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 $x = 5$

The function is defined in three linear pieces. Linear functions are continuous and differentiable everywhere on their own intervals, so any trouble can only happen at the two boundary points $x=-3$ and $x=3$ where the definition changes. The question is simply: does the graph have a break (discontinuity) or a corner (non-differentiability) at either of those points?

Let’s check each point systematically.

1. At $x=-3$ The left-hand piece is $f(x)=3-x$ for $x<-3$. As $x$ approaches $-3$ from the left,

The middle piece gives $f(-3)=6$ directly. The right-hand limit from the middle piece is also $6$ (the function is constant $6$ on $[-3,3]$).

So the left limit, right limit, and function value all equal $6$. The function is continuous at $x=-3$.

For differentiability, compute the left-hand derivative using the left piece:

The right-hand derivative uses the middle piece:

Since $-1 \neq 0$, the derivatives from the two sides do not match. The function has a sharp corner at $x=-3$ — it is not differentiable there.

2. At $x=3$ The middle piece gives $f(3)=6$. Approaching from the left (still on the middle piece),

Approaching from the right, using $f(x)=3+x$ for $x>3$,

$\lim_{x\to 3^+} f(x) = 3+3 = 6.$ …

The key idea: The floor function jumps whenever its input passes an integer. So we find where $x^2 - 3$ equals an integer, then count those $x$ inside $(-1,2)$.

Concept & Intuition

The floor function $\lfloor t \rfloor$ is constant on intervals $[n, n+1)$ and jumps at every integer $n$. So $f(x) = \lfloor g(x) \rfloor$ will be discontinuous at any $x$ where $g(x)$ is an integer and $g$ is crossing that integer (not just touching it). Here $g(x) = x^2 - 3$ is continuous and strictly increasing for $x>0$, strictly decreasing for $x<0$. So each integer value of $g$ will be hit at most twice (once on each side of 0). We just need to list all integers $k$ such that $x^2 - 3 = k$ has a solution in $(-1,2)$, then check which of those $x$ actually cause a jump.

Step-by-step solution

1. Find possible integer values of $x^2 - 3$ on $(-1,2)$.

   Since $x \in (-1,2)$, $x^2$ ranges from $0$ to $4$ (including $0$, excluding $4$ at $x=2$? Actually $2$ is not in the open interval, but $x$ can approach $2$ from below, so $x^2$ approaches $4$ from below). So $x^2 - 3$ ranges from $-3$ to just below $1$.

   The integers in $(-3, 1)$ are: $-2, -1, 0$. Also note $-3$ is attained at $x=0$? $0^2-3 = -3$, but $-3$ is an integer. However, $x=0$ is inside $(-1,2)$, so we must include $k=-3$ as well. So possible integers $k$ are $-3, -2, -1, 0$.

2. Solve $x^2 - 3 = k$ for each $k$.

   • $k = -3$: $x^2 = 0 \Rightarrow x = 0$.
   • $k = -2$: $x^2 = 1 \Rightarrow x = \pm 1$. Both $1$ and $-1$ are in $(-1,2)$? $-1$ is not in the open interval $(-1,2)$ (it's the endpoint), so only $x=1$ counts.
   • $k = -1$: $x^2 = 2 \Rightarrow x = \pm \sqrt{2}$. $\sqrt{2} \approx 1.414$ is inside $(-1,2)$; $-\sqrt{2} \approx -1.414$ is outside. So only $x = \sqrt{2}$.
   • $k = 0$: $x^2 = 3 \Rightarrow x = \pm \sqrt{3}$. $\sqrt{3} \approx 1.732$ is inside; $-\sqrt{3} \approx -1.732$ is outside. So only $x = \sqrt{3}$.

   So candidate points: $0, 1, \sqrt{2}, \sqrt{3}$.

3. Check each candidate for actual discontinuity.

   At each such $x$, $g(x)$ is exactly an integer. Since $g$ is strictly monotonic on each side of $0$ (decreasing for $x<0$, increasing for $x>0$), it crosses that integer transversally — so the floor jumps.

   • At $x=0$: $g(0) = -3$. For $x$ just left or right, $g(x) > -3$ (since $x^2>0$), so floor jumps from $-4$ to $-3$? Wait: For $x$ near $0$, $x^2$ is small positive, so $x^2-3$ is just above $-3$, floor is $-3$. Actually at $x=0$, floor is $-3$. For $x\neq0$, $x^2>0$, so $x^2-3 > -3$, floor is still $-3$ until it reaches $-2$. So is there a jump? Let's check: left of 0, say $x=-0.1$, $x^2=0.01$, $g=-2.99$, floor = $-3$. Right of 0, $x=0.1$, same. So floor is constant $-3$ in a neighborhood of 0. So no discontinuity at $x=0$ — the function touches the integer but doesn't cross it because $x^2$ is minimal at 0. …

Concept & Intuition

A function is differentiable at a point only if it is first continuous there. For a piecewise function, the value at $x=0$ is given as $P$, so we must choose $P$ so that $\lim_{x\to 0} f(x) = f(0)$. Then, the derivative at $0$ is the limit of the difference quotient $\frac{f(x)-f(0)}{x}$ as $x\to 0$. Since the function is defined by a formula for $x\neq 0$, we use series expansions or standard limits to evaluate these limits.

Step-by-step solution

1. Enforce continuity at $x=0$ For $x \neq 0$,

As $x\to 0$, use the standard limits:

Hence

So $\lim_{x\to 0} f(x) = 0$. For continuity, we need $f(0) = P = 0$.

2. Now compute $f'(0)$ using the definition Since $P=0$,

Substitute the expression for $f(x)$:

3. Simplify the limit Write $\tan x = \frac{\sin x}{\cos x}$, so

Now use the standard limits:

Combine carefully:

Concept & Intuition

The question tests how different types of functions behave at specific points or over their domains.

• Absolute value functions create “kinks” but are still continuous everywhere.
• Hyperbolic cosecant $\operatorname{csch} x = 1/\sinh x$ is undefined where $\sinh x = 0$, i.e. at $x=0$.
• Fractional part $x - \lfloor x \rfloor$ jumps at every integer.
• Square root $\sqrt{2-x}$ is only defined for $x \le 2$; at $x=2$ the left‑hand limit exists but the right‑hand limit does not (since the function is not defined for $x>2$).

We examine each function one by one.

Step‑by‑step matching

1. Function A: $f(x) = |x| + |x-2|$

   • This is a sum of absolute values. Each absolute value is continuous everywhere (a V‑shape). The sum of continuous functions is continuous.
   • At $x=2$, $f(2) = |2| + |0| = 2$. The left‑hand limit and right‑hand limit both equal 2.
   • So $f$ is continuous for all real $x$.
   • Match: IV (Continuous for all real values of $x$).

2. Function B: $g(x) = \operatorname{csch} x = \frac{1}{\sinh x}$

   • $\sinh x = \frac{e^x - e^{-x}}{2}$ is zero only at $x=0$.
   • At $x=0$, $g(x)$ is undefined (division by zero). For any $x \neq 0$, $\sinh x \neq 0$, so $g$ is continuous.
   • Thus $g$ is continuous only for non‑zero real $x$.
   • Match: II (Continuous only for non‑zero real values of $x$).

3. Function C: $h(x) = x - \lfloor x \rfloor$ (fractional part)

   • At integer values, $\lfloor x \rfloor$ jumps, so $h(x)$ jumps from nearly 1 to 0.
   • For example, as $x \to 2^-$, $h(x) \to 1$; as $x \to 2^+$, $h(x) \to 0$. The limit does not exist at integers.
   • Hence $h$ is discontinuous at every integer.
   • Match: V (Discontinuous at all integral values of $x$).

4. Function D: $k(x) = \sqrt{2-x}$

   • Domain: $2-x \ge 0 \Rightarrow x \le 2$. …

For distinct real roots the discriminant must be positive:

This must hold for every rational $a$. The quadratic $g(a)=5a^{2}+8a+4K$ attains its minimum at $a = -\dfrac{8}{2\cdot 5} = -\dfrac{4}{5}$, which is itself rational. Evaluating there: …

Concept. A piecewise-defined function can only fail smoothness where its formula changes or where the formula itself misbehaves. Here the only special point is $x = 0$. On any open interval not containing $0$, $f(x) = x^2\cos\dfrac{\pi}{x}$ is a product of the polynomial $x^2$ and the composition $\cos(\pi/x)$, both infinitely differentiable for $x \ne 0$.

Step 1 — locate the point. $x = 2 > 0$, and a whole neighbourhood of $2$ (say $(1,3)$) avoids $x=0$, so on it $f$ is given by the single smooth formula $x^2\cos(\pi/x)$.

Step 2 — differentiate. By the product and chain rules, for $x \ne 0$:

$f'(x) = 2x\cos\frac{\pi}{x} + x^2\left(-\sin\frac{\pi}{x}\right)\left(-\frac{\pi}{x^2}\right) = 2x\cos\frac{\pi}{x} + \pi\sin\frac{\pi}{x}.$ …

The core idea here is that $x$ and $y$ are given in terms of a parameter $t$, so derivatives with respect to $x$ must be computed via the chain rule. The condition $\frac{d^2 y}{dx^2} = -1$ at $t = K$ lets us solve for $K$. Then the limit $\lim_{t \to K} \frac{y}{x}$ is a $0/0$ form (since both $x$ and $y$ vanish at $t=0$, but here $K$ is not zero — we must check), so we use L'Hôpital's rule, which connects back to the first derivative.

Let’s work through it step by step.

1. Find $\frac{dy}{dx}$ in terms of $t$. We have $x = t - \sin t$ and $y = 1 - \cos t$. Differentiate each with respect to $t$:

Hence,

This can be simplified using the half-angle identities: $\sin t = 2\sin(t/2)\cos(t/2)$ and $1 - \cos t = 2\sin^2(t/2)$, so

That’s a neat simplification.

2. Find $\frac{d^2 y}{dx^2}$. The second derivative with respect to $x$ is

Since $\frac{dy}{dx} = \cot(t/2)$, differentiate with respect to $t$:

And $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{1 - \cos t}$. Using $1 - \cos t = 2\sin^2(t/2)$, we get

Therefore,

Since $\csc^2(t/2) = 1/\sin^2(t/2)$, the product gives $-\frac{1}{4\sin^4(t/2)}$.

3. Apply the given condition to find $K$. We are told $\frac{d^2 y}{dx^2} = -1$ at $t = K$, with $K > 0$. So

Hence, $\sin^4(K/2) = \frac{1}{4}$, so $\sin(K/2) = \frac{1}{\sqrt{2}}$ (positive since $K>0$ and we expect $K/2$ in a range where sine is positive). …

The key idea is that a binomial expansion of the form $(a + b)^n$ is only valid when the absolute value of the "ratio term" is less than 1. Here, the exponent is negative and fractional, so we must rewrite the expression to fit the standard binomial form $(1+u)^n$, where the condition for convergence is $|u| < 1$.

Why this works:

The binomial theorem for a general exponent $n$ (not just a positive integer) is an infinite series. It converges only when the term we are expanding in, call it $u$, satisfies $|u| < 1$. So we first factor out the constant term to get a "$1 +$ something" inside the parentheses, then apply the condition.

Step-by-step:

1. Rewrite the expression We have $(7 - 5x)^{-2/5}$. Factor out the 7:

The constant factor $7^{-2/5}$ does not affect convergence; the series behavior depends entirely on the factor $\left(1 - \frac{5}{7}x\right)^{-2/5}$.

2. Identify the expansion variable

   The binomial form is $(1+u)^n$ with $n = -\frac{2}{5}$ and $u = -\frac{5}{7}x$.

   The expansion is valid when $|u| < 1$.

3. Apply the convergence condition

So the set of $x$ for which the expansion is valid is $(-7/5,\; 7/5)$. …

We are asked to find

Direct substitution gives $\cot(\pi/4)=1$ and $\tan(\pi/4)=1$, so numerator $=1-1=0$; denominator $\cos(\pi/4+\pi/4)=\cos(\pi/2)=0$. This is an indeterminate $0/0$ form, so we need to manipulate the expression.

Concept & Intuition

The key is to rewrite everything in terms of sine and cosine, then use algebraic factoring and the known limit $\frac{\sin\theta}{\theta}\to 1$ (or simply cancel a common factor). The denominator $\cos(x+\pi/4)$ suggests using the cosine addition formula, and the numerator $\cot^3 x - \tan x$ can be factored as a difference of cubes after expressing both in terms of $\sin$ and $\cos$.

1. Rewrite numerator in sines and cosines

So

Put over a common denominator $\sin^3 x \cos x$:

2. Factor the numerator $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = (\cos^2 x - \sin^2 x) \cdot 1 = \cos 2x$. So the numerator becomes

3. Rewrite the denominator Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$:

So the whole expression is

4. Simplify $\cos 2x$ in terms of $\cos x - \sin x$ Recall $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$. This is perfect: the factor $(\cos x - \sin x)$ cancels with the same factor in the denominator. After cancellation:

And $\frac{2}{\sqrt{2}} = \sqrt{2}$, so