SATHEE: Differential Equations 1 Question 11

Differential Equations 1 Question 11

11. Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TR10018Adv.)

(a) The curve $y = f (x)$ passes through the point $(1, 2)$

(b) The curve $y = f (x)$ passes through the point $(2, - 1)$

(d) The area of the region $(x, y) \in [0, 1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}$ is $\frac{π - 1}{4}$

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Answer:

Correct Answer: 11. (b)

Solution:

We have,

$f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$

On multiplying $e^{- x}$ both sides, we get

$e^{- x} f (x) = e^{- x} - 2 x e^{- x} + \int_{0}^{x} e^{- t} f (t) d t$

On differentiating both side w.r.t. $x$ , we get

$e^{- x} f^{'} (x) - e^{- x} f (x) = - e^{- x} - 2 e^{- x} + 2 x e^{- x} + e^{- x} f (x)$

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

$\begin{aligned} Let & f (x) & = y \Rightarrow & f^{'} (x) & = \frac{d y}{d x} \end{aligned}$

[dividing both sides by $e^{- x}$ ]

$∴ \frac{d y}{d x} - 2 y = 2 x - 3$

which is linear differential equation of the form $\frac{d y}{d x} + P y = Q$ . Here, $P = - 2$ and $Q = 2 x - 3$ .

Now, IF $= e^{\int P d x} = e^{\int - 2 d x} = e^{- 2 x}$

$∴$ Solution of the given differential equation is

$\begin{aligned} y \cdot e^{- 2 x} = \int \begin{array}{c} (2 x - 3) e^{- 2 x} \\ II \end{array} d x + C \\ y \cdot e^{- 2 x} = \frac{- (2 x - 3) \cdot e^{- 2 x}}{2} + 2 \int \frac{e^{- 2 x}}{2} d x + C \end{aligned}$

[by using integration by parts]

$\Rightarrow y \cdot e^{- 2 x} = \frac{- (2 x - 3) e^{- 2 x}}{2} - \frac{e^{- 2 x}}{2} + C$

$\Rightarrow y = (1 - x) + C e^{2 x}$

On putting $x = 0$ and $y = 1$ , we get

$1 = 1 + C \Rightarrow C = 0$

$∴ y = 1 - x$

$y = 1 - x$ passes through $(2, - 1)$

Now, area of region bounded by curve $y = \sqrt{1 - x^{2}}$ and $y = 1 - x$ is shows as

$∴$ Area of shaded region

$=$ Area of 1 st quadrant of a circle - Area of $△ O A B$

$\begin{aligned} = \frac{π}{4} (1)^{2} - \frac{1}{2} \times 1 \times 1 \\ = \frac{π}{4} - \frac{1}{2} = \frac{π - 2}{4} \end{aligned}$

Hence, options $b$ and $c$ are correct.

Differential Equations 1 Question 11

11. Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TR10018Adv.)

Answer:

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Differential Equations 1 Question 11

11. Let f:[0,∞)→R be a continuous function such that f(x)=1−2x+∫0xex−tf(t)dt for all x∈[0,∞). Then, which of the following statement(s) is (are) TR10018Adv.)

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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11. Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TR10018Adv.)