Question

If A = {2,3,4}, B = {6,8,9,10}. If $\alpha$ represents the number of onto functions, then the value of $\lim_{x \to 0} \frac{(\alpha + 4) \sin^2 x}{x^2 + x^3 \cos x}$ is.

• A. 4

Answer 1

Answer: (A)

4

Answer 2

Let A = {2,3,4} and B = {6,8,9,10}. The number of elements in set A is $|A| = 3$. The number of elements in set B is $|B| = 4$. An onto function (surjective function) from set A to set B exists only if $|A| \ge |B|$. In this case, $|A| = 3$ and $|B| = 4$. Since $|A| < |B|$, it is not possible to define an onto function from A to B. Therefore, the number of onto functions from A to B is 0. Given that $\alpha$ represents the number of onto functions, we have $\alpha = 0$.

Now, we need to evaluate the limit: $L = \lim_{x \to 0} \frac{(\alpha + 4) \sin^2 x}{x^2 + x^3 \cos x}$ Substitute the value of $\alpha = 0$ into the limit expression: $L = \lim_{x \to 0} \frac{(0 + 4) \sin^2 x}{x^2 + x^3 \cos x} = \lim_{x \to 0} \frac{4 \sin^2 x}{x^2 + x^3 \cos x}$ The expression is in the indeterminate form $\frac{0}{0}$ as $x \to 0$. We can factor out $x^2$ from the denominator: $L = \lim_{x \to 0} \frac{4 \sin^2 x}{x^2(1 + x \cos x)}$ We can rewrite the expression as: $L = 4 \lim_{x \to 0} \left( \frac{\sin^2 x}{x^2} \right) \left( \frac{1}{1 + x \cos x} \right)$ Using the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we have $\lim_{x \to 0} \frac{\sin^2 x}{x^2} = \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 1^2 = 1$. For the second part of the expression, as $x \to 0$, $x \cos x \to 0 \cdot \cos(0) = 0 \cdot 1 = 0$. So, $\lim_{x \to 0} (1 + x \cos x) = 1 + 0 = 1$. Therefore, $\lim_{x \to 0} \frac{1}{1 + x \cos x} = \frac{1}{1} = 1$.

Now, substitute these limit values back into the expression for L: $L = 4 \times 1 \times 1 = 4$ The value of the limit is 4.