29Ans 3..f(x) = sinx ... g(x) = cosx
so g'(0) = 0
Q1. Let \begin{Bmatrix} x_{n} \end{Bmatrix} be a sequence of real nos satisfying the recurrence relation
x_{n+1}=\frac{x_{n}\sqrt{3}-1}{\sqrt{3}+x_{n}}, \; n\epsilon N Then the period of the sequence is
got the answer .. but open for practice (ans: 6)
Q2. (MULTI ANSWER)
Let A be a real skew symmetric matrix satisfying A^2+I=0. Then
(a) A is orthogonal
(b) A = [a_{ij}]_{2k \times 2k}, k\epsilon N
(c) A^T = A^{-1}
(d) A must be non-singular
Q3. INTEGER TYPE
Let f and g be two function for which
f(x-y)=f(x)g(y)-f(y)g(x)
g(x-y)=g(x)g(y)+f(x)g(y)
If f'(0^+) exists and f(x) is differentiable at x=0, then g'(0) =
Q4. adding one more though it was not of AIATS .. (source: final test series of Aakash , BBSR)
no. of solutions for
e^x = 2sin(2x), x> 0
-
UP 0 DOWN 0 0 9
9 Answers
1if u dont want to assume function cos x, sin x
put x=0,
f(0)=f(0)g(0)-f(0)g(0)=0
g(0)=g(0)^2
so g(0)=0 or g(0)=1
diffrentiating wrt x we get
f'(x-y)=f'(x)g(y)-g'(x)f(y)
putting x=y=0 we get f'(0)=0
applying similarly to second eq we get g'(0)=0
1Ans for the second one is A B C D .....is it correct??
for the fourth its 2 ...
106@utd4ever and avik: your answers are correct..
can u provide solution?
for the q4. the graphs getting complicated... and wolfram showed how close the two roots are.. in such cases how do get the no. of roots?
@rickde: thanks
13Didn't try to find the roots fr Q4)...Remembered the graph 'coz i had received -1 fr a similar Qn in a test :P
29Ya i also agree with Asish..i too found it tough to say whether 2 roots exist or 0..plz can someone post some alternative method for this one..
1we get A(-A) = I
At is transpose of A
and A is skew symmetric so At=-A
so we get A(At)=I
so A^-1=At and A is orthogonal
as A^-1 exists A is non-singular
n so it will be 2k x 2k matrix
hence a,b,c,d