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.
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Maxima and Minima Exercise Ex. 18.1
Solution 1
Solution 2
Solution 3
![](https://images.topperlearning.com/topper/bookquestions/91881_image004.png)
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
![](https://images.topperlearning.com/topper/bookquestions/91901_image008.png)
Solution 9
![](https://images.topperlearning.com/topper/bookquestions/91911_image010.png)
Maxima and Minima Exercise Ex. 18.2
Solution 1
![](https://images.topperlearning.com/topper/bookquestions/91921_image012.png)
Solution 2
Solution 3
![](https://images.topperlearning.com/topper/bookquestions/91931_18.2-3a.jpg)
Solution 4
![](https://images.topperlearning.com/topper/bookquestions/91941_image016.png)
Solution 5
Solution 6
![](https://images.topperlearning.com/topper/bookquestions/91961_image020.png)
Solution 7
Solution 8
Solution 9
![](https://images.topperlearning.com/topper/bookquestions/91981_image024.png)
Solution 10
![](https://images.topperlearning.com/topper/bookquestions/91991_18.2-10a.jpg)
Solution 11
Solution 12
Solution 13
Solution 14
Maxima and Minima Exercise Ex. 18.3
Solution 1(i)
![](https://images.topperlearning.com/topper/bookquestions/92021_image034.png)
Solution 1(ii)
![](https://images.topperlearning.com/topper/bookquestions/92031_image036.png)
Solution 1(iii)
![](https://images.topperlearning.com/topper/bookquestions/92041_image038.png)
Solution 1(iv)
![](https://images.topperlearning.com/topper/bookquestions/92051_image040.png)
Solution 1(v)
![](https://images.topperlearning.com/topper/bookquestions/92061_image042.png)
Solution 1(vi)
![](https://images.topperlearning.com/topper/bookquestions/92071_image044.png)
Solution 1(vii)
![](https://images.topperlearning.com/topper/bookquestions/92081_image046.png)
Solution 1(viii)
Solution 1(ix)
Solution 1(x)
![](https://images.topperlearning.com/topper/bookquestions/92111_image052.png)
Solution 1(xi)
![](https://images.topperlearning.com/topper/bookquestions/92121_image054.png)
Solution 1(xii)
![](https://images.topperlearning.com/topper/bookquestions/92131_image056.png)
Solution 2(i)
![](https://images.topperlearning.com/topper/bookquestions/92141_image058.png)
Solution 2(ii)
![](https://images.topperlearning.com/topper/bookquestions/92151_image060.png)
Solution 2(iii)
![](https://images.topperlearning.com/topper/bookquestions/92161_image062.png)
Solution 3
Solution 4
Solution 5
![](https://images.topperlearning.com/topper/bookquestions/92181_image066.png)
Solution 6
![](https://images.topperlearning.com/topper/bookquestions/92191_image068.png)
Solution 7
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. 18.4
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
![](https://images.topperlearning.com/topper/bookquestions/104561_image209.gif)
Solution 2
Solution 3
Solution 4
![](https://images.topperlearning.com/topper/bookquestions/104281_18.4.4a.gif)
Solution 5
Maxima and Minima Exercise Ex. 18.5
Solution 1
![](https://images.topperlearning.com/topper/bookquestions/92871_image002.png)
Solution 2
![](https://images.topperlearning.com/topper/bookquestions/92881_image004.png)
Solution 3
![](https://images.topperlearning.com/topper/bookquestions/92891_image006.png)
Solution 4
Solution 6
![](https://images.topperlearning.com/topper/bookquestions/92921_image013.png)
Solution 7
Solution 8
Solution 9
Solution 10
![](https://images.topperlearning.com/topper/bookquestions/93351_image075.png)
Solution 11
Solution 12
![M a x i m u m space v o l u m e space i s space V subscript x equals 3 end subscript equals 3 cross times open parentheses 18 minus 2 cross times 3 close parentheses squared
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Solution 13
Solution 14
Solution 15
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/vimala/page3/ex_18.5_15_1.jpg)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/vimala/page3/ex_18.5_15_2.jpg)
Solution 16
Solution 17
Solution 18
![](https://images.topperlearning.com/topper/bookquestions/93051_image029.png)
![](https://images.topperlearning.com/topper/bookquestions/93051_image030.png)
Solution 19
Solution 20
![](https://images.topperlearning.com/topper/bookquestions/93261_image049.png)
![](https://images.topperlearning.com/topper/bookquestions/93261_image050.png)
![](https://images.topperlearning.com/topper/bookquestions/93261_image051.png)
Solution 21
Solution 22
Solution 24
Solution 25
![](https://images.topperlearning.com/topper/bookquestions/93391_image083.png)
Solution 26
Solution 27
![](https://images.topperlearning.com/topper/bookquestions/93381_image080.png)
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
![](https://images.topperlearning.com/topper/bookquestions/93281_image058.png)
Solution 39
![](https://images.topperlearning.com/topper/bookquestions/93301_image063.png)
![](https://images.topperlearning.com/topper/bookquestions/93301_image064.png)
Solution 40
![](https://images.topperlearning.com/topper/bookquestions/93241_image047.png)
Solution 41
Solution 42
![](https://images.topperlearning.com/topper/bookquestions/93311_image067.png)
![](https://images.topperlearning.com/topper/bookquestions/93311_image068.png)
Solution 43
Solution 44
![](https://images.topperlearning.com/topper/bookquestions/93271_image054.png)
![](https://images.topperlearning.com/topper/bookquestions/93271_image055.png)
Solution 45
Solution 45
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