Class 11 Mathematics Chapter 12 Limits And Derivatives

Key Concept: Basic Derivative Concept

b) $3x^2$

[Solution Description] The derivative of $f(x) = x^3$ is found using the power rule for derivatives, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

$$f'(x) = 3x^{3-1}= 3x^2$$

Therefore, the derivative is $3x^2$.

Key Concept: Constant Function Limit

c) 8

[Solution Description] The limit of a constant function $f(x) = c$ as $x$ approaches any value is simply the constant $c$ itself. Therefore, $$\lim_{x \to 5} 8 = 8$$.

Key Concept: Derivative of Trigonometric Function

c) -1

[Solution Description] The derivative of $\cos x$ is $-\sin x$. Therefore, we have:

$$g'(x) = -\sin x$$

Substituting $$x = \frac{\pi}{2}$$ into $g'(x)$ gives:

$$g’\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right)$$

$$= -1$$

Hence, the derivative at $$x = \frac{\pi}{2}$$ is $-1$.

Key Concept: Limit with Constants

b) 2

[Solution Description] To evaluate this limit, we can use the formula:

$$\lim_{x \to 0} \frac{\sin kx}{kx} = 1$$

We rewrite the given expression as:

$$\lim_{x \to 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \to 0} \frac{4 \cdot \frac{\sin 4x}{4x}}{2 \cdot \frac{\sin 2x}{2x}}$$

Applying the limit for each part gives us:

$$\frac{4 \cdot 1}{2 \cdot 1} = 2$$

Thus, the value of the limit is 2.

Key Concept: Limit with Infinity

c) $\frac{2}{5}$

[Solution Description] As $x$ approaches infinity, the highest degree terms dominate the behavior of the rational function.

The degrees of the numerator and the denominator are both 2, so we focus on the leading coefficients.

Simplifying the expression by dividing each term by $x^2$ yields:

$$\frac{\frac{2x^2}{x^2} + \frac{3x}{x^2} + \frac{1}{x^2}}{\frac{5x^2}{x^2} + \frac{7x}{x^2} + \frac{4}{x^2}} = \frac{2 + \frac{3}{x} + \frac{1}{x^2}}{5 + \frac{7}{x} + \frac{4}{x^2}}$$

As $$x \to \infty$$, the fractional terms ($$\frac{3}{x}$$, $$\frac{1}{x^2}$$ etc.) approach zero:

$$\frac{2 + 0 + 0}{5 + 0 + 0} = \frac{2}{5}$$

Hence, the limit is $$\frac{2}{5}$$.

Key Concept: Complex Rational Limit, Limit with Indeterminate Forms

c) 12

[Solution Description] To find the limit of $\frac{x^3 – 8}{x – 2}$ as $x \to 2$, we first rewrite the numerator using the identity for the difference of cubes: $$a^3 – b^3 = (a-b)(a^2 + ab + b^2)$$. Here, we have:

$$x^3 – 8 = (x – 2)(x^2 + 2x + 4)$$.

Substituting this into our original expression gives:

$$\frac{(x – 2)(x^2 + 2x + 4)}{x – 2}$$

We can cancel $(x – 2)$ in the numerator and denominator, leaving us with:

$$x^2 + 2x + 4$$

Now, substitute $x = 2$:

$$ 2^2 + 2 \cdot 2 + 4$$

$$= 4 + 4 + 4$$

$$ = 12$$

Therefore, the limit is 12.

Key Concept: Distance Function Understanding

b) $s = 4.9t^2$

[Solution Description] The distance function provided is $s = 4.9t^2$. This function shows that the distance $s$ increases quadratically with time $t$. It is expressed as a square of the time variable multiplied by the constant 4.9.

Key Concept: Basic Tangent

d) 8

[Solution Description] To find the slope of the tangent line, we need to compute the derivative of the function and evaluate it at $x = 1$. The derivative of $y = 3x^2 + 2x$ is calculated as follows:

$$\frac{dy}{dx} = \frac{d}{dx}(3x^2 + 2x)$$

$$= 6x + 2$$

Now, substitute $x = 1$ into the derivative:

$$\frac{dy}{dx} = 6(1) + 2 = 8$$

Therefore, the slope of the tangent at $x = 1$ is 8.

Key Concept: Product of Functions

a) $6x^2 – 6x + 2$

[Solution Description] Using the product rule, which states that $$\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$, where $$u(x) = x^2 + 1$$ and $$v(x) = 2x – 3$$.

First, find the derivatives:

$$u'(x) = 2x$$

$$v'(x) = 2$$

Now apply the product rule:

$$g'(x) = (2x)(2x – 3) + (x^2 + 1)(2)$$

Simplifying this gives:

$$g'(x) = 4x^2 – 6x + 2x^2 + 2 = 6x^2 – 6x + 2$$

Key Concept: Derivative at a Point

b) 2

[Solution Description] Using the first principle:

$$f'(1) = \lim_{h \to 0} \frac{(1+h)^2 – 1^2}{h}$$

$$= \lim_{h \to 0} \frac{1 + 2h + h^2 – 1}{h}$$

$$= \lim_{h \to 0} \frac{2h + h^2}{h}$$

$$= \lim_{h \to 0} (2 + h)$$

$$= 2$$

Key Concept: Quotient Rule

a) $\frac{x\cos x – \sin x}{x^2}$

[Solution Description] The quotient rule is $$\left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2}$$ where $u = \sin x$ and $v = x$. Their derivatives are $u’ = \cos x$ and $v’ = 1$. Using the quotient rule:

$$g'(x) = \frac{\cos x \cdot x – \sin x \cdot 1}{x^2} = \frac{x\cos x – \sin x}{x^2}.$$

Key Concept: Derivative of Constant Function

a) 0

[Solution Description] The function $f(x) = 7$ is a constant function. The derivative of any constant function is zero, according to the rule: if $f(x) = c$, then $f'(x) = 0$. Therefore, $f'(x) = 0$.

Key Concept: Constant Function

c) 8

[Solution Description] A constant function remains the same regardless of $x$. Therefore, for the function $h(x) = 8$,

$$\lim_{x \to \infty} h(x) = 8$$

Hence, the limit is simply the constant value itself, which is 8.

Key Concept: Derivative Calculation

b) 5

[Solution Description] To find the derivative using the first principle, we use the formula:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

For $f(x) = x^2 + 3x$, substitute $x+h$ into the function:

$$f(x+h) = (x+h)^2 + 3(x+h)$$

Expanding this gives:

$$f(x+h) = x^2 + 2xh + h^2 + 3x + 3h$$

Subtracting $f(x)$ from $f(x+h)$:

$$f(x+h) – f(x) = (x^2 + 2xh + h^2 + 3x + 3h) – (x^2 + 3x)$$

Simplifying:

$$f(x+h) – f(x) = 2xh + h^2 + 3h$$

Divide by $h$:

$$\frac{f(x+h) – f(x)}{h} = 2x + h + 3$$

Taking the limit as $h \to 0$:

$$f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3$$

Thus, at $x = 1$:

$$f'(1) = 2(1) + 3 = 5$$

Therefore, the derivative at $x = 1$ is 5.

Key Concept: Theorem Application

d) 27

[Solution Description] Here, we use the theorem for evaluating limits of the form $$\lim_{x \to a} \frac{x^n – a^n}{x – a}$$. According to the theorem, this limit equals $na^{n-1}$. In our problem:

$n = 3$, $a = 3$

Therefore, the limit evaluates to:

$$3 \times 3^{3-1} = 3 \times 9 = 27$$

Thus, the limit is 27.

Key Concept: Derivative Calculation

a) $12x^2 – 6$

[Solution Description] To find the derivative of $$f(x) = 4x^3 – 6x + 5$$, we apply the power rule for each term.

The derivative of $4x^3$ is $12x^2$ because:

$$\frac{d}{dx}(4x^3) = 4 \cdot 3x^{3-1} = 12x^2$$

The derivative of $-6x$ is $-6$ because:

$$\frac{d}{dx}(-6x) = -6$$

The derivative of a constant, $5$, is $0$.

Combining these results, the derivative $f'(x) = 12x^2 – 6$.

Key Concept: Complex Function Limit, Limit with Undefined Point

c) The limit does not exist

[Solution Description] To find the limit of $f(x)$ as $x$ approaches 2, we need to determine both the left-hand and right-hand limits and see if they are equal.

For $x < 2$, the function is $f(x) = x^3 - 1$: $$\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (x^3 - 1) = 2^3 - 1 = 7$$ For $$x \geq 2$$, the function is $$f(x) = 3x - 1$$: $$\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (3x - 1) = 3(2) - 1 = 5$$ Since $$\lim_{{x \to 2^-}} f(x) \neq \lim_{{x \to 2^+}} f(x)$$, the limit does not exist at $x = 2$.

Key Concept: Constant Function

c) 5

[Solution Description] The limit of a constant function $f(x) = c$ as $x$ approaches any value is simply the constant itself. So, $$\lim_{x \to a} f(x) = 5$$ for any real number $a$.

Key Concept: Function Analysis

b) 2

[Solution Description] To find the right-hand limit as $x$ approaches 3, use the expression for $x \geq 3$:

$$\lim_{{x \to 3^+}} f(x) = \lim_{{x \to 3^+}} (5 – x) = 5 – 3 = 2$$

Therefore, the right-hand limit of $f(x)$ as $x$ approaches 3 is 2.

Key Concept: Basic Definition

b) To study change in the value of a function as the domain changes

[Solution Description] Calculus involves studying how a function changes as its input value changes. It is primarily concerned with the concept of change, which includes understanding the rate at which quantities evolve over time or space.

Key Concept: Rational Function Simplification

c) 4

[Solution Description] The given rational function has an indeterminate form at $x=2$. We need to factor the expression and then simplify by canceling common terms.

Factor $x^2 – 4$ using difference of squares:

$$x^2 – 4 = (x-2)(x+2)$$

Substitute back into the limit:

$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$$

Cancel the common term $(x-2)$:

$$\lim_{x \to 2} (x+2)$$

Now, substitute $x = 2$:

$$= 2 + 2 = 4$$

Hence, the answer is 4.

Key Concept: Basic Limits

b) 1

[Solution Description] According to the standard limit, we have $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. This is a well-known trigonometric limit derived using L’Hôpital’s rule or by the squeeze theorem involving the inequality $$\sin x < x < \tan x$$ near zero.

Key Concept: Limit with Trigonometric Identities

c) 3

[Solution Description] To find the limit, we use the standard trigonometric limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. Rewrite the expression as:

$$\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 = 3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x}.$$

Since $$\lim_{u \to 0} \frac{\sin(u)}{u} = 1$$, we have $u = 3x$ and as $x \to 0$, $u \to 0$. Therefore:

$$3 \cdot \lim_{u \to 0} \frac{\sin u}{u} = 3 \cdot 1 = 3.$$

So, the limit is 3.

Key Concept: Basic Definition

b) Study of change in the value of a function

[Solution Description] The main focus of calculus is the study of change in functions, which involves understanding how functions behave as their inputs change.

Key Concept: Inequality

c) $\sin x < x < \tan x$ [Solution Description] The inequality $$\sin x < x < \tan x$$ is known to hold true for the interval $$0 < x < \frac{\pi}{2}$$. This can be verified using graphical or calculus-based methods.

Key Concept: Complex Product Rule, Higher Order Derivatives

a) $e^{x}(5x + 2x^2) + \ln(x)e^{x}(3x^2 + 6x)$

[Solution Description] To find the second derivative of $h(x) = f(x)g(x)$, we first apply the product rule to find the first derivative.

The first derivative using product rule:

$$h'(x) = f(x)g'(x) + g(x)f'(x)$$

So,

$$f(x) = x^3e^{x},  g(x) = \ln(x)$$,

$$f'(x) = e^{x}(x^3 + 3x^2),  g'(x) = \frac{1}{x}$$.

Thus,

$$h'(x) = x^3e^{x} \cdot \frac{1}{x} + \ln(x) \cdot e^{x}(x^3 + 3x^2)$$

Simplifying,

$$h'(x) = x^2e^{x} + \ln(x)e^{x}(x^3 + 3x^2)$$

Now for the second derivative $h”(x)$:

Differentiate $h'(x)$:

$$h”(x) = \frac{d}{dx}[x^2e^{x}] + \frac{d}{dx}\left[\ln(x)e^{x}(x^3 + 3x^2)\right]$$

Using the product rule and chain rule:

$$\frac{d}{dx}[x^2e^{x}] = e^{x}(2x + x^2)$$

For the second term:

$$\frac{d}{dx}\left[\ln(x)e^{x}(x^3 + 3x^2)\right] = \ln(x)(e^{x}(3x^2 + 6x)) + \frac{(x^3 + 3x^2)e^{x}}{x}$$

Combine all terms:

$$h”(x) = e^{x}(2x + x^2) + \ln(x)(e^{x}(3x^2 + 6x)) + e^{x}(x^2 + 3x)$$

Key Concept: Derivative of Trigonometric Functions

b) -1

[Solution Description] The derivative of $\cos(x)$ is $-\sin(x)$. Substitute $$x = \frac{\pi}{2}$$:

$$g’\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$$

Key Concept: Quotient Rule Application

b) $\frac{x\cos x + \cos x – \sin x}{(x+1)^2}$

[Solution Description] Let $u(x) = \sin x$ and $v(x) = x+1$. Then $u'(x) = \cos x$ and $v'(x) = 1$. Using the quotient rule:

$$f'(x) = \frac{(\cos x)(x+1) – (\sin x)(1)}{(x+1)^2} = \frac{x\cos x + \cos x – \sin x}{(x+1)^2}$$

Key Concept: Polynomial Limits

c) 6

[Solution Description] Since this is a polynomial function, we can directly substitute the value $x = 1$ into the expression:

$$\lim_{x \to 1} (3x^2 – x + 4) = 3(1)^2 – 1 + 4$$

Calculate each term:

$$= 3 \cdot 1 – 1 + 4 = 3 – 1 + 4 = 6$$

Hence, the answer is $6$.

Key Concept: Simple Comparison

c) $t_1 = 1$ to $t_2 = 3$ seconds

[Solution Description] Calculate the average velocity for each interval using the formula:

$$v_{avg} = \frac{s(t_2) – s(t_1)}{t_2 – t_1}$$

For interval $t_1 = 0$ to $t_2 = 2$:

$$v_{avg} = \frac{4.9(2)^2 – 4.9(0)^2}{2 – 0} = \frac{19.6 – 0}{2} = 9.8 \, \text{m/s}$$

For interval $t_1 = 1$ to $t_2 = 3$:

$$v_{avg} = \frac{4.9(3)^2 – 4.9(1)^2}{3 – 1} = \frac{44.1 – 4.9}{2} = 19.6 \, \text{m/s}$$

Therefore, the interval from $t_1 = 1$ to $t_2 = 3$ seconds has a greater average velocity.

Key Concept: Instantaneous Velocity

b) 19.6 m/s

[Solution Description] To find the instantaneous velocity, calculate the derivative $\frac{ds}{dt}$ and evaluate it at $t = 2$.

$$\frac{ds}{dt} = \frac{d}{dt}(4.9t^2) = 9.8t$$

At $t = 2$:

$$9.8 \times 2 = 19.6 \text{ m/s}$$

Key Concept: Application of First Principle

b) 2

[Solution Description] To find the derivative using the first principle, use the formula:

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) – f(a)}{h}$$

For $$f(x) = x^2 – 4x + 7$$, calculate $f(3+h)$ and $f(3)$:

$$f(3+h) = (3+h)^2 – 4(3+h) + 7$$

$$= 9 + 6h + h^2 – 12 – 4h + 7$$

$$= h^2 + 2h + 4$$

$$f(3) = 3^2 – 4 \cdot 3 + 7$$

$$= 9 – 12 + 7$$

$$= 4$$

Substitute into the limit expression:

$$f'(3) = \lim_{h \to 0} \frac{h^2 + 2h + 4 – 4}{h}$$

$$= \lim_{h \to 0} \frac{h^2 + 2h}{h}$$

$$= \lim_{h \to 0} (h + 2)$$

$$= 2$$

Key Concept: Average Velocity

c) 10 m/s

[Solution Description] The formula for average velocity $v$ is given by:

$$ v = \frac{\Delta s}{\Delta t} $$

where $\Delta s$ is the distance traveled and $\Delta t$ is the time interval.

Here, $\Delta s = 50$ meters and $\Delta t = 5$ seconds.

Substitute these values into the formula:

$$v = \frac{50}{5} = 10 \text{ m/s}$$

Thus, the average velocity is 10 m/s.

Key Concept: Derivative Calculation

b) 1

[Solution Description] First, calculate the derivative of the polynomial function $f(x)$:

$$f'(x) = \frac{d}{dx}(x^4 – 3x^3 + 5x – 7) = 4x^3 – 9x^2 + 5$$

Now, evaluate this derivative at $x = 2$:

$$f'(2) = 4(2)^3 – 9(2)^2 + 5$$

$$= 4(8) – 9(4) + 5$$

$$= 32 – 36 + 5$$

$$= 1$$

Therefore, the derivative of the function at $x = 2$ is 1.

Key Concept: Polynomial Limit

c) 15

[Solution Description] For the polynomial $$f(x) = 2x^2 + 3x + 1$$, substitute $x = 2$ to find the limit:

$$f(2) = 2(2)^2 + 3(2) + 1 = 8 + 6 + 1 = 15$$

Therefore, the limit is 15.

Key Concept: Velocity Estimation

d) 29.4 m/s

[Solution Description] To estimate the instantaneous velocity, we find the derivative of the distance function with respect to time. The distance function is given by $s(t) = 4.9t^2$. We need to calculate $\frac{ds}{dt}$. Using the power rule for differentiation:

$$\frac{ds}{dt} = \frac{d}{dt}(4.9t^2) = 2 \times 4.9 \times t = 9.8t$$

At $t = 3$ seconds:

$$\frac{ds}{dt}\bigg|_{t=3} = 9.8 \times 3 = 29.4 \text{ m/s}$$

Therefore, the estimated instantaneous velocity at $t = 3$ seconds is 29.4 m/s.

Key Concept: Sum and Difference Application

b) $3x^2 – 4$

[Solution Description] To find the derivative, we differentiate term by term:

The derivative of $x^3$ is $3x^2$.

The derivative of $-4x$ is $-4$.

The derivative of the constant $5$ is $0$.

Thus, the derivative of $h(x)$ is:

$$h'(x) = 3x^2 – 4$$

Key Concept: Composite Function Limit

c) 3

[Solution Description] Begin by rewriting the expression using the identity $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. We can manipulate the original limit as follows:

$$\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 = 3 \cdot \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 \times 1 = 3.$$

The effective limit is $3$.

Key Concept: First Derivative Test

c) Neither a local maximum nor a local minimum

[Solution Description] To analyze if $x = 2$ is a local max or min, we need to find the first derivative of the function and evaluate it at $x = 2$. We have:

$$f'(x) = \frac{d}{dx}(x^3 – 6x^2 + 9x + 4) = 3x^2 – 12x + 9$$

Set $f'(x) = 0$ to find critical points:

$$3x^2 – 12x + 9 = 0$$

Dividing all terms by 3 gives:

$$x^2 – 4x + 3 = 0$$

Factoring yields:

$$(x-1)(x-3) = 0$$

So, $x = 1$ and $x = 3$ are critical points. Since $x = 2$ is not a critical point found via the first derivative test, $x = 2$ is neither a local maximum nor a local minimum.

Key Concept: Standard Limits

d) 12

[Solution Description] The expression $$\frac{x^3 – 8}{x – 2}$$ is indeterminate at $x = 2$. Using the standard limit for this form where $a = 2$ and $n = 3$, we have:

$$\lim_{{x \to 2}} \frac{x^3 – 8}{x – 2} = 3 \cdot 2^{2} = 3 \times 4 = 12.$$

Thus, the limit is 12.

Key Concept: Indeterminate Form Simplification

a) 6

[Solution Description] To solve this, note that the expression is initially in the indeterminate form $$\frac{0}{0}$$ when $x = 3$. Factor the numerator:

$$x^2 – 9 = (x + 3)(x – 3)$$

The expression becomes:

$$\frac{(x + 3)(x – 3)}{x – 3}$$

Cancel the $(x – 3)$ terms:

$$\lim_{x \to 3} x + 3$$

Now, substitute $x = 3$:

$$3 + 3 = 6$$

Hence, the limit is 6.

Key Concept: Simple Derivative

c) 24

[Solution Description] The derivative of $f(x) = 4x^2$ with respect to $x$ is calculated using the power rule. The derivative is given by:

$$f'(x) = \frac{d}{dx}(4x^2) = 8x.$$

Evaluating at $x = 3$, we have:

$$f'(3) = 8(3) = 24.$$

Thus, the derivative at $x = 3$ is 24.

Key Concept: Special Limit

c) 27

[Solution Description] Here, $x^3 – 27$ can be expressed as $(x^3 – 3^3)$. By the theorem for limits involving powers:

$$\lim_{x \to a} \frac{x^n – a^n}{x – a} = na^{n-1},$$

with $n = 3$ and $a = 3$, we have:

$$n \cdot a^{n-1} = 3 \cdot 3^{3-1} = 3 \cdot 9 = 27.$$

Therefore, the limit is 27.