આપેલ વિધેયને ધ્યાનમાં લો: \(f\left( x \right) = \left[ x \right] + \left| {1 - x} \right|,\, - 1 \le x \le 3\) કે જ્યાં \([x]\) એ મહતમ પૃણાંક વિધેય છે . વિધેય \(1\) :\(f\) એ \(x = 0, 1, 2\) અને \(3\) આગળ સતત નથી. Statement \(2\) :\(f\left( x \right) = \left( \begin{array}{l}
- x,\,\,\,\,\,\,\,\,\, - 1 \le x < 0\\
1 - x,\,\,\,\,\,\,\,0 \le x < 1\\
1 + x,\,\,\,\,\,\,\,1 \le x < 2\,\\
2 + x,\,\,\,\,\,\,2 \le x \le 3
\end{array} \right.\)

A વિધાન \(- 1\) સાચું છે. વિધાન \(- 2\) ખોટું છે.
B વિધાન \(- 1\) સાચું છે, વિધાન \(- 2\) સાચું છે. વિધાન \(- 2\) એ વિધાન\(- 1\) ની સાચી સમજૂતી નથી.
C વિધાન \(- 1\) સાચું છે, વિધાન \(- 2\) સાચું છે. વિધાન \(- 2\) એ વિધાન\(- 1\) ની સાચી સમજૂતી છે.
D વિધાન \(- 1\) ખોટું છે. વિધાન\(- 2\) સાચું છે.

Verified Solution

Answer & Solution

Correct Answer

(A) વિધાન \(- 1\) સાચું છે. વિધાન \(- 2\) ખોટું છે.

Step-by-step Solution

Detailed explanation

Let \(f\left( x \right) = \left[ x \right] + \left| {1 - x} \right|, - 1 \le x \le 3\) where \([x]=\) greatest integer function. \(f\) is not continous at \(x=0,1,2,3\) But in statement - \(2\) \(f(x)\) is continuous at \(x=3\). Hence, statement - \(1\) is true and \(2\) is…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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