JEE Main 2022 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣, x \in R$ . Consider (S1): $f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = 2$ , (S2): $\int_{- 2}^{2} f (x) d x = 12$ . Then,

Q: Why do we split the integral at $x = -1, 0, 1$?

The function $f(x)$ is defined using modulus terms, which change their algebraic form at their roots. Because the definition of $f(x)$ changes at $x = -1, 0, 1$, the function is piecewise linear. To integrate it correctly, we must integrate each linear piece over its respective domain.

Q: Why is the derivative at the critical points not defined?

You are absolutely right to be cautious! The function $f(x)$ has 'corners' at $x = -1, 0, 1$, meaning it is not differentiable at those specific points. However, the points given in (S1) ($-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}$) are strictly inside the intervals, not at the corners, so the derivative is perfectly well-defined at these values.

Select Answer:

Progressive Hints

Hint 1

The function

f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣

is a piecewise linear function. To analyze it, you must identify the critical points where the expressions inside the modulus change sign:

x = - 1, 0, 1

. These points divide the real number line into four distinct intervals.

Hint 2

In each interval, the modulus terms can be replaced by either the expression itself or its negative, depending on the sign of the argument. For example, if

x < - 1

, then

∣ x ∣ = - x

∣ x - 1∣ = 1 - x

, and

∣ x + 1∣ = - x - 1

. Simplify the function

f (x)

in each of the four intervals:

(- \infty, - 1)

[- 1, 0)

[0, 1)

, and

[1, \infty)

Hint 3

For (S1), calculate the derivative

f^{'} (x)

in each interval by differentiating the simplified linear expression. For (S2), compute the definite integral

\int_{- 2}^{2} f (x) d x

by splitting it into four integrals corresponding to the intervals defined by the critical points:

\int_{- 2}^{- 1} f (x) d x + \int_{- 1}^{0} f (x) d x + \int_{0}^{1} f (x) d x + \int_{1}^{2} f (x) d x

Visualized Solution

Identify Critical Points of $f (x)$

Function: $f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣$
Critical points occur where expressions inside modulus are zero.
$x = 0$ , $x - 1 = 0$ , and $x + 1 = 0$
Critical points: $x = - 1, 0, 1$

Analyze Interval $x < - 1$

For interval $x < - 1$ :
$∣ x ∣ = - x$
$∣ x - 1∣ = - (x - 1) = 1 - x$
$∣ x + 1∣ = - (x + 1) = - x - 1$
$f (x) = 2 + (- x) - (1 - x) + (- x - 1)$
$f (x) = - x$

Analyze Interval $- 1 \leq x < 0$

For interval $- 1 \leq x < 0$ :
$∣ x ∣ = - x$
$∣ x - 1∣ = 1 - x$
$∣ x + 1∣ = x + 1$
$f (x) = 2 + (- x) - (1 - x) + (x + 1)$
$f (x) = x + 2$

Analyze Interval $0 \leq x < 1$

For interval $0 \leq x < 1$ :
$∣ x ∣ = x$
$∣ x - 1∣ = 1 - x$
$∣ x + 1∣ = x + 1$
$f (x) = 2 + x - (1 - x) + (x + 1)$
$f (x) = 3 x + 2$

Analyze Interval $x \geq 1$

For interval $x \geq 1$ :
$∣ x ∣ = x$
$∣ x - 1∣ = x - 1$
$∣ x + 1∣ = x + 1$
$f (x) = 2 + x - (x - 1) + (x + 1)$
$f (x) = x + 4$

Derivatives of $f (x)$

Derivatives for each interval:
$f^{'} (x) = - 1$ for $x < - 1$
$f^{'} (x) = 1$ for $- 1 < x < 0$
$f^{'} (x) = 3$ for $0 < x < 1$
$f^{'} (x) = 1$ for $x > 1$

Evaluate Statement S1

Evaluate Statement (S1):
$f^{'} (- \frac{3}{2}) = - 1$
$f^{'} (- \frac{1}{2}) = 1$
$f^{'} (\frac{1}{2}) = 3$
$f^{'} (\frac{3}{2}) = 1$
Sum $= - 1 + 1 + 3 + 1 = 4 \neq = 2$
Statement (S1) is False.

Setup Integral for S2

Evaluate Statement (S2): $\int_{- 2}^{2} f (x) d x = 12$
Split integral at critical points:
$I = \int_{- 2}^{- 1} f (x) d x + \int_{- 1}^{0} f (x) d x + \int_{0}^{1} f (x) d x + \int_{1}^{2} f (x) d x$

Calculate $I_{1}$ and $I_{2}$

$I_{1} = \int_{- 2}^{- 1} (- x) d x = [- \frac{x ^{2}}{2}]_{- 2}^{- 1} = \frac{3}{2}$
$I_{2} = \int_{- 1}^{0} (x + 2) d x = [\frac{x ^{2}}{2} + 2 x]_{- 1}^{0} = \frac{3}{2}$

Calculate $I_{3}$ and $I_{4}$

$I_{3} = \int_{0}^{1} (3 x + 2) d x = [\frac{3 x ^{2}}{2} + 2 x]_{0}^{1} = \frac{7}{2}$
$I_{4} = \int_{1}^{2} (x + 4) d x = [\frac{x ^{2}}{2} + 4 x]_{1}^{2} = \frac{11}{2}$

Final Conclusion

Total Integral $= \frac{3}{2} + \frac{3}{2} + \frac{7}{2} + \frac{11}{2} = \frac{24}{2} = 12$
Statement (S2) is True.
Conclusion: Only (S2) is correct.

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The function is given by

f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣

. The modulus function acts as a mirror, reflecting negative values to become positive.

To dismantle this, we identify the critical points where the arguments inside the modulus vanish. Setting

x = 0

x - 1 = 0

, and

x + 1 = 0

, we find the critical points:

x = - 1, 0, 1

Mapping the Terrain

We define the function in each of the four intervals created by these points.

For

x < - 1

f (x) = 2 + (- x) - (1 - x) + (- x - 1) = - x

For

- 1 \leq x < 0

f (x) = 2 + (- x) - (1 - x) + (x + 1) = x + 2

For

0 \leq x < 1

f (x) = 2 + (x) - (1 - x) + (x + 1) = 3 x + 2

For

x \geq 1

f (x) = 2 + (x) - (x - 1) + (x + 1) = x + 4

The Derivative Check (S1)

Statement (S1) asks us to sum the derivatives at four specific points. The derivative of a linear function

m x + c

is simply the slope

m

The slopes in the respective intervals are: - For

x < - 1

f^{'} (x) = - 1

. - For

- 1 < x < 0

f^{'} (x) = 1

. - For

0 < x < 1

f^{'} (x) = 3

. - For

x > 1

f^{'} (x) = 1

Evaluating the sum:

f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = - 1 + 1 + 3 + 1 = 4

Since the statement claims this sum is

2

, (S1) is false.

The Area Under the Curve (S2)

We calculate the integral

\int_{- 2}^{2} f (x) d x

by splitting it at the critical points:

I = \int_{- 2}^{- 1} (- x) d x + \int_{- 1}^{0} (x + 2) d x + \int_{0}^{1} (3 x + 2) d x + \int_{1}^{2} (x + 4) d x

Calculating each segment: 1.

\int_{- 2}^{- 1} (- x) d x = [- \frac{x ^{2}}{2}]_{- 2}^{- 1} = - 0.5 - (- 2) = 1.5

\int_{- 1}^{0} (x + 2) d x = [\frac{x ^{2}}{2} + 2 x]_{- 1}^{0} = 0 - (0.5 - 2) = 1.5

\int_{0}^{1} (3 x + 2) d x = [\frac{3 x ^{2}}{2} + 2 x]_{0}^{1} = 1.5 + 2 = 3.5

\int_{1}^{2} (x + 4) d x = [\frac{x ^{2}}{2} + 4 x]_{1}^{2} = (2 + 8) - (0.5 + 4) = 5.5

Summing these values:

1.5 + 1.5 + 3.5 + 5.5 = 12

Statement (S2) is true. Therefore, only (S2) is correct.

Common Traps & FAQs

Why do we split the integral at

x = - 1, 0, 1

The function

f (x)

is defined using modulus terms, which change their algebraic form at their roots. Because the definition of

f (x)

changes at

x = - 1, 0, 1

, the function is piecewise linear. To integrate it correctly, we must integrate each linear piece over its respective domain.

Is there a faster way to find the derivative sum in (S1)?

While you could potentially graph the function and look at the slopes, the most rigorous and reliable method is to find the explicit linear form of

f (x)

in each interval. Since the function is linear in each interval, the derivative is simply the slope of that line. Evaluating the slope at the given points is straightforward once you have the piecewise definition.

Why is the derivative at the critical points not defined?

You are absolutely right to be cautious! The function

f (x)

has 'corners' at

x = - 1, 0, 1

, meaning it is not differentiable at those specific points. However, the points given in (S1) (

- \frac{3}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{3}{2}

) are strictly inside the intervals, not at the corners, so the derivative is perfectly well-defined at these values.

Animated Solution for Mathematics - Definite Integration: Let $f (x) = 2 + ∣ x ∣ - ∣ x - 1∣ + ∣ x + 1∣, x \in R$ . Consider (S1): $f^{'} (- \frac{3}{2}) + f^{'} (- \frac{1}{2}) + f^{'} (\frac{1}{2}) + f^{'} (\frac{3}{2}) = 2$ , (S2): $\int_{- 2}^{2} f (x) d x = 12$ . Then,

Select Answer:

Visualized Solution

Identify Critical Points of $f (x)$

Analyze Interval $x < - 1$

Analyze Interval $- 1 \leq x < 0$

Analyze Interval $0 \leq x < 1$

Analyze Interval $x \geq 1$

Derivatives of $f (x)$

Evaluate Statement S1

Setup Integral for S2

Calculate $I_{1}$ and $I_{2}$

Calculate $I_{3}$ and $I_{4}$

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Mapping the Terrain

The Derivative Check (S1)

The Area Under the Curve (S2)

Similar Questions

Comprehension Passage

The correct statement(s) is(are)

If $\int_{1}^{3} x^{2} F^{'} (x) d x = - 12$ and $\int_{1}^{3} x^{3} F^{''} (x) d x = 40$ , then the correct expression(s) is (are)

Let $f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2$ and $h (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Then $\int_{- 2}^{2} h (x) d x$ is equal to :

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let $f : R \to R$ be a differentiable function such that $f (0) = 0$ , $f^{'} (\frac{π}{2}) = 3$ and $f^{'} (0) = 1$ . If $g (x) = \int_{x}^{π /2} [f^{'} (t) cosec t - cot t cosec t f (t)] d t$ for $x \in (0, \frac{π}{2})$ , then $lim_{x \to 0} g (x) =$

Let $f : R \to R$ be a continuous function such that $f (x) + f (x + 1) = 2$ , for all $x \in R$ . If $I_{1} = \int_{0}^{8} f (x) d x$ and $I_{2} = \int_{- 1}^{3} f (x) d x$ , then the value of $I_{1} + 2 I_{2}$ is equal to ____

Let $f (x)$ be a positive function and $I_{1} = \int_{1/2}^{1} 2 x f (2 x (1 - 2 x)) d x$ and $I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I _{2}}{I _{1}}$ is equal to ________

Let $f : R \to R$ be defined $f (x) = a e^{2 x} + b e^{x} + c x$ . If $f (0) = - 1, f^{'} (lo g_{e} 2) = 21$ and $\int_{0}^{l o g_{e} 4} (f (x) - c x) d x = \frac{39}{2}$ , then the value of $∣ a + b + c ∣$ equals:

Select Answer:

Visualized Solution

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Derivatives of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Evaluate Statement S1

Setup Integral for S2

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Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Mapping the Terrain

The Derivative Check (S1)

The Area Under the Curve (S2)

Similar Questions

Comprehension Passage

The correct statement(s) is(are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫13​x2F′(x)dx=−12 and ∫13​x3F′′(x)dx=40, then the correct expression(s) is (are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={−2,x−2,​−2≤x≤00<x≤2​ and h(x)=f(∣x∣)+∣f(x)∣. Then ∫−22​h(x)dx is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x), for all x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2), f(0)=1 and f(2)=e2. Then the value of ∫02​f(x)dx is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x)∀x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a positive function and I1​=∫1/21​2xf(2x(1−2x))dx and I2​=∫−12​f(x(1−x))dx. Then the value of I1​I2​​ is equal to ________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined f(x)=ae2x+bex+cx. If f(0)=−1,f′(loge​2)=21 and ∫0loge​4​(f(x)−cx)dx=239​, then the value of ∣a+b+c∣ equals:

Identify Critical Points of $f (x)$

Analyze Interval $x < - 1$

Analyze Interval $- 1 \leq x < 0$

Analyze Interval $0 \leq x < 1$

Analyze Interval $x \geq 1$

Derivatives of $f (x)$

Calculate $I_{1}$ and $I_{2}$

Calculate $I_{3}$ and $I_{4}$

If $\int_{1}^{3} x^{2} F^{'} (x) d x = - 12$ and $\int_{1}^{3} x^{3} F^{''} (x) d x = 40$ , then the correct expression(s) is (are)

Let $f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2$ and $h (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Then $\int_{- 2}^{2} h (x) d x$ is equal to :

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let $f (x)$ be a positive function and $I_{1} = \int_{1/2}^{1} 2 x f (2 x (1 - 2 x)) d x$ and $I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I _{2}}{I _{1}}$ is equal to ________

Let $f : R \to R$ be defined $f (x) = a e^{2 x} + b e^{x} + c x$ . If $f (0) = - 1, f^{'} (lo g_{e} 2) = 21$ and $\int_{0}^{l o g_{e} 4} (f (x) - c x) d x = \frac{39}{2}$ , then the value of $∣ a + b + c ∣$ equals: