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

351.

Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

131 Views

352. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is

open parentheses straight a to the power of 2 over 3 end exponent plus straight b to the power of 2 over 3 end exponent close parentheses to the power of 3 over 2 end exponent.

80 Views

353. Find the maximum-area of an isosceles triangle inscribed in the ellipse

straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1

with its vertex at one end of the major axis.

770 Views

Short Answer Type

354.

Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m. BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP² + RQ² is minimum.

80 Views

Long Answer Type

355.

If length of three sides of a trapezium other than base are equal to 10 cm. then find the area of the trapezium when it is maximum.

172 Views

Short Answer Type

356.

Use differentials to approximate $square root of 25.3 end root.$

86 Views

357.

Use differentials to approximate $square root of 0.037 end root$ .

73 Views

358.

Use differentials to approximate $square root of 36.6 end root$

76 Views

359.

Use differentials to approximate:
$square root of 49.5 end root$

79 Views

360.
Use differentials to approximate:
$square root of 0.6 end root$

Take $straight y space equals square root of straight x comma space space space space straight x space space equals 0.64 comma space space space dx space equals space δx space equals space minus 0.04 space so space that space straight x space plus space straight delta space straight x space equals space 0.6$
Now, $straight y plus δy space equals space square root of straight x plus δx end root space space space space space rightwards double arrow space space space δy space equals space square root of straight x plus δx end root space minus space straight y space equals space square root of 0.6 end root space space minus space 0.8$
$rightwards double arrow space space space space space space space space space space space space space square root of 0.6 end root space equals space δy space plus space 0.8$ ...(1)
Now, $δy$ is approximately equal to dy
and $dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx space equals space fraction numerator 1 over denominator 2 square root of 0.64 end root end fraction left parenthesis negative 0.04 right parenthesis$
$equals negative fraction numerator 0.04 over denominator 2 space cross times space 0.8 end fraction space equals space minus fraction numerator 0.04 over denominator 1.6 end fraction space equals space minus 0.025$
$therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space square root of 0.6 end root space equals space minus 0.025 plus 0.8 space equals space 0.775$