# Class 12-science RD SHARMA Solutions Maths Chapter 18 - Maxima and Minima

## Maxima and Minima Exercise MCQ

### Solution 30

Given: is defined as f(x) = 2x + cos x

Differentiating w.r.t x, we get

f'(x) = 2 - sin x

As -sin x ≥ -1

Therefore, 1 - sin x > 0 for all x.

Hence, f(x) is an increasing function.

### Solution 31

Let

Differentiating w.r.t x, we get

f'(x) = 2x - 8

Take f'(x) = 0, we get

2x - 8 = 0

i.e. x = 4

Again differentiating, we get

f"(x) = 2 > 0 for all real x

Therefore, x = 4 is the point of minima.

The minimum value of f(x) is

f(4) = 16 - 32 + 17 = 1

### Solution 32

Let

Differentiating w.r.t x, we get

Take f'(x) = 0, we get

cos 2x = 0

Again differentiating, we get

At

Therefore, is the point of maxima.

The maximum value of f(x) is

### Solution 33

Let

Taking log on both the sides, we get

… (i)

Differentiating w.r.t x, we get

Differentiating w.r.t x, we get

… (iii)

From equation (ii), we get

Take we get

At we have

Therefore, is the point of maxima.

The maximum value of f(x) is

### Solution 34

Let

Taking log on both the sides, we get

… (i)

Differentiating w.r.t x, we get

Take we get

Thus, is the stationary point.

### Solution 35

Given:

Differentiating w.r.t x, we get

which is the slope of the curve

Differentiating w.r.t x, we get

Take

Again differentiating w.r.t x, we get

So, the slope is maximum at x = 1

### Solution 36

Given:

Differentiating w.r.t x, we get

Take f'(x) = 0, we get

Now, let's find f(x) at x = 2 or -1

Therefore, x = -1 is the point of local maxima and the maximum value is 11.

Whereas, x = 2 is the point of local minima and the minimum value is -16.

Hence, f(x) has one maximum and one minimum.

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## Maxima and Minima Exercise Ex. 18.1

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## Maxima and Minima Exercise Ex. 18.2

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## Maxima and Minima Exercise Ex. 18.3

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Given:

Differentiating w.r.t x, we get

Take f'(x) = 0

Differentiating f'(x) w.r.t x, we get

At

Clearly, f"(x) < 0 at

Thus, is the maxima.

Hence, f(x) has maximum value at .

## Maxima and Minima Exercise Ex. 18.4

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## Maxima and Minima Exercise Ex. 18.5

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### Solution 5

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

Volume of cylinder

… (i)

Surface area of cylinder

From (i), we get

### Solution 23

Let be an isosceles triangle with AB = AC.

Let

Here, AO bisects

Taking O as the centre of the circle, join OE, OF and OD such that

OE = OF = OD = r (radius)

Now,

In

Similarly, AF = r cot x

In

As OB bisect we have

In

Similarly, BD = DC = CE =

We have, perimeter of

P = AB + BC + CA

= AE + EC + BD + DC + AF + BF

Differentiating w.r.t x, we get

Taking

As

Therefore, is an equilateral triangle.

Taking second derivative of P, we get

At

Therefore, the perimeter is minimum when

Least value of P

## Maxima and Minima Exercise Ex. 18VSAQ

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