Question

The greatest and least value of $\log_\sqrt2\left(\mathrm{sinx}-\mathrm{cosx}+3\sqrt2\right)$ are respectively-
a.2 and 1
b.5 and 3
c.7 and 5
d.9 and 7

Answer 1

Hint: To find the greatest and least value of the function, it is sufficient to find the greatest and least value of sinx - cosx, because all the other values are constant and only this is the variable part. To find maxima, the differential is 0 and the second differential is less than 0. For minima it is vice versa.

Complete step-by-step answer:
Let us assume f(x) = sinx - cosx,
To find maxima and minima, we can differentiate f(x) with respect to x and equate it with 0.
$\dfrac{\operatorname d\mathrm f\left(\mathrm x\right)}{\operatorname d\mathrm x}=\mathrm{cosx}+\mathrm{sinx}=0\\\dfrac{\mathrm{cosx}}{\sqrt2}+\dfrac{\mathrm{sinx}}{\sqrt2}=0\\\mathrm{cosxsin}\dfrac{\mathrm\pi}4+\mathrm{sinxcos}\dfrac{\mathrm\pi}4=0\\\sin\left(\mathrm x+\dfrac{\mathrm\pi}4\right)=0\\\mathrm x=-\dfrac{\mathrm\pi}4,\dfrac{3\mathrm\pi}4\\\mathrm{At}\;\mathrm x=-\dfrac{\mathrm\pi}4,\\\mathrm f\left(-\dfrac{\mathrm\pi}4\right)=-\left(\dfrac1{\sqrt2}+\dfrac1{\sqrt2}\right)=-\sqrt2\;\left(\mathrm{minima}\right)\\\mathrm f\left(\dfrac{3\mathrm\pi}4\right)=\dfrac1{\sqrt2}+\dfrac1{\sqrt2}=\sqrt2\;\left(\mathrm{maxima}\right)$
Applying the maximum value of f(x)
$\log_\sqrt2\left(\sqrt2+3\sqrt2\right)\\=\log_\sqrt2\left(4\sqrt2\right)\\=5$
Applying the minimum value of f(x)
$\log_\sqrt2\left(-\sqrt2+3\sqrt2\right)\\=\log_\sqrt2\left(2\sqrt2\right)\\=3$

Hence, maximum value is 5 and minimum value is 3. The correct option is B. 5 and 3

Note: In the above question, to find the maximum and minimum values for f(x), we can apply formula for maxima and minima given by-
f(x) = asinx + bcosx
The maximum and minimum values are-
$\pm\sqrt{\mathrm a^2+\mathrm b^2}$