Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 6.4, Q.8,Page 216
Question

If $f\left( x \right) = 3{x^2} + 15x + 5$, then the approximate value of $f\left( {3.02} \right)$ is

(A) $47.66$

(B) $57.66$

(C) $67.66$

(D) $77.66$

Step-by-step Solution

Option D is correct

Given, $f\left( x \right) = 3{x^2} + 15x + 5$
$\Rightarrow f\prime \left( x \right) = 6x + 15$

Also, $f\left( {x + \Delta x} \right) \approx f\left( x \right) + \Delta xf\prime \left( x \right)$ ,

Taking $x = 3$ and $\Delta x = 0.02$ , we get
$f\left( {3.02} \right) \approx 3 \times {3^2} + 15 \times 3 + 5 + 0.02\left( {6 \times 3 + 15} \right) = 77 + 0.66$

$\Rightarrow f\left( {3.02} \right) \approx 77.66$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.