Let f: ℝ → ℝ be a continuous odd function, which vanishes exactly at one point and f(1) = (1)/(2) . Suppose that F(x) = ∫ _ - 1 ^ x f ( t )dt for all x ∈ [- 1, 2] and G(x) = ∫ _ - 1 ^ x t | f(f(t)) | dt for all x ∈ [

Correct answer: 0 f(1) = (1)/(2); f(- x) = - f(x); f(- 1) = - (1)/(2)adf ( x ) is zero only at one point F(x) = ∫ _ - 1 ^ x f(t)dt= ∫ _ 1 ^ x f(t)dtx ∈ [ - 1, 2 ] as it is an odd function, F ^ ′ (x) = f ( x ) G(x) = ∫ _ - 1 ^ x t|f ( f(t) )|dt= ∫ _ 1 ^ x t|f ( f(t) )|dt , as it is an odd function, ⇒ G ^ ′ (x) = x|f ( f(x) )| _ x → 1 (F ( x ))/(G ( x )) = _ x → 1 (F′ ( x ))/(G′ ( x )) = _ x → 1 (f ( x ))/(x|f ( f(x) )|) [ As the limit is in the form of 0/0] = ((1)/(2))/(|f((1)/(2))|) = (1)/(14) ⇒ |f((1)/(2))| = 7 ⇒ f((1)/(2)) = 7asf((1)/(2)) can not be negative

JEE Advanced 2015Paper 2

Exam: JEE Advanced
Year: 2015
Paper: Paper 2
Subject: Mathematics
Chapter: Limits

JEE Advanced 2015 · Paper 2 · Mathematics · Limits

Let f: ℝ → ℝ be a continuous odd function, which vanishes exactly...

Let

f: \mathbb{R} \to \mathbb{R}

be a continuous odd function, which vanishes exactly at one point and

f\left(1\right) = \dfrac{1}{2} .

Suppose that

F\left(x\right) = \mathop{∫}_{- 1}^{x}f ( t )dt

for all

x \in \left[- 1, 2\right]

and

G\left(x\right) = \mathop{∫}_{- 1}^{x}t | f\left(f\left(t\right)\right) | dt

for all

x \in \left[- 1, 2\right] .

\lim_{x \to 1} \dfrac{F ( x )}{G ( x )} = \dfrac{1}{14},

then the value of

f\left(\dfrac{1}{2}\right)

Correct answer

Step-by-step solution

f\left(1\right) = \dfrac{1}{2}; f\left(- x\right) = - f\left(x\right); f\left(- 1\right) = - \dfrac{1}{2}adf ( x )

is zero only at one point

F\left(x\right) = \mathop{∫}_{- 1}^{x}f\left(t\right)dt=\mathop{∫}_{1}^{x}f\left(t\right)dtx \in [ - 1, 2 ]

as it is an odd function,

{F}^{&prime;}\left(x\right) = f ( x )

G\left(x\right) = \mathop{∫}_{- 1}^{x}t\left|f ( f\left(t\right) )\right|dt=\mathop{∫}_{1}^{x}t\left|f ( f\left(t\right) )\right|dt

, as it is an odd function,

⇒ {G}^{&prime;}\left(x\right) = x\left|f ( f\left(x\right) )\right|

\lim_{x \to 1} \dfrac{F ( x )}{G ( x )} = \lim_{x \to 1} \dfrac{F&prime; ( x )}{G&prime; ( x )} = \lim_{x \to 1} \dfrac{f ( x )}{x\left|f ( f\left(x\right) )\right|}

[ As the limit is in the form of 0/0]

= \dfrac{\dfrac{1}{2}}{\left|f\left(\dfrac{1}{2}\right)\right|} = \dfrac{1}{14}

⇒ \left|f\left(\dfrac{1}{2}\right)\right| = 7

⇒ f\left(\dfrac{1}{2}\right) = 7asf\left(\dfrac{1}{2}\right)

can not be negative

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