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Application of Derivatives

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Long Answer Type

261.

Find the maximum and minimum value of x + sin 2 x on $left square bracket 0 comma space 2 straight pi right square bracket$ .

75 Views

Short Answer Type

262.

Find both the maximum value and the minimum value of
3x⁴ – 8x³ + 12x² – 48x + 25

90 Views

263. It is given that at x = 1, function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0, 2], Find the value of a.

88 Views

Long Answer Type

264. $straight y space equals space fraction numerator ax minus straight b over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis end fraction$ has a turning point at P(2, -1). Find the values of a and b and show that y is maximum at P.

Here space straight y space equals space fraction numerator ax minus straight b over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis end fraction space equals space fraction numerator ax minus straight b over denominator straight x squared minus 5 straight x plus 4 end fraction

...(1)

dy over dx space equals space fraction numerator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis. space straight a space minus space left parenthesis ax minus straight b right parenthesis thin space left parenthesis 2 straight x minus 5 right parenthesis over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis squared end fraction

...(2)

Now,

straight P left parenthesis 2 comma space minus 1 right parenthesis

lies on (1)

therefore space space space space space minus 1 space equals space fraction numerator 2 straight a minus straight b over denominator 4 minus 10 plus 4 end fraction space space space rightwards double arrow space space space minus 1 space equals space fraction numerator 2 straight a minus straight b over denominator negative 2 end fraction space space space space rightwards double arrow space space space 2 straight a minus straight b space equals space 2

...(3)

Since P(2, -1) is a turning point at P(2, -1)

therefore space space space space dy over dx space equals space 0 therefore space space space space space space fraction numerator left parenthesis 4 minus 10 plus 4 right parenthesis. space straight a minus left parenthesis 2 straight a minus straight b right parenthesis thin space left parenthesis 4 minus 5 right parenthesis over denominator left parenthesis 4 minus 10 plus 4 right parenthesis squared end fraction space equals space 0 rightwards double arrow space space space space minus 2 straight a plus left parenthesis 2 straight a minus straight b right parenthesis space equals space 0 space space space space space space rightwards double arrow space space space minus 2 straight a plus left parenthesis 2 right parenthesis space equals space 0 space space space space space space space space space space space space open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets rightwards double arrow space space space space space space space space space space 2 straight a space equals space 2 space space space space space rightwards double arrow space space space straight a space equals space 1 therefore space space space space from space left parenthesis 3 right parenthesis comma space space space 2 minus straight b space equals space 2 space space space space space space space space space rightwards double arrow space space straight b space equals space 0 therefore space space space space space space dy over dx equals space fraction numerator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis minus straight x left parenthesis 2 straight x minus 5 right parenthesis over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis end fraction space equals fraction numerator negative straight x squared plus 4 over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis squared end fraction

When space straight x space is space slightly space less than space 2 comma space space space dy over dx space equals space plus ve When space straight x space is space slightly space greater than 2 comma space space space dy over dx space equals space minus ve therefore space space space at space straight x space equals space 2 comma space space dy over dx space changes space sign space from space plus ve space to space minus ve therefore space space space straight y space has space maximum space value space at space straight P left parenthesis 2 comma space minus 1 right parenthesis.

74 Views

265. If y = a log | x | + bx² + x has its extremum values at x = – 1 and x = 2, find the values of a and b. Prove that these extrema are maximum.

74 Views

Short Answer Type

266.

Find the maximum and minimum values of $fraction numerator log space straight x over denominator straight x end fraction space in space 0 less than straight x less than infinity$ .

74 Views

Long Answer Type

267.

Find the shortest distance of the point (0, c) from the parabola y = x² , where 0 ≤ c ≤ 5.

160 Views

Short Answer Type

268. An Apache helicopter of enemy is flying along the curve given by y = x² + 7. A soldier, placed at (3. 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.

139 Views

Multiple Choice Questions

269.

For all real values of x, the minimum value of $fraction numerator 1 minus straight x plus straight x squared over denominator 1 plus straight x plus straight x squared end fraction$ is

85 Views

270.

The maximum value of $open square brackets straight x left parenthesis straight x minus 1 right parenthesis space plus space 1 close square brackets to the power of 1 third end exponent comma space space space 0 space less or equal than space straight x space less or equal than space 1$ is

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