## EXERCISE 6.6

#### Page No 242:

#### Question 1:

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

(a) (b)

#### Answer:

(a) Consider

Then,

Now, *dy* is approximately equal to Δ*y* and is given by,

Hence, the approximate value of = 0.667 + 0.010

= 0.677.

(b) Consider. Let *x* = 32 and Δ*x* = 1.

Then,

Now, *dy* is approximately equal to Δ*y* and is given by,

Hence, the approximate value of

= 0.5 − 0.003 = 0.497.

#### Question 2:

Show that the function given byhas maximum at *x *= *e*.

#### Answer:

Now,

1 − log *x* = 0

#### Question 3:

The two equal sides of an isosceles triangle with fixed base *b* 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?

#### Answer:

Let ΔABC be isosceles where BC is the base of fixed length *b*.

Let the length of the two equal sides of ΔABC be *a*.

Draw AD⊥BC.

Now, in ΔADC, by applying the Pythagoras theorem, we have:

∴ Area of triangle

The rate of change of the area with respect to time (*t*) is given by,

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

∴

Then, when *a *= *b*, we have:

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of.

#### Question 4:

Find the equation of the normal to curve *y*^{2} = 4*x* at the point (1, 2).

#### Answer:

The equation of the given curve is.

Differentiating with respect to *x*, we have:

Now, the slope of the normal at point (1, 2) is

∴Equation of the normal at (1, 2) is* y* − 2 = −1(*x* − 1).

⇒ *y* − 2 = − *x* + 1

⇒ *x* + *y* − 3 = 0

#### Question 5:

Show that the normal at any point *θ* to the curve

is at a constant distance from the origin.

#### Answer:

We have *x* = *a* cos *θ* + *a* *θ* sin *θ**.*

∴ Slope of the normal at any point *θ* is.

The equation of the normal at a given point (*x*, *y*) is given by,

Now, the perpendicular distance of the normal from the origin is

Hence, the perpendicular distance of the normal from the origin is constant.

#### Question 6:

Find the intervals in which the function *f *given by

is (i) increasing (ii) decreasing

#### Answer:

Now,

cos *x* = 0 or cos *x* = 4

But, cos *x* ≠ 4

∴cos *x* = 0

divides (0, 2π) into three disjoint intervals i.e.,

In intervals,

Thus, *f*(*x*) is increasing for

In the interval

Thus, *f*(*x*) is decreasing for.

#### Question 7:

Find the intervals in which the function *f* given byis

(i) increasing (ii) decreasing

#### Answer:

Now, the points *x* = 1 and *x* = −1 divide the real line into three disjoint intervals i.e.,

In intervals i.e., when *x* < −1 and *x* > 1,

Thus, when *x* < −1 and *x* > 1, *f* is increasing.

In interval (−1, 1) i.e., when −1 < *x* < 1,

Thus, when −1 < *x* < 1, *f *is decreasing.

#### Question 8:

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

#### Answer:

The given ellipse is.

Let the major axis be along the *x* −axis.

Let ABC be the triangle inscribed in the ellipse where vertex C is at (*a*, 0).

Since the ellipse is symmetrical with respect to the *x*−axis and *y* −axis, we can assume the coordinates of A to be (−*x*_{1}, *y*_{1}) and the coordinates of B to be (−*x*_{1}, −*y*_{1}).

Now, we have.

∴Coordinates of A are and the coordinates of B are

As the point (*x*_{1}, *y*_{1}) lies on the ellipse, the area of triangle ABC (*A)* is given by,

But, *x*_{1} cannot be equal to *a*.

Also, when, then

Thus, the area is the maximum when

∴ Maximum area of the triangle is given by,

#### Question 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 Rs 70 per sq meters 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* represent the length, breadth, and height of the tank respectively.

Then, we have height (*h)* = 2 m

Volume of the tank = 8m^{3}

Volume of the tank = *l* × *b* × *h*

∴ 8 = *l *× *b* × 2

Now, area of the base = *lb *= 4

Area of the 4 walls (*A)* = 2*h* (*l* + *b*)

However, the length cannot be negative.

Therefore, we have *l* = 4.

Thus, by second derivative test, the area is the minimum when* l* = 2.

We have *l *= *b* = *h* = 2.

∴Cost of building the base = Rs 70 × (*lb*) = Rs 70 (4) = Rs 280

Cost of building the walls = Rs 2*h* (*l* + *b*) × 45 = Rs 90 (2) (2 + 2)

= Rs 8 (90) = Rs 720

Required total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

#### Question 10:

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:

Let *r* be the radius of the circle and *a* be the side of the square.

Then, we have:

The sum of the areas of the circle and the square (*A)* is given by,

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

#### Page No 243:

#### Question 11:

A window is in the form of 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 *x *and *y* be the length and breadth of the rectangular window.

Radius of the semicircular opening

It is given that the perimeter of the window is 10 m.

∴Area of the window (*A)* is given by,

Thus, when

Therefore, by second derivative test, the area is the maximum when length.

Hence, the required dimensions of the window to admit maximum light is given by

#### Question 12:

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

#### Answer:

Let ΔABC be right-angled at B. Let AB = *x* and BC = *y*.

Let P be a point on the hypotenuse of the triangle such that P is at a distance of *a *and *b* from the sides AB and BC respectively.

Let ∠C = *θ*.

We have,

Now,

PC = *b *cosec *θ*

And, AP = *a* sec *θ*

∴AC = AP + PC

⇒ AC = *b* cosec *θ* + *a* sec *θ* … (1)

Therefore, by second derivative test, the length of the hypotenuse is the maximum when

Now, when, we have:

Hence, the maximum length of the hypotenuses is.

#### Question 13:

Find the points at which the function *f *given byhas

(i) local maxima (ii) local minima

(ii) point of inflexion

#### Answer:

The given function is

Now, for values of *x *close toand to the left of Also, for values of *x* close to and to the right of

Thus, is the point of local maxima.

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

Thus, *x* = 2 is the point of local minima.

Now, as the value of *x* varies through −1,does not changes its sign.

Thus, *x* = −1 is the point of inflexion.

#### Question 14:

Find the absolute maximum and minimum values of the function *f* given by

#### Answer:

Now, evaluating the value of *f *at critical pointsand at the end points of the interval (i.e., at *x* = 0 and *x* = π), we have:

Hence, the absolute maximum value of *f* is occurring at and the absolute minimum value of *f* is 1 occurring at

#### Question 15:

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

#### Answer:

A sphere of fixed radius (*r)* is given.

Let *R* and *h* be the radius and the height of the cone respectively.

The volume (*V)* of the cone is given by,

Now, from the right triangle BCD, we have:

∴*h*

∴ The volume is the maximum when

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius *r* is.

#### Question 16:

Let *f* be a function defined on [*a*, *b*] such that *f* ‘(*x*) > 0, for all *x* ∈ (*a*, *b*). Then prove that *f* is an increasing function on (*a*, *b*).

#### Answer:

Let

x1, x2∈(a,b)such that

x1<x2. Consider the sub-interval [

x1, x2]. Since *f* (*x*) is differentiable on (a, b) and

[x1, x2]⊂(a,b). Therefore, f(x) is continous on [

x1, x2] and differentiable on

(x1, x2). By the Lagrange’s mean value theorm, there exists

c∈(x1, x2)such that

f'(c)=f(x2)-f(x1)x2-x1 …(1)Since *f*‘(*x*) > 0 for all

x∈(a,b), so in particular, *f*‘(*c*) > 0

f'(c)>0⇒f(x2)-f(x1)x2-x1>0 [Using (1)]

⇒f(x2)-f(x1)>0 [âˆµ

x2-x1>0 when x1<x2]

⇒f(x2)>f(x1)⇒f(x1)<f(x2)Since

x1, x2are arbitrary points in

(a,b). Therefore,

x1<x2⇒f(x1)<f(x2) for all x1,x2∈(a, b)Hence, *f* (*x*) is increasing on (a,b).

#### Question 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.

#### Answer:

A sphere of fixed radius (*R)* is given.

Let *r* and *h* be the radius and the height of the cylinder respectively.

From the given figure, we have

The volume (*V) *of the cylinder is given by,

Now, it can be observed that at.

∴The volume is the maximum when

When, the height of the cylinder is

Hence, the volume of the cylinder is the maximum when the height of the cylinder is.

#### Question 18:

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 *α* is one-third that of the cone and the greatest volume of cylinder istan^{2}*α*.

#### Answer:

The given right circular cone of fixed height (*h)* and semi-vertical angle (*α**)* can be drawn as:

Here, a cylinder of radius *R* and height *H* is inscribed in the cone.

Then, ∠GAO = *α*, OG =* r*, OA = *h*, OE = *R*, and CE = *H*.

We have,

*r *= *h* tan *α*

Now, since ΔAOG is similar to ΔCEG, we have:

Now, the volume (*V)* of the cylinder is given by,

And, for, we have:

∴By second derivative test, the volume of the cylinder is the greatest when

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as:

Hence, the given result is proved.

#### Question 19:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere 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:

Let *r* be the radius of the cylinder.

Then, volume (*V)* of the cylinder is given by,

Differentiating with respect to time *t*, we have:

The tank is being filled with wheat at the rate of 314 cubic metres per hour.

∴

Thus, we have:

Hence, the depth of wheat is increasing at the rate of 1 m/h.

The correct answer is A.

#### Question 20:

The slope of the tangent to the curveat the point (2, −1) is

(A) (B) (C) (D)

#### Answer:

The given curve is

The given point is (2, −1).

At *x* = 2, we have:

The common value of *t* is 2.

Hence, the slope of the tangent to the given curve at point (2, −1) is

The correct answer is B.

#### Page No 244:

#### Question 21:

The line *y* = *mx* + 1 is a tangent to the curve *y*^{2} = 4*x* if the value of *m* is

(A) 1 (B) 2 (C) 3 (D)

#### Answer:

The equation of the tangent to the given curve is *y* = *mx* + 1.

Now, substituting *y* = *mx* + 1 in *y*^{2} = 4*x*, we get:

Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:

Hence, the required value of *m* is 1.

The correct answer is A.

#### Question 22:

The normal at the point (1, 1) on the curve 2*y* + *x*^{2} = 3 is

(A) *x* + *y* = 0 (B) *x* − *y *= 0

(C) *x* + *y* + 1 = 0 (D) *x *− *y* = 1

#### Answer:

The equation of the given curve is 2*y* + *x*^{2} = 3.

Differentiating with respect to *x*, we have:

The slope of the normal to the given curve at point (1, 1) is

Hence, the equation of the normal to the given curve at (1, 1) is given as:

The correct answer is B.

#### Question 23:

The normal to the curve *x*^{2} = 4*y* passing (1, 2) is

(A) *x* + *y* = 3 (B) *x* − *y* = 3

(C) *x* + *y *= 1 (D) *x* − *y *= 1

#### Answer:

The equation of the given curve is *x*^{2} = 4*y*.

Differentiating with respect to *x*, we have:

The slope of the normal to the given curve at point (*h*, *k*) is given by,

∴Equation of the normal at point (*h*, *k*) is given as:

Now, it is given that the normal passes through the point (1, 2).

Therefore, we have:

Since (*h*, *k*) lies on the curve *x*^{2} = 4*y*, we have *h*^{2} = 4*k.*

From equation (i), we have:

Hence, the equation of the normal is given as:

The correct answer is A.

#### Question 24:

The points on the curve 9*y*^{2} = *x*^{3}, where the normal to the curve makes equal intercepts with the axes are

(A) (B)

(C) (D)

#### Answer:

The equation of the given curve is 9*y*^{2} = *x*^{3}.

Differentiating with respect to *x*, we have:

The slope of the normal to the given curve at point is

∴ The equation of the normal to the curve at is

It is given that the normal makes equal intercepts with the axes.

Therefore, We have:

Also, the pointlies on the curve, so we have

From (i) and (ii), we have:

From (ii), we have:

Hence, the required points are

The correct answer is A.