Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Application of Derivatives

Name: Zigya - For The Curious Learner
Rating: 4.4 (740 reviews)

Short Answer Type

381.

Use differentials to approximate fourth root of $17 over 81$ .

75 Views

382.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 31.9 right parenthesis to the power of 1 fifth end exponent$

76 Views

383.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 0.999 right parenthesis to the power of 1 over 10 end exponent$

85 Views

384.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent$

78 Views

385.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 33 right parenthesis to the power of negative 1 fifth end exponent$

92 Views

386.

If y = x⁴ – 10 and if x changes from 2 to 1.99, what is the approximate change in y?

105 Views

387. If y = x⁴ + 10 and x changes from 2 to 1.99, find the approximate change in y.

Here y = x⁴ + 10
$therefore space space dy over dx space equals space 4 straight x cubed because space space space straight x space changes space from space 2 space to space 1.99 therefore space space straight x space equals space 2 comma space space δx space equals space dx space equals space minus 0.01 Now space δy space is space approximately space equal space to space dy space and space space space space space space space space dy space equals space dy over dx dx space equals space 4 straight x cubed dx space space space space space space space space space space space space space equals space 4 left parenthesis 2 right parenthesis cubed space left parenthesis negative 0.01 right parenthesis space equals space minus 32 space cross times space 0.01 space equals space minus 0.32 when space straight x space equals space 2 comma space space space space space straight y space equals space left parenthesis 2 right parenthesis to the power of 4 plus 10 space equals space 16 plus 10 space equals space 26 therefore space space space space space straight y space plus space δy space equals space 26 space plus left parenthesis negative 0.32 right parenthesis space equals space 25.68 therefore space space space space space space space space space straight y space changes space from space 26 space to space 25.68.$