Question

The complete set of values of $\sin^8x + \cos^8x$ is [a,b]. Then (b – a) is expressed in form of $\frac{m}{n}$ (where m, n are coprime numbers). The value of m + n is -

• A. 15
• B. 1
• C. 23
• D. 9

Answer 1

Answer: (A)

15

Answer 2

To find the complete set of values of $\sin^8x + \cos^8x$, let $y = \sin^8x + \cos^8x$.

We can rewrite the expression using algebraic identities. Recall that $a^4 + b^4 = (a^2+b^2)^2 - 2a^2b^2$. So, $\sin^8x + \cos^8x = (\sin^4x)^2 + (\cos^4x)^2 = (\sin^4x + \cos^4x)^2 - 2\sin^4x \cos^4x$.

First, let's simplify $\sin^4x + \cos^4x$: $\sin^4x + \cos^4x = (\sin^2x + \cos^2x)^2 - 2\sin^2x \cos^2x$ Since $\sin^2x + \cos^2x = 1$, we have: $\sin^4x + \cos^4x = (1)^2 - 2\sin^2x \cos^2x$ We know that $\sin 2x = 2\sin x \cos x$, so $\sin^2 2x = 4\sin^2x \cos^2x$. Therefore, $\sin^2x \cos^2x = \frac{\sin^2 2x}{4}$. Substituting this into the expression for $\sin^4x + \cos^4x$: $\sin^4x + \cos^4x = 1 - 2\left(\frac{\sin^2 2x}{4}\right) = 1 - \frac{\sin^2 2x}{2}$.

Now substitute this back into the expression for $y$: $y = \left(1 - \frac{\sin^2 2x}{2}\right)^2 - 2\sin^4x \cos^4x$ $y = \left(1 - \frac{\sin^2 2x}{2}\right)^2 - 2(\sin x \cos x)^4$ $y = \left(1 - \frac{\sin^2 2x}{2}\right)^2 - 2\left(\frac{\sin 2x}{2}\right)^4$ $y = \left(1 - \frac{\sin^2 2x}{2}\right)^2 - 2\frac{\sin^4 2x}{16}$ $y = \left(1 - \frac{\sin^2 2x}{2}\right)^2 - \frac{\sin^4 2x}{8}$

Let $t = \sin^2 2x$. We know that for any real $x$, $0 \le \sin^2 2x \le 1$. So, $0 \le t \le 1$. Substitute $t$ into the expression for $y$: $y = \left(1 - \frac{t}{2}\right)^2 - \frac{t^2}{8}$ $y = 1 - 2\left(\frac{t}{2}\right) + \left(\frac{t}{2}\right)^2 - \frac{t^2}{8}$ $y = 1 - t + \frac{t^2}{4} - \frac{t^2}{8}$ $y = 1 - t + \frac{2t^2 - t^2}{8}$ $y = 1 - t + \frac{t^2}{8}$

Let $f(t) = \frac{t^2}{8} - t + 1$. We need to find the range of $f(t)$ for $t \in [0, 1]$. This is a quadratic function of $t$, and its graph is a parabola opening upwards (since the coefficient of $t^2$ is positive, $\frac{1}{8} > 0$). The vertex of the parabola occurs at $t = -\frac{\text{coefficient of } t}{2 \times \text{coefficient of } t^2} = -\frac{(-1)}{2 \times \frac{1}{8}} = \frac{1}{\frac{1}{4}} = 4$.

Since the vertex $t=4$ is outside the interval $[0, 1]$ and to the right of it, the function $f(t)$ is monotonically decreasing on the interval $[0, 1]$. Therefore, the maximum value of $f(t)$ occurs at $t=0$, and the minimum value occurs at $t=1$.

Maximum value ($b$): $f_{max} = f(0) = 1 - 0 + \frac{0^2}{8} = 1$. This occurs when $\sin^2 2x = 0$, which means $\sin 2x = 0$. For example, if $x=0$, $y = \sin^8 0 + \cos^8 0 = 0 + 1 = 1$.

Minimum value ($a$): $f_{min} = f(1) = 1 - 1 + \frac{1^2}{8} = \frac{1}{8}$. This occurs when $\sin^2 2x = 1$, which means $\sin 2x = \pm 1$. For example, if $x=\frac{\pi}{4}$, $y = \sin^8(\frac{\pi}{4}) + \cos^8(\frac{\pi}{4}) = \left(\frac{1}{\sqrt{2}}\right)^8 + \left(\frac{1}{\sqrt{2}}\right)^8 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$.

So, the complete set of values of $\sin^8x + \cos^8x$ is $[a, b] = \left[\frac{1}{8}, 1\right]$. We need to find $(b-a)$. $b-a = 1 - \frac{1}{8} = \frac{8-1}{8} = \frac{7}{8}$.

This value is expressed in the form $\frac{m}{n}$, where $m=7$ and $n=8$. $m$ and $n$ are coprime numbers (7 and 8 have no common factors other than 1). We need to find the value of $m+n$. $m+n = 7+8 = 15$.