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

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

Edited By Komal Miglani | Updated on Apr 27, 2025 08:56 AM IST | #CBSE Class 12th

Derivatives, a key concept in calculus, have practical applications in various fields like physics, business, and everyday life. They are used to analyse rates of change, predict future trends, and optimise processes. The "Application of Derivatives" chapter of the NCERT includes all these topics. After studying this chapter, you came across the concept of instantaneous change, increasing and decreasing functions, local maxima and minima, and absolute maxima and minima. In this miscellaneous exercise of Chapter 6, Class 12 Application of Derivatives, you will find various questions from each topic, which will help you understand the concept with clarity. These NCERT Solutions are prepared by the subject matter experts at Careers360, which will help you to learn the concept and approach better.

This Story also Contains
  1. Application of Derivatives: Miscellaneous Exercise
  2. Topics covered in Chapter 6 Application of Derivatives: Miscellaneous Exercise
  3. NCERT Solutions Subject Wise
  4. Subject-wise NCERT Exemplar solutions

Application of Derivatives: Miscellaneous Exercise

Question:1(a) Using differentials, find the approximate value of each of the following:

(17/81)1/4

Answer:

Let y=x14 and x=1681 and Δx=181
Δy=(x+Δx)14x14
=(1681+181)14(1681)14
(1781)14=Δy+23
Now, we know that Δy is approximate equals to dy
So,
dy=dydx.Δx=14x34.181       (y=x14 and Δx=181)=14(1681)34.181=274×8.181=196
Now,
(1781)14=Δy+23=196+23=6596=0.677
Hence, (1781)14 is approximately equal to 0.677

Question:1 Using differentials, find the approximate value of each of the following:
(33)1/5

Answer:

Let y=x15 and x=32 and Δx=1
Δy=(x+Δx)15x15
=(32+1)15(32)15
(33)14=Δy+12
Now, we know that Δy is approximately equals to dy
So,
dy=dydx.Δx=15x65.1       (y=x15 and Δx=1)=15(32)65.1=15×64.1=1320
Now,
(33)15=Δy+12=1320+12=159320=0.497
Hence, (33)15 is approximately equals to 0.497

Question:2. Show that the function given by f(x)=logxx has maximum at x = e.

Answer:

Given function is
f(x)=logxx
f(x)=1x.1x+logx1x2=1x2(1logx)
f(x)=01x2(1logx)=01x20 So logx=1x=e
Hence, x =e is the critical point
Now,
f(x)=2xx3(1logx)+1x2(1x)=1x3(2x+2xlogx1)f(e)=1e3<0
Hence, x = e is the point of maxima



Question:3(i) Find the intervals in which the function f given by f(x)=4sinx2xxcosx2+cosx is

increasing

Answer:

Given function is
f(x)=4sinx2xxcosx2+cosx
f(x)=(4cosx2cosx+xsinx)(2+cosx)(4sinx2xxcosx)(sinx)(2+cosx)2
=4cosxcos2x2+cosx
f(x)=04cosxcos2x2+cosx=0cosx(4cosx)=0cosx=0       and           cosx=4
But cosx4
So,
cosx=0x=π2 and 3π2
Now three ranges are there (0,π2),(π2,3π2) and (3π2,2π)
In interval (0,π2) and (3π2,2π) , f(x)>0

Hence, the given function f(x)=4sinx2xxcosx2+cosx is increasing in the interval (0,π2) and (3π2,2π)
in interval ,(π2,3π2),f(x)<0 so function is decreasing in this inter

Question:3 (ii) Find the intervals in which the function f given by f x is equal to

f(x)=4sinx2xxcosx2+cosx is

decreasing

Answer:

Given function is
f(x)=4sinx2xxcosx2+cosx
f(x)=(4cosx2cosx+xsinx)(2+cosx)(4sinx2xxcosx)(sinx)(2+cosx)2
=4cosxcos2x2+cosx
f(x)=04cosxcos2x2+cosx=0cosx(4cosx)=0cosx=0       and           cosx=4
But cosx4
So,
cosx=0x=π2 and 3π2
Now three ranges are there (0,π2),(π2,3π2) and (3π2,2π)
In interval (0,π2) and (3π2,2π) , f(x)>0

Hence, given function f(x)=4sinx2xxcosx2+cosx is increasing in interval (0,π2) and (3π2,2π)
in interval ,(π2,3π2),f(x)<0
Hence, given function f(x)=4sinx2xxcosx2+cosx is decreasing in interval ,(π2,3π2)

Question:4(i) Find the intervals in which the function f given by f(x)=x3+1x3,x0

Increasing

Answer:

Given function is
f(x)=x3+1x3
f(x)=3x2+3x2x4f(x)=03x2+3x2x4=0x4=1x=±1
Hence, three intervals are their (,1),(1,1) and(1,)
In interval (,1) and (1,),f)x>0
Hence, given function f(x)=x3+1x3 is increasing in interval (,1) and (1,)
In interval (-1,1) , f(x)<0
Hence, given function f(x)=x3+1x3 is decreasing in interval (-1,1)

Question:4(ii) Find the intervals in which the function f given by f(x)=x3+1x3,x0

decreasing

Answer:

Given function is
f(x)=x3+1x3
f(x)=3x2+3x2x4f(x)=03x2+3x2x4=0x4=1x=±1
1654666897895 Hence, three intervals are their (,1),(1,1) and(1,)
In interval (,1) and (1,),f)x>0
Hence, given function f(x)=x3+1x3 is increasing in interval (,1) and (1,)
In interval (-1,1) , f(x)<0
Hence, given function f(x)=x3+1x3 is decreasing in interval (-1,1)

Question:5 Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1 with its vertex at one end of the major axis.

Answer:

1628072034896 Given the equation of the ellipse
x2a2+y2b2=1
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
m=±ba.a2n2
Therefore, Now
Coordinates of A = (n,ba.a2n2)
Coordinates of B = (n,ba.a2n2)
Now,
Length AB(base) = 2ba.a2n2
And height of triangle ABC = (a+n)
Now,
Area of triangle = 12bh
A=12.2ba.a2n2.(a+n)=aba2n2+bna2n2
Now,
dAdn=abna2n2+na2n2bn2a2n2
Now,
dAdn=0abna2n2+na2n2bn2a2n2=0abn+n(a2n2)bn2=0n=a,a2
but n cannot be zero
therefore, n=a2
Now, at n=a2
d2Adn2<0
Therefore, n=a2 is the point of maxima
Now,
b=2ba.a2(a2)2=3b
h=(a+n)=a+a2=3a2
Now,
Therefore, Area (A) =12bh=123b3a2=33ab4

Question:6 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 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Answer:

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8 m3
h = 2m (given)
lb = 4 = l=4b
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
A(b)=4+2h(4b+b)
A(b)=2h(4b2+1)A(b)=02h(4b2+1)=0b2=4b=2
Now,
A(b)=2h(4×2bb3)A(2)=8>0
Hence, b = 2 is the point of minima
l=4b=42=2
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 = 4 m2
building of tank costs Rs 70 per sq metres for the base
Therefore, for 4 m2 Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) = 16 m2
building of tank costs Rs 45 per square metre for sides
Therefore, for 16 m2 Rs = 16 X 45 = 720 Rs
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs

Question:7 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.

Answer:

It is given that
the sum of the perimeter of a circle and square is k = 2πr+4a=ka=k2πr4
Let the sum of the area of a circle and square(A) = πr2+a2
A=πr2+(k2πr4)2
A(r)=2πr+2(k2πr16)(2π)A(r)=02π(8rk2πr8)=0r=k82π
Now,
A(r)=2π(82π8)=0A(k82π)>0
Hence, r=k82π is the point of minima
a=k2πr4=k2πk82π4=2k82π=2r
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Question:8 A window is in the form of a rectangle surmounted 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.

Answer:

Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle (r=l2)
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle

= l+2b+πl2
1628072096595
l+2b+πl2=10l=2(102b)2+π
Area of window id given by (A) = lb+π2(l2)2
=2(102b)2+πb+π2(102b2+π)2
A(b)=208b2+π+π2.2(102b2+π).(2)2+π
=208b2+π2π(102b(2+π)2)A(b)=0208b2+π=2π(102b(2+π)2)40+20π16b8πb=20π4πb40=4b(π+4)b=10π+4
Now,
A(b)=82+π+4π(2+π)2=168π+4π(2+π)2=164π(2+π)2A(10π+4)<0
Hence, b = 5/2 is the point of maxima
l=2(102b)2+π=2(102.104+π)2+π=204+π
r=l2=202(4+π)=104+π
Hence, these are the dimensions of the window to admit maximum light through the whole opening

Question:9 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 (a23+b23)32

Answer:

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle

1628072130108 Let the angle between AC and BC is θ
So, the angle between AD and ED is also θ
Now,
CD = b cosecθ
And
AD = asecθ
AC = H = AD + CD
= asecθ + b cosecθ
dHdθ=asecθtanθbcotθcosecθdHdθ=0asecθtanθbcotθcosecθ=0asecθtanθ=bcotθcosecθasin3θ=bcos3θtan3θ=batanθ=(ba)13
Now,
d2Hdθ2>0
When tanθ=(ba)13
Hence, tanθ=(ba)13 is the point of minima
secθ=aa23+b23a13 and cosecθ=ba23+b23b13

AC = aa23+b23a13+ ba23+b23b13 = (a23+b23)32
Hence proved

Question:10 Find the points at which the function f given by f(x)=(x2)4(x+1)3 has (i) local maxima (ii) local minima (iii) point of inflexion

Answer:

Given function is
f(x)=(x2)4(x+1)3
f(x)=4(x2)3(x+1)3+3(x+1)2(x2)4f(x)=04(x2)3(x+1)3+3(x+1)2(x2)4=0(x2)3(x+1)2(4(x+1)+3(x2))x=2,x=1 and x=27
Now, for value x close to 27 and to the left of 27 , f(x)>0 ,and for value close to 27 and to the right of 27 f(x)<0
Thus, point x = 27 is the point of maxima
Now, for value x close to 2 and to the Right of 2 , f(x)>0 ,and for value close to 2 and to the left of 2 f(x)<0
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point of inflexion

Question:11 Find the absolute maximum and minimum values of the function f given by
f(x)=cos2x+sinx,xϵ[0,π]

Answer: Given function is
f(x)=cos2x+sinx
f(x)=2cosx(sinx)+cosxf(x)=02cosxsinx+cosx=0cosx(12sinx)=0eithercosx=0      and       sinx=12x=π2          and          x=π6     as x ϵ[0,π]
Now,
f(x)=2(sinx)sinx2cosxcosx+(sinx)f(x)=2sin2x2cos2xsinxf(π6)=32<0
Hence, the point x=π6 is the point of maxima and the maximum value is
f(π6)=cos2π6+sinπ6=34+12=54
And
f(π2)=1>0
Hence, the point x=π2 is the point of minima and the minimum value is
f(π2)=cos2π2+sinπ2=0+1=1

Question:12 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

Answer:

1628072169934 The volume of a cone (V) = 13πR2h
The volume of the sphere with radius r = 43πr3
By Pythagoras theorem in ΔADC we ca say that
OD2=r2R2OD=r2R2h=AD=r+OD=r+r2R2
V = 13πR2(r+r2+R2)=13πR2r+13πR2r2+R2
13πR2(r+r2R2)V(R)=23πRr+23πRr2R2+13πR2.2R2r2R2V(R)=013πR(2r+2r2R2R2r2R2)=013πR(2rr2R2+2r22R2R2r2R2)=0R0 So,2rr2R2=3R22r2Square both sides4r44r2R2=9R4+4r412R2r29R48R2r2=0R2(9R28r2)=0R0 So,9R2=8r2R=22r3
Now,
V(R)=23πr+23πr2R2+23πR.2R2r2R23πR2r2R2(1)(2R)(r2+R2)32V(22r3)<0
Hence, the point R=22r3 is the point of maxima
h=r+r2R2=r+r28r29=r+r3=4r3
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r3

Question:13 Let f be a function defined on [a, b] such that f(x)>0 , for all xϵ(a,b) . Then prove that f is an increasing function on (a, b).

Answer:

Let's do this question by taking an example
suppose
f(x)=x3>0,(a.b)
Now, also
f(x)=3x2>0,(a,b)
Hence by this, we can say that f is an increasing function on (a, b)

Question:14 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R3 . Also, find the maximum volume.

Answer:

1628072213939 The volume of the cylinder (V) = πr2h
By Pythagoras theorem in ΔOAB
OA=R2r2
h = 2OA
h=2R2r2
V=2πr2R2r2
V(r)=4πrR2r2+2πr2.2r2R2r2V(r)=04πrR2r22πr3R2r2=04πr(R2r2)2πr3=06πr3=4πrR2r=6R3
Now,
V(r)=4πR2r2+4πr.2r2R2r26πr2R2r2.(1)2r2(R2r2)32V(6R3)<0
Hence, the point r=6R3 is the point of maxima
h=2R2r2==2R22R23=2R3
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R3
and maximum volume is
V=πr2h=π2R23.2R3=4πR333

Question:15 Show that 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

427πh3tan2α

Answer:

1628072251851 Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively

Volume of cylinder = πr2h
Volume of cone = 13πR2h
Now, we have
R=htana
Now, since ΔAOG\andΔCEG are similar
OAOG=CEEG
hR=hRr
h=h(Rr)R
h=h(htanar)htana=htanartana
Now,
V=πr2h=πr2.htanartana=πr2hπr3tana
Now,
dVdr=2πrh3πr2tanadVdr=02πrh3πr2tana=02πrh=3πr2tanar=2htana3
Now,
d2Vdr2=2πh6πrtana
at r=2htana3
d2Vdr2=2πh4πh<0
Hence, r=2htana3 is the point of maxima
h=htanartana=htana2htana3tana=13h
Hence proved
Now, Volume (V) at h=13h and r=2htana3 is
V=πr2h=π(2htana3)2.h3=427.πh3tan2a
hence proved

Question:16 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 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

Answer:

It is given that
dVdt=314 m3/h
Volume of cylinder (V) = πr2h=100πh           (r=10m)
dVdt=100πdhdt314=100πdhdtdhdt=3.14π=1 m/h
Hence, (A) is correct answer


Also Read,

Background wave

Topics covered in Chapter 6 Application of Derivatives: Miscellaneous Exercise

  • If a quantity y varies with another quantity x, satisfying some rule y=f(x), then dydx (or f(x) ) represents the rate of change of y with respect to x and dydxx=x0 (or f(x0)) represents the rate of change of y with respect to x at x=x0.
  • If two variables x and y are varying with respect to another variable t, i.e., if x=f(t) and y=g(t), then by Chain Rule
    dydx=dydt/dxdt, if dxdt0
  • A function f is said to be
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JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

(a) increasing on an interval (a,b) if x1<x2 in (a,b)f(x1)<f(x2) for all x1,x2(a,b). Alternatively, if f(x)0 for each x in (a,b)

(b) decreasing on (a,b) if x1<x2 in (a,b)f(x1)>f(x2) for all x1,x2(a,b).

(c) constant in (a,b), if f(x)=c for all x(a,b), where c is a constant.

  • First Derivative Test: Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then
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(i) If f(x) changes sign from positive to negative as x increases through c , i.e., if f(x)>0 at every point sufficiently close to and to the left of c, and f(x)<0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.

(ii) If f(x) changes sign from negative to positive as x increases through c, i.e., if f(x)<0 at every point sufficiently close to and to the left of c, and f(x)>0 at every point sufficiently close to and to the right of c, then c is a point of local minima.

(iii) If f(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Such a point is called the point of inflexion.

  • Second Derivative Test: Let f be a function defined on an interval I and cI. Let f be twice differentiable at c. Then
JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

(i) x=c is a point of local maxima if f(c)=0 and f(c)<0. The value f(c) is the local maximum value of f.

(ii) x=c is a point of local minima if f(c)=0 and f(c)>0. In this case, f(c) is the local minimum value of f.

(iii) The test fails if f(c)=0 and f(c)=0.

Also, read,

NCERT Solutions Subject Wise

These are links to other subjects' NCERT textbook solutions. Students can check and analyse these well-structured solutions for a deeper understanding.

Subject-wise NCERT Exemplar solutions

Students can check these NCERT exemplar links for further practice purposes.

Frequently Asked Questions (FAQs)

1. What are the concepts to be covered to solve NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise?

Questions related to the tropics rate of change of quantities, approximation, increasing and decreasing functions, tangents and normals and maxima and minima are covered in Class 12 Maths chapter 6 miscellaneous solutions.

2. What is the number of questions covered in the miscellaneous exercise?

24 questions are present in Class 12 Maths chapter 6 miscellaneous exercise solutions

3. How many miscellaneous solved examples are there in NCERT Class 12 Mathematics book?

10 miscellaneous questions are solved in Class 12 NCERT Mathematics book.

4. How many exercises are covered in the NCERT chapter application of derivatives?

Including miscellaneous, there are 6 exercises. For more questions students can use NCERT exemplar.

5. In total how many solved examples are given in class 12 NCERT Maths chapter 6?

51 solved examples are given in the chapter 6 application of derivatives

6. What is covered in exercise 6.5?

The questions regarding maximum and minimum of a function using derivatives are covered in the class 12 maths exercise 6.5

7. In which exercise the questions related to approximation are covered?

Exercise 6.4 and a few questions of Class 12 Maths chapter 6 miscellaneous solutions covers the concept of approximation.

8. What are the two tests discussed in the topic maxima and minima?

The two tests discussed are the first derivative test and the second derivative test.

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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