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,...

Since Volume (V) =    ……….(i)   ……….(ii) It is given that increase in side = 3% =      ……….(iii) Since approximate change in volume V of cube =  =  =    cubic meters Therefore,The correct option is ...

Let   ……….(i)      ……….(ii) Changing  to  and  to  in eq. (i), =   ……….(iii) Here,  and   From eq. (iii),   Since,  and  is approximately equal to  and  respectively.  From eq. (i) and (ii),   =  = ...

Let  be the radius of the sphere.   Surface area of the sphere (S) =           ·  =  square meters

Let  be the radius of the sphere and  be the error in measuring the radius. Then, according to the question,  = 7 m and  = 0.02 m   Volume of sphere (V) =      Approximate error in calculating the...

Since Surface area (S) =     It is given that decrease in side =  of    Since approximate change in surface area S of cube =  =  =    square meters (decreasing)

Since Volume (V) =   ……….(i)   ……….(ii) It is given that increase in side = 1% =       ……….(iii) Since approximate change in volume V of cube =  =  =    cubic...

Let   ……….(i)      ……….(ii) Changing  to  and  to  in eq. (i), =   ……….(iii) Here,  and   From eq. (iii),   Since,  and  is approximately equal to  and  respectively.  From eq. (i) and (ii),   =  = ...

Let   ……….(i)      =   ……….(ii) Changing  to  and  to  in eq. (i), =   ……….(iii) Here,  and   From eq. (iii),   Since,  and  is approximately equal to  and  respectively.  From eq. (i) and (ii),  ...

(xv)  Let  ……….(i)    =     ……….(ii) Now, from eq. (i),  = =   ……….(iii) Here  and  Then  =   =  Since,  and  is approximately equal to  and  respectively.  From eq. (ii),  = 0.001875 Therefore,...

(xii)  Let  ……….(i)    =     ……….(ii) Now, from eq. (i),  = =   ……….(iii) Here  and  Then  =    =  Since,  and  is approximately equal to  and  respectively.  From eq. (ii),   = 0.00462 Therefore,...

(xi)  Let  ……….(i)    =     ……….(ii) Now, from eq. (i),  =  Here,  and , then =    Since,  and  is approximately equal to  and  respectively.  From eq. (ii),   =  Therefore, approximately value...

(ix)  Let  ……….(i)   =     ……….(ii) Now, from eq. (i),  = = ……….(iii) Here  and  Then  =   =  Since,  and  is approximately equal to  and  respectively.  From eq. (ii),   Therefore, approximate...

(vii)  Let  ……….(i)    =      ……….(ii) Now, from eq. (i),  =  Here,  and , then  =    Since,  and  is approximately equal to  and  respectively.  From eq. (ii),   =  Therefore, approximately value...

(v)   Let  ……….(i)    =       ……….(ii) Now, from eq. (i),  =  =  ……….(iii) Here  and  Then  =    =  Since,  and  is approximately equal to  and  respectively.  From eq. (ii), Therefore, approximate...

(iii)  Let  ……….(i)    =     ……….(ii) Now, from eq. (i),  =  Here,  and , then =    Since,  and  is approximately equal to  and  respectively.  From eq. (ii),   =  Therefore, approximately value...

Assume  ……….(i)    =     ……….(ii) Now, from eq. (i),  =  Here,  and , then  =    Since,  and  is approximately equal to  and  respectively.  From eq. (ii),  = 0.03 Therefore, approximately value...

Given:  Equation of the curve  ……….(i) Slope of the tangent at point  is   ……….(ii)  Slope of the line    is ……….(iii) From eq. (ii) and (iii),    From eq. (i),     Therefore, required point is (1,...

Given: Equation of the curve ……….(i)   Slope of the tangent at point  is   Slope of the tangent at  (say)  Slope of the normal =  Therefore,The correct option is option (D)...

fig 1: SOLUTION: Given: Equation of the curve  ……….(i)   Slope of the tangent at point  is    = ……….(ii) Again slope of the line  is ……….(iii) According to the question,  [Parallel lines have same...

Given: Equation of the hyperbola ……….(i)         ……….(ii)  Slope of tangent at  is   Equation of the tangent at  is         ……….(iii) Since  lies on the hyperbola (i), therefore,    From eq....

Given: Equations of the curves are  …..(i) and ……….(ii) Substituting the value of  in eq. (ii), we get      Putting the value of  in eq. (i), we get   Therefore, the point of intersection  is...

Given: Equation of the parabola  ……….(i)   Slope of the tangent at  =       Slope of the tangent at the point  =   Slope of the normal =   Equation of the tangent at the point  =      And Equation...

Equation of the curve  ….(i)   Slope of the tangent at  =   Slope of the normal to the curve at  = ……….(ii) But Slope of the normal (given) =    =            From eq. (i), at  at  Therefore, the...

Equation of the curve ……….(i)        Slope of the tangent at the point  =  =   Slope of the normal at the point  =  Equation of the normal at  =     ...

Equation of the curve  ……….(i)          [tangent is parallel to axis]      From eq. (i),      Therefore, the required points are (1, 2) and 

Given: Equation of the curve  ……….(i)   Slope of the tangent at  passing through origin (0, 0) =  =       Substituting this value of  in eq. (i), we get,       or    or   From eq. (i), at  From eq....

Given: Equation of the curve ………(i)   Slope of tangent at  = ……….(ii) According to question, Slope of the tangent = coordinate of the point          or    or   From eq. (i), at   The point is (0,...

Equation of the curve   Slope of tangent at  =  At the point  Slope of the tangent =  At the point  Slope of the tangent =  Since, the slopes of the two tangents are equal. Therefore, tangents...

Given:  Equation of the curve ……….(i)   Slope of tangent =  …….(ii) (a) Slope of the line  is   Slope of tangent parallel to this line is also = 2  From eq. (ii),      From eq. (i),   Therefore,...

Equation of the curves are    and    Slope of the tangent at  =  (say)  Slope of the normal at  is   Point  =  =  =   Equation of the tangent is       And Equation of the normal is   ...

Equation of the curve  ……….(i)   Now value of  at (1, 1) At   =  (say)  Slope of the normal at (1, 1) is   Equation of the tangent at (1, 1) is       And Equation of the normal at (1, 1) is     ...

(i) Equation of the curve     Now value of  at (0, 5) At  (say)  Slope of the normal at (0, 5) is   Equation of the tangent at (0, 5) is       And Equation of the normal at (0, 5) is     ...

Given: Equation of the curve  ……….(i)          ……….(ii) (i) If tangent is parallel to axis, then Slope of tangent = 0    = 0       From eq. (i),      Therefore, the points on curve (i) where...

Given: Equation of the curve  ……….(i)   =  =  But according to question, slope = 0   = 0     From eq. (i),  Therefore, the point on the curve which tangent has slope 0 is   Equation of the tangent...

Given: Equation of the curve     =  = Slope of the tangent at  But according to question, slope =    = 2   which is not possible. Hence, there is no tangent to the given curve having slope...

Given: Equation of the curve  ……….(i)   =  = Slope of the tangent at  But according to question, slope =    =        or  From eq. (i), when  And when     Points of contact are (2, 1) and  And...

Given: Equation of the curve  ……….(i) Equation of the tangent ……….(ii)    From eq. (i),  = Slope of the tangent at  But from eq. (ii), the slope of tangent =        From eq. (i), when   And when  ...

Let the given points are A (2, 0) and B (4, 4). Slope of the chord AB =  Equation of the curve is    Slope of the tangent at  =  If the tangent is parallel to the chord AB, then Slope of tangent =...

Given: Equation of the curve   ……….(i)    Since, the tangent is parallel to the axis, i.e.,           From eq. (i), when  when   Therefore, the required points...

Given: Equations of the curves are  and    and     and    =   Slope of the tangent at  =  And Slope of the normal at  =  = 

Given: Equations of the curves are    and  =  and     and    =   Slope of the tangent at  =  And Slope of the normal at  =  = 1

Given: Equation of the curve ……….(i)    Slope of the tangent at point  to the curve (i) =  = 27 – 3 = 24

Given: Equation of the curve  ……….(i)    Slope of the tangent at point  to the curve (i) =  = 12 – 1 = 11

Given: Equation of the curve  ……….(i)   =  = ……….(ii)  Slope of the tangent at point  to the curve (i) =  = 

Equation of the curve  ……….(i) Slope of the tangent to the curve = Value of  at the point      Slope of the tangent at point  to the curve (i) =  = 768 – 4 = 764

Given:     is strictly increasing if      i.e., increasing on the interval   is strictly decreasing if      i.e., decreasing on the interval  hence,  is neither strictly increasing nor decreasing on...

(C)  (D)  SOLUTION: (C)    Since     For      Therefore,  is strictly decreasing on  For      Therefore,  is strictly increasing on  Hence,  is neither strictly increasing not strictly decreasing...

 (A)  (B)  SOLUTION: (A)    Since  in  therefore    Therefore,  is strictly decreasing on  (B)    Since      therefore    Therefore,  is strictly decreasing...

(A) (0, 1)  (B)     (C)     (D) None of these SOLUTION: Given:     (A) On (0, 1),    therefore  And for    (0, 1 radian) =  > 0 Therefore,  is strictly increasing on (0, 1). (B) For    = = (1.5,...

     Since  is strictly increasing on (1, 2), therefore  > 0 for all  in (1, 2)  On (1, 2)       Minimum value of  is  and maximum value is  Since  > 0 for all  in (1, 2)   and    and ...

Given:         ……….(i) Here for every  either  or    for ,  (say),  > 0 And for ,  (say),  > 0   > 0 for all  , hence  is strictly increasing on...

Given:    On the interval  i.e., in first quadrant,  > 0 Therefore,  is strictly increasing on . On the interval  i.e., in second quadrant,  < 0 Therefore,  is strictly decreasing...

Given:    On the interval  i.e., in first quadrant,  is positive, thus  < 0 Therefore,  is strictly decreasing on . On the interval  i.e., in second quadrant,  is negative thus  > 0...

Given:      for all  in R. Therefore,  is increasing on R.

(A)     (B)     (C)     (D) (0, 2) SOLUTION:  Given:    =    =    In option (D),  for all  in the interval (0, 2). Therefore,  THE CORRECT OPTION IS option...

Given:    for all  in  Therefore,  is strictly increasing on 

fig 1: SOLUTION: =    =    =  =  Since  and we have , therefore     for  Hence,  is an increasing function of  in 

 Given:       [Applying Product Rule]   =  =    ……….(i)   Therefore, we have  For   taking  (say),     is decreasing. For   taking  (say),     is increasing. For   taking  (say),     is decreasing....

fig 1: solution: Given:     =  =  =     =  ……….(i) Domain of the given function is given to be    Also  and    From eq. (i),  for all  in domain  and  is an increasing...

(e)  SOLUTION: (e) Given:             Here, factors  and  are non-negative for all  Therefore,  is strictly increasing if      And  is strictly decreasing if      Hence,  is strictly increasing...

(c)   (d)   SOLUTION: (c) Given:       =  ……….(i) Now  = 0   or  Therefore, we have three disjoint intervals  and  For interval , from eq. (i),  =  < 0 Therefore,  is strictly decreasing. For...

(a)     (b)     SOLUTION: (a) Given:       ……….(i) Now    Therefore, we have two sub-intervals  and  For interval  taking  (say), from eq. (i),  < 0 Therefore,  is strictly decreasing. For...

(a) Given:     =      ……….(i) Now    or    or  Therefore, we have sub-intervals are  and  For interval    taking  (say), from eq. (i),  > 0 Therefore,  is strictly increasing in  For...

Given:        ……….(i) Now    Therefore, we have two intervals  and  (a) For interval  taking  (say), then from eq. (i),  > 0. Therefore,  is strictly increasing in  (b) For...

 Given:    (a) Since,  > 0, i.e., positive in first quadrant, i.e., in  Therefore,  is strictly increasing in  (b) Since,  < 0, i.e., negative in second quadrant, i.e., in  Therefore,  is...

Given:    =  > 0  i.e., positive for all  R Therefore,  is strictly increasing on R.

Given:     i.e., positive for all  R Therefore,  is strictly increasing on R.

(A)  (B)   (C)   (D)   SOLUTION: Area of circle (A) =  =  Therefore, The correct option is option (B) .

Marginal Revenue (MR) =  =  =  Now, when MR = 26 x 7 + 26 = 208 Therefore, the required marginal revenue is ` 208.

Marginal cost =  =  =  Now, when MC =  = 6.069 – 0.102 + 15 = 20.967 Therefore, required Marginal cost is ` 20.97.

Let the height and radius of the sand-cone formed at time  second be  cm and  cm respectively. According to question,   Volume of cone (V) =  =  =    Now, since       ...

Given: Diameter of the balloon =  Radius of the balloon =   Volume of the balloon =  =  cu. units  Rate of change of volume w.r.t.  =  =  =  = 

Let  cm be the radius of the air bubble at time  According to question,   is positive =  cm/sec  ……….(i) Volume of air bubble  =  =    =  Therefore, required rate of increase of volume of air bubble...

 Given: Equation of the curve  ……….(i) Let  be the required point on curve (i) According to the question,  ……….(ii) From eq. (i),    [From eq. (ii)]     Taking     Required point is (4, 11)....

Let AB be the ladder and C is the junction of wall and ground, AB = 5 m  B Let CA =  meters, CB =  meters According to the equation,  increases,  decreases and  = 2 cm/s In AC2 + BC2 = AB2  [Using...

Since, V =    =  =    Therefore, the volume is increasing at the rate of  cm3/sec.

Given: Rate of decrease of length  of rectangle is 5 cm/minute.   is negative = –5 cm/minute Also, Rate of increase of width  of rectangle is 4 cm/minute   is positive = 4 cm/minute (a) Let  denotes...

 Let  cm be the radius of the circle at time  Rate of increase of radius of circle = 0.7 cm/sec  is positive and = 0.7 cm/sec Let  be the circumference of the circle.   Rate of change of...

 Let  cm be the radius of the circular wave at time  Rate of increase of radius of circular wave = 5 cm/sec  is positive and = 5 cm/sec Let  be the enclosed area of the circular wave.   Rate of...

Let  cm be the edge of variable cube at time  Rate of increase of edge = 3 cm/sec  is positive and = 3 cm/sec Let  be the volume of the cube.   Rate of change of volume of cube =  =  =  =  cm3/sec...

Leave cm alone the range of the circle at time Pace of increment of span of circle = 3 cm/sec dx/dt is positive and = 3 cm/sec Let y be the space of the circle. Pace of progress of space of circle =...

 Let  cm be the edge of the cube. Given: Rate of increase of volume of cube = 8 cm3/sec  is positive = 8     ……….(i) Let  be the surface area of the cube, i.e.,    Rate of change of surface area of...

et  denote the area of the circle of variable radius  Area of circle  Rate of change of area  w.r.t.  =  (a) When  cm, then  sq. cm (b) When  cm, then  sq. cm