IISER Aptitude Test Sample Paper 1 Solution

This page contains solutions to the IISER Aptitude test Sample 1. Currently the solutions are only for Maths and Physics sections. Other sections will be uploaded soon.

  • MATH
  • PHYSICS
  • BIOLOGY
  • CHEMISTRY

 

Some Mnemonics and tricks to solve Permutation and Combination type Problems. Link

                                     MATH

1) If I_m=\int^{\frac{\pi}{4}}_0(\tan{x})^m\,dx;

then I_3+I_5+I_7+I_9 equals ?

Solution-Part1:                                                        Solution-Part2:

 

2) We have a sequence 1^{\frac{1}{\sqrt{1}}} , 3^{\frac{1}{\sqrt{3}}} , 5^{\frac{1}{\sqrt{5}}} , 7^{\frac{1}{\sqrt{7}}} ..... {2n+1}^{{\frac{1}{\sqrt{2n+1}}}} and so on…..

Which is the largest value in the sequence ?

 

Solution-Part-1:                                   Solution-Part-2:                                Solution-Part-3:

 

 

3) Assuming interchange of limit and integration is permisssible, the value of

\lim_{n\to \infty}\int^1_0 \frac{nx^{n-1}}{1+x} \,dx

Note: 0<x<1

Solution-Part-1:                                                                                 Solution-Part-2:

 

4) Let  f: R→R be a function such that |f(x)-f(y)|\leq 6|x-y|^2 for all x,y\in R.

If f(3)=6 then, f(6) equals ?

Solution:

 

5)Let A_n be the area bounded by curve y=x & y=nx^2 in 1st quadrant. Then, the value of \sum\limits_{i=1}^5 \frac{1}{A_n} is.

Solution-Part-1:                                                                         Solution-Part-2: 

 

6) In how many ways can 4 distinguishable pieces be placed on a 8*8 chessboard so that no two pieces are in the same row or column?

Solution:

 

7) A and B are playing a game by alternately rolling a die,with A starting first. Each player’s score is the numbers obtained on his last roll. The ends when the sum of scores of the two players is 7, and the last player to roll the die wins. What is the probability that A wins the game ?

Solution:

8) The binomial coefficient ^{n}C_r, ^{n}C_{r+1}, ^{n}C_{r+2},.....  where 0 \leq r\leq {n-2}

 

Solution:

 

9) The Sum of infinite series arccot{2} + arccot{8} + arccot{18} + ..... + arccot{2n^2} +..... is ?

Solution-Part-1:                                                 Solution-Part-2:                                            Solution-Part-3: 

 

10) The integer values of k for which the equation 7\cos{\theta} + 5\sin{\theta} = 2k+1 has real solutions is ?

Solution-Part-1:                                                   Solution-Part-2:

 

11) The complex solutions of (z+i)^{2011}=z^{2011} lie on :

Solution-Part-1:                                                  Solution-Part-2:                      Solution-Part-3:

Solution-Part-4:                                                  Solution-Part-5: 

 

12) How many 2*2 matrices A satisfy both A^3=I_2 and A^2=A^T, where I_2 denotes the 2*2 identity matix and A^T denotes the transpose of A ?

Solution:    Proof of why the inverse is unique is left to be written but it can be found easily on Google or any Linear Algebra Text Book.

 

13) Let C be the circle that touches the X-Axis and whose centre coincides with the circum-centre of the triangle defined by 4|x| + 3y=12 ; y\geq 0. How many points with both co-ordinates integers are there in the interior of C?

Solution:

 

14) Let P and Q be the centres of the circles that pass through (0,2) and (0,8) and touch the X-Axis. Then the equation of the ellipse with P and Q as focii and touching X-Axis is ?

Solution:

 

15 Let f: R→R be a function such that f(x+y) + f(x-y)= f(xy) for all x,y \in R. Then f is ?

Solution:

Solution:

17 thoughts on “IISER Aptitude Test Sample Paper 1 Solution

    1. Hello Friend sorry. I had made this site in 2014 and totally abandoned it when I did not get good response nor I had much time to do it. Looking at the number of people visiting this month (Actually I also came back to this site after a long time), I will definitely start putting other things up, I have stuff solved and scanned and backed up with me but I did not care to upload, which I will surely do now, maybe on a different and better website. The link to it I will post soon. Thanks for visiting.

  1. Hello, thank you first of all for uploading the answers. See in question no. 10 how can you assume cos{\theta} =3/7 and sin{\theta} = 2/5 ? Doesn’t it violate sin^2{\theta}+cos^2{\theta}=1 ?

    1. Thanks for pointing out the mistake. This is tricky and a tough question to solve on paper. However, I just plotted the LHS on Wolfram

      http://www.wolframalpha.com/input/?i=Plot+7cos(theta)%2B5sin(theta)+for+theta+between+0+to+2*pi

      It seems that there are 16 values of theta between 0 and 2 pi where the RHS is Odd. I am searching for RHS being odd because, k has to be integer, i.e. if k is integer 2k+1 is odd. Visually, check for (-7, -5, -3, -1, 1, 3, 5, 7) on the y-axis. There are 16 thetas (between 0 and 2*pi) which give these values.

      If you have already solved it, be happy to share with others :-D.

      1. The solution for Question 10 is flawed – the correct answer is 8(which is option B).

        Here’s the solution :

        7cos(theta)+5sin(theta)=2k+1. l’m using x in place of theta from now on.

        Subtracting 2k +1 from both sides, we have 7cos(x)+5sin(x)-2k-1=0
        Let f(x) = 7cos(x)+5sin(x) – 2k -1 where f(x)=0

        This basically means that f(x) should intersect/touch the x axis for real solutions.

        Which can be interpreted as 7cos(x)+5sin(x) – 2k – 1 must be 0 at some point(s).

        Which in turn means that 2k+1 shouldn’t exceed the maximum values of 7cos(x)+5sin(x).

        Now, let’s find the maximum values of f(x).
        1st derivative of f(x) = f'(x) = -7sin(x)+5cos(x)

        For an extremum, f'(x) = 0
        => -7sin(x)+5cos(x)=0

        Rearranging makes the equation like this : tan(x) = 5/7
        5/7 = 0.71 which is pretty close to 1. So, we can approximately say that f(x) has extremum near tan(x) = 1 or x = pi/4

        Doing the second derivative test reveals this to be a maximum.

        Now, f(pi/4) = 7cos(pi/4)+5sin(pi/4) – 2k – 1
        Or f(pi/4) = 12/sqrt(2) -2k – 1 Or f(pi/4) = 6sqrt(2) – 2k -1
        6sqrt(2) is pretty close to 7…and 2k – 1 shouldn’t exceed 7 (Initial condition)
        Hence k can take values of 3, 2, 1 or 0

        The period of the function f(x) = 0 is pi (Found through appropriate methods).
        By the nature of circular trigonometric functions, the difference between the abscissa of minima and maxima is the period of the function.

        So, we can say a minima occurs around pi/4+pi. Repeating the above steps. the values of k are found to be -4, -3, -2, and -1.

        Hence total number of values are 8.

        For verification, use this site : https://www.desmos.com/calculator

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