NCERT Solutions class 12 Maths Miscellaneous

NCERT Solutions class 12 Maths Miscellaneous Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

Download NCERT solutions for Applications of Derivatives as PDF.

NCERT Solutions for Class 12 Maths Application of Derivatives

1. Using differentials, find the approximate value of each of the following:

(a)  

(b)  

Ans. (a)

Let  ……….(i)

 

=  ……….(ii)

Changing  to  and  to  in eq. (i), we have

  ……….(iii)

Here  and

 

From eq. (iii),

      0.677

(b)

Let  ……….(i)

 

=  ……….(ii)

Changing  to  and  to  in eq. (i), we have

 …….(iii)

Here  and

From eq. (iii),

 

     0.497

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2. Show that the function given by  has maximum value at  

Ans. Here  ……….(i)

   ……..(ii)

And

=

  ……….(iii)

Now

 

 

 

 

 

From eq. (iii),

 =  =  < 0

  is a point of local maxima and maximum value of  is at .

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NCERT Solutions class 12 Maths Miscellaneous

3. The two equal sides of an isosceles triangle with fixed base  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?

Ans. Let BC =  be the fixed base and AB = AC =  be

the two equal sides of given isosceles triangle.

Since  cm/s  ……….(i)

Area of  x BC x AM

=

=

=  [By chain rule]

=  cm²/s

Now when

  cm²/s

Therefore, the area is decreasing at the rate of  cm²/s.

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NCERT Solutions class 12 Maths Miscellaneous

4. Find the equation of the normal to the curve  at the point (1, 2).

Ans. Equation of the curve is ……….(i)

 

 Slope of the tangent to the curve at the point (1, 2) to curve (i) is  =1

 Slope of the normal to the curve at (1, 2) is

 Equation of the normal to the curve (i) at (1, 2) is  

 

 

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NCERT Solutions class 12 Maths Miscellaneous

5. Show that the normal at any point  to the curve   is at a constant distance from the origin.

Ans. The parametric equations of the curve are

 

 

And

Slope of tangent at point

=

Slope of normal at any point

=

And Equation of normal at any point

i.e., at  =  is

 

 

 

 

Distance of normal from origin (0, 0)

=  which is a constant.

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NCERT Solutions class 12 Maths Miscellaneous

6. Find the intervals in which the function  given by  is (i) increasing (ii) decreasing.

Ans. Given:

=

=

=

=

 

=

=

 

= ……….(i)

Now  for all real  as . Also  > 0

(i)  is increasing if , i.e., from eq. (i),

 lies in I and IV quadrants, i.e.,  is increasing for

and

and (ii)  is decreasing if , i.e., from eq. (i),

  lies in II and III quadrants, i.e.,  is decreasing for

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7. Find the intervals in which the function  given by  is (i) increasing (ii) decreasing.

Ans. (i) Given: 

 

=  =

 

= ……….(i)

Now

 = 0

 = 0

Here,  is positive for all real  

 or  [Turning points]

Therefore,  or  divide the real line into three sub intervals  and

For ,  from eq. (i) at  (say),

 Therefore,  is increasing at

For from eq. (i) at  (say)

Therefore,  is decreasing at

For from eq. (i) at  (say),

Therefore,  is increasing at

Therefore,  is (i) an increasing function for  and for  and (ii) decreasing function for

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NCERT Solutions class 12 Maths Miscellaneous

8. Find the maximum area of an isosceles triangle inscribed in the ellipse  with its vertex at one end of the major axis.

Ans. Equation of the ellipse is  ……….(i)

Comparing eq. (i) with  we have  and

 and

Any point on ellipse is P

Draw PM perpendicular to axis and produce it to meet the ellipse in the point Q.

 OM =  and PM =

We know that the ellipse (i) is symmetrical about axis, therefore, PM = QM and hence triangle APQ is isosceles.

Area of APQ  x Base x Height

=  PQ.AM =  . 2PM.AM = PM (OA – OM)

=

=

 

=

 

Now

 = 0

 

 

 

  or

i.e.,  or

 

 is impossible

At ,

=  [Negative]

 is maximum at

From eq. (i), Maximum area

=

=

=  =

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NCERT Solutions class 12 Maths Miscellaneous

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 8 m³. If building of tank costs ` 70 per sq. meter for the base and ` 45 per square meter for sides. What is the cost of least expensive tank?

Ans. Given: Depth of tank = 2 m

Let  m be the length and  m be the breadth of the base of the tank.

Volume of tank = 8 cubic meters

 

 

 

Cost of building the base of the tank at the rate of ` 70 per sq. meter is

And cost of building the four walls of the tank at the rate of ` 45 per sq. meter is

= `

Let  be the total cost of building the tank.

 and

Now

 

 

 

  [length cannot be negative]

At  [Positive]

 is minimum at .

Minimum cost =

= 280 + 360 + 360 = ` 1000

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NCERT Solutions class 12 Maths Miscellaneous

10. The sum of the perimeter of a circle and square is  where  is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Ans. Let  be the radius of the circle and  be the side of square.

According to question,  Perimeter of circle + Perimeter of square =

 

 ……….(i)

Let  be the sum of areas of circle and square.

 

  [From eq. (i)]

 

=

 and

Now

 

 

 

 

At  [Positive]

 is minimum when

 From eq. (i),

=

=

 

=

=  =

Therefore, sum of areas is minimum when side of the square is double the radius of the circle.

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NCERT Solutions class 12 Maths Miscellaneous

11. 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.

Ans. Let  m be the radius of the semi-circular opening of the window. Then one side of rectangle part of window is  and  m be the other side of rectangle.

Perimeter of window

= Semi-circular arc AB + Length (AD + DC + BC)

 

 

 ……….(i)

Area of window

= Area of semi-circle + Area of rectangle

=

=

 

=

 and  =

Now

 

 

 

 

 

At  [Negative]

 is maximum at

From eq. (i),

=

=  =  m

Therefore, Length of rectangle =  m and Width of rectangle =  m

And Radius of semi-circle =  m

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NCERT Solutions class 12 Maths Miscellaneous

12. A point on the hypotenuse of a triangle is at distances  and  from the sides of the triangle. Show that the maximum length of the hypotenuse is  

Ans. Let P be a point on the hypotenuse AC of a right triangle ABC such that PL AB =  and PM BC =  and let BAC = MPC = , then in right angled  

AP = PL =

And in right angled PMC,

PM = PM

Let AC =  then

 = AP + PC = ……….(i)

Now

 

  

 

   ……….(ii)

And

 

  [ and  is +ve as  )

 is minimum when

 =

 

Also

=

 

Putting these values in eq. (i),

Minimum length of hypotenuse =

=  =

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NCERT Solutions class 12 Maths Miscellaneous

13. Find the points at which the function  given by  has:

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Ans. Given: ……….(i)

=

=

=

Now

= 0

  or  or

  or  or

Now, for values of  close to  and to the left of . Also for values of  close to  and to the right of .

Therefore,  is the point of local maxima.

Now, for values of  close to 2 and to the left of . Also for values of  close to 2 and to the right of .

Therefore,  is the point of local minima.

Now as the values of  varies through  does not change its sign. Therefore,  is the point of inflexion.

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NCERT Solutions class 12 Maths Miscellaneous

14. Find the absolute maximum and minimum values of the function  given by  

Ans. Given:  ……….(i)

=  =

Now

 = 0

  or

  or

 

  [Turning points]

Now

= 0 + 1 = 1

1 + 0 = 1

 = 1

Therefore, absolute maximum is  and absolute minimum is 1.

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NCERT Solutions class 12 Maths Miscellaneous

15. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius  is  

Ans. Let  be the radius of base of cone and  be the height of the cone inscribed in a sphere of radius

OD = AD – AO =

In right angled triangle OBD,

OD² + BD² = OB²

 

  ……….(i)

Volume of cone (V) =  = [From eq. (i)]

 V =

 and

Now

 

 

At  

=  [Negative]

Volume is maximum at

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NCERT Solutions class 12 Maths Miscellaneous

16. Let  be a function defined on  such that  for all  Then prove that  is an increasing function on  

Ans. Let I be the interval

Given:  for all  in an interval I. Let  I with

By Lagrange’s Mean Value Theorem, we have,

 where

 where

Now

  ……….(i)

Also,  for all  in an interval I

 

From eq. (i),

 

Thus, for every pair of points  I with

 

Therefore,  is strictly increasing in I.

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NCERT Solutions class 12 Maths Miscellaneous

17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is  Also find the maximum volume.

Ans. Let  be the radius and  be the height of the cylinder inscribed in a sphere having centre O and radius R.

In right triangle OAM,  AM² + OM² = OA²

 

  ……….(i)

Volume of cylinder (V) =  ……….(ii)

 V =

=  ……….(iii)

and

Now

 

At  

= [Negative]

V is maximum at

From eq. (iii),

Maximum value of cylinder =

=

=

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NCERT Solutions class 12 Maths Miscellaneous

18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height  and having semi-vertical angle  is one-third that of the cone and the greatest volume of the cylinder is  

Ans. Let  be the radius of the right circular cone of height  Let the radius of the inscribed cylinder be  and height

In similar triangles APQ and ARC,

 

 

 

 

Volume of cylinder (V) = ……….(ii)

 V =

= ……….(iii)

 and

Now

 

At  

= [Negative]

V is maximum at

From eq. (iii),

Maximum value of cylinder =

=

=

=

=  

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Choose the correct answer in the Exercises 19 to 24:

NCERT Solutions class 12 Maths Miscellaneous

19. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of 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

Ans. Let  be the depth of the wheat in the cylindrical tank of radius 10 m at time

  V = Volume of wheat in cylindrical tank at time  cu. m

It is given that  = 314 cu. m/hr

 

 

 

  1 m/h

Therefore, option (A) is correct.

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NCERT Solutions class 12 Maths Miscellaneous

20. The slope of the tangent to the curve   at the point  is:

(A)

(B)

(C)

(D)  

Ans. Equation of the curves are …..(i) and …..(ii)

  and

 Slope of the tangent to the given curve at point  =  …..(iii)

At the given point   and

At ,  from eq. (i), 

 

 

 

At , from eq. (ii), 

 

 

 

 

Here, common value of  in the two sets of values is

  From eq. (iii),

Slope of the tangent to the given curve at point  =

Therefore, option (B) is correct.

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NCERT Solutions class 12 Maths Miscellaneous

21. The line  is a tangent to the curve  if the value of  is:

(A) 1

(B) 2

(C) 3

(D)  

Ans. Equation of the curve is ……….(i)

 

 

  Slope of the tangent to the given curve at point  =

 

   ……….(ii)

Now

 

 

   …..(iii)

Putting the values of  and  in eq. (i),

 

 

Therefore, option (A) is correct.

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NCERT Solutions class 12 Maths Miscellaneous

22. The normal at the point (1, 1) on the curve  is:

(A)  

(B)  

(C)

(D)  

Ans. Equation of the given curve is  ……….(i)

 

 

  Slope of the tangent to the given curve at point (1, 1) =  (say)

 Slope of the normal =

 Equation of the normal at (1, 1) is

 

 

Therefore, option (B) is correct.

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NCERT Solutions class 12 Maths Miscellaneous

23. The normal to the curve  passing through (1, 2) is:

(A)  

(B)  

(C)  

(D)

Ans. Equation of the curve is ………..(i)

 

 

 Slope of the normal at  = ……….(ii)

Again slope of normal at given point (1, 2) = ……….(iii)

From eq. (ii) and (iii), we have 

 

 

 

 From eq. (i),

 

 

 

Now, at point (2, 1), slope of the normal from eq. (ii) =

  Equation of the normal is

 

 

Therefore, option (A) is correct.

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NCERT Solutions class 12 Maths Miscellaneous

24. The points on the curve  where the normal to the curve make equal intercepts with axes are:

(A)  

(B)

(C)

(D)

Ans. Equation of the curve is ……….(i)

 

 

 Slope of the tangent to curve (i) at any point  =

 Slope of the normal = negative reciprocal =

  [ Slopes of lines making equal intercepts on the axes are ]

 

Taking positive sign,

  ……….(ii)

From eq. (i) and (ii), we have  and

Taking positive sign,

  ……….(ii)

From eq. (i) and (ii), we have  and

 Required points are

Therefore, option (A) is correct.

NCERT Solutions class 12 Maths Miscellaneous

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