Hello, guys!
This contest is the first JEE contest on Brilliant. The objective of this game is to get well prepared for JEE-Advanced 2016 preparation having fun as well. There is so much to learn from each other, and the contest is an approach to accomplish the same. Each Brilliant user is invited to participate in the contest under the following rules:
- I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
- You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
- Please make a substantial comment.
- Make sure you know how to solve your problem before posting it. In case there is no one can answer it within 24 hours, then you must post the solution, and you have a right to post another problem.
- If the one who solves the last problem does not post his/her problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
- The scope of questions is only what is covered under JEE-Advanced syllabus. If you are unaware of the syllabus cover, You can check out the information brochure.
- You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
- In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
- DO NOT ask the answer to the problem. Just post your detailed solution solution along with the answer. It will be highly helpful to gain confidence in your problem solving and rock in JEE.
- Proof problems are not allowed.
You can post a problem only from Maths section.
- Please write the detailed solutions to the problems.
Format your post is as follows:
SOLUTION OF PROBLEM xxx (number of problem) :
[Post your solution here]
PROBLEM xxx (number of problem) :
[Post your problem here]
Please try to post problems from all the spheres of JEE syllabus and share this note so that maximum users come to know about this contest and participate in it. (>‿◠)✌
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Top NewestHere is the first problem for the inauguration of the contest. It's an easy problem. All the best!
Problem 1:
In a sequence of independent trials, the probability of success in one trial is 41. Find the probability that the second success takes place on or after the fourth trial.
Ritu Roy has solved this problem at the first place and provided the solution. Thanks!
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Solution of problem 1
The required probability is
=13C.41.(43)2.41+14C.41.(43)3.41+15C.41.(43)4.41+...∞=2569.24=3227
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How did we get that 24? I used the reverse probability method 1 - 3C2.1/4.1/4.3/4 - 3C3.1/4.1/4.1/4 = 27/32
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Problem 10 :
In ΔABC, AB=tan−1(sin(2)) and tan(2A)=ln(π)tan(2B)
Then the vertex C lies on
a) Ellipse
b) Parabola
c) Hyperbola
d) Straight Line
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@Aditya Kumar
I see that you've posted the next problem of contest i.e. Problem 11, but there is no solution of 10th problem by you. Can you clarify on this?
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I had posted the solution! Some careless mod has deleted it! I can't believe it.
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Can you please post that again?
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@Aditya Kumar ,your solution was incorrect.
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Problem 3:
Let f be a twice differentiable function, such that f′′(x)=−f(x) and f′(x)=g(x) , h(x)=[f(x)]2+[g(x)]2. Given that h(5)=11, evaluate h(10) .
Vighnesh Shenoy has solved this problem at the first place and provided the solution. Thanks!
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Problem 51
If (a+ω)−1+(b+ω)−1+(c+ω)−1+(d+ω)−1=2ω−1 and (a+ω′)−1+(b+ω′)−1+(c+ω′)−1+(d+ω′)−1=2ω′−1 ; where ω and ω′ are the complex cube roots of unity, then what is the value of (a+1)−1+(b+1)−1+(c+1)−1+(d+1)−1 ?
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Problem 4:
If n∈N, evaluate the value of
(0n)+(4n)+(8n)+…,
where (rn)=r!(n−r)!n!.
Akul Agrawal is the first person to solve this problem and provide the solution. Thanks!
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422n+1cos4nπ+2n
put i, -i,1, -1 in expansion of (1+x)^n and add
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(1+x)n=r=0∑n(rn)xr
1m+(−1)m+im+(−i)m=4,m=4k, else it is equal to 0
Putting x=1,−1,i,−i and adding, we have,
r=0∑∞(4rn)=4(1+i)n+(1−i)n+2n
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Problem 25 :
If p is an odd prime number, then ⌊(2+5)p⌋−2p+1 is always divisible by :
1)2p
2)3p
3)p+1
4)5p
Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!
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(2+5)p+(2−5)p≤(2+5)p≤(2+5)p+(2−5)p+1
Now,
(2+5)p+(2−5)p∈I
(2+5)p+(2−5)p+1∈I
∴⌊(2+5)p⌋=(2+5)p+(2−5)p
S(p)=⌊(2+5)p⌋−2p+1=(2+5)p+(2−5)p−2p+1
∴S=2r=0∑2p−1(2rp)2p−2r5r−2p+1
∴S=2p+1+2r=1∑2p−1(2rp)2p−2r5r−2p+1=2r=1∑2p−1(2rp)2p−2r5r
Clearly, S(p) is divisible by 2 and 5
S(p)=2r=1∑2p−12rp(2r−1p−1)2p−2r5r
Now since p is a prime number, gcd(r,p)=1,r∈[1,p−1]
Therefore, S(p)is divisible by p.
Since S(p) is divisible by 2,p,5 it is divisible by 2p,5p
S(5)=1300
S(5) is not divisible by 15=3⋅5 and is also not divisible by 6=5+1
Thus,
S(p) is not always divisible by p+1 and 3p
The correct options are 1 and 4
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Problem 36:
Let a be a positive real number such that a3=6(a+1) then, find the nature of the roots of x2+ax+a2−6=0.
Both Mayank Chaturvedi and Vighnesh Shenoy have provided complete solution this problem. Thanks!
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A graphical approach.
We have to see the nature of discriminant
a2−(4a2−24)or8−a2Now,graphofa3−6a−6=f(a)takestwoturnsata=2and−2,whichshowsa(rootoffunction)ispositiveandatleast2rememberf(x)iscontinuous.ata=8thef(a)isnegative.Sorootoff(a)>8.Henceforequationa3−6a−6=0,a>8;a2>8Wehave8−a2<0.sotherootsarenonreal
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Δ=(a)2−4(a2−6)=24−3a2=3(8−a2)
Consider,
f(a)=a3−6a−6
f′(a)=3a2−6=3(a2−2)
For 0≤a≤2 , f(a) is decreasing, increasing for a≥2
f(0)=−6<0
f(2)=22−62−6<0
f(8)=88−68−6=28−6<0
Thus, the positive root of f(a) is ≥8
a≥8
Δ<0
Thus, the roots are not real.
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Just completed solving the question and here you post the solution! But I think proof questions are not allowed, isn't it?
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@Vighnesh Shenoy. @Vighnesh Shenoy and @Mayank Chaturvedi,please make changes in your solutions accordingly.
I am changing the wording of the question ,as suggested byLog in to reply
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@Mayank Chaturvedi,please post the next question
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A non-calculus approach.
a3−6a−6=0
Let a=b+2/b
Therefore,(b+2/b)3−6(b+2/b)−6=0
Simplifying we get that, b6−6b3+8=0
Therefore, b3=4 or 2
Substitute these values to get a.
That time we see that only one real solution of a occurs which is, a=21/3+22/3
We see that, a2−6=6/a
Substituting this value in x2+ax+a2−6=0 we get that,
ax2+a2x+6=0
Assume that the roots of these quadratic equation are real, Then using formula for roots for quadratic equations,
x=2−a±a2−24
Then substituting the acquired value of a in this equation we get that x is a complex number. Hence, our assumption was wrong.
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Wondering, how did you thought of that a=b+(2/b).
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ax3+bx+c=0 is to take x=d+y/d and manipulate the value of y to get a solvable 6th degree equation in d
One way of solving cubic equation of the typeLog in to reply
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Problem 8: Find the locus of centers of the circles which touch the two circles
x2+y2=a2 and x2+y2=4ax
externally.
Akul Agrawal has solved this problem in the first place and posted the solution. Thanks!
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Difference of distance of center of touching circle from the two given circles is constant. Hence equation will be hyperbola with (0,0) and (2a,0) as focii. Vertex is (a,0). difference between distance of center from focii is 2a-a=a.
Hence locus is
(2a)2(x−a)2−(32a)2y2=1
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Problem 41
∫tanx.tan2x.tan3x dx
Saarthak Marathe has posted the solution to the problem.Thanks!
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tan3x=1−tanx.tan2xtan2x+tanxtanx.tan2x.tan3x=tan3x−tanx−tan2x
Now integrating we get,
∫tanx.tan2x.tan3x.dx=3ln∣sec3x∣−2ln∣sec2x∣−ln∣secx∣+constant
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Correct!Post the next question.
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Problem 2:
Which real values of k exist such that the following system of equations has no solution? {(k+1)x+8y=4kkx+(k+3)y=3k−1
Nihar Mahajan has solved this problem at the first place and has posted the solution. Thanks!
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For the equations to have no solution, they should be parallel lines, hence their x-y coefficients are in proportion but not in proportion when ratio of the constant terms is included.
kk+1=k+38=3k−14k
First solving the quadratic from kk+1=k+38 we have k=1,3. However when k=3 are substituted in 3k−14k the ratio does not remain consistent. Hence, k=3 is the solution and number of solutions is 1.
(Feel free to correct me)
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Problem 5:
r=1∑7tan2(16rπ)=?
Samuel Jones is the first person to solve this problem and provide the solution. Thanks!
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The sum can be written as −8+r=1∑7sec2(16rπ)
=−8+r=1∑7cos2(16rπ)1
Consider the 15th Chebyshev Polynomials of the Second kind
U15(x)=215k=1∏15(x−cos(nrπ))
Taking logarithm and differentiating two times at x=0, we have
S=−8+43=35
This way, we can generalize to n terms also.
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Problem 6:
Evaluate the sum of the non - real roots of the equation
x4+x3−5x2−12x−6=0.
Aditya Kumar solved the problem at the first place and provided the solution. Thanks!
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The equation can be written as (x2−2x−2)(x2+3x+3)=0
The non- real solutions are: x=2−1(3i+3) and x=21(3i−3)
Hence the sum is: −3
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Just for enlightening, I'm posting my solution (again).
Let x=t−1 the equation converts to
t4−3t3−2t2−3t+1=0
⟹t2−3t−2−t3+t21=0
Now, let y=t+t1, we have,
y2−3y−4=0
⟹(y−4)(y+1)=0
⟹(t2−4t+1)(t2−t+1)=0(∵y=t+t1)
⟹(x2−2x−2)(x2+3x+3)=0(∵t=x+1)
Only the second factor gives non - real roots since it's discriminant is negative, therefore sum is −3
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The equation factors out as,
(x2−2x−2)(x2+3x+3)=0
Now solve x2−2x−2=0 and x2+3x+3=0.
We see that the discriminant of x2+3x+3=0 is negative.
Therefore the sum of the non-real roots are -3
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Problem 7:
If (1+x)n=C0+C1x+C2x2+...+Cnxn.
Then find a closed of: C02+2C12+3C22+...+(n+1)Cn2
Rohit Ner has solved this problem at the first place and posted the solution. Thanks!
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x(1+x)n=C0x+C1x2+C2x3+⋯+Cnxn+1
Differentiating both sides, (1+x)n−1(nx+1+x)=C0+2C1x+3C2x2+⋯+(n+1)Cnxn
Also (1+x1)n=C0+xC1+x2C2+⋯+xnCn
The given series is sum of all constant terms obtained in the expansion of
(1+x)n−1((n+1)x+1).(1+x1)n
which is same as coefficient of xn in (1+x)2n−1 i.e. (n2n−1)
plus coefficient of xn−1 in (n+1)(1+x)2n−1 i.e. (n+1)(n−12n−1)
So the closed form is (n2n−1)+(n+1)(n−12n−1)
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Problem 9:
(On behalf of Akul Agarwal)
If f(x)=x+∫01t(x+t)f(t).dt, then find the value of the definite integral,
∫01f(x)dx
Ishan Singh has provided the complete solution to the problem. Thanks!
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f(x)=x(1+∫01tf(t) dt)+∫01t2f(t) dt
=Ax+B (say)
⟹f(x)=x(1+∫01t(At+B) dt)+∫01t2(At+B) dt
=x(1+3A+2B)+4A+3B
⟹Ax+B=x(1+3A+2B)+4A+3B
Comparing coefficients, we have,
A=2365
⟹∫01f(t)dt=A−1=2342
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Problem 11:
If the medians of a ΔABC make angles α,β,γ with each other, then find the value of:cot(α)+cot(β)+cot(γ)+cot(A)+cot(B)+cot(C)
Consider α,β,γ to be acute.
Saarthak Marathe is the first person to solve this problem and provide the solution. Thanks!
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The answer is 0.
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Problem 12:
Evaluate:
x→0lim{xsinx}{xtanx}1
where,{.} is fractional part function.
Rishabh Cool has provided a complete solution to this problem. Thanks!
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Problem 13:
f2(n+1)f(n)+2f(n+1)=f(n)
Given f(0)=1
Find n→∞lim2nf(n)
Ashu Dablo has solved this problem at the first place and has posted the solution. Thanks!
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@Akul Agrawal Simplify the given equation to the form:
f(n) = 1−f2(n+1)2(f(n+1))
Now we say that for f(n)=tanθ, f(n+1)=tan2θ
As f(1) =1, from above relation, we can say that f(n)=tan2(n+1)π which for n −>∞,−>0
multiply divide by tan2(n+1)π
now for x−>0 xtanx=1
So answer is 4π
Sorry for my latex!
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The equation becomes inconsistent for n=0 . Please check .
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Problem 14 :
If the sum of the first n terms of an AP is cn2, then find the sum of the cubes of these n terms.
Saakshi Singh has posted the solution first. Thanks!
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Using sum formula of AP, we get 2cn=2a+(n−1)d
⇒2cn−a=a+(n−1)d⇒(2cn−a)3=(a+(n−1)d)3.
We have to find n=1∑n(a+(n−1)d)3=n=1∑n(2cn−a)3. Nos putting n=1, we get a=c.
So we have sum of the cubes of n terms
=c3⋅n=1∑n(2n−1)3
=c3(n2(2n2−1)).
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I've cleaned up your solution for now. But it's very easy to type an equation using latex which you can learn from this note. @Saakshi Singh
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This one is a fairly easy one:
Problem 16:
The maximum value of the function f(x)=2x3−15x2+36x−48
for {x∣x2+20≤9x}
Vignesh Shenoy has posted the correct solution. Thanks!
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x2−9x+20≤0
(x−4)(x−5)≤0→x∈[4,5]
f(x)=2x3−15x2+36x−48
Differentiating,
f(x)=6x2−30x+36
f′(x)=6(x2−5x+6)
f′(x)=6(x−2)(x−3)
For x∈[4,5] , f′(x)>0→f(x) is increasing.
f(x)max=f(5)=7
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Problem 18:
Consider the ellipse :
25x2+16y2=1
Let L be the length of perpendicular drawn from the origin to any normal of the ellipse. Find the maximum value of L.
Bonus : Solve it without calculus and generalize for any ellipse
Aditya Kumar has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!
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Solution to Problem 18:
Consider the point (5cos(θ),4sin(θ)).
The equation of normal would be: 5xsec(θ)−4ycsc(θ)
Hence, distance d=∣∣∣∣25sec2(θ)+16csc2(θ)9∣∣∣∣
To maximise d, we have to minimise the denominator. By minimising the denominator(did that by calculus), the denominator comes out to be 9.
Hence dmax=1
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Absolutely Correct.
Generalisation for any ellipse x^2 / a^2 + y^2/b^2 = 1
Maximum distance of normal from origin = Modulus (b-a)
Also you could have avoided calculus by using AM Gm in last step by writing sec and cosec in terms of tan and cot
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Problem 26 :
Let na=ntimesaa⋅⋅a
f(x)=r=1∑x(rr)
Find the last digit of f(15)
Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!
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The last digit is 8. Short methods are always welcome. I went to find out last digit of individual terms. Yes!. This looks lengthy but when we note down last terms, last digits of 1,2,4,5,6,10,11,14,15 are easy to find. For rest terms, dividing exponents by 4, using modular arithmetic would do.
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Correct.
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Problem 39:
Solve
x3dxdy=y3+y2y2−x2
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The answer is xy=c(y−y2−x2) The above equation is a homogeneous equation and can be written of the form dxdy=xy3+xy2∗y2−x2 Making the substitution y=vx and we see that dxdy=(v+dxdv∗x). On simplification we get v(v2−1+vv2−1)dv=xdx. Taking v2−1 and then Multiplying and dividing by v−v2−1 we have v2−1dv−vdv=xdx. On integrating both sides we have lnvv−v2−1=lnx+c. This on substituting back v gives us the above answer.
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Problem 43 :
Find the remainder when 323232 is divided by 7.
Mayank Chaturvedi has solved this problem with the correct solution.
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32≡−3(mod7)33≡−1(mod7)323∗10+2≡(−3)3∗10+2≡(1)∗9≡9(mod7)323232≡(9)3∗10+2≡(1)∗81≡4(mod7)
Hence the answer is 4
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The answer is right, however you used
323232≡(3232)32≡932≡4(mod7)
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@Saarthak Marathe the answer is a coincidence this time.
Yes i agree with you. Have a checkLet me try once.
Reference-euler's theorem
φ(7)=6andgcd(32,7)=1326≡1mod(7)............me−1Nowwefind3232mod63232≡232≡44≡4mod(6)......me−2Usingme−1andme−2323232≡326x+4≡324mod(7)324≡44≡256=4mod(7)So4istheremainder:)
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Problem 44:
Find the locus of point of intersection of tangents to an ellipse a2x2+b2y2=1 at two points,whose eccentric angles differ by a constant angle α .
Kunal Verma has provided a complete solution to this problem.Thanks!
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Point of intersection of tangents at points who's eccentric angles are i and j :-
x= a×cos2i−jcos2i+j and y= b×cos2i−jsin2i+j
Thus a2x2 +b2y2 =sec22i−j
a2x2 +b2y2 =sec22α
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Correct! But prove your 1st statement in your solution and post the next question
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Problem 45
The coefficient of xn−6 in the expansion:
n!× [ x −(0n)(0n) +(1n) ] [2x −(1n)(1n) +(2n) ] .... [nx −(n−1n)(n−1n) +(