1. ⇒ (MHT CET 2023 12th May Evening Shift )

The value of $\int_{0}^{π} | \sin x - \frac{2 x}{π} | d x$ is

A. $\frac{π}{4}$

B. $\frac{π}{2}$

C. $π$

D. $2 π$

Correct Option is (B)

$\begin{aligned} \int_{0}^{π} | \sin x - \frac{2 x}{π} | d x \\ = \int_{0}^{\frac{π}{2}} (\sin x - \frac{2 x}{π}) d x + \int_{\frac{π}{2}}^{π} (\frac{2 x}{π} - \sin x) d x \\ = [- \cos x]_{0}^{\frac{π}{2}} - {[\frac{x^{2}}{π}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2}}{π}]}_{\frac{π}{2}}^{π} + [\cos x]_{\frac{π}{2}}^{π} \\ = (0 + 1) - (\frac{π}{4} - 0) + (π - \frac{π}{4}) + (- 1 - 0) \\ = \frac{π}{2} \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

$\int_{0}^{4} | 2 x - 5 | d x =$

A. $\frac{13}{2}$

B. $\frac{15}{2}$

C. $\frac{17}{4}$

D. $\frac{17}{2}$

Correct Option is (D)

$\begin{aligned} \int_{0}^{4} | 2 x - 5 | d x & = \int_{0}^{\frac{5}{2}} (5 - 2 x) d x + \int_{\frac{5}{2}}^{4} (2 x - 5) d x \\ = {[5 x - x^{2}]}_{0}^{\frac{5}{2}} + {[x^{2} - 5 x]}_{\frac{5}{2}}^{4} \\ = \frac{25}{4} + \frac{9}{4} \\ = \frac{34}{4} \\ = \frac{17}{2} \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

Let $f : [- 1, 2] \to [0, \infty)$ be a continuous function such that $f (x) = f (1 - x), \forall x \in [- 1, 2]$

Let $R_{1} = \int_{- 1}^{2} x f (x) d x$ and $R_{2}$ be the area of the region bounded by $y = f (x), x = - 1, x = 2$ and the $X$ -axis, then $R_{2}$ is

A. $\frac{1}{2} R_{1}$

B. $2 R_{1}$

C. $3 R_{1}$

D. $\frac{1}{3} R_{1}$

Correct Option is (B)

Given that $f (x) = f (1 - x)$ and $R_{1} = \int_{- 1}^{2} x f (x) d x$

$\begin{array}{ll} ∴ R_{1} = \int_{- 1}^{2} (1 - x) f (1 - x) d x \dots [∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] \\ ∴ R_{1} = \int_{- 1}^{2} f (x) d x - \int_{- 1}^{2} x f (x) d x \dots [∵ f (x) = f (1 - x)] \\ ∴ R_{1} = \int_{- 1}^{2} f (x) d x - R_{1} \\ ∴ 2 R_{1} = \int_{- 1}^{2} f (x) d x \end{array}$

Note that $R_{2} = \int_{- 1}^{2} f (x) d x = 2 R_{1}$

4. ⇒ (MHT CET 2023 11th May Morning Shift )

Let $f (x)$ be positive for all real $x$ . If $I_{1} = \int_{1 - h}^{h} x f (x (1 - x)) d x$ and $I_{2} = \int_{1 - h}^{h} f (x (1 - x)) d x$ , where $(2 h - 1) > 0$ , then $\frac{I_{1}}{I_{2}}$ is

A. 2

B. $h$

C. $\frac{1}{2}$

D. 1

Correct Option is (C)

$\begin{aligned} I_{1} = \int_{1 - h}^{h} x f (x (1 - x)) d x and . . . (i) \\ I_{2} = \int_{1 - h}^{h} f (x (1 - x)) d x . . . (ii) \\ I_{1} = \int_{1 - h}^{h} (1 - x) f [(1 - x) (1 - 1 + x)] d x \\ \dots [∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] \end{aligned}$

$\begin{aligned} ∴ I_{1} & = \int_{1 - h}^{h} (1 - x) f (x (1 - x)) d x \\ = \int_{1 - h}^{h} f (x (1 - x)) d x - \int_{1 - h}^{h} x f (x (1 - x)) d x \\ \Rightarrow I_{1} = I_{2} - I_{1} ....[From (i) and (ii)] \\ \Rightarrow 2 I_{1} = I_{2} \\ \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{2} \end{aligned}$

5. ⇒ (MHT CET 2023 10th May Evening Shift )

Let $f : R \to R$ and $g : R \to R$ be continuous functions. Then the value of the integral $\int_{\frac{- π}{2}}^{\frac{π}{2}} [f (x) + f (- x)] [g (x) - g (- x)] d x$ is

A. $π$

B. 1

C. $- 1$

D. 0

Correct Option is (D)

Let $h (x) = [f (x) + f (- x)] [g (x) - g (- x)]$

$\begin{aligned} ∴ h (- x) & = [f (- x) + f (x)] [g (- x) - g (x)] \\ = - [f (x) + f (- x)] [g (x) - g (- x)] \\ = - h (x) \end{aligned}$

$∴ h (x)$ is an odd function.

$∴ \int_{\frac{- π}{2}}^{\frac{π}{2}} h (x) = 0$

⇒ Definite Integration