IITPrep
Ques. Paper code : MALG2
Solutions Code : SMALG2
MATHEMATICS ALGEBRA
SOLUTIONS
- The given expression can be written as ax -
Ö(ax-2)2.
The transformations are quite simple. The second part of the question demands us to solve ax-½
ax-2½
=1. If ax ³
2, then this equation obviously has no solution. If ax < 2, then we have ax=3/2. Noting that the condition ax < 2 for this value of x, we find the desired value to be x=loga3/2.
- Since the number equals 1/2 + log23, it suffices to demonstrate that log23 is irrational. Assume the contrary, that this number is rational which means log23 = p/q. Since log23 > 0, both numbers p and q may be regarded as natural. Rewrite as 2p = 3q. But this is a contradiction since LHS is even but RHS is odd! This completes the proof.
- The number 1 + Ö3 i a root of the eqn. if 3(1+Ö3)3+a(1+Ö3)2+b(1+Ö3)+12 = 0. That is 4a+b+42+Ö3(2a+b+18) = 0. We are interested only in integer a and b therefore p = 4a+b+42, q = 2a+b+18 will also be integer. Thus we have to solve p + Ö3q = 0. Now the "obvious" conclusion drawn is p = q = 0. But can you prove it? Lets do it. The above holds for certain p and q (¹0) for which Ö3 = -p/q, but this is not true since Ö3 is irrational. Therefore q = 0, it follows p = 0. Thus it now remains to solve the system:
4a + b + 42 = 0
2a + b + 18 = 0.
This has a unique solution namely a = -12 and b = 6.
- From first equation we have z3 = -(w7)* and from second z5 = 1/w11, hence we obtain z15 = -(w35)* and z15 = 1/w33 and hence w33(w35)* = -1. From this equation it follows that ½w33(w35)*½ = 1. Using properties of moduli and conjugate numbers, we obtain ½w33(w35)*½ = ½w33½½(w35)*½ = ½w68½ = 1. Reverting to the equation w33(w35)* = -1, we rewrite its left hand member: (w33(w33)*)(w2)* = (w2)*. We have thus arrived at the equation (w2)* = -1 or (w1)* = i and (w2)* = -i or vice-versa. Now one can compute Further the value of z as z1 = -i and z2 = i.
- By hypothesis both sides of the inequality are positive and so both sides can be squared and one can proceed. The final result turns out to be (a.3y - b.2x)2 + (2x - a)2 + (3y - b)2 ³ 0 which is true. Therefore the original inequality is true.
- The given inequality is equivalent to (1 + 1/n)n < n. It suffices to prove that (1 + 1/n)n < 3.
- Note that a, b, c are all factors of the det. Also if we subsitute -(b+c) for a we get 3 identical columns. Therefore (a+b+c)2 is a factor. Hence, since det is of 6th degree, D = Nabc(a+b+c)3. To evaluate N put a = b = c = 1 in det whence N=2. Hence the required result.
- Denoting sum by S, we have
S(1-x) = 1 + 3x + 5x2 + ... + (2n-1)xn-1 - n2xn
S’ = 1 + 3x + 5x2 + ... + (2n-1)xn-1, then
S’(1-x) = 1 + 2x + 2x2 + ... + 2xn-1-(2n-1)xn
= 2(1-xn)/(1-x)-1-(2n-1)xn
Therefore S = 2(1-xn)/(1-x)3 - (1+(2n-1)xn)/(1-x)2 - n2xn/(1-x)
- We have to obtain a system of three equations in the common root, a
(say). The following system can be obtained: aa2
+ ba
+ c = 0; ba2
+ ca
+ ak = 0; ca2
+ aka
+ bk = 0. Thus solving for a2
and a
we get the reqd det zero.
- Quite straight forward.
- The coefficient = å
(5!/(a
!.b
!.g
!)).1a
.2b
.3g
where a
,b
,g
have any values 0,1,2, ... ,5 such that a
+b
+g
=5 and b
+2g
=4. The solutions of the 2nd eqn are (4,0), (2,1), (0,2),
and the possible values of a
are correspondingly 1, 2, 3. Therefore the coefficient = (5!/1!.4!.0!)24 + (5!/2!.2!.1!)22.3
+ (5!/3!0!2!)32 = 80+360+90 = 530.
- Straight forward we obtain a quadratic in terms of the common ratio which is equal to a2 and it boils down to get the limits of this quadratic.
- We get tr+1 = r.tr/(r+5)
Þ r.tr - (r+1)tr+1 = 4tr+1
So, we have
1t1 - 2t2 = 4t2
2t2 - 3t3 = 4t3
...........................
(n-1)tn-1 - ntn = 4tn
Þ t1 - ntn = 4(Sn - t1)
Þ Sn = 1/96 - (n!)/(n+4)!
- Given f(x) = ax/(ax + Öa).
Now f(1-x) = Öa/(Öa + ax).
Therefore f(x) + f(1-x) = 1
Þ f(r/2n) + f( (2n-r)/2n ) = 1
Þ r=1å2n-1 {f(r/2n) + f( (2n-r)/2n )} = 2n-1
Þ r=1å2n-1 f(r/2n) + t=1å2n-1f(t/2n) = 2n-1
Þ r=1å2n-1 f(r/2n) = (2n-1)/2