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x 2 +bx+c = 0 x 2 +bx = -c ---------------------------(1) x 2 +bx+c(x+a)(2x+b)=0 --------------------------(2) put (1) in (2) we get, -c + c[2x^2 +(2a +b)x + ab -1] = 0 c[2x^2 + (2a +b)x + ab -1] = 0 ...

Triangle CDO / triangle APO = CD^2/ AB^2 q/p = CD^2/AB^2 CD / AB = root q / root p then, CD = root q and AB = root p area of triangle ABO = 1/2bh p = 1/2 root p.h 2p/root p = h now, h = 2 root p...

Given equation x 5 – 2x 2 + 7 = 0 = f let then f’ = 5x^4 -4x = 0 x(5x^3 -4) = 0 plot the wavy curve and u will get for x > .983 slope increasing and for x and 0 just after plotting the grapgh u will...

Find the determinant first by arranging the termss in a 3x3 matrix form, Then find Dx by replacing first column of the marix by [0 0 0] T similarly find Dy and Dz by replacing the suitable columns...

Let the first term be a and common difference = d. The given relation yields a+5d = 3......1 Since a is not an integer, d cannot be an integer, but its rational Let the fractional part of a be f 0...

Consider the feet D, E, F from the Orthocentre H on AB, BC and AC respectively. In triangle DBE, we have BD = a cos B and BE = c cos B (from rt angled triangles BDC and AEB resp.) Now use cosine rule ...

Denote the points as O (0,0), A(1,0), B(0,1) and C (t,t). Note that OAB is a right triangle and hence the circle passing through these three points is the one with AB as diameter. which is (x-1/2) 2...

Write the given determinant as and note that A T = -A (i.e. the matrix is a skew symmetric matrix) and hence the determinant is zero. i.e.

Yes

Logarithm is strictly defined for postive real numbers. It must not take any negative number. U can also see the graph. not defined for -ve x values. If u are defining this then x must be negative...

We check the derivative and equate it to 0 we get x 4 -1 = 0 => maxima at x=-1 and minima at x=1 (check with 2 nd derivative) so now we can vizulize the curve and if there is only 2 peaks => it ...

A = (a+b)/2 A1, A2 , … An be the n AM’s between a and b then the common difference d is b = a + (n+2-1)d => d = (b-a)/(n+1) now evaluating the sum from A1 to An is S = (n/2)(A1+An) A1...

Writing n = 2k, the given expression = 8k 3 +40k = 8(k 3 -k)+48 k. It remains to prove that k 3 -k = (k-1) k (k+1) is divisible by 6. But this is true as it is the product of 3 consecutive integers...

Note that 3x 2 +p is the derivative of x 3 +px+r and hence the common root ‘a’ is at least a double root of the cubic. If a is a triple root then since sum of roots is 0, a=0=p=r and the statement is ...

Calculus is the most important part in Jee point of view as it comprises mostly 40% of the Jee Paper. After this, Analytical geometry is the most crucial one especially from parabola and circle...

S 2n =3S n [n + (2a + (2n -1)d)] = [3n/2 + 6a + 3nd -3d] after solving , 2a – d = -n/4 – nd/2.......................................(1) S3n = [3n/2 + 2a + 3nd -d] after substituting from eqn. (1) S3n ...

distance between two parallel lines is given by the formulae: m = |9 – 15|/|(root(36 + 64)| = 6/10 = 3/5 is the answer.

but dontneglect the other subject you prepare them also becaz they are also used in the qualification but if u concentrate in one subject and score good and qualify in other without of lot of strain...

if we are asked to bring same colors pens posibility=4 but total posibility =24 therefore, probability is 4/24=1/6 therefore the solution is 1/6

lets check the domain as the term inside squareroot must be >= 0 => term inside ln be >=1 thus now at x=3 and x=4 ,f(3)= f(4) = 0 hence it cannot be strictly increasing and it’s...

Take Cases for x5 Case 1 when x (x-3)(x-5)^2010/(x+3) after pltting wavy curve x belongs to (-infinity, 0] – {1} Similary take the cases plot the wavy curves and then wite the x contribution in...

let x^2 = t then t >=0 then t^2 – 2at + a^2 – a =0 For Real Roots, D>=0 4a^2 -4(a^2 -a) >=0 a>=0 The roots in terms of a are: t1 = a + root(a) > 0 and t2 = a – root(a) > 0

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