Mathematics Times – July 2019

(III)

Area of

Area of

MZN

ZMW



 (R)

5

4

(IV)  (S)

21

5

(T) 2 6

(U)

10

3

17. Which of the following is the only correct
    combination?
    (a) (II) - (T) (b) (I) - (S)
    (c) (II) - (Q) (d) (I) - (U)
    18.Which of the following is the only Incorrect
    combination?
    (a) (IV) - (S) (b) (I) - (P)
    (c) (III) - (R) (d) (IV) - (U)

SECTION-I

1. (a,b,d) 2. (a,d) 3. (b,d) 4. (a,b,c)


2. (a,c) 6. (a,b) 7. (a,b,c) 8. (a,c,d)

SECTION-II

9. (6.20) 10. (0.00) 11. (18.00)


10. (30.00) 13. (0.50) 14. (422)

SECTION-III

15. d 16. c 17. c 18. a

SECTION-I

SECTION-II

SECTION-III

SECTION-I

1.Sol: We can evaluate

SECTION-I

1.Sol: F x  explicitly. But for

the options (A) to (C), we shall be dealing

with F x'  which equals f x  by the

Fundamental Theorem of Calculus.

Thus F x'  has three zeros, 1, 2 and 5. For

x 1 , all the three factors x x1, 2 , and

x 5 are negative and so F x' ^0. For

x1,2 F x'  0 as two factors are

negative and one positive. So, F changes from

decreasing to increasing at x 1. Hence there

is a local minimum at x 1. By a similar

reasoning, there is a local maximum at x 2.

So, both (A) and (B) are true.

For x2,5 F x'  0 as two factors are

positive and the third negative. For x 5 ,

F x'  0. So there is a local minimum as 5.

As there is already a local minimum atx 1 ,

(C) is false.

For (D), suppose there is some b0,5such

that F b  0. Since F0 0 too, by Rolle’ss

theorem there is somec b0, such that

F c'  0 , i.e. c c c (^1)   (^2)   5 0. But
c 5 and so c 5 0. So c must be either 1
or 2. However, this is not a sufficient ground
to prove (D). So here we actually calculate
F x by integrating f. Since t t (^1)   (^2) 
t  ^5  t t t^3 8 17 10^2   ,
F x  comes out as
  0 ^3 8 17 10^2 
x

F x  t t t dt  

4

(^8317210)
4 3 2
x
  x  x x
On the interval [0,5], F x can attain its
maximum either at 2 or at one of the end points
0 and 5. Since F is decreasing on [2,5] we
need not consider the end point 5. As between
0 and 2, F0 0 while  
64
2 4 34
3
F   
64
20 18 0
3
   .
So, the maximum of F x on [0,5] is
F0 0. Therefore there cannot be any