Ans. (i) Given:   

Since  for all  R

Adding 3 both sides,     

Therefore, the minimum value of  is 3 when , i.e., 

This function does not have a maximum value.

(ii) Given:  

 

 

= 

        ……….(i)

Since  for all  R

Subtracting 2 from both sides, 

 

Therefore, minimum value of  is  and is obtained when , i.e., 

And this function does not have a maximum value.

(iii) Given:  f(x)=−(x−1)2+10f(x)=−(x−1)2+10  ……….(i)

Since  for all  R

Multiplying both sides by  and adding 10 both sides,

     [Using eq. (i)]

Therefore, maximum value of  is 10 which is obtained when  i.e., 

And therefore, minimum value of  does not exist.

(iv) Given:  

As    

As    

Therefore, maximum value and minimum value of  do not exist.

Ans. (i) Given:         ……….(i) 

Since  for all  R

Subtracting 1 from both sides, 

  

Therefore, minimum value of  is  which is obtained when  i.e., 

From eq. (i), maximum value of  hence it does not exist.

(ii) Given:  

Since  for all  R

Multiplying by  both sides and adding 3 both sides,

|x+2|−1≥−1|x+2|−1≥−1

  

Therefore, maximum value of  is  which is obtained when  i.e., 

From eq. (i), minimum value of  hence it does not exist.

(iii) Given:         ……….(i)

Since   for all  R

Adding 5 to all sides,  

 

Therefore, minimum value of  is 4 and maximum value is 6.

(iv) Given:  

Since   for all  R

Adding 3 to all sides,  −1+3≤sin4x+3≤1+3−1+3≤sin⁡4x+3≤1+3

 

Therefore, minimum value of  is 2 and maximum value is 4.

(v) Given:        ……….(i)

Since  

Adding 1 to both sides, 

 

Therefore, neither minimum value not maximum value of  exists.

Ans. (i) Given:   

    and 

Now 

      [Turning point]

Again, when ,    [Positive]

Therefore,  is a point of local minima and local minimum value = 

(ii) Given:  

    and 

Now 

  

 

 

  or  [Turning points]

Again, when ,

 [Negative]

   is a point of local maxima and local maximum value 

And when    [Positive]

   is a point of local minima and local minimum value 

(iii) Given:         ……….(i)

    and 

Now 

 

 

 

    [Positive]

   can have values in both I and III quadrant.

But,  therefore,  is only in I quadrant.

  = tan

x= [Turning point]

At  

  = 

=  =  =   [Negative]

   is a point of local maxima and local maximum value

= 

= 

(iv) Given:         ……….(i)

    and 

Now 

 

 

 

    [Negative]

   can have values in both II and IV quadrant.

   = 

=  or 

  or 

   and    [Turning point]

At   = 

  = 

= 

= 

=  =   [Negative]

   is a point of local maxima and local maximum value = 

= 

=  = 

At   

= 

   = 

= 

= 

=  =      [Positive]

   is a point of local maxima and local maximum value = 

= 

=  = 

(v) Given:  

    and 

Now 

 

 

 

  or    [Turning points]

At   [Negative]

   is a point of local maxima and local maximum value is 

At     [Positive]

   is a point of local minima and local minimum value is 

(vi) Given:  

  

= 

=   and 

Now 

  = 0

 

  or 

But  therefore  is only the turning point.

   is a point of local minima and local minimum value is 

(vii) Given:  

    and

h′′(x)=(x2+2)2(−2)−(−2x)(2)(x2+2)2x(x2+2)4h″(x)=(x2+2)2(−2)−(−2x)(2)(x2+2)2x(x2+2)4

=(x2+2)[−2(x2+2)+8x2](x2+2)4=(x2+2)[−2(x2+2)+8×2](x2+2)4

=−2x2−4+8x2(x2+2)3=−2×2−4+8×2(x2+2)3

 

= 

Now 

 

  [Turning point]

At    [Negative]

   is a point of local maxima and local maximum value is 

(viii) Given:  

  

= 

=  = 

And 

= 

= 

Now 

  = 0

 

     turning point

Again 

= 

   is a point of local maxima and local maximum value is 

  has local maximum value at 

Ans. 

1. We have,
   f(x) = e^x
   ∴∴f"(x) = e^x
   Now, if f"(x) = 0, then e^x = 0. But the exponential function can never assume 0 for any value of x.
   Therefore, there does not exist c ∈∈R such that f"(c) = 0
   Hence, function f does not have maxima or minima.
2. We have,
   g(x) = log x
   ∴g′(x)=1x∴g′(x)=1x
   Since, log x is defined for a positive number x, g"(x)>0 for any x.
   Therefore, there does not exist c ∈∈R such that g"(c) = 0
   Hence, function g does not have maxima or minima.
3. We have,
   h(x) = x³ + x² + x + 1
   ∴∴ h"(x) = 3x² + 2x + 1
   Now, h(x) = 0,
   ∴3x2+2x+1=0∴3×2+2x+1=0
   ∴x=−2±22√i6=−1±2√i3∉R∴x=−2±22i6=−1±2i3∉R
   Therefore, there does not exist c ∈∈R such that h"(c) = 0
   Hence, function h does not have maxima or minima.

Ans. 

1. The given function is f(x) = x³.
   ∴∴f"(x) = 3x²
   Now,
   f"(x) = 0 ⇒x=0⇒x=0
   Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [-2, 2].
   f(0) = 0, f(-2) = (-2)³ = -8, f(2) = (2)³ = 8
   Hence, we conclude that the absolute maximum value of f on [-2, 2] is 8 occurring at x = 2.
   Also, the absolute minimum value of f on [-2, 2] is -8 occurring at x = -2.
2. The given function is f(x) = sin x + cos x.
   ∴∴f"(x) = cosx – sin x
   Now,
   f"(x) = 0 ⇒sinx=cosx⇒sin⁡x=cos⁡x⇒tanx=1⇒x=π4⇒tan⁡x=1⇒x=π4
   Then, we evaluate the value of f at critical point x = π4π4 and at end points of the interval [0, ππ].
   f(π4)=sinπ4+cosπ4=f(π4)=sin⁡π4+cos⁡π4=12√+12√=22√=2–√12+12=22=2
   f(0) = sin 0 + cos 0 = 1
   f(ππ) = sinπ+cosπsin⁡π+cos⁡π =  0 – 1 = -1
   Hence, we conclude that the absolute maximum value of f on [0, ππ] is 2–√2 occurring at x = π4π4.
   Also, the absolute minimum value of f on [0, ππ] is -1 occurring at x = ππ.
3. The given function is f(x) =4x−12x2,x∈[−2,92]4x−12×2,x∈[−2,92]
   ∴∴f"(x) =4−12(2x)=4−x4−12(2x)=4−x
   Now,
   f"(x) = 0 ⇒x=4⇒x=4
   Then, we evaluate the value of f at critical point x = 4 and at end points of the interval [−2,92][−2,92].
   f(4)=16−12(16)f(4)=16−12(16)= 16 – 8 = 8
   f(-2) = −8−12(4)−8−12(4)= -8 – 2= -10
   f(92)=4(92)−12(92)2=18−818f(92)=4(92)−12(92)2=18−818= 18 – 10.125 = 7.875
   Hence, we conclude that the absolute maximum value of f on [−2,92][−2,92] is 8 occurring at x = 4.
   Also, the absolute minimum value of f on [−2,92][−2,92] is -10 occurring at x = -2.
4. The given function is f(x) =(x – 1)² + 3
   ∴∴f"(x) = 2(x – 1)
   Now,
   f"(x) = 0 ⇒⇒2(x – 1) = 0
   ∴∴ x = 1
   Then, we evaluate the value of f at critical point x = 1 and at end points of the interval [-3, 1].
   f(1) = (1 – 1)² + 3 = 0 + 3 = 3
   f(-3) = (-3 – 1)² + 3 = 16 + 3 = 19
   Hence, we conclude that the absolute maximum value of f on [-3, 1] is 19 occurring at x = -3.
   Also, the absolute minimum value of f on [-3, 1] is 3 occurring at x = 1.

Ans. Given: Profit function  

   and 

Now 

 

 

At ,  [Negative]

   has a local maximum value at .

  At , Maximum profit

= 

= 41 + 16 – 8 = 49

Ans. Let  on [0, 3] 

  

Now 

 

 

 

  or 

Since  is imaginary, therefore it is rejected.

   is turning point.

  At  = 

At  

At  

Therefore, absolute minimum value is  and absolute maximum value is 25.

Ans. Let  

  

Now 

 

 

 

Putting  

Now 

  

= 

=  = 

Putting 

Also   and 

Since  attains its maximum value 1 at  and 

Therefore, the required points are  and 

Ans. Let  

 

Now 

 

 

 = 

   [Turning point]

  

= 

= 

= 

= 

If  is even, then 

If  is odd, then 

Therefore, maximum value of  is  and minimum value of  is 

Ans. Let  

 

Now 

 

 

 

  or   [Turning points]

For Interval [1, 3],  is turning point.

At  

At  

At  

Therefore, maximum value of  is 89.

For Interval    is turning point.

At  

At  

At  

Therefore, maximum value of  is 139.

Ans. Let  

 

Since,  attains its maximum value at  in the interval [0, 2], therefore 

  

 

 

Ans. Let  

 

Now 

 

 

 

=  = 

= 

  where  Z   

For   But  , therefore 

For   =  and 

For  

But , therefore 

Therefore, it is clear that the only turning point of  given by  which belong to given closed interval  are, 

At 

 nearly

At 

 nearly

At 

 nearly

At 

 nearly

At  

At 

 nearly

Therefore, Maximum value =  and minimum value = 0

Ans. Let the two numbers be  and  

According to the question, 

   …….(i)

And let  is the product of  and 

 

  [From eq. (i)]

 

  and 

Now to find turning point, 

    

At ,  [Negative]

   is a point of local maxima and  is maximum at .

  From eq. (i),  

Therefore, the two required numbers are 12 and 12.

Ans. Given:         ……….(i) 

Let  P =   [To be maximized]      ……….(ii)

Putting from eq. (i),  in eq. (ii),

P = 

  …..(iii)

Now 

 

 

It is clear that  changes sign from positive to negative as  increases through 45.

Therefore, P is maximum when 

Hence,  is maximum when  and 

Ans. Given:  

      ……….(i)

Let 

    [From eq. (i)]

 

 

 

 

  ……….(ii)

Now 

 

  or  or 

  or  or 

Now  is rejected because according to question,  is a positive number.

Also  is rejected because from eq. (i),  = 35 – 35 = 0, but  is positive.

Therefore,  is only the turning point.

 

At , 

= 

  By second derivative test,  will be maximum at  when .

Therefore, the required numbers are 10 and 25.

Ans. Let the two positive numbers are  and  

  

     ……….(i)

Let 

  [From eq. (i)]

 

= 

  and 

Now  = 0

 

 

At    is positive.

   is a point of local minima and  is minimum when .

  

Therefore, the required numbers are 8 and 8.

Ans. Given: Each side of square piece of tin is 18 cm. 

Let  cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are  and  cm.

Let  denotes the volume of the open box.

  

  

  

= 

    and 

Now  = 0

 

 

  or 

 is rejected because at  length =  which is impossible.

   is the turning point.

At ,   [Negative]

·  is minimum at  i.e., side of each square to be cut off from each corner for maximum volume is 3 cm.

Ans. Given: Dimensions of rectangular sheet are 45 cm and 24 cm. 

Let  cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are  and  cm.

Let  denotes the volume of the open box.

  

  

= 

   and 

Now  = 0

 

 

 

  or 

 is rejected because at  length =  which is impossible.

   is the turning point.

At ,   [Negative]

·  is minimum at  i.e., side of each square to be cut off from each corner for maximum volume is 5 cm.

Ans. Let PQRS be the rectangle inscribed in a given circle with centre O and radius  

Let  and  be the length and breadth of the rectangle, i.e.,  and 

In right angled triangle PQR, using Pythagoras theorem,

PQ² + QR² = PR²

  

 

   …..(i)

Let A be the area of the rectangle, then A =  = 

 

= 

And 

= 

 

= 

Now 

  = 0

 

  At ,   [Negative]

  At , area of rectangle is maximum.

And from eq. (i), ,

i.e., 

Therefore, the area of inscribed rectangle is maximum when it is square.

Ans. Let  be the radius of the circular base and  be the height of closed right circular cylinder. 

  Total surface area (S) = 

   =  (say)

 

   …..(i)

Volume of cylinder 

=  [From eq. (i)]

 

  and 

Now 

 

 

At   [Negative]

   is maximum at .

  From eq. (i), 

=  = 

  Height = Diameter

Therefore, the volume of cylinder is maximum when its height is equal to the diameter of its base.

Ans. Let  be the radius of the circular base and  be the height of closed right circular cylinder. 

According to the question, Volume of the cylinder 

  …(i)

  Total surface area (S) = 

= 

=   [From eq. (i)]

 S = 

= 

  and 

Now 

 

 

 

 

 

At  

=  =    [Positive]

  S is minimum when radius  cm

  From eq. (i)

=  = 

Ans. Let  meters be the side of square and  meters be the radius of the circle. 

Length of the wire = Perimeter of square + Circumference of circle

   = 28

 

   ……….(i)

Area of square =  and Area of circle = 

Combined area (A) =  +  =  + 

= 

  and 

Now 

 

 

 

 

 

And   [Positive]

  A is minimum when 

Therefore, the wire should be cut at a distance  from one end.

Ans. Let O be the centre and R be the radius of the given sphere, BM =  and AM =  

In right angled triangle OMB, using Pythagoras theorem,

OM² + BM² = OB²

  

 

 

   ……….(i)

Volume of a cone inscribed in the given sphere

 = 

           ……….(ii)

   and 

Now 

 

 

 

 

At  

= 

=   [Negative]

   is maximum at 

  From eq. (i) 

= 

  Maximum volume of the cone

= 

=  (Volume of the sphere)

Ans. Let  be the radius and  be the height of the cone. 

  Volume of the cone (V) = 

   (say) ……….(i)

And Surface area of the cone (S) = 

  (say) ….(ii)

 

= 

= 

   and 

Now 

 

 

 

    ……….(iii)

At   [Positive]

   is minimum when 

  From eq. (i), 

=  [From eq. (iii)]

 

 

Therefore, Surface area is minimum when height =  (radius of base)

Ans. Let  be the radius,  be the height,  be the slant height of given cone and  be the semi-vertical angle of cone. 

  

        ……….(i)

  Volume of the cone (V) =        ……….(ii)

  V = 

= 

  and 

Now 

 

 

 

 

At  

=    [Negative]

  V is maximum at 

  From eq. (i), 

 

  Semi-vertical angle, 

= 

  

Ans. Let  be the radius and  be the height of the cone and semi-vertical angle be  

  Total Surface area of cone (S) = 

   (say)

 

 

 

 

    ……….(i)

Volume of cone (V) = 

= 

= 

 

=   [Using quotient rule]

        ……….(ii)

Now 

  

 

 

 

   [height can’t be negative]

   is the turning point.

Since, , therefore, Volume is maximum at 

 From eq. (i), 

  

Now Semi-vertical angle of the cone = 

 

Ans. Equation of the curve is                                                                                 ……….(i) 

Let P be any point on the curve (i), then according to question,

Distance between given point (0, 5) and P =   (say)

 

=             [From eq. (i)]

 

  = Z (say)

  and 

Now 

 

 

At 

 [Positive]

 Z is minimum and  is minimum at 

 From eq. (i)

 

  and  are two points on curve (i) which are nearest to (0, 5).

Therefore, option (A) is correct.

Ans. Given:  ……….(i) 

 

 

 

  = 

Now 

  = 0

 

 

 

 

  and  [Turning points]

At ,

from eq. (i),

At ,

from eq. (i),

 [Minimum value]

Therefore, option (D) is correct.

Ans. Let  

= ,      ……….(i)

  

= 

Now 

   = 0

 
    [Turning point] and it belongs to the given enclosed interval  i.e., [0, 1].
At , from eq. (i),

At   from eq. (i),

At ,  from eq. (i),

  Maximum value of  is 1.
Therefore, option (C) is correct.