Mathematics Times – July 2019

5. If

5

is equal to : [2016]

(a) 4 (b) 13 (c)

(c)

51 63

84 72

 

 

 

(d)

51 84

63 72

 

 

 

3 2

a b

A

  

 

 

and A adjA AA T, then 5 a b

 1 (d) 5

6. The system of linear equations

has a non-trival solution for : [2016]

(a) Exactly two values of

7. The set of all values of

x y z   0

x y z   0

x y z   0

.

(b) Exactly three values of .

(c) Infinitely many values of .

(d) Exactly one value of .

 for which the system of

linear equations : [2015]

2 2x x x x 1  2   3  1

2 3 2x x x x 1  2  3  2

 x x x 1 22  3 has a non-trivial solution,

(a) Contains two elements

(b) Contains more than two elements

(c) Is an empty set.

(d) Is a singleton

8. If  , 0, and f n( ) n n and

3 1 (1) 1 (2)

1 (1) 1 (2) 1 (3)

1 (2) 1 (3) 1 (4)

f f

f f f

f f f

 

  

  

k(1 ) (1 ) (^2   ^2  ) ,^2 then k is equal to

[2014]

(a)  (b)

1

 (c) 1 (d) -1

9. The number of values of k for which the linear
   equations [2013]
   4 x ky z  2 0
   kx y z   4 0
   2 2x y z   0
   possesses a non-zero solution is
   (a) zero (b) 3 (c) 2 (d) 1

10.The number of values of k, for which the system

of equations: [2013]

has no solution, is

(a) Infinite (b) 1 (c) 2 (d) 3

CONTINUITY & DIFFERENTIABILITY

[ONLINE QUESTIONS]

1. If the function f defined as

( 1) 8 4k x y k  

kx k y k ( 3) 3 1 

2

( )^11 , 0

x 1

f x k x

x e

   

 , is continuous at x^0 ,

then the ordered pair ( , (0))k f is equal to: [2018]

(a) (2,1) (b) (3,1) (c) (3,2) (d)

2. Let

1

,2

3

 

 

S{ , }   R R f t: ( ) ( et).sin(2 ),t t R ,

is a differentiable function}. Then S is a subset

of: [2018]

(a)  ,0 R (b)R [0, )

(c) [0, ) R (d) R ( ,0)

3. The value of k for which the function

 

tan 4

4 tan5

, 0

5 2

(^2) ,
5 2
x
x
x
f x
k x


 
   
 

  

is continuous at ,
2
x is : [2017]
(a)

4. If

2

5 (b)

2

5

 (c)

3

5 (d)

17

20

1 1

2 x y y ^5 ^5 and

 

2 2

1 2 0,

d y dy

x x ky

dxdx

    then k is equal

to [2017]

(a) -23 (b) -24 (c) 26 (d) -26

5. Let f be a polynomial function such that

f x f x f x x R (^3)  '  '' ,   . Then : [2017]
(a)
5.
[2016]
6.
[2016]
7.
[2015]
8.
[2014]
9.
[2013]
10.
[2013]
CONTINUITY & DIFFERENTIABILITY
[ONLINE QUESTIONS]

1. [2018]


2. 
   

   [2018]

   
   


3. 
   

   [2017]

   
   


4. 
   

   [2017]

   
   


5. 
   

   [2017]
   f 2 f' 2 28 