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Long Answer Type

261.

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

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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.

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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.

straight y space equals space straight a space log space open vertical bar straight x close vertical bar space plus space bx squared plus straight x

therefore space space space space dy over dx space equals space straight a fraction numerator 1 over denominator open vertical bar straight x close vertical bar end fraction. space straight d over dx open parentheses open vertical bar straight x close vertical bar close parentheses space plus space 2 bx plus 1

equals space straight a. space fraction numerator 1 over denominator open vertical bar straight x close vertical bar end fraction space open parentheses fraction numerator straight x over denominator open vertical bar straight x close vertical bar end fraction. space 1 close parentheses space plus space 2 bx plus 1 space equals space fraction numerator straight a space straight x over denominator straight x squared end fraction plus 2 bx plus 1

open square brackets because space space space open vertical bar straight x close vertical bar squared space equals space straight x squared close square brackets

equals straight a over straight x plus 2 bx plus 1

Since y has its extremum values at x = -1 and x = 2

therefore space space space space open square brackets dy over dx close square brackets subscript straight x equals negative 1 end subscript space space equals space 0 comma space space space space space space open square brackets dy over dx close square brackets subscript straight x equals 2 end subscript space space equals 0 therefore space space space minus straight a minus 2 straight b plus 1 space equals 0 comma space space space straight a over 2 plus 4 straight b plus 1 space equals space 0 therefore space space space space space space space straight a plus 2 straight b space equals space 1 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 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 and space space space space space space straight a plus 8 straight b space equals space minus 2 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 space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis 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

Subtracting (1) from (2),

6 straight b space equals negative 3 space space space space space space rightwards double arrow space space space straight b space equals space minus 1 half

therefore space space space space space space from space left parenthesis 1 right parenthesis comma space space space straight a minus 1 space equals 1 space space space rightwards double arrow space space space straight a space equals space 2 therefore space space space space space space dy over dx space equals space 2 over straight x minus straight x plus 1 space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 2 over straight x squared minus 1

At space space straight x space space equals negative 1 comma space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 2 over 1 minus 1 space equals space minus 3 space less than space 0 At space space space straight x minus 2 comma space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 1 half minus 1 space equals space minus 3 over 2 less than 0 therefore space space space space both space the space points space straight x space equals negative 1 comma space space space straight x space equals space 2 space of space extrema space are space maxima

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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