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New answer posted
4 months agoContributor-Level 10
Draw g(t) = t3 – 3t
g'(t) = 3(t2 – 1)
g(1) is maximum in (-2, 2)
So, maximum (t3 – 3t) =
=
I =
again rewrite the f(x)
So f(x) is not differentiable at x = 2, 3, 4, 5
so m = 4
New answer posted
4 months agoContributor-Level 10
Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is
Let
Now according to the question
(a, b, c) =
Now
=
= 28 + 22 [Use ]
= 6
New answer posted
4 months agoContributor-Level 10
Put y =
differentiable worst = x
V +
apply variable separable method
Given y(1) = 3 C = 8
Now at x = 2
y2 + 64 = (32 – y)2
y = 15
New answer posted
4 months agoContributor-Level 10
Set of first 10 prime numbers
= {2,3,5,7,11,13,17,19,23,29,31}
So sample space = 104.
Favourable cases
So required probability
New answer posted
4 months agoNew answer posted
4 months agoContributor-Level 10
so f (x) is decreasing in
Tangent at y2 = 2x is y = mx + it is passing through (4, 3) therefore we get m =
So tangent may be passes through (-2, 0) so rejected.
Equation of normal
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