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Maximum value of \(\rm sinx \ . cosx\)  is 
1. 1
2. \(\frac{1}{\sqrt2}\)
3. 0 
4. \(\frac{1}{2}\)
5. -1

1 Answer

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Correct Answer - Option 4 : \(\frac{1}{2}\)

Concept:

Following steps to finding maxima and minima using derivatives.

Find the derivative of the function.

Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.

Now we have to find the second derivative.

  • f"(x) is less than 0 then the given function is said to be maxima
  • If f"(x) Is greater than 0 then the function is said to be minima

 

\(\rm sin2 x =2sinx \ . cosx\)

 

Calculation:

Let, f(x) = \(\rm sinx \ . cosx\) 

\(\rm\frac12\times sin2 x \)            (∵ \(\rm sin2 x =2sinx \ . cosx\)

f'(x) = \(\rm\frac12\times (2cos2 x)\)

= cos 2x

Now, f'(x) = 0 ⇒ cos 2x = 0

We know, \(\rm cos(\fracπ2)=0\)

∴ cos 2x = \(\rm cos(\fracπ2)\)

⇒ 2x = \(\fracπ2\)

⇒ x = \(\fracπ4\)

Also, f''(x) = \(\rm-(4sin2 x)\)

⇒ f''(\(\fracπ4\)) = - 4 < 0

At x = \(\fracπ4\) , f(x) is maximum.

∴ f(\(\fracπ4\)) = \(\rm\frac12\times sin (\frac{\pi}{2})\) 

\(\frac12\)

Hence, option (4) is correct.

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