How Would I Solve This Maths Question

How do I solve this math question?

First let's try to develop a general theory for these kinds of questions. So lets consider a general equation of  Conic section which is [math]ax^2+2hxy+by^2+2fx+2gy+c = 0[/math]Rearrange this equation as [math]ax^2+2x(hy+f)+(by^2+2gy+c) = 0 [/math]Solving for the [math] x[/math] we have [math]x= \dfrac{-2(hy+f)\pm2\sqrt{(hy+f)^2-a(by^2+2gy+c)}}{2a} [/math]Now if we want that given equation should have linear factors then this also means that above expression should have linear relationship between [math]x[/math]  and [math]y[/math]. Now for this to happen the term under square-root should be a perfect square.[math]\implies (hy+f)^2-a(by^2+2gy+c)[/math] should be perfect square Simplifying this equation and we will get [math](h^2-ab)y^2+2y(gh-af)+(g^2-ac) = 0 [/math]Now this equation will be a perfect square if it has only one root i.e it's discriminant is zero.[math]\implies 4(gh-af)^2 - 4(g^2-ac)(h^2-ab) = 0[/math] Simplify this and we will get [math]\boxed{af^2+ch^2+bg^2-2fgh-abc = 0}[/math] Hence this condition should be true if you want to factorize any such equation into two linear factor. More neat and aesthetic form to express this condition is [math] \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c  \end{vmatrix} = 0  [/math]. This is also called discriminant of Conic section . (It decides whether the given equation represent an ellipse or parabola or circle or point or hyperbola or pair of straight line. Here it represents the pair of straight line).Now come to this particular equation. (I am changing [math]a[/math] to [math]P[/math] and [math]b[/math] to [math]Q[/math] to avoid confusion).So given equation is [math]3x^2 + 2Qxy + 2y^2 + 2Px - 4y + 1[/math]. If you compare this to standard form you will find that ,[math]a= 3 , h = Q , g = P , b = 2 , f = -2[/math] and [math]c = 1[/math] So discriminant of this equation is [math] \begin{vmatrix} 3 & P & Q \\ Q & 2 & -2 \\ P & -2 & 1 \end{vmatrix} = 0[/math]Solving and rearranging this we have ,[math]Q^2+4PQ + 2P^2 + 6 = 0 [/math]Clearly [math]q[/math] is a root of quadratic equation given by [math]x^2+4Px+2P^2+6[/math]

How do I solve this mensuration maths question?

First of all the question is not complete.The equation of the curve shown is not stated.I am answering it ‘assuming that it is a circle.’NOTE: area of sector is proportional to its angle.above area of circle will be 36*pi for 360 degreeso area of sector which includes 90 degree angle will be( 36/4) *pisimilarly, area of sector which includes 60 degree angle will be (36/6)*pir=6, so area of sector APC=(36/4)*pi=9*piFinding area of DPC :Draw DP and PC , DP=PC=r=6, as they are radii of same circle.area of equilateral triangle DPC=1/2*6*(3*sqrt(3))=9*sqrt(3)area of sector DPC=(36/6)*pi=6*piarea of DPC=((area of sector DPC)+(area of sector CDP))-(area of triangleDPC)=[6*pi]+[6*pi]-[9*sqrt(3)]=[12*pi]-[9*sqrt(3)]Finding area of APD:area of APD=area of sector APC- area of DPC=9*pi-12*pi+9*sqrt(3)=9*sqrt(3)-3*pi=6.168required area=2*6.168=12.336.

What are some tips/tricks to solve math questions of the IIT JEE quickly?

Maths questions are easy but you can apply few tricks in order to solve the questions just by a simple method.If you get long multiplication questions like 855 * 142*311It seems to be a long calculation. But you have a trick to solve it . Just take a look at the unit digits. Yes, you got it right.5*2*1 =10 i.e unit digit is 0. If the options are in different unit digits than it is just easy for you.When you get the question to find a series than you simply need to apply numbers n upto 3. Than you would get the answer.The sum of the first n terms of the series1^2+2 . 2^2 + 3^2 +2 . 4^2 +5^2+2 . 6^2 + …………. is:n(n+1)^2⁄2 when n is even.When n is odd the sum is :A.3n(n+1)/2B.[n(n+1)/2]^2C.n(n+1)^2⁄4D.n^2*(n+1)/2Without even bothering to see what the question actually is let us put a value for n,ie; n=1 (note that this value should be odd)Now the required sum = 1^2 (the first term of the series only)= 1Now lets check in the optionsA.3n(n+1)/2 =31(1+1)/2=3 CANT BE THE ANSWERB.[n(n+1)/2]^2=[1(2)/2]^2= 1C.n(n+1)^2⁄4 =1D.n^2(n+1)/2=1Now we see that the options B,C,D all give us the correct answer.But only one of them is correct.Hence we need to put another value for n.Lets put n=3 and check.The required sum is 1+8+9=18A.3n(n+1)/2 Already eliminatedB.[n(n+1)/2]^2=[3(4)/2]^2= 36 therefore not the answerC.n(n+1)^2⁄4=36 therefore not the answerD.n^2*(n+1)/2=18Thus D is the correct answer.Make Algebra your weapon. Create the ability to picturize functions as graphs and learn to apply vertical and horizontal origin shifts carefully.Integral Calculus can be scoring if you use tricks and some basic varieties of integral functions. You can save a lot of time by using properties and applying them wisely.Work on these aspects and it would be easy for you to solve questions. Even if you need further help, than you can apply for online courses like Kaysons as they provide you all kinds of study materials and also give guidance to solve using tricks.