Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Choose the correct answer The anti derivative of (Q22 )

Q22   Choose the correct answer   The anti derivative of  

           If   \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}  such that f (2) = 0. Then f (x) is

       A ) x ^ 4 + \frac{1}{x^3} - \frac{129 }{8} \\\\ B ) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8} \\\\ C ) x ^ 4 + \frac{1}{x^3} + \frac{129 }{8}\\\\ D) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8}

Answers (1)

best_answer

Given that the anti derivative of \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}

So, \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}

f(x) = \int 4 x ^3 - \frac{3}{x^4}\ dx

f(x) = 4\int x ^3 - 3\int {x^{-4}}\ dx

f(x) = 4\left ( \frac{x^4}{4} \right ) -3\left ( \frac{x^{-3}}{-3} \right )+C

f(x) = x^4+\frac{1}{x^3} +C

Now, to find the constant C;

we will put the condition given, f (2) = 0

f(2) = 2^4+\frac{1}{2^3} +C = 0

16+\frac{1}{8} +C = 0

or C = \frac{-129}{8}  

\Rightarrow f(x) = x^4+\frac{1}{x^3}-\frac{129}{8}

Therefore the correct answer is A.

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 Chemistry Exam Analysis 2025: Paper 'moderately difficult' but well balanced
anam-khan

CBSE Class 12 board exam still once a year; 20 lakh expected to appear in 2026 from February 17
anam-khan

CBSE board exams twice a year from 2025-26; 2 subject groups, 'better of two marks' in draft policy

Latest Question