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NCERT Exemplar · Q76

Q.State whether True or False: If ff is continuous on its domain DD, then ∣f∣|f| is also continuous on DD.

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Concept understanding — Continuity of Compositions

Continuity of Composite Functions

Many functions we meet are really one function fed into another: sin⁡(x2)\sin(x^2) is "square, then take sine"; x2+1\sqrt{x^2 + 1} is "add one to the square, then take the root." A natural question is: if each piece is continuous, is the combined function continuous? The answer is a reassuring yes, and it saves an enormous amount of work.


The Composition Rule

Important

If gg is continuous at x=ax = a, and ff is continuous at the point g(a)g(a), then the composite (f∘g)(x)=f(g(x))(f \circ g)(x) = f\big(g(x)\big) is continuous at x=ax = a.

Why it works, in plain terms: as x→ax \to a, continuity of gg pushes g(x)→g(a)g(x) \to g(a). Then continuity of ff at that landing point pushes f(g(x))→f(g(a))f\big(g(x)\big) \to f\big(g(a)\big). The limit slides cleanly through both functions:

lim⁡x→af(g(x))=f(lim⁡x→ag(x))=f(g(a)).\lim_{x \to a} f\big(g(x)\big) = f\Big(\lim_{x \to a} g(x)\Big) = f\big(g(a)\big).

That last equality is the definition of continuity for f∘gf \circ g at aa.


Using It in Practice

Most "is this function continuous?" problems become one-liners:

  • sin⁡(x2)\sin(x^2) — x2x^2 is continuous everywhere and sin⁡\sin is continuous everywhere, so the composite is continuous for all xx.
  • x2+1\sqrt{x^2 + 1} — the inside is continuous and always ≥1>0\ge 1 > 0, and  \sqrt{\ } is continuous on (0,∞)(0, \infty), so the composite is continuous everywhere.
  • ecos⁡xe^{\cos x} — continuous on all of R\mathbb{R}, being a composition of two everywhere-continuous functions.

The Trap to Watch

Watch out

The outer function must be continuous at the value g(a)g(a), not merely somewhere. For f(u)=uf(u) = \sqrt{u} with g(x)=x−4g(x) = x - 4, the composite x−4\sqrt{x-4} needs g(a)≥0g(a) \ge 0; at a=2a = 2, g(2)=−2g(2) = -2 lands outside the domain of  \sqrt{\ }, so continuity there simply doesn't apply. …

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