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Install NowNCERT 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

**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 .

### 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]

= cm^{2}/s

Now when

cm^{2}/s

Therefore, the area is decreasing at the rate of cm^{2}/s.

### 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

### 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.

### 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

**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

### 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

=

=

= =

### 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 m**^{3}. 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?

^{3}. 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

### 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.

### 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

### 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 =

= =

### 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.

### 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.

### 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,

OD^{2} + BD^{2} = OB^{2}

……….(i)

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

V =

and

Now

At

= [Negative]

Volume is maximum at

### 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.

### 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, AM^{2} + OM^{2} = OA^{2}

……….(i)

Volume of cylinder (V) = ……….(ii)

V =

= ……….(iii)

and

Now

At

= [Negative]

V is maximum at

From eq. (iii),

Maximum value of cylinder =

=

=

### 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 =

=

=

=

=

**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.

### 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.

### 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.

### 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.

### 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.

### 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

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 20 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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