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y= tan –1[{√(1 + x^2) – √(1 – x^2)}/{√(1 + x^2) + √(1 – x2)}]Put x^2= cos θ, theny = tan –1[{√(1 + cos θ) – √(1 – cos θ)}/{√(1 + cos θ) + √(1 – cos θ)}]= tan-1[(√2 cos θ/2 – √2 sin θ/2)/(√2 cos θ/2 + √2 sin θ/2)]= tan-1[(1 – tan θ/2)/(1 + tan θ/2)]= tan-1[tan(π/4 – θ/2)]= π/4 – θ/2 = π/4 – 1/2 cos –1 x^2.Differentiating w.r.t. x we getdy/dx = 0 – 1/2. d/dx(cos-1x^2)= – 1/2. – 1/√{1 – (x2)2}.d/dx(x^2)= 1/2 .1/√(1 – x^4). 2x= x/√(1 – x^4)

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3x-2x-6x= -4-8-3-5x= -15x= 3

Since 2 is one root of the quadratic equation,

We substitute X=2

k(2)^2 - 14(2) +8= 0

4k-28+8=0

4k-20=0

4k=20

k=5

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5x*x - 15x - 20 = 5(x*x - 3x - 4) = 5(x*x - 4x + x - 4) = 5(x(x-4)+1(x-4)) = 5(x+1)(x-4)

Answer

(x*x-3x+2)/(x*x-5x+4) = (x*x-6x+8)/(x*x-9x+14) (x-2)(x-1)/((x-4)(x-1)) = (x-4)(x-2)/((x-7)(x-2)) (x-2)/(x-4) = (x-4)/(x-7) (x-2)(x-7) = (x-4)(x-4) x*x-9x+14 = x*x-8x+16 8x - 9x = 16 - 14 -x = 2 x = -2

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x=3(3)^2-2k(3)-6=09-6k-6=03-6k=0k=1/2

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3x/5-2=x+1/3

Multiply by LCM=15 to get rid of all fractions:

9x - 30 = 15x + 5

-6x - 30 = 5 [subtract 15x from both sides]

-6x = 35 [add 30 to both sides]

x = -35/6 [divide both sides by (-6)]

x = -5 5/6

If -1/2 is a solution of given quadratic equation the it would satisfy the condition ;3(-1/2)^2 + 2×k×(-1/2) + 3 = 0

-3/4 - k + 3 = 0k = (-3+12)/4k = 9/4

yes because there are two variables

One of the roots is -1/2. So it will satisfy the equation,3(1/4)+2k(-1/2)-3=03/4-k-3=0 k=3/4-3=3-12/4=-9/4

x²-2kx-6= 0(3) ²-2k(3) -6= 09-6k-6=03-6k=0k= 1/2

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When roots are equal then discrimanat is 0

(-2k)² - 4 × 12 × 3 = 0

4k² = 144

k² = 36

k = + / - 6

Given :x=3 is one of the root of equation x^2-2kx-6+0Solution:Substituting x=3 in the equation3^2-2k3-6=09-6k-6=06k=3k=1/2Therefore the value of k=1/2

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[(64)^2/3 * 2^-2 ÷ 8^0]^1/2= [(4)^2/2 * 2^-2/2 ÷ 1]= (4)*(2)-1 ÷1= 4/2= 2

( b ) 412( e ) 6081( h) 96260( k) 12458.175

(b) 412(e) 6081(h) 96260(k) 123458.17 are the following questions answer

412,6081,96260,123458.17 is the correct answer of the given question

b. 412e. 6081h. 96260k. 12458.175

Let there be X 10Rs noteno of Rs 5 note= 2X+1no of Rs 20 note= X-5Sum of the notes= 3X-4value of notes= 10*X+ 5(2x+1)+20(X-5)=185 ⇒ 10X +10X+5+20X-100=185 ⇒ 40X= 280 ⇒ X=7Rs 10 Qty 7Rs 5 Qty 15Rs20 Qty 2

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x+8+3x+4+2x=180; 6x+12=180; 6x=180-12=168; x=168/6=28;

sum of all internal angle of ∆ = 180x+8+3x+4+2x= 1806x+12=1806x= 180-12= 168x= 168/6= 28°

x+8+3x+4+2x=1806x+12=1806x=180-126x=168x=168/6x=28

X = is correct answer in this questions

28 yar the value of x

Let theta = xThen,tan x = 3/4 = P/B

Using pythagoras theoram H^2 = P^2 + B^2 H^2 = 3*3 + 4*4H^2 = 9 + 16H^2 = 25H = 5

sin x = P/H = 3/5cos x = B/H = 4/5

Thus,(4sin x - cosx + 1)/(4sin x + cosx - 1)

= (4*3/5 - 4/5 + 1)/ (4*3/5 + 4/5 - 1)

= (12/5 - 4/5 +1)/ (12/5 + 4/5 - 1)

= [(12 - 4 + 5)/5]/ [(12 + 4 - 5)/5]

= (13/5)/(11/5)

= 13/11

the right answer is 20

8/32 = 1/4

15/45 = 1/3

44/22 = 240/76 = 10/1952/1176/1372/8576/133

i) Base = 6 and Power =8

ii)Base = -3 and Power =4

iii)Base = -52 and Power =3

iv)Base =2*2=4 and Power =6

thanks

Polynomial=x³-5x²+7x+mdivided by x-2 Remainder=52So, x³-5x²+7x+m+52 should be divisible by x-2

Put x=2

2³-5(2)²+7(2)+m+52=08-20+14+m+52=08-6+m+52=02+m+52=0m=-54

52*(-8)+(-52)52*(-8-52)52*-60-3120 is answer

sorry mistake ho gae mana na 2 nahi dakha

52*(-8)+(-52)*2-416-[(52)*2]-416-104-105 is answer

your 1st questions answer is 15