JAMIA 2026 — Mathematics PYQ

Let us test each option individually by substituting $-x$ for $x$:

• Checking Option (a): $f(x)=1-x^2$

  $$f(-x)=1-(-x{)}^2$$

  $$f(-x)=1-x^2=f(x)$$

  Since $f(-x)=f(x)$, this is an even function.

• Checking Option (b): $f(x)=x-{\cos }^{3}x$

  $$f(-x)=(-x)-[\cos (-x){]}^3$$

  $$f(-x)=-x-[\cos x{]}^3$$

  $$f(-x)=-x-{\cos }^{3}x=-(x+{\cos }^{3}x)$$

  Since $f(-x)\ne -f(x)$ and $f(-x)\ne f(x)$, this function is neither even nor odd.

• Checking Option (c): $f(x)=x-\sin x$

  $$f(-x)=(-x)-\sin (-x)$$

  Since $\sin (-x)=-\sin x$:

  $$f(-x)=-x-(-\sin x)$$

  $$f(-x)=-x+\sin x$$

  Now, factor out a negative sign:

  $$f(-x)=-(x-\sin x)$$

  $$f(-x)=-f(x)$$

  Since $f(-x)=-f(x)$, this is an odd function.

• Checking Option (d): $f(x)=x^3-|x|$

  $$f(-x)=(-x{)}^3-|-x|$$

  $$f(-x)=-x^3-|x|=-(x^3+|x|)$$

  Since $f(-x)\ne -f(x)$ and $f(-x)\ne f(x)$, this function is neither even nor odd.

Conclusion

The function that satisfies the algebraic condition for being an odd function ($f(-x)=-f(x)$) is $f(x)=x-\sin x$.

Therefore, the correct option is (c).