SATHEE: Differential Equations 1 Question 23

Differential Equations 1 Question 23

23. Consider the statements.

I. There exists some $x \in R$ such that, $f (x) + 2 x = 2 (1 + x^{2})$

II. There exists some $x \in R$ such that,

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

(a) Both I and II are true

(b) I is true and II is false

(d) Both I and II are false

Show Answer

Solution:

Here, $f (x) + 2 x = (1 - x)^{2} \cdot \sin^{2} x + x^{2} + 2 x$

where, I: $f (x) + 2 x = 2 (1 + x)^{2}$

$\begin{aligned} ∴ 2 (1 + x^{2}) = (1 - x)^{2} \sin^{2} x + x^{2} + 2 x \\ \Rightarrow (1 - x)^{2} \sin^{2} x = x^{2} - 2 x + 2 \\ \Rightarrow (1 - x)^{2} \sin^{2} x = (1 - x)^{2} + 1 \\ \Rightarrow (1 - x)^{2} \cos^{2} x = - 1 \end{aligned}$

which is never possible.

$∴ I$ is false.

Again, let $h (x) = 2 f (x) + 1 - 2 x (1 + x)$

where,

$\begin{aligned} h (0) = 2 f (0) + 1 - 0 = 1 \\ h (1) = 2 (1) + 1 - 4 = - 3 as [h (0) h (1) < 0] \end{aligned}$

$\Rightarrow h (x)$ must have a solution.

$∴$ II is true.

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Differential Equations 1 Question 23

23. Consider the statements.

Solution:

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