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)

Long Answer Type

161.

Prove that $straight y equals space fraction numerator 4 space sin space straight theta over denominator 2 plus cos space straight theta end fraction minus straight theta$ is an increasing function of $straight theta$ in $open square brackets 0 comma space straight pi over 2 close square brackets$ .

78 Views

162.

Find the intervals in which the function f is given by
$straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 space sin space straight x space minus space 2 straight x minus space straight x space cosx over denominator 2 plus cos space straight x end fraction$
is (i) increasing (ii) decreasing

90 Views

163.

Determine the values of x for which the function $straight f left parenthesis straight x right parenthesis space equals fraction numerator straight x over denominator straight x squared plus 1 end fraction$ is increasing and for which it is decreasing.

73 Views

Short Answer Type

164. Prove that the function x² – x + 1 is neither increasing nor decreasing on (0, 1).

100 Views

165.

Let I be any interval disjoint from (– 1, 1). Prove that the function f given by $straight f apostrophe left parenthesis straight x right parenthesis space equals space straight x plus 1 over straight x$ is strictly increasing on I.

107 Views

166.

Find the intervals in which the following function is decreasing:
f (x) = x³– 12x

88 Views

167.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing (b) strictly decreasing

f(x) = 2x² – 3x $rightwards double arrow space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 straight x minus 3$
Now $straight f apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space space space space space space space space space space space space space rightwards double arrow space space space space 4 straight x minus 3 space equals space 0 space space space space space rightwards double arrow straight x space equals space 3 over 4$
The point $straight x equals 3 over 4$ divides the real line into two disjoint intervals $open parentheses negative infinity comma space 3 over 4 close parentheses space and space open parentheses 3 over 4 comma space infinity close parentheses$ .
$In space space open parentheses negative infinity comma space space space 3 over 4 close parentheses comma space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 straight x minus 3 space space less than space space 0 therefore space space space space space space space space space straight f space is space strictly space decreasing space in space open parentheses negative infinity comma space space 3 over 4 close parentheses. In space open parentheses 3 over 4 comma space infinity space close parentheses comma space space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 straight x minus 3 space greater than space 0 therefore space space space space straight f space is space strictly space increasing space in space open parentheses 3 over 4 comma space infinity close parentheses.$