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Application of Derivatives

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Long Answer Type

351.

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

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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.

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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.

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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.

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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.

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Short Answer Type

356.

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

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357.
Use differentials to approximate $square root of 0.037 end root$ .

Take $straight y space equals space square root of straight x comma space space space space straight x space equals space 0.0361 comma space space space space dx space equals space δx space equals space 0.0009 space space so space that space space straight x plus δx space equals space 0.037$
Now $straight y plus δy space equals space square root of straight x plus δx end root$
$rightwards double arrow space space space space space δy space equals square root of straight x plus δx end root space minus space square root of straight x space equals space square root of 0.037 end root space minus space square root of 0.0361 end root space equals space square root of 0.037 end root space minus 0.19 rightwards double arrow space space space space square root of 0.037 end root space equals space δy space plus 0.19 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space δy space is space approximately space equal space to space dy$
$and space space dy space equals dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx space equals space fraction numerator 1 over denominator 2 square root of 0.0361 end root end fraction left parenthesis 0.0009 right parenthesis space space space space space space equals space fraction numerator 0.0009 over denominator 2 left parenthesis 0.19 right parenthesis end fraction space equals space fraction numerator 0.0009 over denominator 0.38 end fraction space equals 0.023 therefore space space space space space from space left parenthesis 1 right parenthesis comma space space square root of 0.037 end root space equals space 0.0023 plus space 0.19 space equals space 0.1923.$