Page 26 - Engineering Mathematics Workbook_Final
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Calculus

            102.   A real function                                           tan x                  tan x
                                   x +  2    , x   for x   0         (a)                   (b)
                            
                    f  ( ) x =                                              2y − 1                2y − 1
                               x +    x +  5sin , x  x   0
                                        2
                                 3
                            
                                                                                                      2
                       f
                   . If  ( ) x  is twice differentiable then             (c)   sec x           (d)   sec x
                                                                             2y − 1                2y − 1
                   (a)    1, =   =  0   (b)   1, =   =  5                                       dy
                                                                                            +
                   (c)    5, =  = − 10   (d)   5, =  =  5     106.   If  x y = a  b  (x +  ) y  a b  then   dx  =

                                            2x                             x                     1
            103.   The derivative of sin − 1     2         with      (a)                 (b)
                                             +
                                            1 x                            y                     y
                                     2x   
                   respect to  tan − 1     2         is equal to       1                     y
                                      −
                                     1 x                               (c)                   (d)
                                                                             x                     x
                   (a) 0                 (b) 1
                                                                  107.   By applying, Rolle’s theorem for
                        2x                                                        sin x
                                                                                                
                   (c)                   (d) 2                            f  ( ) x =     in 0, , the value of
                        −
                       1 x  2                                                      e x
                                                                                   )
                                                                         c  (0,  is
            104.   If  x =  ( a   sin −  ) ,

                                             2
                                            d y                                                  
                    y    ( a  =  cos −  ) then   =                     (a)                   (b)
                                            dx 2                             6                     4

                            1                                                                    
                   (a)  −                                               (c)                   (d)
                        a sin 2                                              2                     3
                              2
                                                                  108.   Which of the following function satisfied
                         1                                              all the conditions of Rolle’s Theorem in
                   (b)  −   cosec 4
                         4a        2                                     the interval [0,1]

                         1                                                   f x =       
                   (c)  −  sec   2  / 2 cosec     4  / 2              (a)  ( ) tan x
                        4a
                                                                                                1
                        1                                                             x , 0   x 
                                                                                    
                   (d)    sec   2  / 2 cosec     4  / 2                                      2
                                                                             f
                       4a                                                (b)  ( ) x = 
                                                                                    
                                                                                     −   ,  1    x  1
                                                                                     1 x
                                                                                          2
                                                                                    
            105.   If  y =  tan x +  tan x +  tan x + .....
                        dy                                        109.  By applying Lagranges mean value for
                   then     =                                            the function
                        dx




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