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प्रश्न
The curves y = 4x2 + 2x – 8 and y = x3 – x + 13 touch each other at the point ______.
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उत्तर
The curves y = 4x2 + 2x – 8 and y = x3 – x + 13 touch each other at the point `(- 1/3, (-74)/9)`.
Explanation:
We have y = 4x2 + 2x – 8 .....(i)
And y = x3 – x + 13 .....(ii)
Differentiating eq. (i) w.r.t. x, we have
`"dy"/'dx"` = 8x + 2
⇒ m1 = 8x + 2 .....[m is the slope of curve (i)]
Differentiating eq. (ii) w.r.t. x, we get
`"dy"/"dx"` = 3x2 – 1
⇒ m2 = 3x2 – 1 ......[m2 is the slope of curve (ii)]
If the two curves touch each other, then m1 = m2
∴ 8x + 2 = 3x2 – 1
⇒ 3x2 – 8x – 3 = 0
⇒ 3x2 – 9x + x – 3 = 0
⇒ 3x(x – 3) + 1(x – 3) = 0
⇒ (x – 3)(3x + 1) = 0
∴ x = 3, `(-1)/3`
Putting x = 3 in equation (i), we get
y = 4(3)2 + 2(3) – 8
= 36 + 6 – 8
= 34
So, the required point is (3, 34)
Now for x = `- 1/3`
y = `4((-1)/3)^2 + 2((-1)/3) - 8`
= `4 xx 1/9 - 2/3 - 8`
= `4/9 - 2/3 - 8`
= `(4 - 6 - 72)/9`
= `(-74)/9`
∴ Other required point is `(- 1/3, (-74)/9)`.
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