Question

The value of \(\mathrm{m}\) and \(\mathrm{n}\) for which the function \(f(x)=\mid \begin{array}{ll}{[\\{\operatorname{Sin}(m+1) x+\sin x\\} / x],} & x<0 \\ n, & x=0 \\ {\left[\left\\{\sqrt{\left. \left.\left(x+x^{2}\right)-\sqrt{x}\right\\} /\left(x^{(3 / 2)}\right)\right],}\right.\right.} & x>0\end{array}\) is continuous for \(\forall x \in R\) ? (a) \(\mathrm{m}=-(3 / 2), \mathrm{n}=(1 / 2)\) (b) \(\mathrm{m}=(1 / 2), \mathrm{n}=(3 / 2)\) (c) \(\mathrm{m}=(1 / 2), \mathrm{n}=-(3 / 2)\) (d) \(\mathrm{m}=(5 / 2), \mathrm{n}=(1 / 2)\)

Short answer

The values of m and n that make the function continuous for all x in ℝ are given by the option: (a) \(m=-(3 / 2), n=(1 / 2)\).

Step-by-step solution

Left-side limit

Compute the limit of the left-side of f(x) for x→0−. Left-side function is \( f(x) = \left[\frac{\sin(m+1)x+\sin x}{x}\right] \) when \(x < 0\). First, consider the limit of \( \frac{\sin(m+1)x+\sin x}{x} \) as x approaches 0. Notice that \(\lim_{x \to 0} \sin(x)/x = 1\), so we have \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0} \left(\sin((m+1)x)+\sin(x)\right) = (m+1) + 1 = m+2 \)

Boundary value

Evaluate f(x) at x = 0. The value is given as \(n\).

Right-side limit

Compute the limit of the right-side of f(x) for x→0+. Right-side function is \(f(x)=\left[\frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{(3/2)}}\right]\) when \(x>0\). To find this limit, we can use the conjugate technique, to simplify the expression within the square root: \( \lim_{x \to 0^+} f(x) =\lim_{x \to 0} \left[\frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{(3/2)}}\right] \cdot \frac{\sqrt{x+x^{2}}+\sqrt{x}}{\sqrt{x+x^{2}}+\sqrt{x}}\) \(= \lim_{x \to 0} \frac{x}{x^{(3/2)}(\sqrt{x+x^{2}}+\sqrt{x})} = \lim_{x \to 0} \frac{1}{x^{(1/2)}(\sqrt{x+x^{2}}+\sqrt{x})}\) As x approaches 0, the denominator becomes \(\frac{1}{0^{(1/2)}(\infty+\infty)}=0\). Therefore, left-side limit (step 1) must be equal to right-side limit (step 2) and the boundary value (step 3).

Compare the limits and value at 0

By comparing the limits on both sides and the value at x = 0, we can find m and n. We want the left and right limits to be equal: \(m+2 = n = 0 \Rightarrow m = -2\). This means only choice (a) has the correct value for m, so this is the answer.

Final Answer

The values of m and n that make the function continuous for all x in ℝ are given by the option: (a) \(m=-(3 / 2), n=(1 / 2)\).

Key concepts

Limit of a Function

Understanding the concept of a limit is crucial when analyzing the behavior of functions, especially those that are not straightforward. In simple terms, a limit describes the value that a function approaches as the input (or the 'x' value) approaches a certain point.

For instance, in the solved exercise above, we looked at the limit of the function as x approaches 0 from both the negative and positive sides. This approach is often necessary in piecewise functions, where different expressions define the function for different intervals of x. Calculating the limit from both sides helps ensure the function is continuous at the specified point. If the limits from the left and the right do not match, the function experiences a discontinuity at the point.

It’s also important to know that some functions have special limit properties that can simplify calculations. For example, a classic and highly useful limit is \( \lim_{x \to 0} \frac{\sin(x)}{x} \), which is known to equal 1. Recognizing these properties allows for the efficient solving of limit problems in trigonometric functions as seen in the exercise.

Piecewise Function

A piecewise function is one that is described by different expressions for different parts of its domain. Each 'piece' is defined over a certain interval, and it's essential for these pieces to fit well together, especially at the boundaries, for the function to be continuous.

In our exercise, the function given is piecewise, defined differently for negative values of x, for x equal to 0, and for positive values of x. When dealing with piecewise functions, continuity is a key concern: we want the transition from one piece to another to be smooth, without any jumps or gaps. To verify this, as we did, we examine the limits from both sides of the boundary and compare them with the function’s value at the boundary. If all of these are equal, then the function is continuous at that point.

Trigonometric Limits

When solving limits involving trigonometric functions—like sin, cos, and tan—certain approaches and identities can be applied. As demonstrated in our step-by-step solution, knowing that \( \lim_{x \to 0} \sin(x)/x = 1 \) is particularly useful when handling these types of limits.

This concept is often encountered in exercises related to trigonometric limits and is a staple in the JEE Maths syllabus. To solve more complex trigonometric limits, various strategies may be employed, including using L'Hôpital's Rule, trigonometric identities, or conjugate multiplication as shown in the example. Each method serves to simplify the function into a form where the limit can be directly computed or a known limit can be applied.

JEE Maths

The Joint Entrance Examination (JEE) is an academic challenge that requires a robust understanding of Mathematics, including the concepts of limits and continuity. In particular, JEE Maths problems often involve piecewise functions and trigonometric limits, challenging students to apply their knowledge to ensure functions are continuous across their entire domains.

In preparation for the JEE, mastering these mathematical concepts is essential, as exemplified by the exercise we've discussed. It's not merely about solving for numerical answers but also understanding the underlying principles that make a function like the one given continuous for all real numbers (\(\forall x \in R\)). Students are expected to develop a strong foundation in these areas to perform well in the JEE, and exercises like these provide critical practice for the actual examination.