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QJEEAdv18Ma1c3b: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma1c3b Solution: Let f : (0, ) R be a twice differentiable function such that…

Let f : (0, π\pi ) →\to R be a twice differentiable function such that lim⁡t→xf(x)sin⁡t−f(t)sin⁡xt−x=sin⁡2x\mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}} = {\sin ^2}x for all x ∈\in (0, π\pi ).

If f(π6)=−π12f\left( {{\pi \over 6}} \right) = - {\pi \over {12}} , then which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A f(π4)=π42f\left( {{\pi \over 4}} \right) = {\pi \over {4\sqrt 2 }}
  • B f(x)<x46−x2f(x) < {{{x^4}} \over 6} - {x^2} for all x ∈\in (0, π\pi )
  • C There exists α\alpha ∈\in (0, π\pi ) such that f'( α\alpha ) = 0
  • D f′′(π2)+f(π2)=0f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0

Step-by-Step Verified Solution

Given,

lim⁡t→xf(x)sin⁡t−f(t)sin⁡xt−x=sin⁡2x\mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}} = {\sin ^2}x

Using L' Hospital rules

lim⁡t→xf(x)cos⁡t−f′(t)sin⁡x1=sin⁡2x\mathop {\lim }\limits_{t \to x} {{f(x)\cos t - f'(t)\sin x} \over 1} = {\sin ^2}x

⇒f(x)cos⁡x−f′(x)sin⁡x=sin⁡2x\Rightarrow f(x)\cos x - f'(x)\sin x = {\sin ^2}x

⇒f′(x)sin⁡x−f(x)cos⁡x=−sin⁡2x\Rightarrow f'(x)\sin x - f(x)\cos x = - {\sin ^2}x

⇒f′(x)sin⁡x−f(x)cos⁡xsin⁡2x=−1\Rightarrow {{f'(x)\sin x - f(x)\cos x} \over {{{\sin }^2}x}} = - 1

⇒d(f(x)sin⁡x)=−1\Rightarrow d\left( {{{f(x)} \over {\sin x}}} \right) = - 1

On integrating, we get

f(x)sin⁡x=−x+C{{f(x)} \over {\sin x}} = - x + C

⇒f(x)=−xsin⁡x+Csin⁡x\Rightarrow f(x) = - x\sin x + C\sin x

It is given that x=π6, f(π6)=−π12x = {\pi \over 6},\,f\left( {{\pi \over 6}} \right) = - {\pi \over {12}}

∴\therefore f(π6)=−π6sin⁡π6+Csin⁡π6f\left( {{\pi \over 6}} \right) = - {\pi \over 6}\sin {\pi \over 6} + C\sin {\pi \over 6}

=−π12=−π12+12C= - {\pi \over {12}} = - {\pi \over {12}} + {1 \over 2}C

⇒C=0\Rightarrow C = 0

∴\therefore f(x)=−xsin⁡xf(x) = - x\sin x

(a) f(x)=−xsin⁡xf(x) = - x\sin x

f(π4)=−π4sin⁡π4=−π42f\left( {{\pi \over 4}} \right) = - {\pi \over 4}\sin {\pi \over 4} = - {\pi \over {4\sqrt 2 }} false

(b) f(x) = −- x sin x

\sin x &gt; x - {{{x^3}} \over 6},\,\forall x \in (0,\pi )

\Rightarrow - x\sin x &lt; - {x^2} + {{{x^4}} \over 6}

\Rightarrow f(x) &lt; {{{x^4}} \over 6} - {x^2},\,\forall x \in (0,\,\pi )

It is true.

(c) f(x) = −- x sin x

f'(x) = −- sin x −- x cos x

f'(x) = 0

⇒\Rightarrow −- sin x −- x cos x = 0

tan x = −- x

JEE Advanced 2018 Paper 2 Offline Mathematics - Limits, Continuity and Differentiability Question 38 English Explanation

⇒\Rightarrow Their exists α\alpha ∈\in (0, π\pi ) for which f'( α\alpha ) = 0

It is true.

(d) f(x)=−xsin⁡xf(x) = - x\sin x

f′(x)=−sin⁡x−xcos⁡xf'(x) = - \sin x - x\cos x

f′′(x)=−2cos⁡x+xsin⁡xf''(x) = - 2\cos x + x\sin x

f′′(π2)=π2, f(π2)=−π2f''\left( {{\pi \over 2}} \right) = {\pi \over 2},\,f\left( {{\pi \over 2}} \right) = - {\pi \over 2}

∴\therefore f′′(π2)+f(π2)=0f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0

It is true.

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