NIMCET 2009 — Mathematics PYQ

First, let's simplify the term ${\sin }^{4}x+{\cos }^{4}x$ using the identity $a^2+b^2=(a+b{)}^2-2ab$:

Since ${\sin }^{2}x+{\cos }^{2}x=1$:

Using the double angle identity $\sin 2x=2\sin x\cos x$, we can write $2{\sin }^{2}x{\cos }^{2}x$ as $\frac{1}{2}(2\sin x\cos x{)}^2=\frac{1}{2}{\sin }^{2}2x$.

So, ${\sin }^{4}x+{\cos }^{4}x=1-\frac{1}{2}{\sin }^{2}2x$.

2. Substitute back into the Original Equation

Replace the simplified expression in the equation:

Rearrange to solve for $\alpha$:

3. Analyze as a Quadratic in $\sin 2x$

Let $y=\sin 2x$. We know that for real $x$, $-1\le y\le 1$.

The equation becomes:

To find the range of $\alpha$, we find the minimum and maximum values of the function $f(y)=\frac{1}{2}{y}^2-y-1$ on the interval $[-1,1]$.

• Check the Vertex: The vertex occurs at $y=-\frac{b}{2a}=-\frac{-1}{2(1/2)}=1$.

• Evaluate at Endpoints:

  • At $y=1$: $f(1)=\frac{1}{2}(1{)}^2-1-1=\frac{1}{2}-2=-\frac{3}{2}$ (Minimum value)

  • At $y=-1$: $f(-1)=\frac{1}{2}(-1{)}^2-(-1)-1=\frac{1}{2}+1-1=\frac{1}{2}$ (Maximum value)

4. Conclusion

The range of $\alpha$ for which the equation is solvable is the set of values $f(y)$ can take, which is:

Final Answer:

The correct option is (c) $-\frac{3}{2}\le \alpha \le \frac{1}{2}$.