Let $$cos$$−$$1(4x3$$−$$3x)=a+bcos$$−$$1xand$$ x∈$$(12$$,$$1$$] then the value of limy→abcosy is

Let $$x=$$$$cos$$θ. Now x∈$$(12$$,$$1$$]⇒$$12$$

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Let $\cos^{- 1} (4 x^{3} - 3 x) = a + b \cos^{- 1} x$ and $x \in (\frac{1}{2}, 1]$ then the value of $\lim_{y \to a}$ $b \cos y$ is

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Solution

$L e t x = \cos θ . N o w x \in (\frac{1}{2}, 1] \Rightarrow \frac{1}{2} < \cos θ \leq 1 \Rightarrow 0 \leq θ < \frac{π}{3} \Rightarrow 0 \leq 3 θ < π ∴ \cos^{- 1} (4 x^{3} - 3 x) = \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) = \cos^{- 1} (\cos 3 θ) = 3 θ = 3 \cos^{- 1} x = a + b \cos^{- 1} x \Rightarrow a = 0, b = 3 . ∴ \lim_{y \to a} b \cos y = \lim_{y \to 0} 3 \cos y = 3$

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Properties of ITF

Remember concepts with our Masterclasses.

Let cos−1(4x3−3x)=a+bcos−1xand x∈(12,1] then the value of limy→abcosy is

Let x=cosθ. Now x∈(12,1]⇒12<cosθ≤1 ⇒0≤θ<π3⇒0≤3θ<π ∴cos-14x3-3x=cos-14cos3θ-3cosθ =cos-1cos3θ =3θ =3cos-1x =a+bcos-1x ⇒a=0, b=3. ∴limy→abcosy=limy→03cosy =3

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Let $\cos^{- 1} (4 x^{3} - 3 x) = a + b \cos^{- 1} x$ and $x \in (\frac{1}{2}, 1]$ then the value of $\lim_{y \to a}$ $b \cos y$ is