Question

Let $f(x)=1+2\cos x+3\sin x$. If real numbers a, b, c are such that $a f(x)+b f(c-x)=1$ holds for any $x\in R$

then $\frac{b\cos c}{a}=$

• A. 1
• B. -1
• C. $\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

$\frac{5}{13}$

Answer 2

We are given

$$f(x)=1+2\cos x+3\sin x,$$

and

$$a\,f(x)+b\,f(c-x)=1\quad\text{for all }x\in\mathbb{R}.$$

Since the identity holds for all $x$, the coefficients of $\cos x$ and $\sin x$ must vanish and the constant term must match.

1. Write

$$f(c-x)=1+2\cos(c-x)+3\sin(c-x).$$

Using the cosine and sine angle‐difference formulas:

$$\cos(c-x)=\cos c\cos x+\sin c\sin x,$$ $$\sin(c-x)=\sin c\cos x-\cos c\sin x,$$

we obtain

$$f(c-x)=1+ (2\cos c+3\sin c)\cos x + (2\sin c-3\cos c)\sin x.$$

2. Now,

$$a\,f(x)+b\,f(c-x)= a\,[1+2\cos x+3\sin x] + b\,[1+(2\cos c+3\sin c)\cos x+(2\sin c-3\cos c)\sin x].$$

Collect coefficients:

• Constant term: $a+b$
• Coefficient of $\cos x$: $2a+ b(2\cos c+3\sin c)$
• Coefficient of $\sin x$: $3a+ b(2\sin c-3\cos c)$

For the identity to be $1$ (with no $\cos x$ or $\sin x$ terms) we must have:

$$a+b=1,\quad 2a+b(2\cos c+3\sin c)=0,\quad 3a+b(2\sin c-3\cos c)=0.$$

3. From the second equation:

$$2a = -b (2\cos c+3\sin c) \quad\Longrightarrow\quad a = -\frac{b}{2}(2\cos c+3\sin c).$$

Substitute this into the third equation:

$$3\Bigl[-\frac{b}{2}(2\cos c+3\sin c)\Bigr] + b(2\sin c-3\cos c)=0.$$

Multiply by 2:

$$-3b(2\cos c+3\sin c)+2b(2\sin c-3\cos c)=0.$$

Factor $b$ (noting $b\ne0$):

$$-3(2\cos c+3\sin c)+2(2\sin c-3\cos c)=0.$$

Expanding:

$$-6\cos c -9\sin c+4\sin c-6\cos c=0\quad\Longrightarrow\quad -12\cos c -5\sin c=0.$$

Thus,

$$12\cos c+5\sin c =0\quad\Longrightarrow\quad \sin c = -\frac{12}{5}\cos c.$$

4. Now substitute $\sin c = -\frac{12}{5}\cos c$ back into the equation for $a$ from step 3:

$$a = -\frac{b}{2}\Bigl[2\cos c+3\Bigl(-\frac{12}{5}\cos c\Bigr)\Bigr] = -\frac{b}{2}\left[2\cos c-\frac{36}{5}\cos c\right] = -\frac{b}{2}\left[-\frac{26}{5}\cos c\right] = \frac{13}{5}\,b\,\cos c.$$

5. We need the value of $\dfrac{b\cos c}{a}$:

$$\frac{b\cos c}{a}=\frac{b\cos c}{\frac{13}{5}\,b\,\cos c}=\frac{5}{13}.$$

Thus the required ratio is $\frac{5}{13}$.