Question

Solution to equation is ordinate of a coordinate point
3-5x-(5-x)=1/2(3-x). find the ordinate

Answer 1

Solution : Comparing the given polynomial with ax3 + bx2 + cx + d, we get

a = 3, b = – 5, c = –11, d = – 3. Further

p(3) = 3 × 33 – (5 × 32) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)3 – 5 × (–1)2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

p(-1/3) = 3* (-1/3)3 - 5*(-1/3)2 - 11*(-1/3) - 3,

= -1/9-5/9+11/3-3 = 0

Therefore, 3, -1 and -1/3 are the zeroes of 3x3 - 5x2 -11x -3

So, we take α = 3, β = -1, γ = -1/3,

Now,

α + β + γ = 5/3 = -(-5)/3 = -b/a,

αβ + βγ + γα = -11/3 = c/a,

αβγ = 1 = -(-3)/3 = -d/a

Answer 2

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f(x)=

3/x^3−2/5x^2+7x−4

since, tangent have intercepts equal in magnitude but

opposite in sign

then let equation of tangent be x−y=a

Slope of tangent=1=f

′

(x)=x^2−5x+7

⇒x^2−5x+6=0⇒x=2,3

Therefore, co-ordinates are (2,f(2))=(2,8/3) & (3,f(3))=(3,7/2)

Ans: A,B