Let fx= (1+cosxπ−x^2⋅sin^2x)/(ln1+π^2−2πx+x^2) x≠π λ x=π is continuous at x=π, then λ is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (C) 1/2. For fx to be continuous at x=π, limx→π⁡fx=fπ Now, limx→π⁡(1+cos⁡xπ-x^2·sin^2⁡x)/(ln⁡1+π-x^2)=λ Putting x=π+t, we get, ∴ lim t→0 (1−cost)/(t^2) ⋅(sin^2 t)/(ln 1+t^2) =λ lim t→0 1/2 ⋅(sin^2 t)/(t^2) ⋅(t^2)/(ln 1+t^2) =λ 1/2×1×1=λ⇒λ=1/2

Let $f (x) = \{\begin{cases} \frac{1 + \cos x}{{(π - x)}^{2}} \cdot \frac{\sin^{2} x}{\ln (1 + π^{2} - 2 π x + x^{2})} & x \neq π \\ λ & x = π \end{cases}$ is continuous at $x = π$ , then $λ$ is equal to

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More Continuity and Differentiability Questions

Let fx=1+cosxπ−x2⋅sin2xln1+π2−2πx+x2 x≠πλ x=π is continuous at x=π, then λ is equal to

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More Continuity and Differentiability Questions

Let $f (x) = \{\begin{cases} \frac{1 + \cos x}{{(π - x)}^{2}} \cdot \frac{\sin^{2} x}{\ln (1 + π^{2} - 2 π x + x^{2})} & x \neq π \\ λ & x = π \end{cases}$ is continuous at $x = π$ , then $λ$ is equal to