NIMCET 2007 — Mathematics PYQ

Step 1: Simplify $a$, $b$, and $c$

We know that $1={\log }_{x}x$. Substituting this into the equation for $a$:

$$a={\log }_{x}x+{\log }_x(yz)$$

Using the logarithmic product rule (${\log }_{m}p+{\log }_{m}q={\log }_m(pq)$):

$$a={\log }_x(xyz)$$

Similarly, we can simplify $b$ and $c$:

$$b={\log }_{y}y+{\log }_y(xz)={\log }_y(xyz)$$

$$c={\log }_{z}z+{\log }_z(xy)={\log }_z(xyz)$$

Step 2: Take the reciprocal of each term

Using the base change property (${\log }_{m}n=\frac{1}{{\log }_{n}m}$), we get:

$$\frac{1}{a}={\log }_{xyz}x$$

$$\frac{1}{b}={\log }_{xyz}y$$

$$\frac{1}{c}={\log }_{xyz}z$$

Step 3: Add the reciprocals

Now, let's add $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ together:

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}={\log }_{xyz}x+{\log }_{xyz}y+{\log }_{xyz}z$$

Using the product rule again:

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}={\log }_{xyz}(xyz)$$

Since ${\log }_{m}m=1$, we have:

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$$

Step 4: Find the value of $ab+bc+ca$

To get the required algebraic expression, take the common denominator on the left side:

$$\frac{bc+ac+ab}{abc}=1$$

Multiplying both sides by $abc$:

$$ab+bc+ca=abc$$

Correct Answer

The correct option is (c) abc.