Q.Find the values of $k$ so that the function $f$ is continuous at the indicated point, where $f$ is defined by $f(x) = \begin{cases} kx+1, & \text{if } x \leq 5 \\ 3x-5, & \text{if } x > 5 \end{cases}$ at $x = 5$

Concept: continuity at x = 0 requires LHL = RHL = f(0). (The stem is truncated but the standard question asks for k that makes f continuous at 0.)

Right-hand limit (x -> 0+): f(x) = sin x / x + cos x -> 1 + 1 = 2.

Left-hand limit (x -> 0-): f(x) = 4(1 - sqrt(1 - x))/x. Rationalise: …

The key idea is that continuity at a point means the function’s value and the limits from both sides all agree. Here, the function is defined by two different linear expressions meeting at $x = 3$. For it to be continuous, the y-value computed from the left piece (at $x = 3$) must equal the y-value computed from the right piece (also at $x = 3$). That gives a simple equation linking $a$ and $b$.

1. Identify the left-hand value. For $x \leq 3$, we use $f(x) = a x + 1$. At $x = 3$, this gives

2. Identify the right-hand limit. For $x > 3$, we use $f(x) = b x + 3$. As $x$ approaches 3 from the right, the value approaches

3. Set them equal for continuity. Continuity at $x = 3$ requires

so

4. Solve for the relationship. Subtract 1 from both sides: $3a = 3b + 2.$ …

Concept: a piecewise polynomial function is continuous everywhere except possibly at the break point; there the left limit, right limit and the value must agree.

Both pieces (ax + 3 and a^2 x - 1) are polynomials, hence continuous on their own domains. Continuity at x = 2 requires

lim(x -> 2-) f(x) = f(2) = a(2) + 3 = 2a + 3 …

Concept & Intuition

Continuity at $x=0$ means $\lim_{x\to 0} f(x) = f(0) = k$. The given expression for $x \neq 0$ is a $0/0$ form, so we need to simplify it. The classic trick: rationalize the numerator to eliminate the square roots, then use $\lim_{x\to 0} \frac{\sin x}{x} = 1$.

Step-by-step solution

1. Set up the continuity condition Since $f$ is continuous at $x=0$,

2. Rationalize the numerator Multiply numerator and denominator by the conjugate $\sqrt{1+x} + \sqrt{1-x}$:

The numerator simplifies to $2x$.

3. Rewrite the limit

4. Separate into known limits

We know $\lim_{x\to 0} \frac{x}{\sin x} = 1$, so $\lim_{x\to 0} \frac{2x}{\sin x} = 2 \cdot 1 = 2$.

5. Evaluate the remaining factor …

Check $x=1$ in the numerator: $1-5+7-3=0$, so $(x-1)$ is a factor. Dividing:

Thus for $x\neq1$, $f(x)=\dfrac{(x-1)^2(x-3)}{(x-1)^2}=x-3$. …

The key idea is that a piecewise function is continuous at the boundary point if the limit from both sides matches the defined value. Here, the limit of the rational expression as $x\to 1$ is a standard derivative-like result: $\frac{1-x^m}{1-x}$ is the sum of a geometric series when $m$ is a positive integer, but the limit formula holds for any real $m$. So we compare that limit to $2m-1$; if they are unequal, the function is discontinuous.

1. Write the continuity condition For $f$ to be continuous at $x=1$, we need

Since for $x\neq 1$, $f(x)=\frac{1-x^m}{1-x}$, we compute

2. Evaluate the limit Recognize that $\frac{1-x^m}{1-x} = 1 + x + x^2 + \cdots + x^{m-1}$ when $m$ is a positive integer, but the limit formula works for any real $m$ using L’Hôpital’s rule or the definition of derivative:

So $L = m$.

3. Set equality for continuity Continuity requires $m = 2m-1$. Solving:

Thus the function is continuous only when $m=1$.

4. Discontinuity condition …

The standard form of this function is

For $f$ to be continuous the pieces must agree at the joining points.

At $x=-\tfrac{\pi}{2}$: left piece $=2\sin(-\tfrac\pi2)=2(-1)=-2$; middle piece $=a\sin(-\tfrac\pi2)+b=-a+b$. So

$-a+b=-2.\qquad(1)$ …

The key idea: continuity at a point means the function’s value equals its limit there. Here, $f$ is defined piecewise, so we compute the limit of the expression for $x\neq \pi/2$ as $x$ approaches $\pi/2$, and set $\lambda$ equal to that limit.

Why this approach works:

Direct substitution gives $0/0$, an indeterminate form. We need to simplify algebraically or use a substitution to reveal a known limit. The presence of $\sin x$ and $(\pi-2x)^2$ suggests letting $t = \pi/2 - x$, which turns the denominator into $(2t)^2$ and the numerator into $1 - \cos t$, a classic limit pair.

Step-by-step:

1. Set up the limit for continuity For $f$ to be continuous at $x = \pi/2$, we require

2. Substitute to simplify Let $t = \frac{\pi}{2} - x$. Then as $x \to \pi/2$, $t \to 0$. Also:

so $(\pi - 2x)^2 = 4t^2$.

For the numerator: $\sin x = \sin\left(\frac{\pi}{2} - t\right) = \cos t$, so $1 - \sin x = 1 - \cos t$.

The limit becomes:

3. Use the half-angle identity Recall $1 - \cos t = 2\sin^2(t/2)$. Then:

4. Rewrite to use $\lim_{u\to 0} \frac{\sin u}{u} = 1$ Let $u = t/2$, so $t = 2u$ and $t^2 = 4u^2$. Then: $$$$

Step 1 — Identify which branch each one-sided limit uses

• $x\to 2^{-}$ means $x$ approaches $2$ from below, so $x$ lies in $(0,2)$ → use $f(x)=x^2-1$.
• $x\to 2^{+}$ means $x$ approaches $2$ from above, so $x$ lies in $[2,3)$ → use $f(x)=2x+3$.

This is the whole point of a one-sided limit on a piecewise function: the side selects the formula.

Step 2 — Compute the limits

Both branches are polynomials, hence continuous, so the limits are found by direct substitution:

(As an aside, $3\neq 7$, so $f$ is discontinuous at $x=2$ — but that does not affect the one-sided values.) …

The key idea is that a complex number is purely real when its imaginary part is zero. Here we have a fraction involving $i \sin \alpha$, so we need to manipulate it into a standard form $x + iy$ and then set $y = 0$.

1. Rationalise the denominator Multiply numerator and denominator by the conjugate of the denominator:

The denominator becomes:

The numerator expands as:

Since $i^2 = -1$, the last term is $-2 \sin^2 \alpha$. So:

2. Write in $x + iy$ form The whole expression becomes:

The imaginary part is $\displaystyle \frac{-3 \sin \alpha}{1 + 4 \sin^2 \alpha}$.

3. Set the imaginary part to zero For the number to be purely real:

The denominator is always positive (since $1 + 4 \sin^2 \alpha \ge 1$), so this reduces to:

4. Solve $\sin \alpha = 0$ …

Step 1 — Find $f(1)$

The middle piece gives $f(1) = 2(1) - 3 = -1$. This is the target value both one-sided limits must reach.

Step 2 — Left-hand limit fixes $a$

Setting this equal to $-1$: $a + 7 = -1 \implies a = -8$. …