Prove that f ( x ) = sin x + 3 cos x has maximum value at x = π 6 .

Given that, f ( x ) = sin x + 3 cos x Differentiating with respect to x we get, f ' ( x ) = cos x + 3 ( - sin x ) = cos x - 3 sin x For f ' ( x ) = 0 , cos x = 3 sin x ⇒ tan x = 1 3 = tan π 6 ⇒ x = π 6 Again differentiating f ( x ) with respect to x we get, f ' ' ( x ) = - sin x - 3 cos x At x = π 6 , f ' ' ( x ) = - sin π 6 - 3 cos π 6 = - 1 2 - 3 · 3 2 = - 1 2 - 3 2 = - 2 < 0 Hence, at x = π 6 , f ( x ) has maximum value and π 6 is the point of local maxima.

If the straight line x cos α + y sin α = p touches the curve x 2 a 2 + y 2 b 2 = 1 , then prove that a 2 cos 2 α + b 2 sin 2 α = p 2 .

Given that, equation of starlight line is x cos α + y sin α = p . . . i and curve is x 2 a 2 + y 2 b 2 = 1 ⇒ b 2 x 2 + a 2 y 2 = a 2 b 2 . . . i i Differentiating E q i i with respect to x we get, b 2 · 2 x + a 2 · 2 y · d y d x = 0 ⇒ d y d x = - 2 b 2 x 2 a 2 y = - x b 2 y a 2 . . . i i i From E q i . y sin α = p - x cos α ⇒ y = - x cos α sin α + p sin α So, the slope of the line is ( - cot α ) Thus, the given equation of line will be tangent to the E q . i i if - x y × b 2 a 2 = - cos α sin α ⇒ x a 2 cos α = y b 2 sin α = k [say] ⇒ x = k a 2 cos α and y = b 2 k sin α Thus, the line x cos a + y sin α = p will touch the curve x 2 a 2 + y 2 b 2 at point k a 2 cos α , k b 2 sin α . Substituting the values of x , y in E q . i we get, ⇒ k a 2 cos 2 α + k b 2 sin 2 α = p ⇒ a 2 cos 2 α + b 2 sin 2 α = p k Squaring both sides we get, a 2 cos 2 α + b 2 sin 2 α 2 = p 2 k 2 Substituting the values of x , y in E q . i i we get, ⇒ b 2 k 2 a 4 cos 2 α + a 2 k 2 b 4 sin 2 α = a 2 b 2 Dividing a 2 b 2 from both sides we get, ⇒ k 2 a 2 cos 2 α + b 2 sin 2 α = 1 ⇒ a 2 cos 2 α + b 2 sin 2 α = 1 k 2 . . . v Dividing E q . i v by E q . v we get, a 2 cos 2 α + b 2 sin 2 α = p 2 Hence, proved.

E M B I B E

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 23

MEDIUM

12th CBSE

IMPORTANT

Earn 100

At what point, the slope of the curve $y = - x^{3} + 3 x^{2} + 9 x - 27$ is maximum? Also find the maximum slope.

Important Questions on Application of Derivatives

EASY

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 24

Prove that

f (x) = \sin x + \sqrt{3} \cos x

has maximum value at

x = \frac{π}{6}

.

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 25

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is

\frac{π}{3}

.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 26

Find the points of local maxima, local minima and the points of inflection of the function

f (x) = x^{5} - 5 x^{4} + 5 x^{3} - 1

. Also find the corresponding local maximum and local minimum values.

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 27

A telephone company in a town has

500

subscribers on its list and collects fixed charges of

Rs . 300 / -

per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of

R s . 1 / -

one subscriber will discontinue the service. Find what increase will bring maximum profit?

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 28

If the straight line

x \cos α + y \sin α = p

touches the curve

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

, then prove that

a^{2} \cos^{2} α + b^{2} \sin^{2} α = p^{2}

.

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 29

An open box with square base is to be made of a given quantity of cardboard of area

c^{2}

. Show that the maximum volume of the box is

\frac{c^{3}}{6 \sqrt{3}}

cubic units.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 30

Find the dimensions of the rectangle of perimeter

36 cm

which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 6 - Application of Derivatives>EXERCISE>Q 31

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?