GATE | Quiz for Sudo GATE 2021 | Question 30

Which of the following option(s) is/are correct ?
• (I) Function f(x) = x(x2 – 27) where x is a real number. Then the function has one minima and one maxima.
• (II) The Integral value of f(x) = x * sinx within the limits [0, Ï€] is Ï€/2.
• (III) The mean value for the function f(x) = âˆš(x2 – 9) in the interval [3, 6] is 3/âˆš2.
Note – This question is multiple select questions (MSQ). (A) Statement I is correct (B) Statement II is correct (C) Statement III is correct (D) Only I and II are correct

Explanation: Statement I : Given function,
```f(x) = x(x2-27)
f'(x) = (x2-27)  + x.2x
3x2 - 27 = 0
x2 - 9 = 0
x = Â±3 ```
And,
`f''(x) = 6x `
Check minima and maxima at both the values,
```f''(3)
= 6(3)
= 18 > 0
So, minima at x = 3 ```
Also,
```f''(-3) = 6(-3) = -18 < 0
So, maxima at x = -3 ```
Statement II : Let,
```I = âˆ«x*sinx -----------> 1
I = âˆ«(Ï€-x)*sin(Ï€-x) = âˆ«(( Ï€-x)*sin(x) ------ 2 ```
```=> I + I
= âˆ« Ï€*sinx = Ï€*[-cosx]
= - Ï€*[cos Ï€ â€“ cos0]

=> 2I = - Ï€*-2
=> I = Ï€ ```
The value of Integral is Ï€. Statement III : f(x) being a polynomial function is continuous in [2, 4], and f(x) being a polynomial function is derivable in (2, 4). Thus, both conditions of Lagrange's mean value theorem is satisfied. Therefore there exists atleast one real number c in (3, 6) such that,
```fâ€™(c) = f(6) - f(3)/6 - 3
â†’ fâ€™(x) = x / âˆš(x2 - 9) ```
Then,
```fâ€™(c) = âˆš(62 - 9) - âˆš(32 - 9) / 6 - 3
â†’ c/âˆš (c2 - 9) =  âˆš27 / 3
â†’ c2 = 3(c2 - 9)
â†’ 2c2 = 9
â†’ c = 3/âˆš2 ```
So, option (A) and (C) are correct.

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