NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

Mathematics (Maths) Class 12

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Q.1. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.
Ans.
Ball of salt is spherical
∴ Volume of ball,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev where r = radius of the ball
As per the question,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev where S = surface area of the ball
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev[∵ S = 4pr2]
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev(K = Constant of proportionality)
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the radius of the ball is decreasing at constant rate.

Q.2. If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.
Ans.
We know that:
Area of circle, A = πr2, where r = radius of the circle.
and perimeter = 2πr
As per the question,
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= K, where K = constant
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(1)
Now Perimeter c = 2πr
Differentiating both sides w.r.t., t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev[From (1)]
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the perimeter of the circle varies inversely as the radius of the circle.

Q.3. A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.
Ans.
Given that height of the kite (h) = 151.5 m
Speed of the kite(V) = 10 m/s
Let FD be the height of the kite and AB be the height of the boy.
Let AF = x m
∴ BG = AF = x
m andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
From the figure, we get that
GD = DF – GF ⇒ DF - AB
= (151.5 – 1.5) m = 150 m [∵ AB = GF]
Now in ΔBGD,
BG2 + GD2 = BD2 (By Pythagoras Theorem)
⇒ x2 + (150)2 = (250)2
⇒ x2 + 22500 = 62500 ⇒ x2 = 62500 – 22500
⇒ x2 = 40000 ⇒ x = 200 m
Let initially the length of the string be y m
∴ In ΔBGD
BG2 + GD2 = BD2 ⇒ x2 + (150)2 = y2
Differentiating both sides w.r.t., t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 2 x 200 x 10 = 2 x 250NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the rate of change of the length of the string is 8 m/s.

Q.4. Two  men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated.
Ans.
Let P be any point at which the two roads are inclined at an angle of 45°.
Two men A and B are moving along the roads PA and PB respectively with the same speed ‘V’.
Let A and B be their final positions such that AB = y
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∠APB = 45° and they move with the same speed.
∴ ∠APB is an isosceles triangle. Draw NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev 
AB = y ∴NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevand PA = PB = x (let)
∠APQ = ∠BPQ =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
[∵ In an isosceles D, the altitude drawn from the vertex, bisects the base]
Now in right ΔAPQ ,
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t, t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the rate of their separation isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevunit/s.

Q.5. Find an angle θ, 0 < θ <NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevwhich increases twice as fast as its sine.
Ans.
As per the given condition,
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 2 cosNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev⇒ 1 = 2 cos θ
∴ cos θ =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev⇒ cos θ =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the required angle isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

Q.6. Find the approximate value of (1.999)5.
Ans.
(1.999)5 = (2 – 0.001)5 
Let x = 2 and Δx = – 0.001
Let y = x5
Differentiating both sides w.r.t, x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 5x= 5(2)4= 80
Now Δy =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 80 × (- 0.001) = - 0.080

∴ (1.999)5 = y + Dy
= x5 – 0.080 = (2)5 – 0.080 = 32 – 0.080 = 31.92
Hence, approximate value of (1.999)5 is 31.92.

Q.7. Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.
Ans.
Internal radius r = 3 cm
and external radius R = r + Δr = 3.0005 cm
∴ Δr = 3.0005 – 3 = 0.0005 cm
Let y = r3 ⇒ y + Δy = (r + Δr)3 = R3 = (3.0005)3 ...(i)
Differentiating both sides w.r.t., r, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 3r2
∴ Δy =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 3rx 0.0005
= 3 x (3)2 x 0.0005 = 27 x 0.0005 = 0.0135
∴ (3.0005)3 = y + Δy [From eq. (i)]
= (3)3 + 0.0135 = 27 + 0.0135 = 27.0135
Volume of the shell =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
= 4π × 0.005 = 4 x 3.14 x 0.0045 = 0.018 π cm3
Hence, the approximate volume of the metal in the shell is 0.018 π cm3.

Q.8. A man, 2m tall, walks at the rate ofNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevtowards a street light which is NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevabove the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevfrom the base of the light?
Ans.
Let AB is the height of street light post and CD is the height of the man such that
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevand CD 2 m
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Let  BC = x length (the distance of the man from the lamp post) and CE = y is the length of the shadow of the man at any instant.
From the figure, we see that
ΔABE ~ ΔDCE [by AAA Similarity]
∴ Taking ratio of their corresponding sides, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 8y = 3x + 3y ⇒ 8y - 3y = 3x ⇒ 5y = 3x
Differentiating both sides w.r.t, t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
[∵ man is moving in opposite direction]
= - 1 m/s
Hence, the length of shadow is decreasing at the rate of 1 m/s.
Now let u = x + y
(u = distance of the tip of shadow from the light post)
Differentiating both sides w.r.t. t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the tip of the shadow is moving at the rate ofNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
towards the light post and the length of shadow decreasing at the rate of 1 m/s.

Q.9. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?
Ans.
Given that L = 200(10 – t)2
where L represents the number of litres of water in the pool.
Differentiating both sides w.r.t, t, we get∴
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 200 x 2(10 - t) (-1) = - 400(10 - t)
But the rate at which the water is running out
=NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev 400(10 - t)...(1)
Rate at which the water is running after 5 seconds
= 400 x (10 - 5) = 2000 L/s (final rate)
For initial rate put t = 0 = 400(10 - 0) = 4000 L/s
The average rate at which the water is running out
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the required rate = 3000 L/s.

Q.10. The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.
Ans.
Let x  be the length of the cube
∴ Volume of the cube V = x3 ...(1)
Given thatNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating Eq. (1) w.r.t. t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now surface area of the cube, S = 6x2 
Differentiating both sides w.r.t. t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev(4K = constant)
Hence, the surface area of the cube varies inversely as the length of the side.

Q.11. x and y are the sides of two squares such that y = x – x2 . Find the rate of change of the area of second square with respect to the area of first square.
Ans.
Let area of the first square A1 = x2 
and area of the second square A2 = y2 
Now A1= x2 and A2 = y2  = (x – x2)
Differentiating both A1 and A2 w.r.t. t, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the rate of change of area of the second square with respect to first is
2x2 – 3x + 1.

Q.12. Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally.
Ans.
The two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90°.
Equation of the two circles are given as
2x = y2...(i)
and 2 = k ...(ii)
Differentiating eq. (i) and (ii) w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
(m1 = slope of the tangent)
⇒ 2xy = k
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
[m2 = slope of the other tangent]
If the two tangents are perpendicular to each other,
then m1 x m2 = – 1
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now solving 2x = y2 [From (i)]
and 2xy = k [From (ii)]
From eq. (ii) NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Putting the value of y in eq. (i)
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 8x3 = k2 ⇒ 8(1)3 = k2 ⇒ 8 = k2
Hence, the required condition is k= 8.

Q.13. Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.
Ans.
Given circles are xy = 4 ...(i)
and x2 + y2 = 8 ...(ii)
Differentiating eq. (i) w.r.t., x
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(iii)
where, m1 is the slope of the tangent to the curve.
Differentiating eq. (ii) w.r.t. x
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
where, m2 is the slope of the tangent to the circle.
To find the point of contact of the two circles
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Putting the value of y2 in eq. (ii)
x2 + x2 = 8 ⇒ 2x2 = 8 ⇒ x2 = 4
∴ x = ±    2
∵ x= y2 ⇒ y = ± 2
∴ The point of contact of the two circles are (2, 2) and ( - 2, 2).

Q.14. Find the co-ordinates of the point on the curveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev at which tangent is equally inclined to the axes.

Ans.
Equation of curve is given byNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Let (x1, y1) be the required point on the curve   
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x1, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Since the tangent to the given curve at (x1, y1) is equally inclined to the axes.
∴ Slope of the tangentNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, from eq. (i) we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Putting the value of y1 in the given equation of the curve.
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Since y1 = x1 
∴ y1 = 4
Hence, the required point is (4, 4).

Q.15. Find the angle of intersection of the curves y = 4 – x2 and y = x2.
Ans.
We know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.
The given curves are y = 4 – x2 ... (i)  and    y = x2 ...(ii)
Differentiating eq. (i) and (ii) with respect to x, we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
mis the slope of the tangent to the curve (i).
andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
m2 is the slope of the tangent to the curve (ii).
So, m1 = – 2x and m2 = 2x
Now solving eq. (i) and (ii) we get
⇒ 4 – x2 = x2 ⇒ 2x2 = 4 ⇒ x2 = 2 ⇒ x = ± √2
So, m1 = - 2 x =- 2√2 and m2 = 2x = 2√2
Let θ be the  angle of intersection of two curves
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the required angle is tan-1NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

Q.16. Prove that the curves y= 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2).
Ans.
Given that the equation of the two curves are y2 = 4x ...(i)
and x2 + y2 – 6x + 1 = 0 ...(ii)
Differentiating (i) w.r.t. x, we getNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Slope of the tangent at (1, 2),NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev = 1
Differentiating (ii) w.r.t. x ⇒ 2x + 2yNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Slope of the tangent at the same point (1, 2)
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
We see that m1 = m2 = 1 at the point (1, 2).
Hence, the given circles touch each other at the same point (1, 2).

Q.17. Find the equation of the normal lines to the curve 3x2 – y2 = 8 which are parallel to the line x + 3y = 4.
Ans.
We have equation of the curve 3x2 – y2 = 8
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Slope of the tangent to the given curve =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Slope of the normal to the curve =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now differentiating both sides the given line x + 3y = 4
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Since the normal to the curve is parallel to the given line x + 3y = 4.
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Putting the value of y in 3x2 – y2 = 8, we get
3x2 – x2 = 8 ⇒ 2x2 = 8 ⇒ x2 = 4 ⇒ x = ±    2
∴ y = ± 2
∴ The points on the curve are (2, 2) and (– 2, – 2).
Now equation of the normal to the curve at (2, 2) is
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 3y – 6 = – x + 2 ⇒ x + 3y = 8
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 3y + 6 = – x – 2 ⇒ x + 3y = – 8
Hence, the required equations are x + 3y = 8 and x + 3y = – 8 or x + 3y = ± 8.

Q.18. At what points on the curve x2 + y2 – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis?
Ans.
Given that the equation of the curve is
x2 + y2 – 2x – 4y + 1 = 0 ...(i)
Differentiating both sides w.r.t. x, we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Since the tangent to the curve is parallel to the y-axis.
∴ SlopeNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, from eq. (ii) we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now putting the value of y in eq. (i), we get
⇒ x2 + (2)2 – 2x – 8 + 1 = 0
⇒ x2 – 2x + 4 – 8 + 1 = 0
⇒ x2 – 2x – 3 = 0 ⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0 ⇒ (x - 3) (x + 1) = 0
⇒ x = – 1 or 3
Hence, the required points are (– 1, 2) and (3, 2).

Q.19. Show that the lineNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevtouches the curve y = b.

NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevat the point where the curve intersects the axis of y.
Ans.
Given that y = b × e– x/a, the equation of curve
andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevthe equation of line.
Let the coordinates of the point where the curve intersects the y-axis be (0, y1) Now differentiating y = b × e– x/a both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, the slope of the tangent, m=NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevboth sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, the slope of the line, m=NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
If the line touches the curve, then m1 = m2
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev(Taking log on both sides)
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev⇒ x = 0
Putting x = 0 in equation y = b × e – x/a 
⇒ y = b × e0 = b
Hence, the given equation of curve intersect at (0, b) i.e. on y-axis.

Q.20. Show that f (x) = 2x + cot–1x + logNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevis increasing in R.
Ans.
Given that f (x) = 2x + cot–1x + logNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
For increasing function, f ′(x) ≠ 0
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get 4x4 + 1 + 4x2 ≥ 1 + x2
⇒ 4x4 + 4x2 - x2 ≥ 0 ⇒ 4x4 + 3x2 ≥ 0 ⇒ x2(4x2 + 3) ≠ 0
which is true for any value of x ∈ R.
Hence, the given function is an increasing function over R.

Q.21. Show that for a ≥ 1, f (x) = √3 sinx – cosx – 2ax + b is decreasing in R.
Ans.
Given that: f (x) = √3 sin x - cos x - 2ax + b, a≥ 1
Differentiating both sides w.r.t. x, we get
f ′(x) = √3 cos x + sin x- 2a
For decreasing function, f ′(x) < 0
∴ √3 cos x + sin x- 2a < 0
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Since cos x ∈ [ - 1, 1] and a ≥ 1
∴ f ′(x) < 0
Hence, the given function is decreasing in R.

Q.22. Show that f (x) = tan–1(sinx + cosx) is an increasing function in
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Ans.
Given that: f (x) = tan–1(sin x + cos x) inNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get  
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
For an increasing function f ′(x) ≥ 0
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ cos x – sin x ≥ 0 NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ cos x ≥ sin x, which is true forNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the given function f(x) is an increasing function inNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

Q.23. At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.
Ans.
Given that: y = – x3 + 3x2 + 9x – 27
Differentiating both sides w.r.t. x, we getNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= – 3x2 + 6x + 9
Let slope of the curveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ z = -3x2 + 6x + 9
Differentiating both sides w.r.t. x, we getNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= – 6x + 6
For local maxima and local minima,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= 0
∴ – 6x + 6 = 0 ⇒ x = 1
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Put x = 1 in equation of the curve y = (– 1)3 + 3(1)2 + 9(1) – 27
= – 1 + 3 + 9 – 27 = – 16
Maximum slope = – 3(1)2 + 6(1) + 9 = 12
Hence, (1, – 16) is the point at which the
slope of the given curve is maximum and maximum slope = 12.

Q.24. Prove that f (x) = sinx + √3 cosx has maximum value at x =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Ans.
We have: f (x) = sin x + √3 cosx =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev(Maxima)
Maximum value of the function atNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev   
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the given function has maximum value atNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevand     the maximum value is 2.

Long Answer (L.A.)
Q.25. If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Ans.
Let ΔABC be the right angled triangle in which ∠B = 90°
Let AC = x, BC = y
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∠ACB = θ
Let Z  = x + y (given)
Now area of ΔABC,A =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. y we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(i)
For local maxima and local minima,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ yZ ≠ 0 ( ∵ y ≠ 0 and Z ≠ 0)
∴ Z - 3y = 0
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev( ∵ Z = x + y)
⇒ 3y = x + y ⇒ 3y - y = x ⇒ 2y = x
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating eq. (i) w.r.t. y, we haveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the area of the given triangle is maximum when the angle
between its hypotenuse and a side isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

Q.26. Find the points of local maxima, local minima and the points of inflection of the function f (x) = x5 – 5x4 + 5x3 – 1.  Also find the corresponding local maximum and local minimum values.
Ans.
We have f (x) = x5 – 5x4 + 5x3 – 1
⇒ f ′(x) = 5x4 - 20x3 + 15x2
For local maxima and local minima, f ′(x) = 0
⇒ 5x4 – 20x3 + 15x2 = 0 ⇒ 5x2(x2 - 4x + 3) = 0
⇒ 5x2(x2 - 3x - x + 3) = 0 ⇒ x2(x - 3) (x - 1) = 0
∴ x = 0, x = 1 and x = 3
Now f ″(x) = 20x3 - 60x2 + 30x
⇒ f ″(x)at x = 0 = 20(0)3 - 60(0)2 + 30(0) = 0 which is neither maxima nor minima.
∴ f (x) has the point of inflection at x = 0
f ″(x)at x = 1 = 20(1)3 - 60(1)2 + 30(1)
= 20 - 60 + 30 = - 10 < 0 Maxima
f ″(x)at x = 3 = 20(3)3 - 60(3)2 + 30(3)
= 540 - 540 + 90 = 90 > 0 Minima
The maximum value of the function at x = 1
f(x) = (1)5 – 5(1)4 + 5(1)3 – 1 = 1 – 5 + 5 – 1 = 0
The minimum value at x = 3 is
f (x) = (3)5 – 5(3)4 + 5(3)3 – 1
= 243 – 405 + 135 – 1 = 378 – 406 = – 28
Hence, the function has its maxima at x = 1 and the maximum value = 0 and it has minimum value at x = 3 and its minimum value is – 28.
x = 0 is the point of inflection.

Q.27. A telephone company in a town has 500 subscribers on its list and collects fixed charges of ₹ 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of ₹ 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?
Ans.
Let us consider that the company increases the annual subscription by ₹ x.
So, x is the number of subscribers who discontinue the services.
∴ Total revenue, R(x) = (500 – x) (300 + x)
= 150000 + 500x – 300x – x2
= – x2 + 200x + 150000
Differentiating both sides w.r.t. x, we get R ′(x) = - 2x + 200
For local maxima and local minima, R ′(x) = 0
2x + 200 = 0 ⇒ x = 100
R ″(x) =  - 2 < 0 Maxima
So, R(x) is maximum at x = 100
Hence, in order to get maximum profit, the company should increase its annual subscription by ₹ 100.

Q.28. If the straight line x cosα + y sinα = p touches the curveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev then prove that a2 cos2α + b2 sin2α = p2.
Ans.
The given curve isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev ...(i)
and the straight line x cos α + y sin α = p ...(ii)
Differentiating eq. (i) w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So the slope of the curve =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now differentiating eq. (ii) w.r.t. x, we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, the slope of the straight line = – cot α
If the line is the tangent to the curve, then
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now from eq. (ii) we have x cos α + y sin α = p
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ a2 cos2 αy + b2 sin2 αy = b2 sin αp
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

⇒ a2 cos2 α + b2 sin2 α = p × pNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, a2 cos2 α + bsin2 α = p2
Alternate method: 
We know that y = mx + c will touch the ellipse
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Here equation of straight line is x cos α + y sin α = p and
that of ellipse isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
x cos α + y sin α = p
⇒ y sin α = – x cos α + p
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Comparing with y = mx + c, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So, according to the condition, we get c2 = a2m+ b2
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, a2 cos2 α + b2 sin2 α = p2
Hence proved.

Q.29. An open box with square base is to be made of a given quantity of card board of area c2. Show that the maximum volume of the box isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevcubic units.
Ans.
Let x  be the length of the side of the square base of the cubical open box and y be its height.
∴ Surface area of the open box
c2 = x2 + 4xyNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev ...(i)
Now volume of the box, V = x x x x y
⇒ V = x2y
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(ii)
For local maxima and local minima,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now again differentiating eq. (ii) w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Volume of the cubical box (V) = x2 y
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the maximum volume of the open box isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevcubic units.

Q.30. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.
Ans.
Let x and y be the length and breadth of a given rectangle ABCD as per question, the rectangle be revolved about side AD which will make a cylinder with radius x and height y.
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Volume of the cylinder V = πr2h
 ⇒ V = πx2y ...(i)
Now perimeter of rectangle P = 2(x + y) ⇒ 36 = 2(x + y)
⇒ x + y = 18 ⇒ y = 18 – x ...(ii)
Putting the value of y in eq. (i) we get
V = πx2(18 – x)
⇒ V = π(18x2 – x3)
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(iii)
For local maxima and local minimaNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ p(36x – 3x2) = 0 ⇒ 36x – 3x2 = 0
⇒ 3x(12 – x) = 0
⇒ x ≠ 0 ∴ 12 – x = 0 ⇒  x = 12
From eq. (ii) y = 18 – 12 = 6
Differentiating eq. (iii) w.r.t. x, we getNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
at x = 12 NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev= π(36 – 6 x 12)
= π(36 - 72) = - 36π < 0 maxima
Now volume of the cylinder so formed = πx2y
= π x (12)2 x 6  = π x 144 x 6 = 864π cm3
Hence, the required dimensions are 12 cm and 6 cm  and the maximum volume is 864π cm3.

Q.31. If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?
Ans.
Let x  be the edge of the cube and r be the radius of the sphere.
Surface area of cube = 6x2
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
and surface area of the sphere = 4πr2
∴ 6x2 + 4πr2 = K(constant) ⇒NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(i)
Volume of the cube = x3 and the volume of sphere =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Sum of their volumes (V) = Volume of cube + Volume of sphere
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(ii)
For local maxima and local minima,NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
x ≠ 0 ∴ x -NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now putting the value of K in eq. (i), we get
6x2 + 4πr2 = x2(p + 6)
⇒ 6x2 + 4πr2 = πx2 + 6x2 ⇒ 4πr2 = πx2 ⇒ 4r2 = x2
∴2r = x
∴ x : 2r = 1 : 1
Now differentiating eq. (ii) w.r.t x, we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
PutNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So it is minima.
Hence, the required ratio is 1 : 1 when the combined volume is minimum.

Q.32. AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ ABC is maximum, when it is isosceles.
Ans.
Let AB be the diameter and C be any point on the circle with radius r.
∠ACB = 90° [angle in the semi circle is 90°]
Let AC = x
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(i)
Now area ofNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Let A2 = Z
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(ii)
For local maxima and local minimaNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev

x ≠ 0 ∴ 2r2 - x2 = 0 

⇒ x2 = 2r2 ⇒ x = √2r = AC
Now from eq. (i) we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
So AC = BC
Hence, ΔABC is an isosceles triangle.
Differentiating eq. (ii) w.r.t. x, we getNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Put x = √2r
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the area of ΔABC is maximum when it is an isosceles triangle.

Q.33. A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5/cm2 and the material for the sides costs Rs 2.50/cm2 . Find the least cost of the box.
Ans.
Let x  be the side of the square base and y be the length of the vertical sides.
Area of the base and bottom = 2x2 cm2 
∴Cost of the material required = ₹ 5 x 2x2 = ₹ 10x2
Area of the 4 sides = 4xy cm2
∴ Cost of the material for the four sides
= ₹ 2.50 x 4xy = ₹ 10xy
Total cost C = 10x2 + 10xy ...(i)
New volume of the box = x x x x y
⇒ 1024 = x2y
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(ii)
Putting the value of y in eq. (i) we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(iii)
For local maxima and local minimaNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
⇒ 20x3 – 10240 = 0 ⇒ x= 512 ⇒ x = 8 cm
Now from eq. (ii)
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Cost of material used C = 10x2 + 10xy
= 10 x 8 x 8 + 10 x 8 x 16  = 640 + 1280 = 1920
Now differentiating eq. (iii) we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Put x = 8
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the required cost is ₹1920 which is the minimum.

Q.34. The sum of the surface areas of a rectangular parallelopiped with sides x, 2x andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevand a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Ans.
Let ‘r ’ be the radius of the sphere.
∴ Surface area of the sphere = 4πr2
Volume of the sphere =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
The sides of the parallelopiped are x, 2x andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Its surface area =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Volume of the parallelopiped =NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
As per the conditions of the question,
Surface area of the parallelopiped + Surface area of the sphere = constant
⇒ 6x2 + 4πr2 = K (constant) ⇒ 4πr2 = K - 6x2
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev...(i)
Now let V = Volume of parallelopiped + Volume of the sphere
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev[from eq. (i)]
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
For local maxima and local minima, we haveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Here x ≠ 0 andNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Squaring both sides, we get
4πx2 = 9(K – 6x2)  ⇒ 4πx2 = 9K - 54x2
⇒ 4πx2 + 54x2 = 9K
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev ...(ii)

⇒ 2x2(2π + 27) = 9K
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now from eq. (i) we have
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Now we haveNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Differentiating both sides w.r.t. x, we get
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Put NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
[ ∵ 27 - 2π > 0]
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevso, it is minima.

Hence, the sum of volume is minimum forNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
∴ Minimum volume,
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
NCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRev
Hence, the required minimum volume isNCERT Exemplar- Application of Derivatives(Part-1) Notes | EduRevand x = 3r. 

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