Let f :R→R be a continuous odd function, which vanishes exactly at one point and f1=12. Suppose that Fx= ∫-1xf(t)dt for all x∈-1, 2 and Gx= ∫-1xt|fft|dt for all x∈-1, 2. If limx→1⁡F(x)G(x)=114, then the value of f12 is

JEE Advanced · Mathematics · 16. Limits

Let $f : R \to R$ be a continuous odd function, which vanishes exactly at one point and $f (1) = \frac{1}{2} .$ Suppose that $F (x) = \int_{- 1}^{x} f (t) d t$ for all $x \in [- 1, 2]$ and $G (x) = \int_{- 1}^{x} t | f (f (t)) | d t$ for all $x \in [- 1, 2] .$ If $\lim_{x \to 1} \frac{F (x)}{G (x)} = \frac{1}{14},$ then the value of $f (\frac{1}{2})$ is

A 7
B 9
C 10
D 15

Verified Solution

Answer & Solution

Correct Answer

(A) 7

Step-by-step Solution

Detailed explanation

f (1) = \frac{1}{2}; f (- x) = - f (x); f (- 1) = - \frac{1}{2} a d f (x)

is zero only at one point

F (x) = \int_{- 1}^{x} f (t) d t

\int_{1}^{x} f (t) d t x \in [- 1, 2]

as it is an odd function,

F^{'} (x) = f (x)

G (x) = \int_{- 1}^{x} t |f (f (t))| d t

\int_{1}^{x} t |f (f (t))| d t

, as it is an odd function,

\Rightarrow G^{'} (x) = x |f (f (x))|

\lim_{x \to 1} \frac{F (x)}{G (x)} = \lim_{x \to 1} \frac{F' (x)}{G' (x)} = \lim_{x \to 1} \frac{f (x)}{x |f (f (x))|}

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

= \frac{\frac{1}{2}}{|f (\frac{1}{2})|} = \frac{1}{14}

\Rightarrow |f (\frac{1}{2})| = 7

\Rightarrow f (\frac{1}{2}) = 7 a s f (\frac{1}{2})

can not be negative

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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