NCERT Solutions class 12 Maths Exercise Miscellaneous (Ex Misc) Chapter 6 Application of Derivatives


NCERT Solutions for Class 12 Maths Exercise Miscellaneous Chapter 6 Application of Derivatives – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise Miscellaneous (Ex Misc) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise Miscellaneous Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex Misc) Exercise Miscellaneous



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),

(33)1/51215(2)6(33)−1/5∼12−15(2)6     0.497


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 .


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.


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  
 
  


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. 


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 


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 disjoint 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 


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

 = 


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.

=70x(4/x)+180x+180(4/x)70x(4/x)+180x+180(4/x)

280+180x+720x280+180x+720x

 

 and 

Now 

 

 

 

  [length cannot be negative]

At  [Positive]

 is minimum at .

Minimum cost = 

= 280 + 360 + 360 = Rs   1000


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.


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), 

y=10π+4010π202(π+4)y=10π+40−10π−202(π+4)

 =  m

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

And Radius of semi-circle =  m


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 = 

 = 


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

 

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Ans. Given: ……….(i) 

f(x)=(x2)4ddx(x+1)3+(x+1)3ddx(x2)4f′(x)=(x−2)4ddx(x+1)3+(x+1)3ddx(x−2)4

3(x2)4(x+1)2+4(x+1)3(x2)33(x−2)4(x+1)2+4(x+1)3(x−2)3

(x2)3(x+1)2[3(x2)+4(x+1)](x−2)3(x+1)2[3(x−2)+4(x+1)]

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.


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

 

Ans. Given:  ……….(i) 

 = 

Now 

 = 0

  or 

  or 

 

  [Turning points]

thus two turning points are π/2andπ/6π/2andπ/6

Now 

= 0 + 1 = 1

1 + 0 = 1

 = 1

Therefore, absolute maximum is  and absolute minimum is 1.


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 

   and y0y≠0

 

At  

 [Negative]

Volume is maximum at 


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

 

Ans. Let II be the interval  

Given:  for all  in an interval  II. Let  II 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 II.


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 = 


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)

dVdx=πhr(2rx3x2)dVdx=πhr(2rx−3×2) and d2Vdx2=πhr(2r6x)d2Vdx2=πhr(2r−6x)

Now dVdx=0dVdx=0

   and  x0x≠0

At       ,    d2Vdx2=πhr(2r12r3)d2Vdx2=πhr(2r−12r3)

[Negative]

V is maximum at 

From eq. (iii),

Maximum value of cylinder = πhr[r4r298r327]πhr[r4r29−8r327]

 


Choose the correct answer in the Exercises 19 to 24:

 

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   cu m/h

(B) 0.1  cu m/h

(C) 1.1   cu m/h

(D) 0.5   cu  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

 

 100πdydt=314100πdydt=314

 100(3.14)dydt=314100(3.14)dydt=314

  1   cu  m/h

Therefore, option (A) is correct.


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.


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.


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.


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.


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),

      9(x26)2=x39(−x26)2=x3

      x=4    and substituting this value of x in (i)

    

we have  and 

Taking positive sign, 

  ……….(ii)

From eq. (i) and (ii),

9(x26)2=x39(x26)2=x3

    x=4 and substituting in 

we have  and 

 Required points are 

Therefore, option (A) is correct.