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CBSE Sample Paper 2026-27 · Section A · 1 mark

Q19: ∫−11∣x−2∣ dx=4\int_{-1}^{1}|x - 2|\,dx = 4 and the even-function property

Sample 2026-27Ch 7 · Integrals
The question

Select the correct answer from the options (A), (B), (C) and (D) as given below:

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(C) Assertion (A) is true but Reason (R) is false.

(D) Assertion (A) is false but Reason (R) is true.

Assertion: The value of the integral ∫−11∣x−2∣ dx\int_{-1}^{1}|x - 2|\,dx is 4

Reason: If f(−x)=f(x)f(-x) = f(x) then ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx

Exam answer

For −1≤x≤1-1 \le x \le 1⁠, x−2<0x - 2 < 0⁠, so ∣x−2∣=2−x|x - 2| = 2 - x⁠.

Why?

∣t∣=t|t| = t when t≥0t \ge 0⁠, and ∣t∣=−t|t| = -t when t<0t < 0⁠.

For xx from −1-1 to 11⁠, x−2x - 2 runs from −3-3 to −1-1⁠, so it is negative all the way and ∣x−2∣=−(x−2)=2−x|x - 2| = -(x - 2) = 2 - x⁠. No splitting of the integral is needed.

∫−11∣x−2∣ dx=∫−11(2−x) dx=[2x−x22]−11=(2−12)−(−2−12)=32−(−52)=4\begin{aligned} &\int_{-1}^{1}|x - 2|\,dx \\ &= \int_{-1}^{1}(2 - x)\,dx \\ &= \left[2x - \frac{x^2}{2}\right]_{-1}^{1} \\ &= \left(2 - \frac{1}{2}\right) - \left(-2 - \frac{1}{2}\right) \\ &= \frac{3}{2} - \left(-\frac{5}{2}\right) = 4 \end{aligned}
Why?

∫2 dx=2x\int 2\,dx = 2x and ∫x dx=x22\int x\,dx = \frac{x^2}{2}⁠. Upper limit minus lower limit: at x=1x = 1 the bracket is 32\frac{3}{2}⁠, and at x=−1x = -1 it is −52-\frac{5}{2}⁠.

x=1:2(1)−(1)22=2−12=42−12=32x=−1:2(−1)−(−1)22=−2−12=−42−12=−5232−(−52)=32+52=82=4\begin{aligned} x = 1&: 2(1) - \frac{(1)^2}{2} \\ &= 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \\[4pt] x = -1&: 2(-1) - \frac{(-1)^2}{2} \\ &= -2 - \frac{1}{2} = -\frac{4}{2} - \frac{1}{2} \\ &= -\frac{5}{2} \\[4pt] \frac{3}{2} - \left(-\frac{5}{2}\right) &= \frac{3}{2} + \frac{5}{2} \\ &= \frac{8}{2} = 4 \end{aligned}

Check by area: the region under y=2−xy = 2 - x from −1-1 to 11 is a trapezium with parallel sides 3 and 1 and width 2, so its area is

x=−1:y=2−(−1)=3x=1:y=2−1=1width=1−(−1)=2\begin{aligned} x = -1&: y = 2 - (-1) = 3 \\ x = 1&: y = 2 - 1 = 1 \\ \text{width} &= 1 - (-1) = 2 \end{aligned} 12(3+1)(2)=4\frac{1}{2}(3 + 1)(2) = 4

So Assertion (A) is true.

Reason (R) is a true property of even functions.

Why?

Split the integral at 0:

∫−aaf(x) dx=∫−a0f(x) dx+∫0af(x) dx\begin{aligned} \int_{-a}^{a} f(x)\,dx &= \int_{-a}^{0} f(x)\,dx \\ &\quad + \int_{0}^{a} f(x)\,dx \end{aligned}

Put x=−tx = -t⁠, dx=−dtdx = -dt in the first part. It becomes ∫0af(−t) dt\int_0^a f(-t)\,dt⁠, which equals ∫0af(t) dt\int_0^a f(t)\,dt when f(−x)=f(x)f(-x) = f(x)⁠. So the two parts are equal.

But ∣x−2∣|x - 2| is not even (its values at x=1x = 1 and x=−1x = -1 are 1 and 3), so the Reason does not apply to this integral, and Reason (R) does not explain Assertion (A).

Why?

An even function has the same value at xx and −x-x⁠. Here f(1)=∣1−2∣=1f(1) = |1 - 2| = 1 but f(−1)=∣−1−2∣=3f(-1) = \lvert -1 - 2\rvert = 3⁠.

Watch out: using the Reason anyway gives 2∫01(2−x) dx=32\int_0^1 (2 - x)\,dx = 3⁠, not 4. The Assertion is proved by direct integration, so the code is (B).

2∫01(2−x) dx=2[2x−x22]01=2(32−0)=3\begin{aligned} &2\int_0^1 (2 - x)\,dx \\ &= 2\left[2x - \frac{x^2}{2}\right]_0^1 \\ &= 2\left(\frac{3}{2} - 0\right) = 3 \end{aligned}

Answer: (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Where this comes from

Chapter 7, Integrals: Definite integrals: properties (even functions).

Question type: Definite integrals + Limits from −a-a to aa (combines 2 types)

How to spot it: click to see

Type 10. Definite integrals: integrate, then top minus bottom

Spot it: numbers on the integral sign. The answer is a number, with no + C+\,C⁠.

Here: On [−1,1][-1, 1]⁠, x−2x - 2 stays negative, so ∣x−2∣=2−x|x - 2| = 2 - x with no split. Integrate to 2x−x222x - \frac{x^2}{2} and take top minus bottom: 32−(−52)=4\frac32 - \left(-\frac52\right) = 4⁠, so the Assertion is true.

Type 11. Limits from −a-a to aa⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: The limits −1-1 and 11 call for the even test, but ∣x−2∣|x - 2| is 1 at x=1x = 1 and 3 at x=−1x = -1⁠. It is not even, so the Reason, though true, does not explain the Assertion.

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