106haan 36
Q1. For what values of d is the product of two numbers of the form x2-dy2 and u2-dv2 is also of the same form (d is not a perfect square)
Q2. How many solutions does the equation
xax = axa, 0<a<1 have in positive real nos.?
Q3. Let x,y,z satisfy
\frac{x^2}{2^2-1^2}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1
\frac{x^2}{4^2-1^2}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1
\frac{x^2}{6^2-1^2}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1
\frac{x^2}{8^2-1^2}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1
then find the value of x2+y2+z2+w2
Q3. For how many values of a are a, a+4, a+14 primes?
Q4. Consider the numbers A=4125+1254 and B = 4245+2454. Then A and B are primes or composites ??
Q5. Find the value to which \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+...}}}}} converges
Q6. The value of 1^2C_{1} + 2^{2}C_{2}+3^2C_{3}+...+n^2C_{n} is
Q7. If a=\sum_{r=0}^{n}{\frac{r}{^{n}C_{r}}} then find the value of \sum_{r=0}^{n}{\frac{1}{^{n}C_{r}}} in terms of a and n
Q8. For a positive number n, let Sn denote the sum of the digits of n. The number of integers for which n-Sn = 1234 equals?
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18 Answers
62Q3. For how many values of a are a, a+4, a+14 primes?
Hint: think of divisibility by 3
a gives a remainder r,
then a+4 gives a remainder r+1 and a+14 gives a remainder r+2
so the only possibility fo these three numbers being prime is when a=2 or 3
a=2 will not be possible. so the answer is for a=3
1708(5) Yes Nishant Sir You are saying Right.
x =\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4...........}}}}
and put 4=a
then equation become
x=\sqrt{a+\sqrt{a-x}}
square both side
(x^{2}-a)^{2}=(a-x)
x^{4}+a^{2}-2ax^{2}=a-x
x^{4}-2ax^{2}+x+a^{2}-a=0
now this is a biquadratic in x but quadratic in a
so solve for a
a^{2}-a(2x^{2}+1)+x^{4}-x=0
2a-(2x^{2}+1)=\pm (2x+1)
and put a=4 in this equation and solve for x.........
118) No solutions.....FOT QSN.
Think of dividing both sides by 3.
Number 2 was once done in an AOPS MATH-JAM.....
And if i remember the Pell's EQN properly, I'd rather think of applying it for the 1st one....
Number 4 was an old GOIIT QSN.....I rememebred abt the lost legacy of goiit after seeing that qsn.
14.
A = (2125)2 + (1252)2 + 2(2125)(1252) - 2(2125)(1252)
A = (2125 + 1252)2 - (263 125)2
A = (2125 + 1252 + 263 125)(2125 + 1252 - 263 125)
62Q7. If a=\sum_{r=0}^{n}{\frac{r}{^{n}C_{r}}} then find the value of in terms of a and n \sum_{r=0}^{n}{\frac{1}{^{n}C_{r}}}
Keep in mind: \sum_{r=a}^{b}{f(r)}=\sum_{r=a}^{b}{f(a+b-r)}
Hence a=\sum_{r=0}^{n}{\frac{r}{^{n}C_{r}}} = \sum_{r=0}^{n}{\frac{n-r}{^{n}C_{n-r}}} = \sum_{r=0}^{n}{\frac{n-r}{^{n}C_{r}}}
adding the first with the last result, we get
\\a=\sum_{r=0}^{n}{\frac{r}{^{n}C_{r}}} = \sum_{r=0}^{n}{\frac{n-r}{^{n}C_{n-r}}} \\a= \sum_{r=0}^{n}{\frac{n-r}{^{n}C_{r}}} \\ Thus, \\ 2a=\sum_{r=0}^{n}{\frac{n-r+r}{^{n}C_{r}}} \\2a=\sum_{r=0}^{n}{\frac{n}{^{n}C_{r}}} \\\sum_{r=0}^{n}{\frac{1}{^{n}C_{r}}}= 2a/n
62x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+...}}}}}
y=\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+...}}}}
so we have x=√4+y
while y=√4-x
I remember a question on this one where we had 7 instead of 4 and the observation that y=2 and x=3 worked well there.....
But does not seem to work here....
Solving the biquadratic to get the answer would not be a very prudent thing.. (i guess!)
21Q2.....
take loga both sides
tat will yield axlogax=xa
consider functions from R+→R
f(x)=ax
g(x)=logax
h(x)=xa
then both functions f and g are decreasing and h is increasing
it follows that f(x)g(x)=h(x) has a unique soln at x=a
btw from wer u got des ones from.............
24arre nahin..maine to solve bhi nahin kiya tha..subah se bino hi akr raha thaa...yaad ho gaye the answers:P
106kk .. thanks got it ..
main kahaan poora solution maangta hoon .. ? just hints are enuf (most of the time)
24(1+x)^n=C_0+C_1x+C_2x^2+...C_nx^n
diff it n(1+x)^{n-1}=C_1+2C_2x+...+nC_nx^{n-1}
multiply by x nx(1+x)^{n-1}=C_1x+2C_2x^2+...+nC_nx^{n}
diff again
RHS will be C_1+2^2C_2x+...+n^2C_nx^{n-1}
lhs thoda lamba hai isliye type nahin kiya..
ab x=1 aur ans