Q.Differentiate w.r.t. x: tan−1(1+cosx1−cosx), −4π<x<4π.
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Domain of a Composite Function
Picture a building with two doors: the first opens with a blue pass, the second with a red pass. Walking through f(g(x)) means passing the inner door g first, then the outer door f. The domain of the composite is simply: which inputs make it through both doors?
The intuition
Take f(x)=x and g(x)=x−5, so f(g(x))=x−5.
- The inner g(x)=x−5 accepts every real number.
- The outer f accepts only non-negative inputs.
So the real question is: which x make g(x) land inside the domain of f? The domain of the composite is not just the domain of g, nor just the domain of f — it is the overlap seen through g.
The precise statement
Domain(f∘g)={x∈Domain(g)∣g(x)∈Domain(f)}.
Two steps, in order:
- Keep only the x that g can handle.
- Among those, keep only the x for which g(x) is something f can handle.
A frequent mistake is to restrict x using the domain of f directly. The restriction comes from g(x) lying in Domain(f), not from x itself.
A worked check
For f(x)=x, g(x)=x−11:
- Domain of g: x=1.
- Outer condition: g(x)≥0⇒x−11≥0⇒x−1>0⇒x>1.
So Domain(f∘g)=(1,∞) — the condition from f already excludes x=1.
Order matters …
Concept: Chain rule after simplifying the argument.
With the half-angle identities 1−cosx=2sin22x and 1+cosx=2cos22x,
1+cosx1−cosx=tan22x=tan2x.
The absolute value matters: the square root is never negative, but tan2x is negative when x<0. On −4π<x<4π, 2x∈(−8π,8π), where tan2x has the same sign as x. Hence
y=tan−1tan2x=2∣x∣. …
Half-angle identities turn the argument into tan2x, so y=2∣x∣ on the interval; hence dxdy=21 for x>0, −21 for x<0, and it fails to exist at x=0.
Set up
Let
y=tan−1(1+cosx1−cosx),−4π<x<4π.
Direct differentiation would be ugly; simplifying the inside first makes it easy.
Simplify the argument
Using 1−cosx=2sin22x and 1+cosx=2cos22x,
1+cosx1−cosx=tan22x ⇒ 1+cosx1−cosx=tan2x.
The absolute value is essential: a principal square root cannot be negative, but tan2x is negative for x<0.
Resolve the sign
On −4π<x<4π we have 2x∈(−8π,8π), where tan2x has the same sign as x: …
Method: Half-Angle Substitution With Explicit Sign (Absolute Value) Handling
Use this method whenever a square root of a ratio of (1±cosx) appears inside an inverse trig function — the half-angle identities collapse it to a single tangent, but you must track the sign carefully because a square root is never negative while tan(x/2) can be.
Steps
Step 1: Apply the half-angle identities
1−cosx=2sin22x,1+cosx=2cos22x
so the ratio under the root becomes tan22x.
Step 2: Take the square root carefully — introduce the absolute value
tan22x=tan2x
A principal square root is never negative, but tan(x/2) is negative whenever x<0, so this absolute value is not optional.
Step 3: Resolve the sign on the given interval
Determine the sign of tan(x/2) throughout the given domain. Here x/2 stays in a small interval around 0 where tan(x/2) has the same sign as x, so the problem splits into two cases: x>0 and x<0. …
Common Mistakes
Mistake 1: Writing tan2(x/2)=tan(x/2) without the absolute value
Why it's wrong: a square root is defined to be non-negative, but tan(x/2) is negative for x<0 — skipping the absolute value silently produces the wrong sign (and hence the wrong derivative) on half the domain. Correct approach: always write u2=∣u∣, then resolve the sign using the specific interval given.
Mistake 2: Treating the function as differentiable everywhere on the given interval
Why it's wrong: y=∣x∣/2 has a corner at x=0 — the left- and right-hand derivatives disagree (−21 vs 21), so the derivative genuinely does not exist there, even though x=0 lies inside the stated domain. Correct approach: report the derivative piecewise and explicitly note the point where it fails to exist, rather than quoting a single formula for the whole interval. …
Showing the 12 most recent of 13 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.Let f(x)=sin−1x and g(x)=x−2. To define the composite function f∘g, the largest domain of g(x) has to be (A) [2,5] (B) [1,3] (C) [0,2] (D) [−1,3] (E) [−3,3]
›Reveal solutionSolution
sin−1 needs its argument in [−1,1]; with g(x)=x−2 that forces x∈[1,3].
The composite is f∘g(x)=f(g(x))=sin−1(x−2).
The domain of sin−1 is [−1,1], so we require
−1≤x−2≤1.
Add 2 throughout:
1≤x≤3. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.The domain of the function f(x)=2sin−1(2x−1)−4π is (A) [−1,1] (B) [0,1] (C) [0,2] (D) [2,5] (E) [−2,2]
›Reveal solutionSolution
The domain comes entirely from the sin−1 argument constraint −1≤2x−1≤1.
The added constant −4π and the factor 2 do not restrict the domain. We need …
- KEAM 2026Set eng-2026-04194 marksMCQQ.The domain of the function f(x)=x2+3x−4log2(x−5), is (A) (1,∞) (B) (10,∞) (C) (5,∞) (D) R∖{−4} (E) R∖{−4,1}
›Reveal solutionSolution
Logarithm forces x>5; that alone avoids the denominator's zeros, so domain =(5,∞).
Log argument. log2(x−5) requires x−5>0⇒x>5. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.The domain of the function f(x)=sin−1(5−4x2)+cos−1(5−4x2) is (A) [1,23]∪[−23,−1] (B) [1,25]∪[−23,−1] (C) [1,23]∪[−25,−1] (D) [2,23]∪[−23,−1] (E) [1,23]∪[−23,−2]
›Reveal solutionSolution
The argument 5−4x2 must lie in [−1,1]. Solving −1≤5−4x2≤1 gives 1≤x2≤23, i.e. 1≤∣x∣≤3/2, which is [1,3/2]∪[−3/2,−1].
Both sin−1(t) and cos−1(t) require −1≤t≤1 with t=5−4x2.
Upper bound:
5−4x2≤1⇒4x2≥4⇒x2≥1.
Lower bound:
5−4x2≥−1⇒4x2≤6⇒x2≤23. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.Let f(x)=log5x (x>0) and g(x)=cos−1x (−1≤x≤1). Then the domain of g∘f is (A) (0,1] (B) [−1,α) (C) [0,α) (D) [51,5] (E) [−1,5]
›Reveal solutionSolution
The domain of g∘f is [1/5, 5].
Concept and Intuition
For a composition g(f(x)), x must be in the domain of f and f(x) must lie in the domain of g. Here f=log5 x (needs x>0) feeds into g=cos⁻¹ (needs input in [−1,1]).
Step-by-Step Solution
- Require x>0 (domain of log5).
- Require −1≤log5x≤1.
- Exponentiate base 5: 5−1≤x≤51.
- Domain =[51,5] (already within x>0). …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The domain of the function f(x)=7−11x is (A) (−∞,−1] (B) [117,∞) (C) (−∞,117] (D) [117,1] (E) [−1,1]
›Reveal solutionSolution
The radicand must be non-negative: 7−11x≥0⇒x≤7/11.
For f(x)=7−11x to be real, the expression under the square root must be ≥0: …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The domain of the function f(x)=(8x−x2−7)3/2 is (A) [1,7] (B) [−3,3] (C) [−7,−1] (D) [3,7] (E) [1,4]
›Reveal solutionSolution
The inner square root requires 8x−x2−7≥0, which gives 1≤x≤7.
f(x)=(8x−x2−7)3/2 is defined when the radicand is non-negative:
8x−x2−7≥0⇒x2−8x+7≤0⇒(x−1)(x−7)≤0. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The domain of the function f(x)=x2+x−2 is (A) (−∞,−2)∪[1,∞) (B) (−∞,−2]∪(1,∞) (C) (−∞,−2)∪(1,∞) (D) (−∞,−2]∪[1,∞) (E) (−∞,1)∪[0,∞)
›Reveal solutionSolution
The domain is where (x+2)(x−1)≥0, namely (−∞,−2]∪[1,∞).
For f(x)=x2+x−2 the radicand must be non-negative:
x2+x−2=(x+2)(x−1)≥0. …
- KEAM 2025Set eng-2025-04294 marksMCQQ.The domain of the function f(x)=cos−1([x]) (where [x] denotes the greatest integer function) is (A) [−1,2] (B) [−1,2) (C) (−2,2) (D) (−2,1) (E) (−1,1)
›Reveal solutionSolution
cos−1 requires its argument in [−1,1]; since [x] is an integer, [x]∈{−1,0,1}, which corresponds to x∈[−1,2). …
- KEAM 2025Set eng-2025-04294 marksMCQQ.The domain of f(x)=∣x∣−1+4−∣x∣ is (A) [−4,−1]∪(1,4) (B) (−4,−1)∪(1,4) (C) [−4,−1] (D) [−4,−1)∪(1,4) (E) [−4,−1]∪[1,4]
›Reveal solutionSolution
Both radicands must be ≥0: ∣x∣≥1 and ∣x∣≤4, so 1≤∣x∣≤4, i.e. [−4,−1]∪[1,4].
For ∣x∣−1 to be real: ∣x∣−1≥0⇒∣x∣≥1.
For 4−∣x∣ to be real: 4−∣x∣≥0⇒∣x∣≤4. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.Let f(x)=loge(x) and let g(x)=x2+1x−2. Then the domain of the composite function f∘g is (A) (2,∞) (B) (−1,∞) (C) (0,∞) (D) (1,∞) (E) (1,0)
›Reveal solutionSolution
The composite f(g(x))=loge(x2+1x−2) is defined only where its argument is positive. As x2+1>0 for all real x, positivity depends solely on x−2.
Solving x2+1x−2>0: …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The domain of the function f(x)=9−x2sin−1(x−3) is (A) [1,2] (B) [2,3] (C) [2,3) (D) [1,2) (E) (1,2)
›Reveal solutionSolution
Combine 2≤x≤4 (from the arcsine) with −3<x<3 (denominator strictly positive) to get [2,3).
For sin−1(x−3) to be defined: −1≤x−3≤1, i.e. 2≤x≤4. …
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