nice one (but easy)

yes sir, could you just show how you solved the integral? I solved it by using hit n trial .. (time was short in hand)

Q3. MULTI ANSWER CORRECT

If a,b,c,d,c₁,c₂,c₃....,c_n are arbitrary constants, then the order of the differential equation whose solution is given by

y= (asin(x+b) + csin(x+d) + xtan^{-1}(\frac{c_{1}-c_{2}}{1+c_{1}c_{2}}) + 2xtan^{-1}(\frac{c_{2}-c_{3}}{1+c_{2}c_{3}}) + ... + (n-2)xtan^{-1}(\frac{c_{n-2}-c_{n-1}}{1+c_{n-1}c_{n-2}}) + c_{n}e^{x+c_{n-1}}

where n is a natural no. is

(a) undefined (c) 3
(b) 2 (d) 4

Q4. MULTI ANSWER

Let a₁,a₂,b₁,b₂,c₁,c₂ be selected from the set A= {1,2,3,.....,100}. If the roots of the equations a₁x²+b₁x+c₁ =0 and a₂x²+b₂x+c₂ =0 are x₁, x₂ and 2x₁, 3x₂ respectively, then the probability that (b₁²-4a₁c₁)(b₂²-4a₂c₂) < 0 is always less than

(a) 1/2 (b) 1/3 (c) 1/4 (d) 3/4

hey in the first is it y=1-x or y=f(1-x) ???

Q1.
Let \int_0^1f(x)\ \mathrm dx =a and \int_0^2f(x)\ \mathrm dx =b
Then
f(x)=ax+b-5
Hence
a=\int_0^1f(x)\ \mathrm dx =\int_0^1(ax+b-5)\ \mathrm dx=\dfrac{a}{2}+b-5
which gives
2b - a = 10 --- (1)

Again
b=\int_0^2f(x)\ \mathrm dx =\int_0^2(ax+b-5)\ \mathrm dx=2a+2b-10
which gives
2a +b =10 ---- (2)
Solving for a and b gives a=2, b=6
Hence f(x)=2x+1
Hence A = 3/2 and so 2A = 3

Let g(x) = x^x

Then g'(x) = x^x(1+\log x)

and \log g(x) = x \log x

The integrand is x^x(1+\log x) x \log x = g'(x) \log g(x)

Hence \int x^x(1+\log x) x \log x \ dx = \int g'(x) \log g(x) \dx = \int \log y \ dy

where y = \log g(x)

Hence the integral evaluates to g(x) (\log g(x) -1)+C

Then evaluate at x =e

http://targetiit.com/iit-jee-forum/posts/sum-of-coefficients-13190.html

Q3,4 unsolved [2]