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Integrals Ncert Solutions Maths class 12th Maths Class 12th

Let f, g : R ® R be functions defined by

$f (x) = {\begin{matrix} [x] & , x < 0 \\ | 1 - x | & , x \geq 0 \end{matrix} a n d g (x) = {\begin{matrix} e^{x} - x & , x < 0 \\ {(x - 1)}^{2} - 1 & , x \geq 0 \end{matrix}$

where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly:

Option 1 -

One point

Option 2 -

Two points

Option 3 -

Three points

Option 4 -

Four points

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Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$f (x) = {\begin{matrix} [x], x < 0 & | 1 - x |, x \geq 0 \end{matrix}$ $g (x) = {\begin{matrix} c^{x} - x, x < 0 & {(x - 1)}^{2} - 1, x \geq 0 \end{matrix}$

fog

$f o g (\begin{matrix} \end{matrix} [e^{x} - x], (e^{x} - x < 0) (x < 0) [{(x - 1)}^{2} - 1] {(x - 1)}^{2} - 1 < 0 x \geq 0 | 1 - e^{x} + x |, e^{x} - x \geq 0 x < 0 | 1 - {(x + 1)}^{2} + 1 |, {(x - 1)}^{2} - 1 \geq 0 x \geq 0, x \in (2, \infty) ($

Not possible as of inequalities give $? .$

e^x – x < 0, x < 0 (x – 1)² – 1 < 0

Not possible (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x $\in (0, 2)$

continuous $\forall$ x < 0

$f o g {\begin{matrix} | 1 - e^{x} + x |, x < 0 & | 1 - {(x - 1)}^{2} + 1 |, x = 0 & | 1 - {(x - 1)}^{2} + 1 |, x \geq 2 \end{matrix}$ Discontinuous at 0

continuous $\forall x$

$∴$ fog is discontinuous at 0

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The value of $\int_{0}^{π / 4} \frac{x d x}{s i n^{4} (2 x) + c o s^{4} (2 x)}$ equal to

alok kumar singh

$I = \int_{0}^{π / 4} \frac{x d x}{s i n^{4} (2 x) + c o s^{4} (2 x)}$

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The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
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I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
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Kindly considwer the following

alok kumar singh

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

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Let f, g : R ® R be functions defined by f ( x ) = { [ x ] , x < 0 | 1 − x | , x ≥ 0 a n d g ( x ) = { e x − x , x < 0 ( x − 1 ) 2 − 1 , x ≥ 0 where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly:

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