161 - Mathematics Questions Answers
The number of solutions of : 4x²-4x+2 = sin² y & x²+y² ≤ 3 are ? (please solve it graphically i am getting one solution by using graph but the answer is 2)
Given eq. can be written as (2x-1)2 + 1 = sin2y
and max. of siny is 1 so only possible value of x = 1/2
and possible values of y = ±nπ/2 where n is odd integer
but by drawing the circle of 2nd eq. we find that only y = ±π/2 are possible values so answer will be 2
What is the range of the function y = (e^-x)/ (1+[x]) is ?
Domain of the given function is (-∞, -1) υ [0, ∞)
Now at x=0, y=1
and at x=∞, y=0
similarly left limiting value of y = -e
and for x=-∞, y=-∞
so range is (-∞, -e] υ (0, 1]
line AB intersects CD internally in the ratio K : 1 where K is undefined , these line may be ( 1) concurrent , (2) intersecting , (3) skew < (4) all of these . explain plese ?
I think it will be skew because skew lines do not intersect in real points
If the lines represented by 2x²-5xy+2y² =0 , be the two sides of a || gm and the line 5x+2y=1 be one of teh diagonals. then the equation of the other diagonal is ?
2x²-5xy+2y² =0
=> (x-2y)(2x-y) = 0
these 2 lines pass through origin so one of the diagonal will definitely pass through origin. Given diagonal does not contain origin so second diagonal will pass through origin.
For finding any other point in the second diagonal first solve the two lines with given diagonal one by one. After calculating the two vertax use the formula for middle point of given diagonal. This point will be present in the second diagonal also.
The AM of two positive numbers a & b exceeds their GM by 3/2 & the GM exceeds the HM by 6/5 such that a+b= α , |a-b|=β , then which of them are correct,
1) α + β² = 96 2) α + β² =74 3)α²+β = 234 4)α²+β = 84
(a+b)/2 = √ab + 3/2 (1) or a+b = 2√ab + 3
√ab = 2ab/(a+b) + 6/5 (2) or 5(a+b)√ab = 10ab + 6(a+b)
on solving these 2 we get 5[2√ab + 3]√ab = 10ab + 6[2√ab + 3]
or 10ab + 15√ab = 10ab + 12√ab + 18
or 3√ab = 18 or ab = 36 and a+b = 15 = α
so a-b = √(225 - 144) = √81 = 9 = β
if x,y,z are acute and cos x= tan y , cos y = tanz , cos z = tan x, then the value of sin x is ;
by 1st equation sec y = √(1+cos2x) so cos y = 1/ √(1+ cos2x)
by 3rd equation sec z = cot x so tan z = √(cot2x - 1)
on putting these 2 values in 2nd equation we get
1/ √(1+ cos2x) = √(cot2x - 1)
so sin2x = (1+cos2x)(cos2x-sin2x)
or sin2x = (2-sin2x)(1-2sin2x)
or sin2x = 2-4sin2x-sin2x+2sin4x
or 2sin4x-6sin2x+2 = 0
now solve it for sinx
∫0π [cot-1x]dx
i think the question will be ∫0π [cotx]dx because limits are in angular terms.
by graph given below we get answer as -π/2
here in the graph same coloured shaded region are dx*(-1) always, because for every similar pair [+k]+[-k] = -1 , here k is a real number, so complete area = -1(π/2)
(α,β);(β,γ) &(γ,α) are respectively the roots of x²-2px+2=0, x²-2qx+3=0,x²-2rx+6=0. if α,β,γ are all positive , then the value of p+q+r is
According to the given conditions
α+β=2p (1) α*β=2 (2)
β+γ=2q (3) β*γ=3 (4)
γ+α=2r (5) γ*α=6 (6)
so α+β+γ = p+q+r (7)
and α*β*γ=6 (8)
dividing eq. (8) by (2), (4) , (6) one by one we will get α, β, γ hence we will get p+q+r
If a sinx + b cos(x+θ) + b cos(x-θ) = d, then the minimum value of |cos θ| is equal to :
sir i have expanded cos(x+θ) & cos(x-θ) and after that i got the equation: (d sec x - a tanx)/2b = cos θ. now , can i write minimum value of (d sec x -a tanx) = (d²-a²)^(1/2).. as all the options contain (d²-a²)^(1/2).......or do tell the other hint to solve it..
Dear amit your starting steps are correct
now let us consider that d secx - a tanx = y
on differentiating we will get d secx tanx - a sec2x = dy/dx
on comparing with 0 we get d tanx = a secx or sinx = a/d so secx = d/√(d2-a2) and tanx = a/√(d2-a2)
so minimum of given expression = [d2/ √(d2-a2)] - [a2/√(d2-a2)] = √(d2-a2)
left hand derivative of f(x) = [x]sin( pie x) at x=k where k is an integer,
Left hand derivative of given function at x = k (k is an integer) is
limh -> 0 {f(k-h) - f(k)}/ {k-h-k}
= limh -> 0 {[k-h] sin π(k-h) - [k] sin πk}/{-h} now 2 cases arise for even and odd k
first case
= limh ->0 (k-1) sin (-πh) - 0 / (-h) if k is even , here we used sin πk = 0 and sin (2π-x) = sin (-x)
= π(k-1) here we used sin (-πx)/(-πx) = 1
second case
= limh->0 (k-1) sin (πh) - 0 / (-h) if k is odd, here we used sin πk = 0 and sin (3π-x) = sin x
= -π(k-1)
Left hand derivative of given function at x = k (k is an integer) is
limh -> 0 {f(k-h) - f(k)}/ {k-h-k}
= limh -> 0 {[k-h] sin π(k-h) - [k] sin πk}/{-h} now 2 cases arise for even and odd k
first case
= limh ->0 (k-1) sin (-πh) - 0 / (-h) if k is even , here we used sin πk = 0 and sin (2π-x) = sin (-x)
= π(k-1) here we used sin (-πx)/(-πx) = 1
second case
= limh->0 (k-1) sin (πh) - 0 / (-h) if k is odd, here we used sin πk = 0 and sin (3π-x) = sin x
= -π(k-1)
so for even and odd k answer will be different.