Content of this article has been merged with Chapter 6 Application of Derivatives – Miscellaneous Exercise as per the revised syllabus of NCERT.
Question 12. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is [Tex](a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}[/Tex]
Solution:
Given, a triangle ABC
Let, PE = a & PD = b
In the △ABC, ∠B = 90
Let ∠C = θ, so, ∠DPA = θ
DP|| BC.
Now in â–³ADDP,
cosθ = DP/AP = b/AP
AP = b/cosθ
In â–³EPC,
sinθ = EP/CP = a/CP
CP = a/sin θ
Now AC = h = PA + PC
h = [Tex]\frac{b}{\cos θ}+\frac{a}{\sin θ}[/Tex]
h(θ) = b sec θ + a cosec θ
Put h'(θ) = [Tex]\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}[/Tex]
[Tex]\frac{b}{\cos θ}.\frac{\sin θ}{\cos θ}=\frac{a}{\sin θ}.\frac{\cos θ}{\sin θ}[/Tex]
b sin3θ = a cos 3θ
tan3θ = a/b
tanθ = (a/b)1/3
secθ = [Tex]\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}[/Tex]
cosecθ = [Tex]\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}[/Tex]
hmax = [Tex]b.\frac{\sqrt{b^{2/3}+a^{2/3}}}{b^{1/3}}+a.\frac{\sqrt{b^{2/3}+a^{2/3}}}{a^{1/3}}[/Tex]
hmax = (b2/3+a2/3)3/2
Question 13. Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution:
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
Question 14. Find the absolute maximum and minimum values of the function f given by f(x) = cos2 x + sin x, x ∈ [0, π]
Solution:
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”(Ï€/2) = -1 + 2 = 1 > 0
x = π/2 is a point of local minima f(π/2) = 1
f”(Ï€/6) = [Tex]\frac{-1}{2}-2.\frac{1}{2}=\frac{-3}{2}<0[/Tex]
x = π/6 is a point of local maxima f(π/6) = 5/4
[Tex]f”(\frac{5Ï€}{6})=\frac{-1}{2}-2.(\frac{-1}{2})>0[/Tex]
x = 5π/6​ is a point of local minima f(5π/6) = 5/4
Global/Absolute maxima = ma{f(0), f(Ï€/6), f(Ï€)}
= max{1, 5/4, 1}
= 5/4 = Absolute maxima value
Global/Absolute minima = min{f(0), f(Ï€/2), f(Ï€/6), f(Ï€)}
= min{1, 1, 5/4, 1}
= 1 = Absolute minima value
Question 15. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4Ï€/3
Solution:
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 â–³ 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 = [Tex]\frac{1}{3}Ï€r^2h[/Tex]
v(θ) = [Tex]\frac{1}{3}.πR^2\sin{-2}θ.R(1+\cosθ)[/Tex]
[Tex]v'(θ)=\frac{}{}[\sin{-2}θ(-\sinθ)+(1+\cosθ)(2\sinθ\cosθ)][/Tex]
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([Tex]1+\frac{1}{3} [/Tex])
h = 4r/3
Hence proved
Question 16. Let f be a function defined on [a, b] such that f′(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).
Solution:
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
Question 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also, find the maximum volume.
Solution:
In â–³ABC,
AC2 = BC2 + AB2
4R2 = 4r2 + h2
r2 = R2-[Tex]\frac{h^2}{4} [/Tex] ……….(1)
Now, volume of cylinder = πr2h
Put the value ov r2 from eq(1), we get
V = π([Tex]R^2\frac{-h^2}{4} [/Tex]).h
V(h) = [Tex]Ï€R^2h-\frac{Ï€h^3}{4}[/Tex]
On differentiating both side we get
V ‘(h) = [Tex]Ï€R^2h-\frac{3Ï€h^3}{4}[/Tex]
Now, put V'(h) = 0
Ï€R2 = [Tex]\frac{3}{4}Ï€h^2[/Tex]
[Tex]h=\frac{2R}{\sqrt{3}}[/Tex]
Now the maximum volume of cylinder = Ï€[R2. 2R/√3 – 1/4.4R2/3.2R/√3]
= 4πR3/ 3√3
Question 18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α is one-third that of the cone and the greatest volume of the cylinder is 4/27πh3tan2α.
Solution:
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 = Ï€r’2h’
v = Ï€r’2(h(r – r’)/r)
= Ï€(h(rr’2 – r’3)/r)
On differentiating we get
v’ = Ï€h/r(2rr’ – 3r’2)
Again on differentiating we get
v” = Ï€h/r(2r – 6r’) ………(1)
Now put v’ = 0
Ï€h/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 = Ï€h/r[r. 4r2/9 – 8r2/27]
= πhr2[4/27]
= 4/27πh(h tanα)2
= 4/27πh3 tan2α
Question 19. A cylindrical tank of a radius 10 m is being filled with wheat at the rate of 314 cubic meters per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h
Solution:
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 = 100Ï€h
On differentiating w.r.t t, we get
dv/dt = 100Ï€[Tex]\frac{dh}{dt}[/Tex]
[Tex]\frac{dh}{dt}=\frac{1}{100Ï€}.\frac{dv}{dt}=\frac{1}{314}.314[/Tex]
[Tex]\frac{dh}{dt}=1m/h [/Tex]
So option A is correct
Question 20. The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) is
(A) 22/7 (B) 6/7 (C) 7/6 (D) -6/7
Solution:
Given that the slope of the tangent to the curve x = t2 + 2t – 8 and y = 2t2 – 2t – 5
On differentiating we get
[Tex]\frac{dy}{dx}=2t+3 ;\frac{dy}{dt}=4t-2[/Tex]
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, [Tex]\frac{dy}{dx}=slope=\frac{dy}{dt}[/Tex]
[Tex]\frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}[/Tex]
So, option B is the correct.
Question 21.The line y = mx + 1 is a tangent to the curve y2 = 4x if the value of m is
(A) 1 (B) 2 (C) 3 (D)1/2
Solution:
The curve if y2 = 4x …….(1)
On differentiating we get
[Tex]2y\frac{dy}{dx}=4[/Tex]
[Tex]\frac{dy}{dx}=\frac{2}{y}[/Tex]
The slope of the tangent to the given curve at point(x, y)
[Tex]\frac{dy}{dx}=\frac{2}{y}[/Tex]
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
Question 22. The normal at the point (1, 1) on the curve 2y + x2 = 3 is
(A) x + y = 0 (B) x – y = 0
(C) x + y +1 = 0 (D) x – y = 1
Solution:
The equation of curve 2y + x2 = 3
On differentiating w.r.t x, we get
2[Tex]\frac{dy}{dx}+2x=0[/Tex]
dy/dx = -x
The slope of the tangent to the given curve at point(1, 1)
dy/dx = -x = -1
m = -1
And slope of normal = 1
Now the equation of normal
(y -1) = 1(x – 1)
x – y = 0
So, B option is correct
Question 23. The normal to the curve x2 = 4y passing (1, 2) is
(A) x + y = 3 (B) x – y = 3 (C) x + y = 1 (D) x – y = 1
Solution:
The equation of curve is x2 = 4y …….(1)
On differentiating w.r.t x, we get
2x = [Tex]4\frac{dy}{dx}[/Tex]
[Tex]\frac{dy}{dx}=\frac{x}{2}[/Tex]
The slope of normal at (x, y)
-dx/dy = -2/x = m
The slope 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
Question 24. The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are
(A) [Tex](4,±\frac{8}{3}) [/Tex] (B) [Tex](4,\frac{-8}{3})[/Tex]
(C)[Tex](4,±\frac{3}{8}) [/Tex] (D)[Tex] (±4,\frac{3}{8})[/Tex]
Solution:
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
[Tex]-1=\frac{6y_1}{x^2_1}[/Tex]
Hence, the equation of the normal to the curve at point (x1, y1) is
[Tex]y-y_1=\frac{-6y_1}{x^2_1}(x-x_1)[/Tex]
[Tex]x^2_1y-x_1^2y=-6xy_1+6x_1y_1[/Tex]
[Tex]\frac{6xy_1}{6x_1y_1+x_1^2y_1}+\frac{x_1^2y}{6x_1y_1+x^2_1y_1}[/Tex]
[Tex]\frac{\frac{x}{x_1(6+x_1)}}{6}+\frac{\frac{y}{y_1(6+x_1)}}{x_1}=1[/Tex]
According to the question it is given that the normal
make equal intercepts with the axes.
So,
[Tex]\frac{x_1(6+x_1)}{6}+\frac{y_1(6+x_1)}{x_1}[/Tex]
[Tex]x_1^2=6y_1 [/Tex] …………(1)
The point (x1, y1)lie on the curve,
[Tex]9_1^2=x_1^22 [/Tex] …………(2)
From eq(1) and (2), we get
[Tex]9(\frac{x_1^2}{6})^2=x_1^3=\frac{x_1^4}{4}=x^3_1=x_1=4[/Tex]
From eq(2), we get
[Tex]9y_1^2=(4)^3=64[/Tex]
[Tex]y_1^2=\frac{64}{9}[/Tex]
[Tex]y_1=±\frac{8}{3}[/Tex]
Hence, the required points are [Tex](4,±\frac{8}{3}) [/Tex]
So, option A is correct.
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