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...

Given:  ……….(i)         =  Now    = 0           and  [Turning points] At , from eq. (i), At , from eq. (i),  [Minimum value] Therefore, The correct option is option...

Equation of the curve is                                                                                 ……….(i) Let P be any point on the curve (i), then according to question, Distance between...

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]...

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   ...

 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)...

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...

 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...

 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....

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 ...

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...

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...

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...

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...

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....

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...

 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...

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...

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

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,...

Let    Now       =     [Turning point]    =  =  =  =  If  is even, then  If  is odd, then  Therefore, maximum value of  is  and minimum value...

Let     Now        Putting   Now     =  =  =  Putting  Also   and  Since  attains its maximum value 1 at  and  Therefore, the required points...

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...

Profit function     and  Now      At ,  [Negative]    has a local maximum value at .   At , Maximum profit =  = 41 + 16 – 8 = 49

(iii) Given:     Now       At   At   At   Therefore, absolute minimum value is  and absolute maximum value is 8. (iv) Given:     Now      At   At   Therefore, absolute minimum value is  and absolute...

 (i) Given:     Now      At   At   At   Therefore, absolute minimum value of  is  and absolute maximum value is 8. (ii) Given:     Now          [Positive]   is in I quadrant. [ ]       Therefore,...

(iii) Given:     Now      =  =  Here, values of  are imaginary.    does not have maxima or minima.

(i) Given:     Now    But this gives no real value of  Therefore, there is no turning point.    does not have maxima or minima. (ii) Given:     Now     1 = 0 But this is not possible. Therefore,...

(vii) Given:       and =  Now      [Turning point] At    [Negative]    is a point of local maxima and local maximum value is  (viii) Given:      =  =  =  And  =  =  Now    = 0        is a point of...

(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...

(iii) Given:         ……….(i)     and  Now            [Positive]    can have values in both I and III quadrant. But,  therefore,  is only in I quadrant.   =  [Turning point] At     = ...

(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...

(v) Given:        ……….(i) Since   Adding 1 to both sides,    Therefore, neither minimum value not maximum value of  exists.

(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,    ...

(iii) Given:    ……….(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,...