160 - Mathematics Questions Answers

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]

I think it will be skew because skew lines do not intersect in real points 

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. 

(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

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 

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)

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

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 given function at x = k (k is an integer) is

   lim_{h -> 0} {f(k-h) - f(k)}/ {k-h-k}  

= lim_{h -> 0} {[k-h] sin π(k-h) - [k] sin πk}/{-h}                  now 2 cases arise for even and odd k

first case

= lim_{h ->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

= lim_h->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

   lim_{h -> 0} {f(k-h) - f(k)}/ {k-h-k}  

= lim_{h -> 0} {[k-h] sin π(k-h) - [k] sin πk}/{-h}                  now 2 cases arise for even and odd k

first case

= lim_{h ->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

= lim_h->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.

I think the right question is  

f(x)  = ₀∫^x ( t²-t+2 )²⁰⁰⁵( t²-t-2 )²⁰⁰⁷( t²-t-6 )²⁰⁰⁹( t²-t-12 )²⁰¹¹( t²-3t+2 )²⁰¹²  dt    

then sum of values of x,  where maxima of f(x) will be obtained is

for maxima f '(x) = 0

by using Newton Lebnitz formula we get

f ^'(x) =  (x²-x+2 )²⁰⁰⁵( x²-x-2 )²⁰⁰⁷( x²-x-6 )²⁰⁰⁹( x²-x-12 )²⁰¹¹( x²-3x+2 )²⁰¹²

comparing it with 0 and making factors we will get x = -1, 2, 3, -2, 4, -3, 1

now obtain the maxima point then sum these values