If the tangent to the curve at the point (a, b) is also tangent to the curve y = 5x2 + 2x – 25 at the point (2, 1), then is equal to
If the tangent to the curve at the point (a, b) is also tangent to the curve y = 5x2 + 2x – 25 at the point (2, 1), then is equal to
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1 Answer
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y = 5x2 + 2x – 25
P(2, -1)
T(p) : T = 0
Þ y – 1 = 10x(2) + 2(x + 2) – 50
is also tangent to y = x3 – x2 + x at point (a, b)
For y = x3 -x2 + x
y = 22x which is not tangent to the curve.
(slope of tangent)
b = 27 – 9 + 3 = 21
tangent : y – 21 = 22(x – 3)
⇒ y = 22x – 45
a = 3, b = 21
2a + 9b = 6 + 189 = 195
Also,
For a =
b =
=
Similar Questions for you
y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|
f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)

Area
3x2 = 10
x = k
3k2 = 10
By truth table
So F1 (A, B, C) is not a tautology
Now again by truth table
So F2 (A, B) be a tautology.
From option let it be isosceles where AB = AC then
=
Now ar
then
So .
Hence be equilateral having each side of length
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