# GATE | Quiz for Sudo GATE 2021 | Question 30

• Last Updated : 16 Dec, 2020

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, π] is π/2.
• (III) The mean value for the function f(x) = √(x2 – 9) in the interval [3, 6] is 3/√2.

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 = ±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 = ∫x*sinx -----------> 1
I = ∫(π-x)*sin(π-x) = ∫(( π-x)*sin(x) ------ 2 ```

```=> I + I
= ∫ π*sinx = π*[-cosx]
= - π*[cos π – cos0]

=> 2I = - π*-2
=> I = π ```

The value of Integral is π.

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,

```f’(c) = f(6) - f(3)/6 - 3
→ f’(x) = x / √(x2 - 9) ```

Then,

```f’(c) = √(62 - 9) - √(32 - 9) / 6 - 3
→ c/√ (c2 - 9) =  √27 / 3
→ c2 = 3(c2 - 9)
→ 2c2 = 9
→ c = 3/√2 ```

So, option (A) and (C) are correct.

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