Mathematics Times – July 2019

   2 0 1 0 cos 2sin sin cos 0 1 1 0 1 sin cos x x x x x x       cos 2sin 0x x or sin cosx x i.e., tan^1 2 x  or t ...

Now            1 1 2 2 1 3 2 1 2 1 1 3 4 1 1 2 2 n r r r r r n n a n n n n n                We know  ...

For non-trivial solution, we have i.e., 1 1 1 1 1 1 0 1 1 1 a b c           R R R 2   2 3 1 1 1 0 0 1 1 1 a b c c  ...

y az  0 z ax  0 1 0 0 1 0 1 a A a a          Now, | | [1 ( )] 1A  a a^2   a^30 So, system has only trivial ...

now, 2 4 2 2 2 4 2 ( 4)( 4) 2 2 4 x x x x x x Bx x x x x       1 1 ;^22 R R R R x x   and R^3 R 3 x  2 4 1 2 2 4 4 4 2 ...

^2  51 63 3 12 84 72 adj A A        5.Sol: Given A adjA AA  T A A adjA A AA^1   ^1 T adjA A T 2 5 3 3 5 2 b ...

2.Sol:  It has only one common root i.e., it has only one solution. CONTINUITY & DIFFERENTIABILITY [ONLINE QUESTIONS] 1.Sol ...

Now  ^3  ^2 3 9 ' 2 2 f x x f x x   and f x x''  9 i.e.,  ^2 9 ' 2 2 18 2 f   and f'' 2 18   f' 2 f'' 2  ...

Now   2 0 1 lim 12 sin log 1 4 x x e x x k               2 0 1 lim. sin x x x e k k x x k           ...

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i.e.,    '  1 g f x' f x  1 g'(f (x)) f '(x)  1 g'(f (2)) f '(2)   1 g'(7) 3   Sol: Given   2 2 sin , 0 sin , 0 ...

f0 0 and f0 0 When x y0, 0 y is continuous at x0. Clearly at all other points y is continuous. Therefore, the set of ...

SECTION-I 1.Let f R R:  be given by f x x x x( ) ( 1)( 2)( 5)   . Define 0 ( ) ( ) , 0 x F x f t dt x . Then which of th ...

(c) 1 lim ( ) n 2 f n  (d) If tan(cos (6)), then^1 f ^2 2 1  0 5.Let f R R:  be a function we say that f has Propert ...

in the interval 3 , 4 4        equals 11.Let a i j k   2 ˆ ˆ ˆ and b i j k ˆ 2 ˆˆ  be two vectors. Consider a v ...

(III) Area of Area of MZN ZMW   (R) 5 4 (IV)  (S) 21 5 (T) 2 6 (U) 10 3 Which of the following is the only correct c ...

x0,5 for which F x  0. Hence (D) is correct. 2.Sol: If P, Q and R are collinear, then 1 1 1 PQ kPR             ...

likewise option (b) is true i.e., 3 (4) cos 6 2 f    and option (c) is limcosn n 2 1          finally option (d) i ...

(^113) I x dx 1  0 (3) and   1 2 0 2 1 I dx a x    (4) Both the integrals are easy to evaluate. Indeed, 4 1 3 1 0 3 3 [ ] ...