CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Application of derivatives

Q.8. At what points will the tangent to the curve y = 2x³ – 15x² + 36x – 21 be parallel to x-axis? Also, find the equations of tangents to the curve at those points.

Solution :

We have, y = 2x³ – 15x² + 36x – 21
dy/dx = 4x² – 30x + 36
= 6(x² – 5x + 6)
= 6(x – 2)(x – 3).
As tangent is parallel to x-axis, dy/dx = 0 => 6(x – 2)(x – 3) = 0 => x = 2, 3.
When x = 2, y = 2(8) – 15(4) + 36(2) – 21 = 16 – 60 + 72 – 21 = 7.
Therefore, point is (2, 7).
When x = 3, y = 2(27) – 15(9) + 36(3) – 21 = 54 – 135 + 108 – 21 = 6.
Therefore, point is (3, 6).
Therefore, required points are (2, 7) and (3, 6). [Ans.]
Equation of tangent at (2, 7) is [ Using : y – y1 = m (x – x1)]
y – 7 = 0 (x – 2)
=> y – 7 = 0. [Ans.]
Equation of tangent at (3, 6) is
y – 6 = 0 (x – 3)
=> y – 6 = 0. [Ans.]

6.4. Approximation.

Q.1. If f(x) = 3x² + 15x + 5, then find the approximate value of f(3.02), using differentials.

Solution :

We have, f(x) = 3x² + 15x + 5.
Let x = 3 and ∆x = 0.02, then f(3.02) = f(x + ∆x) = 3(x + ∆x)² + 15(x + ∆x) + 5.
As, ∆y = f(x + ∆x) – f(x)
Therefore, f(x + ∆x) = f(x) + ∆y = f(x) + f’(x).∆x
And f(3.02) = 3x2 + 15x + 5 + (6x + 15).∆x
= 3(3)² + 15(3) + 5 + [6(3) + 15] (0.02)
= 27 + 45 + 5 + (33)(0.02)
= 77 + 0.66
= 77.66. [Ans.]

Paper By Mr. M.P.Keshari
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