Q.If $f(x) = x^2$, find $\dfrac{f(1.1) - f(1)}{(1.1 - 1)}$.

Step 1 — Continuity at $x=0$ (necessary for differentiability).

A differentiable function is always continuous, so these must be equal:

Step 2 — Left-hand derivative at $x=0$. For $x<0$, $f'(x)=e^{x}+a$, so

Step 3 — Right-hand derivative at $x=0$. For $x\ge 0$, $f'(x)=2b(x-1)$, so

Step 4 — Equate them. Differentiability at $0$ means $\mathrm{LHD}=\mathrm{RHD}$:

Step 5 — Check. With $a=-3,\ b=1$: left branch $e^{x}-3x$ has value $1$ and slope $1-3=-2$ at $0$; right branch $(x-1)^2$ has value $1$ and slope $2(0-1)=-2$. Value and slope both match. ✓

ANSWER: C

The core idea is that a limit of the form

is actually twice the derivative of $f$ at 0, provided the derivative exists. Here $f(t)=\sqrt[3]{t}=t^{1/3}$, so we can avoid messy algebra by using the derivative rule.

1. Recognize the derivative structure Write the limit as

Let $f(t)=\sqrt[3]{t}$. Then the numerator is $f(1+x)-f(1-x)$. For small $x$,

so their difference is approximately $2f'(1)x$. Dividing by $x$ gives $2f'(1)$.

2. Compute the derivative $f(t)=t^{1/3}$ so $f'(t)=\frac{1}{3}t^{-2/3}$. At $t=1$,

Hence the limit is $2\cdot\frac{1}{3}=\frac{2}{3}$.

3. Verify with algebraic manipulation (optional) Use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=\sqrt[3]{1+x},\;b=\sqrt[3]{1-x}$. Multiply numerator and denominator by $a^2+ab+b^2$:

As $x\to0$, $a\to1$ and $b\to1$, so the denominator tends to $1+1+1=3$. Thus the limit is $\frac{2}{3}$.

The key insight is recognizing that the expression

is exactly the definition of the derivative of the function $f(x)=x^2\sin x$ evaluated at $x=a$.

So instead of manipulating the limit directly, we can differentiate $f(x)$ and then plug in $x=a$.

1. Identify the function and the derivative definition The general definition of the derivative is

Here, $f(x)=x^2\sin x$, so $f(a+h)=(a+h)^2\sin(a+h)$ and $f(a)=a^2\sin a$.

Thus the given limit is simply $f'(a)$.

2. Differentiate $f(x)=x^2\sin x$ Use the product rule:

3. Evaluate at $x=a$

4. Match with the options The expression $a^2\cos a + 2a\sin a$ is exactly option (D).

The key insight is that the numerator $\sin(a+x) - \sin(a-x)$ is a difference of two sines. Instead of memorizing formulas, think: as $x \to 0$, both $\sin(a+x)$ and $\sin(a-x)$ approach $\sin a$, so we have a $\frac{0}{0}$ form. The natural tool is either the sine difference identity or recognizing this as a derivative.

1. Use the sine difference identity Recall: $\sin P - \sin Q = 2 \cos\left(\frac{P+Q}{2}\right) \sin\left(\frac{P-Q}{2}\right)$. Here $P = a+x$, $Q = a-x$. Then

Simplify:

2. Rewrite the limit The original limit becomes

3. Apply the fundamental limit We know $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$. Therefore,

We are asked to evaluate

The direct substitution $x=0$ gives $\frac{0}{0}$, an indeterminate form. The presence of square roots suggests rationalizing the numerator — a classic trick that turns a difference of square roots into an expression where the $x$ cancels cleanly.

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

2. Simplify the numerator The numerator becomes $(a+x) - a = x$. So we have:

3. Cancel the common factor $x$ (valid for $x \neq 0$, which is fine since we take the limit):

4. Take the limit as $x \to 0$ Now substitute $x = 0$:

Since $a > 0$ (implied by the square roots in the original expression), $\sqrt{a^2} = a$. Thus:

Step 1 — What we are given.

To find $f'$ we must first know $f$ as a function of its own argument — so we must express $x^3 - \frac{1}{x^3}$ purely in terms of $\left(x - \frac1x\right)$.

Step 2 — Use the algebraic identity.

Recall $\;(a-b)^3 = a^3 - b^3 - 3ab(a-b)$. Put $a = x$, $b = \dfrac1x$ (so $ab = 1$):

Rearranging,

Step 3 — Read off $f$.

The right-hand side is written entirely in terms of $g(x)$:

Since this holds for every $x$ (and $g$ takes all real values), the rule of $f$ is

Step 4 — Differentiate.

Step 5 — Verify with the chain rule (independent check).

$\dfrac{d}{dx}\left(x^3 - \dfrac1{x^3}\right) = 3x^2 + \dfrac{3}{x^4}$, and $g'(x) = 1 + \dfrac{1}{x^2}$. The chain rule demands $f'(g(x))\,g'(x) = 3x^2 + \dfrac{3}{x^4}$, i.e.

Simplifying, $\dfrac{3(x^6+1)}{x^2(x^2+1)} = 3\left(x^2 - 1 + \dfrac{1}{x^2}\right) + 3 \cdot 1 = 3\left(x - \dfrac1x\right)^2 + 3$, which is exactly $3\big[g(x)\big]^2 + 3$ ✓ — confirming $f'(t) = 3t^2 + 3$.

(Option (A) $3x^2 + \frac{3}{x^4}$ is the trap: that is $\dfrac{d}{dx}(f\circ g)$, not $f'$.)

ANSWER: D

Step 1 — Identify the function.

The coefficients $1,\; n,\; \dfrac{n(n-1)}{2},\; \dfrac{n(n-1)(n-2)}{6},\ldots,1$ are precisely $\binom{n}{0},\binom{n}{1},\binom{n}{2},\binom{n}{3},\ldots,\binom{n}{n}$, because

So by the binomial theorem

This is the key move: instead of differentiating a long polynomial term by term, we collapse it to a closed form.

Step 2 — Differentiate twice.

Using the power/chain rule,

Step 3 — Substitute $x=1$.

(Sanity check with $n=2$: $f(x)=1+2x+x^2$, so $f''(x)=2$ for all $x$. The formula gives $2\cdot 1\cdot 2^{0}=2$. ✓)

ANSWER: A

Write

As $x\to0$, $\frac{a^x-1}{x}\to\log a$ and $\frac{b^x-1}{x}\to\log b$, so

Step 1 — Note the key fact: $x=1$ is a fixed point

We are given $f(1)=1$. Therefore

So whenever we substitute $x=1$, every inner argument collapses to $1$, and every derivative factor becomes $f'(1)=3$. This is what makes the problem tractable without knowing $f$.

Step 2 — Differentiate the composite $f(f(f(x)))$

Apply the chain rule from the outside in:

Evaluate at $x=1$:

Step 3 — Differentiate $(f(x))^{2}$

Power rule + chain rule:

At $x=1$:

Step 4 — Add (derivative is linear)

ANSWER: A

Step 1 — Recognise the indeterminate form

As $y\to 0$, the numerator $\to \sqrt{3}-\sqrt{3}=0$ and the denominator $y^{3}\to 0$: the form is $\dfrac{0}{0}$, so direct substitution is not allowed.

Step 2 — Substitute to simplify

Notice that $y$ appears only as $y^{3}$. Put

Step 3 — Rationalise the numerator

The standard tool for a surd difference is multiplication by the conjugate, because $(\sqrt A-\sqrt B)(\sqrt A+\sqrt B)=A-B$ removes the radicals from the numerator:

Since $t\neq 0$ in the limiting process, cancel $t$:

Step 4 — Now substitute $t=0$

The expression is continuous at $t=0$, so

Step 5 — Cross‑check with the derivative definition

$\displaystyle\lim_{t\to0}\frac{f(3+t)-f(3)}{t}=f'(3)$ with $f(x)=\sqrt{x}$. Since $f'(x)=\dfrac{1}{2\sqrt{x}}$, we get $f'(3)=\dfrac{1}{2\sqrt3}$. ✓ Same value.

ANSWER: D

The key idea here is that continuity at a point means the function's value at that point equals the limit from both sides. For a piecewise function, we must check the boundary where the definition changes — here, $x=0$.

The left-hand piece ($-1 \leq x < 0$) involves a difference of square roots, which is a classic indeterminate form $\frac{0}{0}$ when $x \to 0$. The right-hand piece ($0 \leq x \leq 1$) is a rational function, and at $x=0$ it gives a finite value directly.

Let's work through it step by step.

1. Find $f(0)$ from the right-hand definition. Since $0 \leq x \leq 1$ includes $x=0$, we use the second piece: