=\frac{-1}{2}-2.(\frac{-1}{2})>0$

x = 5π/6​ is a point of local minima f(5π/6) = 5/4

Global/Absolute maxima = ma{f(0), f(π/6), f(π)}

= max{1, 5/4, 1}

= 5/4 = Absolute maxima value

Global/Absolute minima = min{f(0), f(π/2), f(π/6), f(π)}

= min{1, 1, 5/4, 1}

= 1 = Absolute minima value

问题15：证明可以在半径为r的球体内刻出的最大体积的右圆锥的高度为4π/ 3

解决方案：

Let ABC be the cone

and o is the centre of the sphere.

AO = BO = CO = R

AO = h = height of cone

BD = CD = r = radius of cone.

∠DOC = θ                  -(Properties of circle)

In △ DOC,

OD = R cosθ & CD = Rsinθ,

r = R sin θ 

AD = AO + OD = R + Rcosθ 

h = R(1 + cosθ)

Now, the volume of the cone is 

V = $\frac{1}{3}πr^2h$

v(θ) = $\frac{1}{3}.πR^2\sin{-2}θ.R(1+\cosθ)$

$v'(θ)=\frac{}{}[\sin{-2}θ(-\sinθ)+(1+\cosθ)(2\sinθ\cosθ)]$

Put v(θ) = 0

sinθ[2cosθ + 2cos²θ − sin2θ] = 0

sinθ[2cosθ + 2cos²θ − 1] = 0

sinθ(3cosθ − 1)(1 + cosθ) = 0

sinθ = 0, cos = 1/3​, cosθ = −1

If sinθ = 0, then volume will be 0.

If cosθ = -1, then sinθ = 0 & again volume will be 0.

But if cosθ = 1/3; sinθ = 2√2/3 and 

Volume, v = 32/81​πR³, which is maximum.

Height, h = R(1 + cosθ) = R($1+\frac{1}{3}$)

h = 4r/3

Hence proved

问题16。令f是在[a，b]上定义的函数，对于所有x∈(a，b)，f’(x)> 0。然后证明f是(a，b)上的一个递增函数。

解决方案：

Given that on [a, b] f'(x) > 0, for all x in interval I. 

So let us considered x1, x2 belongs to I with x1 < x2 

To prove: f(x) is increasing in (a, b)

According to the Lagrange’s Mean theorem

f(x2) – f(x1)/ x2 – x1 = f'(c)

f(x2) – f(x1) = f'(c)(x2 – x1)

Where x1 < c < x2 

As we know that x1 < x2 

so x1 < x2 > 0

It is given that f'(x) > 0

so, f'(c) > 0

Hence, f(x2) – f(x1) > 0

f(x2) < f(x1)

Therefore, for every pair of points x1, x2 belongs to I with x1 < x2 

f(x2) < f(x1)

f(x) is strictly increasing in I

问题17：证明可以在半径为R的球体内切出的最大容积圆柱体的高度为2R /√3。另外，找到最大音量。

解决方案：

In △ABC, 

AC² = BC² + AB²

4R² = 4r² + h²

r² = R²–$\frac{h^2}{4}$          ……….(1)

Now, volume of cylinder = πr²h

Put the value ov r² from eq(1), we get

V = π($R^2\frac{-h^2}{4}$).h      

V(h) = $πR^2h-\frac{πh^3}{4}$

On differentiating both side we get

V ‘(h) = $πR^2h-\frac{3πh^3}{4}$

Now, put V'(h) = 0

πR² = $\frac{3}{4}πh^2$

$h=\frac{2R}{\sqrt{3}}$

Now the maximum volume of cylinder = π[R². 2R/√3 – 1/4.4R²/3.2R/√3]

= 4πR³/ 3√3

问题18。表明最大容积的圆柱体的高度(可以刻在高度为h和半垂直角α的右圆锥形中)是圆锥体的三分之一，最大圆柱体的体积为4/27 ^πH3个黄褐色^2个α。

解决方案：

Let, 

XQ = r

XO = h’

AO = h

OC = r’

∠XAQ = α

In triangle AXQ and AOC = XQ/OC = AX/AO

So, r’/r = h-h’/h

hr’ = r(h-h’)

hr’ = rh – rh’

rh’ = rh – hr’

rh’ = h(r – r’)

h’ = h(r – r’)/r

The volume of cylinder = πr’²h’

v = πr’²(h(r – r’)/r)

= π(h(rr’² – r’³)/r)

On differentiating we get

v’ = πh/r(2rr’ – 3r’²)

Again on differentiating we get

v” = πh/r(2r – 6r’) ………(1)

Now put v’ = 0

 πh/r(2rr’ – 3r’²) = 0

(2rr’ – 3r’²) = 0

2r’r = 3r’²

r’ = 2r/3

So, v is maximum at r’ = 2r/3

The maximum volume of cylinder = πh/r[r. 4r²/9 – 8r²/27]

= πhr²[4/27]

= 4/27πh(h tanα)²

= 4/27πh³ tan²α

问题19：一个半径为10 m的圆柱形罐正在以每小时314立方米的速度装满小麦。然后小麦的深度以

(A)1立方米/小时(B)0.1立方米/小时(C)1.1立方米/小时(D)0.5立方米/小时

解决方案：

Given,

Radius of cylinder = 10m   [radius is fixed]

Rate of increase of volume = 314m³/h

ie   dv/dt = 314m³/h

Now, the volume of cylinder = πr²h

v = π.(10)².h

v = 100πh

On differentiating w.r.t t, we get

dv/dt = 100π$\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{100π}.\frac{dv}{dt}=\frac{1}{314}.314$

$\frac{dh}{dt}=1m/h$  

So option A is correct

问题20.在点(2，-1)上，曲线x的切线的斜率x = t ² + 3t – 8，y = 2t ² – 2t – 5为

(A)22/7(B)6/7(C)7/6(D)-6/7

解决方案：

Given that the slope of the tangent to the curve x = t² + 2t – 8 and y = 2t² – 2t – 5

On differentiating we get

$\frac{dy}{dx}=2t+3 ;\frac{dy}{dt}=4t-2$

Now, when x = 2,

t² + 3 – 8 = 2

t² + 3 – 10 = 0

t² – 2t + 5t – 10 = 0

(t – 2)(t + 5) = 0

Here, t = 2, t = -5   ……….(1)

When y = -1

2t² – 2t – 5 = -1

2t² – 2t – 4 = 0

t² – t – 2 = 0

(t + 1)(t – 2) = 0

t = -1 or t = 2  ……….(2)

From eq(1) & eq(2) satisfies both,

Now, $\frac{dy}{dx}=slope=\frac{dy}{dt}$

$\frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}$

So, option B is the correct.

问题21：如果m的值是y，则线y = mx +1是曲线y ² = 4x的切线

(A)1(B)2(C)3(D)1/2

解决方案：

The curve if y² = 4x  …….(1)

On differentiating we get

$2y\frac{dy}{dx}=4$

$\frac{dy}{dx}=\frac{2}{y}$

The slope of the tangent to the given curve at point(x, y)

$\frac{dy}{dx}=\frac{2}{y}$

m = 2/y

y = 2/m

The equation of line is y = mx + 1

Now put the value of y, we get the value of x

2/m = mx + 1

x = 2 – m/m

Now put the value of y and x in eq(1), we get

(2/m)² = 4(2 – m/m)

m = 1

Hence, the option A is correct

问题22。曲线2y + x ² = 3上的点(1、1)的法线为

(A)x + y = 0(B)x – y = 0

(C)x + y +1 = 0(D)x – y = 1

解决方案：

The equation of curve 2y + x² = 3

On differentiating w.r.t x, we get

2$\frac{dy}{dx}+2x=0$

dy/dx = -x

The slope of the tangent to the given curve at point(1, 1)

dy/dx = -x = -1

m = -1

And slop of normal = 1

Now the equation of normal 

(y -1) = 1(x – 1)

x – y = 0

So, B option is correct

问题23.曲线x ² = 4y通过(1、2)的法线是

(A)x + y = 3(B)x – y = 3(C)x + y = 1(D)x – y = 1

解决方案：

The equation of curve is x² = 4y …….(1)

On differentiating w.r.t x, we get

2x = $4\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{x}{2}$

The slop of normal at (x, y)

-dx/dy = -2/x = m

The slop at given point(1, 2)

m = (y – 2)/(x – 1)

-2/x = (y – 2)/(x – 1)

y = 2/x

Now put the value of y in eq(1)

x² = 4(2/x)

x = 2

and y = 1

So the point is (2, 1)

Now the slope of normal at point(2, 1) = -2/2 = -1

The equation of the normal is

(y – 1) = -1(x – 2)

x + y = 3

So option A is correct

问题24.曲线9y ² = x ³上的点，曲线的法线与轴的交点相等，分别为

(一种) $(4,±\frac{8}{3})$ (B) $(4,\frac{-8}{3})$

(C) $(4,±\frac{3}{8})$ (D) $(±4,\frac{3}{8})$

解决方案：

Given equation 9y² = x³

On differentiating w.r.t x, we get

18y dy/dx = 3x²

dy/dx = 3x²/18y 

dy/dx = x²/6y 

Now, the slope of the normal to the given curve at point (x₁, y₁) is

$-1=\frac{6y_1}{x^2_1}$

Hence, the equation of the normal to the curve at point (x₁, y₁) is

$y-y_1=\frac{-6y_1}{x^2_1}(x-x_1)$

$x^2_1y-x_1^2y=-6xy_1+6x_1y_1$

$\frac{6xy_1}{6x_1y_1+x_1^2y_1}+\frac{x_1^2y}{6x_1y_1+x^2_1y_1}$

$\frac{\frac{x}{x_1(6+x_1)}}{6}+\frac{\frac{y}{y_1(6+x_1)}}{x_1}=1$

 According to the question it is given that the normal 

make equal intercepts with the axes.

So,

$\frac{x_1(6+x_1)}{6}+\frac{y_1(6+x_1)}{x_1}$

$x_1^2=6y_1$          …………(1)

The point (x₁, y₁)lie on the curve,

$9_1^2=x_1^22$      …………(2)

From eq(1) and (2), we get

$9(\frac{x_1^2}{6})^2=x_1^3=\frac{x_1^4}{4}=x^3_1=x_1=4$

From eq(2), we get

$9y_1^2=(4)^3=64$

$y_1^2=\frac{64}{9}$

$y_1=±\frac{8}{3}$

Hence, the required points are $(4,±\frac{8}{3})$

So, option A is correct.