Q244 / 270 · NDA 2026 (Phase 1) · Mathematics

Passage: Let $f(x)=\sin x$ and $g(x)-f(x)=f(4-x)$.

Question: What is $\displaystyle\int_0^4\dfrac{f(4-x)}{g(4-x)}\,dx$ equal to ?

  • A
    $0$
  • B
    $1$
  • Correct answer
    C
    $2$
  • D
    $4$

Solution

Given $g(x) - f(x) = f(4-x)$

$g(x) = f(x) + f(4-x)$

Replacing $x$ with $4-x$, we get:

$g(4-x) = f(4-x) + f(x) = g(x)$

Let $I = \displaystyle\int_0^4\dfrac{f(4-x)}{g(4-x)}\,dx$

Using the property $\displaystyle\int_0^a h(x)\,dx = \displaystyle\int_0^a h(a-x)\,dx$, we have:

$I = \displaystyle\int_0^4\dfrac{f(x)}{g(x)}\,dx$

Adding the two expressions for $I$:

$2I = \displaystyle\int_0^4\dfrac{f(4-x)}{g(4-x)}\,dx + \displaystyle\int_0^4\dfrac{f(x)}{g(x)}\,dx$

Since $g(4-x) = g(x)$, this becomes:

$2I = \displaystyle\int_0^4\dfrac{f(4-x) + f(x)}{g(x)}\,dx$

Substituting $f(x) + f(4-x) = g(x)$:

$2I = \displaystyle\int_0^4\dfrac{g(x)}{g(x)}\,dx = \displaystyle\int_0^4 1\,dx$

$2I = [x]_0^4 = 4$

$I = 2$

Answer: $2$

Hence, the correct answer is option: 'C'.

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