Question

If \(\lim\limits_{x\rightarrow0}\frac{e^{ax}-\cos(bx)-\frac{cxe^{-ce}}{2}}{1-\cos(2x)}=17\) ,then 5a² + b² is equal to

• A. 68
• B. 64
• C. 72
• D. 76

Answer 1

The Correct Option is A

Step 1: Use the given condition. \[ \lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17 \] On expanding the terms: \[ \left( 1 + ax + \frac{(ax)^2}{2!} + \cdots \right) - \left( 1 - \frac{(bx)^2}{2!} + \cdots \right) - \frac{cx}{2} \left( 1 - cx + \frac{(cx)^2}{2!} \right) \] \[ \lim_{x \to 0} \frac{(1 - \cos(2x))}{(2x)^2} \times \left( \frac{x}{4x^2} \right) = 17 \] Step 2: For the limit to exist: \[ a - \frac{c}{2} = 0 \quad \Rightarrow \quad c = 2a \] Substituting this into the equation: \[ \frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2} = 17 \quad \Rightarrow \quad \frac{a^2}{2} + \frac{b^2}{2} + \frac{4a^2}{2} = 34 \] \[ \Rightarrow \quad 5a^2 + b^2 = 68 \]