Q.State whether True or False: Trigonometric and inverse-trigonometric functions are differentiable in their respective domains.

Concept and Intuition

The oscillating factor $\sin(1/x)$ is always bounded between $-1$ and $1$, so whether $x^\alpha\sin(1/x)\to0$ (continuity) or whether the difference quotient $x^{\alpha-1}\sin(1/x)$ converges (differentiability) depends entirely on the power of $x$ multiplying the bounded oscillation — a higher power tames the oscillation faster.

Step-by-Step Solution

1. Continuity at 0: $|f(x)-f(0)|=|x^\alpha\sin(1/x)|\le|x|^\alpha$. This $\to0$ as $x\to0$ whenever $\alpha>0$. So $f$ is continuous at $0$ for every $\alpha>0$ — this already contradicts option (A) (which wrongly restricts continuity+differentiability to $0\le\alpha<1$, an interval where differentiability actually fails) and option (B) (which wrongly claims discontinuity throughout $0\le\alpha<1$, whereas $f$ is continuous there for $\alpha>0$).
2. Differentiability at 0: $f'(0)=\lim_{x\to0}\dfrac{x^\alpha\sin(1/x)-0}{x}=\lim_{x\to0}x^{\alpha-1}\sin(1/x)$.
   • If $\alpha>1$: exponent $\alpha-1>0$, so $x^{\alpha-1}\to0$ and the bounded oscillation is killed — limit is 0, so $f'(0)$ exists: continuous AND differentiable.
   • If $0<\alpha\le1$: exponent $\alpha-1\le0$, so $x^{\alpha-1}$ does not decay (it's bounded-below-1 or blows up) while $\sin(1/x)$ keeps oscillating — the limit does not exist: continuous but NOT differentiable. …

Concept and Intuition

$\min\{x,x^2\}$ switches which of the two "candidate" functions is smaller depending on the sign of $x(x-1) = x^2-x$. Since both candidate functions are individually continuous (and even differentiable) polynomials, the only way the minimum can fail to be smooth is exactly at the points where the two candidates cross over — i.e. where $x=x^2$.

Step-by-Step Solution

1. Compare $x$ and $x^2$: $x^2-x = x(x-1)$, which is $\ge0$ outside $[0,1]$ and $\le0$ inside $[0,1]$.
   • For $x\le0$ or $x\ge1$: $x^2\ge x$, so $\min=x$.
   • For $0\le x\le1$: $x^2\le x$, so $\min=x^2$.
2. So $f(x)=\begin{cases}x, & x\le0\\ x^2, & 0\le x\le1\\ x, & x\ge1\end{cases}$.
3. Continuity: at $x=0$, both pieces give $0$; at $x=1$, both give $1$. Each piece is a polynomial (continuous on its own), so $f$ is continuous everywhere — option (A) is true.
4. Differentiability at $x=0$: left derivative (of $x$) is $1$; right derivative (of $x^2$, i.e. $2x$) at $0$ is $0$. These differ, so $f$ is not differentiable at $x=0$. …

Concept and Intuition

A piecewise function differentiable everywhere must, at the junction point, have matching function values (continuity, otherwise the derivative can't exist) and matching one-sided derivatives (the slopes must agree, else there's a corner).

Step-by-Step Solution

1. Continuity at $x=1$: left value $=2(1)+3=5$. Right value $=a(1)^2+b(1)=a+b$. So $a+b=5$.
2. Differentiability at $x=1$: derivative of $2x+3$ is $2$ (constant). Derivative of $ax^2+bx$ is $2ax+b$, at $x=1$: $2a+b$. Equate: $2a+b=2$.
3. Subtract the two equations: $(2a+b)-(a+b)=2-5 \Rightarrow a=-3$.
4. Then $b=5-a=5-(-3)=8$. …

Concept and Intuition

Differentiability at a breakpoint requires both continuity (matching values from both sides) and matching one-sided derivatives. With a piecewise cubic/quadratic/linear function, we get one continuity equation and one derivative-matching equation at each breakpoint — four equations total for two unknowns, which must be mutually consistent for the function to be genuinely differentiable everywhere.

Step-by-Step Solution

1. Continuity at $x=1$: left value $= a(1)^2+b(1)-\frac{13}{8}=a+b-\frac{13}{8}$; right value (from the middle piece as $x\to1^+$) $=3(1)-3=0$.

2. Continuity at $x=2$: left value (middle piece) $=3(2)-3=3$; right value (as $x\to2^+$) $= b(2)^3+1=8b+1$.

3. From (i): $a = \tfrac{13}{8}-\tfrac14 = \tfrac{13}{8}-\tfrac28=\tfrac{11}{8}$.
4. Check differentiability at $x=1$: left derivative $=2ax+b|_{x=1}=2a+b = 2(\tfrac{11}{8})+\tfrac14=\tfrac{22}{8}+\tfrac28=\tfrac{24}{8}=3$; right derivative (middle piece, constant slope) $=3$. Matches. …

Concept and Intuition

If $f'(x)=m$ (a fixed constant) for every real $x$, then integrating (or simply recognizing this as the defining property of a straight line) gives $f(x)=mx+k$ for some constant $k$. Polynomials — including linear functions — are continuous everywhere on $\mathbb R$, so there's no room for any of the "fails at some points" options to be correct; differentiability everywhere is actually a stronger condition than continuity everywhere, and it's given here as a hypothesis (constant derivative existing at every real $x$), so continuity everywhere follows immediately.

Step-by-Step Solution

1. $f'(x)=m$ (constant) for all $x\in\mathbb R$ means $f(x)=mx+k$ for some constants $m,k$ (this is the unique family of functions with constant derivative everywhere).
2. $f'(0)=1\Rightarrow m=1$.
3. $f(0)=2\Rightarrow k=2$.
4. So $f(x)=x+2$ — a polynomial, hence continuous (and differentiable) at every real number.
5. Since $f$ is differentiable everywhere (given, as $f'(x)$ exists and is constant for all $x$), it must also be continuous everywhere — differentiability at a point always implies continuity there.

Step-by-Step Solution (continued reasoning on the options)

• Option (A) claims discontinuity somewhere — contradicts having a derivative everywhere.
• Option (B) claims continuity only at isolated points $0,1,2,3$ — but a function differentiable at every real $x$ can't be discontinuous at, say, $0.5$. …

Concept and Intuition

When a function has a jump discontinuity but stays bounded near the point, multiplying by $(x-x_0)$ (here $x_0=0$) forces continuity there (since bounded $\times\to0 \to 0$). But that trick doesn't automatically restore differentiability — checking the derivative of $xf(x)$ at $0$ from first principles brings back exactly the one-sided limits of $f$ itself.

Step-by-Step Solution

1. As $x\to0^+$: $\frac1x\to+\infty$, so $e^{1/x}\to\infty$. Divide num/denom by $e^{1/x}$: $f(x)=\dfrac{5+2e^{-1/x}}{3e^{-1/x}-1}\to\dfrac{5+0}{0-1}=-5$.
2. As $x\to0^-$: $\frac1x\to-\infty$, so $e^{1/x}\to0$. $f(x)\to\dfrac{5(0)+2}{3-0}=\dfrac23$.
3. Since $-5\ne\dfrac23$, $\lim_{x\to0}f(x)$ doesn't exist, so $f$ is discontinuous at $x=0$, hence automatically not differentiable there.
4. For $g(x)=xf(x)$: as $x\to0$ (either side), $f(x)$ stays bounded (approaching $-5$ or $2/3$), so $x\cdot f(x)\to0$. Also $g(0)=0\cdot f(0)=0$ (given $f(0)=0$). So $\lim_{x\to0}g(x)=g(0)=0$ — $g$ is continuous at $0$. …

Concept and Intuition

A piecewise function built from two polynomial pieces is differentiable at the junction point only if both the function values AND the derivatives agree there — continuity alone (matching values) is not enough for differentiability.

Step-by-Step Solution

1. Continuity at $x=3$: $9a-3b+2 = 9b-3 \Rightarrow 9a-12b=-5$. — (i)
2. Derivatives: left $f'(x)=2ax-b$, right $f'(x)=2bx$. At $x=3$: $6a-b=6b \Rightarrow 6a=7b \Rightarrow a=\dfrac{7b}{6}$. — (ii)
3. Substitute (ii) into (i): $9\left(\dfrac{7b}6\right)-12b=-5 \Rightarrow \dfrac{63b}6-12b=-5 \Rightarrow 10.5b-12b=-5 \Rightarrow -1.5b=-5 \Rightarrow b=\dfrac{10}3$.
4. $a=\dfrac{7}{6}\cdot\dfrac{10}{3}=\dfrac{70}{18}=\dfrac{35}{9}$.
5. Line $\dfrac{x}{a}+\dfrac{y}{b}=1$ has x-intercept $=a=\tfrac{35}9$ and y-intercept $=b=\tfrac{10}3$. …

Concept and Intuition

The theorem "differentiable $\Rightarrow$ continuous" is standard and true. Its logical contrapositive — "not continuous $\Rightarrow$ not differentiable" — is automatically true whenever the original implication is true, and is a direct restatement/explanation of it.

Step-by-Step Solution

1. (R): differentiable at a point $\Rightarrow$ continuous at that point. This is a fundamental, true calculus theorem.
2. (A): not continuous at $x=a$ $\Rightarrow$ not differentiable at $x=a$. This is logically the contrapositive of (R): "$P\Rightarrow Q$" is equivalent to "$\lnot Q\Rightarrow\lnot P$".
3. Since a statement and its contrapositive are logically equivalent, (A) being the contrapositive of the true statement (R) is itself true. …

Concept and Intuition

When checking continuity/differentiability of a piecewise-defined function at the "seam" point, replace the transcendental factors by their leading-order Taylor approximations near that point — this converts a hard limit into simple algebra with powers of $x$.

Step-by-Step Solution

1. For small $x$: $\cos x \approx 1-\dfrac{x^2}2$, so $\log(\cos x)\approx\log\left(1-\dfrac{x^2}2\right)\approx-\dfrac{x^2}2$.
2. Also $\log(1+x)\approx x-\dfrac{x^2}2+\cdots\approx x$ for the leading order.
3. So for $x\ne0$: $f(x) = \dfrac{x^2\log(\cos x)}{\log(1+x)} \approx \dfrac{x^2\cdot(-x^2/2)}{x} = -\dfrac{x^3}{2}$.
4. Continuity at $0$: $\displaystyle\lim_{x\to0}f(x) = \lim_{x\to0}\left(-\dfrac{x^3}2\right)=0=f(0)$. So $f$ is continuous at $0$. …

Concept and Intuition

A piecewise function can only fail to be differentiable at the points where the pieces are stitched together. At such a point, the function must first be continuous (otherwise differentiability is moot), and then the left- and right-hand derivatives must be checked to see if they agree — a "kink" (unequal slopes) is exactly what breaks differentiability even when the function itself is continuous.

Step-by-Step Solution

1. Check continuity at $x=-\sqrt5$: left piece value $=4$; right piece $x^2-1$ at $x=-\sqrt5$ gives $5-1=4$. Equal — continuous there.
2. Check continuity at $x=\sqrt5$: middle piece gives $5-1=4$; right piece constant $4$. Equal — continuous there too. So $f$ is continuous everywhere.
3. Derivative of the middle piece $x^2-1$ is $2x$. At $x=-\sqrt5$: slope from inside $=2(-\sqrt5)=-2\sqrt5$; slope from the constant piece outside $=0$. Since $-2\sqrt5\neq0$, not differentiable at $x=-\sqrt5$. …