IIT JEE 1982 Maths - 'Adapted' - FIB to Single-Digit Integer Q3

Calculus Level 3

If y = f ( 2 x − 1 x 2 + 1 ) y=f \left(\dfrac{2x-1}{x^2+1}\right) and f ′ ( x ) = sin ⁡ ( x 2 ) f'(x)=\sin \left(x^2\right) , then find ⌊ ∣ d y d x ∣ ⌋ \left \lfloor \left| \dfrac {dy}{dx} \right| \right \rfloor at x = 0 x=0 .

Notations:

Assumptions

All angles are in radians.

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The answer is 1.

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

y = f ( 2 x − 1 x 2 + 1 ) d y d x = d d x ( 2 x − 1 x 2 + 1 ) ⋅ sin ⁡ ( 2 x − 1 x 2 + 1 ) 2 = ( 2 x 2 + 1 − ( 2 x − 1 ) ( 2 x ) ( x 2 + 1 ) 2 ) ⋅ sin ⁡ ( 2 x − 1 x 2 + 1 ) 2 d y d x ∣ x = 0 = ( 2 − 0 ) ⋅ sin ⁡ ( − 1 ) 2 = 2 sin ⁡ 1 ≈ 1.683 \begin{aligned} y & = f \left(\frac {2x-1}{x^2+1} \right) \\ \frac {dy}{dx} & = \frac d{dx} \left(\frac {2x-1}{x^2+1} \right) \cdot \sin \left(\frac {2x-1}{x^2+1} \right)^2 \\ & = \left(\frac 2{x^2+1} - \frac {(2x-1)(2x)}{(x^2+1)^2} \right) \cdot \sin \left(\frac {2x-1}{x^2+1} \right)^2 \\ \frac {dy}{dx} \bigg|_{x=0} & = \left(2 - 0 \right) \cdot \sin \left(-1 \right)^2 = 2 \sin 1 \approx 1.683 \end{aligned}

⟹ ⌊ ∣ d y d x ∣ ⌋ x = 0 = ⌊ ∣ 1.683 ∣ ⌋ = 1 \begin{aligned} \implies \left \lfloor \left| \frac {dy}{dx} \right| \right \rfloor _{x=0} & = \lfloor \left| 1.683 \right| \rfloor = \boxed{1} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

@Shubhamkar Ayare , you don't need to provide the hint "Enter your answer of digit from 0 to 9." Can just write a complete sentence such as: "then find ..." instead of "then ... is _ ."

Chew-Seong Cheong - 4 years, 4 months ago

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Alright! I will follow it next time; as also correct previously written questions.

Shubhamkar Ayare - 4 years, 4 months ago

@Shubhamkar Ayare i think you have to specify if x x is in degree or radian, the answer is 0 if x x is in degrees

Anirudh Sreekumar - 4 years, 4 months ago

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Added the assumption. Thanks for pointing out!

Shubhamkar Ayare - 4 years, 4 months ago

I feel that we can assume it is in radians when it is calculus, unless it is stated otherwise.

Chew-Seong Cheong - 4 years, 4 months ago

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Yp, sorry about that,just learned about this now

apparently,

lim ⁡ x → 0 sin ⁡ ( x ) x = 1 \lim_{x \to 0} \dfrac{\sin(x)}{x} = 1 ,only if x is in radians

if x is in degrees,

lim ⁡ x → 0 sin ⁡ ( x ∘ ) x = π 180 \lim_{x \to 0} \dfrac{\sin(x^\circ)}{x} = \dfrac{\pi}{180}

which induces coefficients like these

d d x sin ⁡ ( x ∘ ) = π 180 cos ⁡ ( x ∘ ) \dfrac{d}{dx}\sin(x^\circ)=\dfrac{\pi}{180}\cos (x^\circ)

so yeah i think there is no need to specify radians.Thanks for pointing it out- :)

Anirudh Sreekumar - 4 years, 4 months ago

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@Anirudh Sreekumar – Also arc length d s = r d θ ds = r \ d \theta , area d A = 1 2 r 2 d θ dA = \frac 12 r^2 \ d \theta and volume.

Chew-Seong Cheong - 4 years, 4 months ago

@Anirudh Sreekumar – d d x sin ⁡ ( x ∘ ) = π 180 cos ⁡ ( x ∘ ) \dfrac{d}{dx}\sin(x^\circ)=\dfrac{\pi}{180}\cos (x^\circ) - that makes a good question! Has it been already asked?

Shubhamkar Ayare - 4 years, 4 months ago

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