Let a, b, c be three real numbers. If the function fx=cos2x+πifx≤0ax2+bif0<x<1cx+4if1≤x≤23a+1ifx≥2 is continuous everywhere, then b2-bc+c2=

TS EAMCET · Maths · Continuity and Differentiability

Let $a, b, c$ be three real numbers. If the function
$f (x) = \{\begin{cases} \cos (2 x + π) & if & x \leq 0 \\ a x^{2} + b & if & 0 < x < 1 \\ c x + 4 & if & 1 \leq x \leq 2 \\ 3 a + 1 & if & x \geq 2 \end{cases}$
is continuous everywhere, then $b^{2} - b c + c^{2} =$

A $133$
B $157$
C $43$
D $31$

Verified Solution

Answer & Solution

Correct Answer

(A) $133$

Step-by-step Solution

Detailed explanation

Given, fx=cos2x+πifx≤0ax2+bif0<x<1cx+4if1≤x≤23a+1ifx≥2 Is continuous everywhere. Now, ∵fx is continuous at x=0. ∴f0=f0+=f0- f0-=limx→0-cos2x+π=cosπ=-1 And f0=-1 f0+=limx→0+ax2+b=b ∴b=-1................i…

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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Let a, b, c be three real numbers. If the functionfx=cos2x+πifx≤0ax2+bif0<x<1cx+4if1≤x≤23a+1ifx≥2is continuous everywhere, then b2-bc+c2=

Step-by-step Solution

Let $a, b, c$ be three real numbers. If the function
$f (x) = \{\begin{cases} \cos (2 x + π) & if & x \leq 0 \\ a x^{2} + b & if & 0 < x < 1 \\ c x + 4 & if & 1 \leq x \leq 2 \\ 3 a + 1 & if & x \geq 2 \end{cases}$
is continuous everywhere, then $b^{2} - b c + c^{2} =$