If f (x) = 3x² + 15x + 5, then the approximate value of f (3.02) is
(A) 47.66  (B) 57.66  (C) 67.66  (D) 77.66

Solution:

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

Let x = 3 and Δx = 0.02

Then,

On replacing x with x + Δx,

f (3.02) = f (x + Δx) = 3(x + Δx)² + 15(x + Δx) + 5

Δy = f (x + Δx) - f (x)

f (x + Δx) = f (x) + Δy

≈ f (x) + f' (x).Δx

(∵ dx = Δx)

f (3.02) ≈ (3x² + 15x + 5) + (6x + 15) Δx

= [3(3)² + 15(3) + 5] + [6(3) + 15] (0.02)

[∵ x = 3, Δx = 0.02]

= (27 + 45 + 5) + (18 + 15) (0.02)

= 77 + (33)(0.02)

= 77 + 0.66

= 77.66

Hence, the approximate value of f (3.02)

= 77.66.

Thus, the correct option is D

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 8

If f (x) = 3x² + 15x + 5, then the approximate value of f (3.02) is (A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Summary:

Given that f (x) = 3x² + 15x + 5,the approximate value of f (3.02) is 77.66. Thus, the correct option is D