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# Please elaborate the answer.Need complete solution.

Arun
25763 Points
3 years ago
Dear student

Minimum value of expression is obtained when denominator maximizes.
hence
maximum value of 3 sin@ – 4 cos@ +7 = 5 + 7 = 12
Hence minimum value of original expression is 1/12
Soumendu Majumdar
159 Points
3 years ago
Dear Student,
$-\sqrt{a^2 +b^2}\leq asin\Theta + bcos\Theta \leq \sqrt{a^2 +b^2}$
Here a = 3, b = 4 or – 4 (doesn’t matter since squaring it will give the same result)
Now since the expression is in the denominator so we need to find the maximum value of $3sin\Theta -4cos\Theta$ since greater is the denominator lesser will be the value of the original expression.
So maximum value = $\sqrt{3^2+4^2 } = \sqrt{5^2} = 5$
so now maximum value of denominator = 5 + 7 = 12
Hence, minimum value of the expression = 1/12
Hope it helps!