QJEEM23Maf6a0: JEE Main 2023 Mathematics - Class 12 JEEM23Maf6a0 Solution: If lim x → 0 e a x − cos ⁡ ( b x ) − c x e − c x 2 1 − cos ⁡ (…

The Taylor series for $e^x$ , $\cos x$ , and $e^{-x}$ around $x=0$ are:

$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$

$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$

$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$

Substituting these into the limit and simplifying:

$\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.$

This simplifies to:

$\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.$

Let's evaluate this expression:

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.$

We know that $1 - \cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x = 1 - 2\sin^2x$ . So, $1 - \cos 2x = 2\sin^2x$ .

Substituting this into our expression we get:

$(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.$

This simplifies to:

${1 \over 2}$ $(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.$

So,

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.$

$\therefore$ $\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.$

Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:

$a-\frac{c}{2} = 0 \Rightarrow c=2a.$

Then, for the limit to equal $17$ , the coefficient of $x^2$ in the numerator must equal $17 \times 2 = 34$ . This gives us the equation:

$\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.$

Since $c=2a$ , we can substitute $c$ in the equation to get:

$a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.$