Mathematics Times – July 2019

SECTION-III

15.Sol: So g x cos 2 sin xvanishes only

when 2 sin x is an odd multiple of

2



, i.e.

2 sin xis an odd multiple of

1

2

, i.e. when

sinx is an odd multiple of

1

4

. For this to

be possible,

1

sin

4

x  or

3

sin

4

x . So (U)

fails since

1

sin

6 2



 is an even and not an

odd multiple of

1

4

. That rules out both (A)

and (C), without bothering to check anything

else.

Now the choice of the correct option is

narrowed down to (B) and (D), both of which

are about W i.e. the set of zeros of g x' .

By a direct calculation

g x'  2 sin 2 sin cos   x x

Hence for g x' to vanish, either cosx 0 or

sin2 sin x 0. The first possibility gives

xas an odd multiple of

2



. The second gives

2 sin xas in integral multiple of , i.e.

2sinx as an integer, which is possible only

when

1

sin 0, , 1

2

x  . Hence x is an odd

multiple of

2



which is already included (in the

zeros of cosx ), or xis an angle of the form

k, or k 6



 for some integer k. AllAll

possibilities put together show that (P) and (S)

hold for W. But

3

W



. So W is not an A.P..

Hence (Q) is false and (R) is true. So (B) is

the only correct option.

16.Sol: Clearly, f x sin cos x 0 if and

only if cosxis an integral multiple of ,

which can happen only when cosx 0 or  1.

Hence X consists of all multiples of

2



. Hence

only (P_) and (Q) are correct. That rules out

the options (A) and (C). To choose between

(B) and (D), we need to identify Y, the set of

zeros of f'. Since f x'  cos cos x

sinx, f x' can vanish only when sinx 0

or cos cos x 0. the first possibility gives

all integral multiples of . The second gives

all values of x for which cosx is an odd

multiple of

2



, i.e. cosxis an odd multiple of

1

2

. This is possible only when
cos^1
2

x  , that

is when^2

3

x k



  or

2

2

3

x k



  for

some integer k. Combined with all integral

multiples of , the set Y is now all integral

multiples of

3



. These for an A.P. Hence (Q)

is true and (R) false. So withoutany further

checking, (B), if all, must be true. Still, it is

easy to verify that Y contain

SECTION-III

15.Sol:
16.Sol:

17,18.Sol:

2

,

3 3

 

and .

So (T) is true too.

17,18.Sol: c to d