NIMCET 2023 — Mathematics PYQ

Step 1: Write down the expression

$$E={\cos }^2\theta -6\sin \theta \cos \theta +3{\sin }^2\theta +2$$

Step 2: Convert standard terms into double angles ($2\theta$)

We will use the following standard trigonometric identities:

1. ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

2. ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

3. $2\sin \theta \cos \theta =\sin 2\theta ⟹6\sin \theta \cos \theta =3\sin 2\theta$

Step 3: Substitute the identities into the expression

$$E=\left(\frac{1+\cos 2\theta }{2}\right)-3\sin 2\theta +3\left(\frac{1-\cos 2\theta }{2}\right)+2$$

Multiply out the brackets to separate the terms:

$$E=\frac{1}{2}+\frac{\cos 2\theta }{2}-3\sin 2\theta +\frac{3}{2}-\frac{3\cos 2\theta }{2}+2$$

Step 4: Combine the constant terms and variable terms

• Constants: $\frac{1}{2}+\frac{3}{2}+2=2+2=4$

• $\cos 2\theta$ terms: $\frac{1}{2}\cos 2\theta -\frac{3}{2}\cos 2\theta =-\cos 2\theta$

• $\sin 2\theta$ terms: $-3\sin 2\theta$

Now, combine everything:

$$E=4-\cos 2\theta -3\sin 2\theta$$

$$E=4-(\cos 2\theta +3\sin 2\theta )$$

Step 5: Apply the Extreme Value Formula

We know that for any expression of the form $a\cos x+b\sin x$, its range is:

$$-\sqrt{a^2+b^2}\le a\cos x+b\sin x\le \sqrt{a^2+b^2}$$

For our internal term $(\cos 2\theta +3\sin 2\theta )$, here $a=1$ and $b=3$:

$$\text{Maximum/Minimum value}=\pm \sqrt{1^2+3^2}=\pm \sqrt{1+9}=\pm \sqrt{10}$$

Therefore:

$$-\sqrt{10}\le (\cos 2\theta +3\sin 2\theta )\le \sqrt{10}$$

Step 6: Calculate the largest value

To make $E=4-(\cos 2\theta +3\sin 2\theta )$ as large as possible, we must subtract the minimum value ($-\sqrt{10}$):

$$E_{\text{max}}=4-(-\sqrt{10})$$

$$E_{\text{max}}=4+\sqrt{10}$$