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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.

Maxima and Minima Exercise Ex. 18.3

Solution 8

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. 18.5

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