NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 1.
Using differentials, find the approximate value of each of the following.
a. \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\)
b. \(\text { (33) }^{-\frac{1}{5}}\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 1
∆y is approximately equal to dy
∴ (1) ⇒ (33)\(\frac { -1 }{ 5 }\) ≈ 2-1 – 0.003
= 0.5 – 0.003 = 0.497
Hence (33)\(\frac { -1 }{ 5 }\) ≈ 0.497

Question 2.
Show that the function given by f(x) = \(\frac { log x }{ x }\) has maximum at x = e.
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 2
For maxima, f'(x) = 0
i.e., \(\frac{1-\log x}{x^{2}}\) = 0 ⇒ 1 – log x = 0
⇒ log x = 1 ⇒ x = e
At x = e, f”(e) = \(\frac{-3+2 \log e}{e^{3}}\) < 0,
∴ f has a maximum at x = e

Question 3.
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 3
∆ ABC be an isosceles triangle with AB = AC
Let AB = x, BC = b, \(\frac { dx }{ dt }\) = – 3cm/sec at
Draw AD ⊥ BC. Then D is the midpoint of BC.
i.e., BD = \(\frac { b }{ 2 }\)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 3a
Hence area is decreasing at the rate of \(\sqrt{3}\) bcm²/sec.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 4.
Find the equation of the normal to curve y² = 4x at the point (1,2).
Solution:
y² = 4x
Differentiating w.r.t. x,
2y\(\frac { dy }{ dx }\) = 4
\(\frac { dy }{ dx }\) = \(\frac { 2 }{ y }\)
At(1, 2), \(\frac { dy }{ dx }\) = \(\frac { 2 }{ 2 }\) = 1
Slope of the tangent at (1, 2) = \(\frac { -1 }{ 1 }\) = – 1
Slope of the normal at(1, 2)
Equation of the tangent at (1, 2) is
y – 2 = 1(x – 1)
y – x – 1 = 0
Equation of the normal at (1, 2) is
y – 2 = 1(x – 1)
x + y – 3 = 0

Question 5.
Show that the normal at any point θ to the curve x = a cosθ + aθ sinθ, y = a sinθ – aθ cosθ is at a constant distance from the origin.
Solution:
x = acosθ + a0sinθ
y = asinθ – aθ cosθ
Differentiating w.r.t θ, dx
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 4
Slope of the tangent at the point 0 is \(\frac { sinθ }{ cosθ }\)
∴ Slope of the normal at the point θ is \(\frac { – cosθ }{ sinθ }\)
∴ Equation of the normal at the point θ is
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 5
i.e., xcosθ + ysin θ = a, which is the equation of the straight line in the normal form. Hence ‘a’ is the distance of the normal line from the origin. Thus the normal at any point 0 to the curve is at a constant distance from the origin.

Question 6.
Find the intervals in which the function f given by
f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}, 0 \leq x \leq 2 \pi\) (i) increasing (ii) decreasing.
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 6
Since f is continuous
i. f is increasing: 0 ≤ x ≤ \(\frac { π }{ 2 }\) and \(\frac { 3π }{ 2 }\) ≤ 2π
ii. is decreasing: \(\frac { π }{ 2 }\) ≤ x ≤ \(\frac { 3π }{ 2 }\)

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 7.
Find the intervals in which the function f given by f(x) = x³ + \(\frac{1}{x^{3}}\), x ≠ 0 is
i. increasing
ii. decreasing
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 7
The critical points are x = – 1, x = 0 and x =1. x = – 1, x = 0, x = 1 divide Rinto disjoint open intervals as
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 8
(-∞, -1), (-1, 0),(0, 1) and (1, ∞)
i. f is increasing: x < – 1 and x > 1
ii. f is decreasing: – 1 < x < 0 and 0 < x < 1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 9

Question 8.
Find the maximum area of an isosceles tri-angle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) = 1 with its vertex at one end of the major axis.
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 10
Consider the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) = 1 , a > b
Let A be one end point of the major axis and P(x, y) be a point on the ellipse. Draw PM perpendicular to x axis and extend it to meet the ellipse at Q.
Then the coordinate of Q is (x, – y)
∆APQ is an isosceles triangle
Let S be the area of ∆APQ
S = \(\frac { 1 }{ 2 }\)PQ.AM = \(\frac { 1 }{ 2 }\)(2y)(x + a)
S = y(x + a) = \(\frac { b }{ 2 }\)\(\sqrt{a^{2}-x^{2}}\)(x + a)
Since S is positive, S² is maximum when S is maximum
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 11
When x = – a, the point P coincides with A which is not possible.
∴ x = – \(\frac { a }{ 2 }\)
When x = \(\frac { a }{ 2 }\), \(\frac{d^{2}}{d x^{2}}\left(\mathrm{~S}^{2}\right)\)
= \(\frac{-12 b^{2}}{a^{2}} \frac{a}{2}\left(\frac{a}{2}+a\right)\) < 0
∴ S² is maximum when x = \(\frac { a }{ 2 }\)
i.e., S is maximum when x = \(\frac { a }{ 2 }\)
The maximum value of S
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 12

Question 9.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8m³. If building of tank costs ₹ 70 per sq.metre for the base and ₹ 45 per square metre for sides, what is the cost of least expensive tank?
Solution:
Let x and y be the length and breadth of the base of the tank (in metre). Let C be the cost of building the tank.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 13
Volume of the tank = 8m³ (x)(y).(2) = 8 ⇒ xy = 4
∴ y = \(\frac { 4 }{ x }\)
Area of base = xy
Area of 4 sides = 2(x + y)(2) = 4(x + y)
Cost of construction
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 14
But x = – 2 is not possible, since length of the base cannot be negative
∴ x = 2 When x = 2,
\(\frac{d^{2} \mathrm{C}}{d x^{2}}=\frac{180 \times 8}{8}\) > 0
The minimum cost of construction,
C = 280 + 180(2 + \(\frac { 4 }{ 2 }\)) = ₹ 1000

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 10.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution:
Let r be the radius and x be the side of the square.
Then 2πr + 4x = k
∴ r = \(\frac{k-4 x}{2 \pi}\) … (1)
Let S be the sum of areas of circle and square
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 15

Question 11.
A window is in the form of a rectangle sur-mounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Solution:
Length of the window = 2x,
Height of the rectangular portion = y
i. Perimeter of the window = 10 m
2x + 2y + πx = 10
(π + 2)x + 2y = 10

ii. Let A be the area of the window
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 16

iii. Differentiating A w.r.t. x,
∴ A’ = 10 – 2 πx – 4x + π x
A’ = 10 – π x – 4x
A” = – π – 4 = – (π + 4)
For maxima, A’ = 0
10 – πx – 4x = 0
x(π + 4) = 10
x = \(\frac { 10 }{ π + 4 }\)
When x = \(\frac { 10 }{ π + 4 }\), A” = – (π + 4) < 0
∴ A is maximum when x = \(\frac { 10 }{ π + 4 }\)
2y = 10 – (π + 2)x
= 10 – (π + 2)\(\frac { 10 }{ π + 4 }\)
= \(\frac{10 \pi+40-10 \pi-20}{\pi+4}\) = \(\frac { 20 }{ π + 4 }\)
∴ y = \(\frac { 10 }{ π + 4 }\)
∴ Length of the window = 2x = \(\frac { 20 }{ π + 4 }\)
Length of larger side of window = y = \(\frac { 10 }{ π + 4 }\).

Question 12.
A point on the hypotenuse of a triangle is at distance a and b from the side of the tri¬angle. Show that the minimum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 17
Let ∆ABC be right angled at B and P be a point on the hypotenuse AC at a distance of a from BC and b from AB.
∴ PQ = a and PR = b
Let ∠A = θ then ∠C = 90 – θ
From right triangles ARP and PQC, we get
AP = \(\frac { b }{ sin θ }\) = b cosec θ
PC = \(\frac { a }{ sin(90 – θ) }\) = \(\frac { a }{ cos θ }\) = asec θ
Let l be the length of the hypotenuse
∴ l = AP + PC = asecθ + bcosecθ,
0 < θ < \(\frac { π }{ 2 }\)
\(\frac { dl }{ dθ }\) = asecθ tanθ – bcosecθ cotθ
\(\frac{d^{2} l}{d \theta^{2}}\) = a[secθ.sec²θ + tanθ.secθ tanθ] – b[cosecθ(-cosec²θ) + cot θ(-cosecθ cotθ)] = a[sec³θ + secθtan²θ] + b[cosec³θ + cosecθ cot²θ]
For minima
\(\frac { dl }{ dθ }\) = 0 ⇒ asecθ tanθ – bcosecθ cotθ = 0
⇒ a secθ tanθ = b cosecθ cotθ
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 18
Minimum length of the hypotenuse = \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 13.
Find the points at which the function given by f(x) = (x – 2)4 (x + 1)³ has
i. local maxima
ii. local minima
iii. point of in flexion
Solution:
f(x) = (x – 2)4 (x + 1)³
f’(x) = (x – 2)4[3(x + 1)²] + (x+ 1)³[4(x – 2)³]
= (x – 2)³(x + 1)²[3x – 6 + 4x + 4]
= (x – 2)³(x + 1)²(7x – 2)
f’(x) = 0 ⇒ (x – 2)³(x + 1)²(7x – 2) = 0
⇒ x = – 1, and x = \(\frac { 2 }{ 7 }\) x = 2
Let us apply first derivative test.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 19
i. local maxima is at x = \(\frac { 2 }{ 7 }\)
ii local minima ¡s at x = 2
iii. point of inflexion is at x = – 1

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) = cos² x + sin x, x ∈ [0, π]
f'(x) = – 2cosx sinx + cosx = cosx (1 – 2sinx)
f'(x) = 0 ⇒ cosx (1 – 2sinx) = 0
⇒ cosx = 0 or sin x = \(\frac { 1 }{ 2 }\)
⇒ x = \(\frac { π }{ 2 }\) or x = \(\frac { π }{ 6 }\) and x = \(\frac { 5π }{ 6 }\)
∴ x = \(\frac { π }{ 6 }\), \(\frac { π }{ 2 }\), \(\frac { 5π }{ 6 }\)
f(0) = cos²0 + sin0 = 1² + 0 = 1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 20

Question 15.
Show that the altitude of the right circular cone of maximum volume that can be in-scribed in a sphere of radius r is \(\frac { 4r }{ 3 }\)
Solution:
Let R be the radius and A be the height of the right circular cone inscribed in a sphere of radius r.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 21
i.e., AB = h, OA = OC = r and BC = R
In right triangle OBC, OB = h – r
and OC² = OB² + BC²
i.e., r² = (h – r)² + R²
r² = h² – 2 hr + r² + R²
∴ R² = 2hr – h²
Volume of the cone V = \(\frac { 1 }{ 3 }\) πR²h
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 22
∴ When h = \(\frac { 4r }{ 3 }\), the volume is maximum.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

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:
Let c1, c2 ∈ [a , b\ such that c1, c2.
Since f'(x) > 0 in (a, b), f is continuous in [a, b] and differentiable in (a, b).
∴ By mean value theorem, there exists c ∈ (c1, c2) such that
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 23
Since c1 and c2 are arbitrary, we get f(x) is increasing on (a, b).

Question 17.
Show that the height of the cylinder of maxi-mum volume that can be inscribed in a sphere of radius R is \(\frac{2 \mathrm{R}}{\sqrt{3}}[/latex . Also find the maximum volume.
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 24
Let y be the radius and x be the height of the cylinder.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 25
∴ V is maximum when x = [latex]\frac{2 \mathrm{R}}{\sqrt{3}}\)
The height of cylinder with maximum volume is \(\frac{2 \mathrm{R}}{\sqrt{3}}\)
∴ (1) → Maximum volume,
V = \(\frac{\pi}{4}\left(4 \mathrm{R}^{2} \cdot \frac{2 \mathrm{R}}{\sqrt{3}}-\frac{8 \mathrm{R}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{R}^{3}}{3 \sqrt{3}}\)

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 a is one-third that of the cone and the greatest volume of cylinder is \(\frac { 4 }{ 27 }\)πh³ tan²α.
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 26
Let y be the radius and x be the height of the right circular cylinder inscribed in a cone of height h and semivertical angle α.
Then AB h, CD = y, BC = x
∴ AC = h – x
From right triangle ACD, tan α = \(\frac { CD }{ AC }\)
= \(\frac { y }{ h – x }\)
⇒ y = (h – x)tan α
Volume of the cylinder V = πy²x
= π(h – x)² tan² α.x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 27
∴ V is maximum when the height of the cylinder is one third of the height of the cone.
(1) → Maximum volume,
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 28

Question 19.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
a. 1 m³/h
b. 0.1 m³/h
c. 1.1 m³/h
d. 0.5 m³/h
Solution:
a. 1 m³/h
Let h be the height of the wheat at time t and V be the volume at that instant
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 29

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. \(\frac { 22 }{ 7 }\)
b. \(\frac { 6 }{ 7 }\)
c. \(\frac { 7 }{ 6 }\)
d. \(\frac { – 6 }{ 7 }\)
Solution:
b. \(\frac { 6 }{ 7 }\)
x = t² + 3t – 8 and y = 2t² – 2t – 5
Put x = 2 and y = – 1, we get
t² + 3t – 8 = 2 and 2t² – 2t – 5 = – 1
i.e., t² + 3t – 10 = 0 and 2t² – 2t – 4 = 0
i.e, (t + 5) (t – 2) = 0 and 2(t – 2) (t+ 1) = 0
i.e., t = -5 or t = 2 and t = 2 or t = – 1
Hence t = 2
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 30

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

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. \(\frac { 1 }{ 2 }\)
Solution:
y² = 4x
Differentiating w.r.t x,
2y\(\frac { dy }{ dx }\) = 4 ⇒ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ y }\)
∴ Slope of the tangent, m = \(\frac { 2 }{ y }\)
∴ y = mx + 1 becomes y = \(\frac { 2 }{ y }\) x + 1
y² = 2x + y
i.e., y² = \(\frac { y² }{ 2 }\) + y since y² = 4x
\(\frac { y² }{ 2 }\) = y ⇒ y² = 2y
⇒ y(y – 2) = 0
y ≠ 0 since slope = \(\frac { 2 }{ y }\)
∴ y = 2
∴ slope m = \(\frac { 2 }{ y }\) = \(\frac { 2 }{ 2 }\) = 1

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 = 0
Solution:
b. x – y = 0
2y + x² = 3
Differentiating w.r.t x,
2\(\frac { dy }{ dx }\) + 2x = 0
\(\frac { dy }{ dx }\) = – x
Slope of the tangent at (1, 1)
= \(\frac { dy }{ dx }\) at (1, 1) = – 1
∴ Slope of the normal at (1, 1) = \(\frac { – 1 }{ – 1 }\) = 1
Equation of the normal at (1, 1) is
y – 1 = 1(x – 1)
⇒ x – y = 0

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 23.
The normal to the curve x² = 4y passing through (1, 2) is
a. x + y = 3
b. x – y = 3
c. x + y = 1
d. x – y = 1
Solution:
a. x + y = 3
x² = Ay
Differentiating w.r.t. x,
2x = 4\(\frac { dy }{ dx }\)
∴\(\frac { dy }{ dx }\) = \(\frac { x }{ 2 }\)
Slope of the tangent = \(\frac { x }{ 2 }\)
∴ Slope of the normal = \(\frac { – 2 }{ x }\)
Let (x1, y1) be a point on the curve
∴ Slope of normal at (x1, y1) = \(\frac{-2}{x_{1}}\)
Equation of the normal at (x1, y1) is
y – y1 = \(\frac{-2}{x_{1}}\)(x – x1)
The normal passes through the point (1, 2)
∴ (2 – y1) = \(\frac{-2}{x_{1}}\)(1 – x1)
Equation of the normal at(x1, y1) is
y – y1 = \(\frac{-2}{x_{1}}\)(1 – x1)
2 – y1 = \(\frac{2}{x_{1}}\) + 2
∴ y1 = \(\frac{2}{x_{1}}\)
Since (x1, y1) is a point on x² = 4y, we get
x1² = 4y1
i.e., x1³ = \(\frac{4 \times 2}{x_{1}}\), sjnce y1 = \(\frac{2}{x_{1}}\)
x1² = 8 ⇒ x, = 2
When x1 = 2, y1 = \(\frac { 2 }{ 2 }\) = 1
∴ Equation of the normal at (2, 1) is
y – 1 = \(\frac { – 2 }{ 2 }\)(x – 2)
y – 1 = – (x – 2)
i.e., x + y = 3

Question 24
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are
a. \(\left(4, \pm \frac{8}{3}\right)\)
b. \(\left(4, \pm \frac{-8}{3}\right)\)
c. \(\left(4, \pm \frac{3}{8}\right)\)
d. \(\left(\pm 4, \frac{3}{8}\right)\)
Solution:
9y² = x³
Let (x1, y1) be the point at which the nor¬mal make equal intercepts on the coordinate axes.
Differentiating 9y³ = x³ w.r.t. x, we get
18y\(\frac { dy }{ dx }\) = 3x²
\(\frac { dy }{ dx }\) = \(\frac{x^{2}}{6 y}\)
Slope of the tangent at (x1, y1) = \(\frac{x_{1}^{2}}{6 y_{1}}\)
∴ Slope of normal at (x1, y1) = \(\frac{-6 y_{1}}{x_{1}^{2}}\)
Equation of the normal at (x1, y1) is
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 31
which is in the intercept form.
Since the intercepts of the normal are equal,we get
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 32
But x1 ≠ 0 since x1 = 0 makes the intercepts meaningless.
∴ x1 = 4
When x1, 9y²1 = x³1 becomes
9y²1 = 4³ = 64
1 = \(\frac { 64 }{ 9 }\) ⇒ y1 = ≠ \(\frac { 8 }{ 3 }\)
∴ The required points are (4, \(\frac { 8 }{ 3 }\))

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