No links please Q (iii) If f (x) = 1-sin xπ-2x2, when x≠π2 K, when x=π2 Find the value - Maths - Inverse Trigonometric Functions

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Q (iii) If f (x) = 1 - sin   x π - 2 x 2 ,   w h e n
  x ≠ π 2   K ,   w h e n   x =
π 2
Find the value of K for which f (x) is continuous at x = π
2
Ans : 1 2

Varun Rawat · Accepted Answer

LHL = limx→π2- fx=limx→π2- 1 - sin xπ - 2x2Put x = π2 - has, x → π2-, h→0LHL = limh→01 - sinπ2 - hπ - 2π2 - h= limh→01 - cos h4h2=14 limh→02 sin2h/2h2=12 limh→0sinh/2h/2×12×limh→0sinh/2h/2×12=12×1×12×1×12=18RHL = limx→π2+ fx=limx→π2+ 1 - sin xπ - 2x2Put x = π2 + has, x → π2+, h→0RHL = limh→01 - sinπ2 + hπ - 2π2 + h= limh→01 - cos h4h2=14 limh→02 sin2h/2h2=12 limh→0sinh/2h/2×12×limh→0sinh/2h/2×12=12×1×12×1×12=18Now, fπ2 = kSince, f is continuous at x = π2, thenLHL = RHL = fπ2⇒k = 18

No links please Q. (iii) If f (x) = 1 - sin x π - 2 x 2 , w h e n x ≠ π 2 K , w h e n x = π 2 Find the value of K for which f (x) is continuous at x = π 2 . Ans : 1 2

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Q. (iii) If f (x) = $\{\begin{cases} \frac{1 - \sin x}{{(π - 2 x)}^{2}}, w h e n x \neq \frac{π}{2} \\ K, w h e n x = \frac{π}{2} \end{cases}$
Find the value of K for which f (x) is continuous at x = $\frac{π}{2}$ .
Ans : $\frac{1}{2}$