CAT 2020

Slot 1

QA

Question & Solution

Given:

$(x^{2}-7x+11)^{(x^{2}-13x+42)} = 1$

This equation holds under the following conditions:

1. $(x^{2}-13x+42) = 0$
2. $(x^{2}-7x+11) = 1$
3. $(x^{2}-7x+11) = -1$ and $(x^{2}-13x+42)$ is even.

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Step 1: Solve for conditions where $(x^{2}-13x+42) = 0$

We need to solve:

$$ x^{2} - 13x + 42 = 0 $$

Factoring the quadratic expression:

$$ (x - 6)(x - 7) = 0 $$

Thus, $x = 6$ or $7$.

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Step 2: Solve for conditions where $(x^{2}-7x+11) = 1$

We need to solve:

$$ x^{2} - 7x + 11 = 1 $$

Simplifying the equation:

$$ x^{2} - 7x + 10 = 0 $$

Factoring:

$$ (x - 5)(x - 2) = 0 $$

Thus, $x = 5$ or $2$.

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Step 3: Solve for conditions where $(x^{2}-7x+11) = -1$ and $(x^{2}-13x+42)$ is even

We need to solve:

$$ x^{2} - 7x + 11 = -1 $$

Simplifying the equation:

$$ x^{2} - 7x + 12 = 0 $$

Factoring:

$$ (x - 3)(x - 4) = 0 $$

Thus, $x = 3$ or $4$.

Also, for $x = 3$ or $4$, we need to check if $(x^{2} - 13x + 42)$ is even.

For $x = 3$:

$$ x^{2} - 13x + 42 = 3^{2} - 13(3) + 42 = 9 - 39 + 42 = 12 $$ (which is even)

For $x = 4$:

$$ x^{2} - 13x + 42 = 4^{2} - 13(4) + 42 = 16 - 52 + 42 = 6 $$ (which is also even)

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Step 4: Final Solution Set

From the above steps, the values of $x$ that satisfy the given equation are:

$$ x = 2, 3, 4, 5, 6, 7 $$

Final Answer: The solution set is $\boxed{{2, 3, 4, 5, 6, 7}}$ Hence the answer is 6.