Q.Let f(x)=∫ex(x−1)(x−2)dx. Then write the interval in which f(x) decreases.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Derivative Sign Analysis
Derivative Sign Analysis: What the Slope Tells You
Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.
Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.
The Intuition First
If f′(x) is positive, the function is increasing — the graph rises as you move right. If f′(x) is negative, it is decreasing. If f′(x)=0, there is a horizontal tangent — a potential peak, valley, or flat inflection.
The key: a single point tells you little; you look at intervals. If f′(x)>0 for all x in (a,b), the function is strictly increasing on that whole interval. Same logic for negative.
The analysis is local — it describes behaviour on intervals, not isolated points. A zero derivative at a single point doesn't guarantee a max or min; check the sign change across that point.
The Precise Statement
Let f be differentiable on an open interval I. Then:
- If f′(x)>0 for all x in I, then f is strictly increasing on I.
- If f′(x)<0 for all x in I, then f is strictly decreasing on I.
- If f′(x)=0 for all x in I, then f is constant on I.
Points where f′(x)=0 (or where f′ does not exist) are critical points — the candidates for local maxima and minima.
If f′(x)>0 on (a,b)⟹f increasing on (a,b)
If f′(x)<0 on (a,b)⟹f decreasing on (a,b)
How to Perform It (Step-by-Step)
- Find the derivative f′(x).
- Find critical points: solve f′(x)=0 and check where f′(x) is undefined (but f is defined).
- Plot these on a number line — they split the domain into intervals.
- Pick a test point inside each interval and evaluate f′; only the sign matters.
- Record the sign in each interval and interpret: + means increasing, – means decreasing.
A Concrete Example
Take f(x)=x3−3x.
Step 1: f′(x)=3x2−3=3(x−1)(x+1).
Step 2: Critical points: x=−1 and x=1.
Step 3: Intervals: (−∞,−1), (−1,1), (1,∞).
Step 4: Test points:
- x=−2: f′(−2)=3(4−1)=9>0.
- x=0: f′(0)=−3<0.
- x=2: f′(2)=9>0.
Step 5: So f increases on (−∞,−1), decreases on (−1,1), increases on (1,∞). Thus x=−1 is a local maximum (sign changes + to –), and x=1 is a local minimum (– to +). …
By the Fundamental Theorem, f′(x)=ex(x−1)(x−2), and since ex>0 the sign is set by (x−1)(x−2), which is negative only between its roots 1 and 2. …
Since f(x)=∫ex(x−1)(x−2)dx, we get f′(x)=ex(x−1)(x−2); f decreases where f′(x)<0, i.e. on (1,2).
By the Fundamental Theorem of Calculus, if f(x)=∫ex(x−1)(x−2)dx, then
f′(x)=ex(x−1)(x−2)
A function decreases on an interval where its derivative is negative: f′(x)<0.
Since ex>0 for every real x, the sign of f′(x) is entirely determined by the sign of (x−1)(x−2):
- For x<1: both factors negative ⇒ product positive ⇒f′(x)>0. …
- CBSE 2026Set V11 markMCQQ.Statement I : The function f(x)=x2 is decreasing in the interval (0,∞) Statement II : Any function y=f(x) is decreasing if dxdy<0. Which of the following is correct?(a) Both the Statements I and II are true(b) Both the Statements I and II are false(c) Statement I is true and Statement II is false(d) Statement I is false and Statement II is true
›Reveal solutionSolution
Statement I is false and Statement II is true, so the answer is (d).
Statement I: For f(x)=x2, f′(x)=2x. On (0,∞) we have f′(x)=2x>0, so f is increasing there, not decreasing. False. …
- CBSE 2026Set ANNUAL1 markMCQQ.Let f(x)=∫ex(x−1)(x−2)dx. Then write the interval in which f(x) decreases.(a) (−∞,−2)(b) (−2,−1)(c) (1,2)(d) (2,+∞)
›Reveal solutionSolution
Since f(x)=∫ex(x−1)(x−2)dx, we get f′(x)=ex(x−1)(x−2); f decreases where f′(x)<0, i.e. on (1,2).
By the Fundamental Theorem of Calculus, if f(x)=∫ex(x−1)(x−2)dx, then
f′(x)=ex(x−1)(x−2)
A function decreases on an interval where its derivative is negative: f′(x)<0.
Since ex>0 for every real x, the sign of f′(x) is entirely determined by the sign of (x−1)(x−2):
- For x<1: both factors negative ⇒ product positive ⇒f′(x)>0. …
- CBSE 2025Set 65/4/11 markMCQQ.The values of λ so that f(x)=sinx−cosx−λx+C decreases for all real values of x are : (A) 1<λ<2 (B) λ≥1 (C) λ≥2 (D) λ<1
›Reveal solutionSolution
A function decreases everywhere when its derivative is non-positive for all x. Here f′(x)=cosx+sinx−λ must satisfy cosx+sinx≤λ for all x, which requires λ≥2 (the maximum of cosx+sinx).
A function decreases for all real x when its rate of change is never positive. This translates to the condition f′(x)≤0 for all x∈R. The question asks us to find which values of the parameter λ enforce this condition.
The key insight is that we need to understand the range of the trigonometric expression in the derivative, then choose λ large enough to dominate it everywhere.
Finding the derivative
- Differentiate f(x)=sinx−cosx−λx+C:
f′(x)=cosx+sinx−λ
- For f to be decreasing everywhere, we need:
f′(x)≤0for all x∈R
This means:
cosx+sinx−λ≤0
cosx+sinx≤λfor all x
Finding the maximum of cosx+sinx
- The condition cosx+sinx≤λ for all x is equivalent to requiring:
λ≥maxx∈R(cosx+sinx)
- To find this maximum, we can express the sum as a single sinusoid. Using the identity:
cosx+sinx=2sin(x+4π)
›Proof
Derivation of the identity:
We write cosx+sinx=Rsin(x+ϕ) for some amplitude R and phase ϕ.
Expanding: Rsin(x+ϕ)=R(sinxcosϕ+cosxsinϕ)=Rcosϕ⋅sinx+Rsinϕ⋅cosx
Comparing coefficients:
- Coefficient of sinx: Rcosϕ=1
- Coefficient of cosx: Rsinϕ=1
Squaring and adding: R2(cos2ϕ+sin2ϕ)=1+1=2, so R=2.
…
- CBSE 2024Set 65/2/11 markMCQQ.The function f(x)=x3−3x2+12x−18 is: (A) strictly decreasing on R (B) strictly increasing on R (C) neither strictly increasing nor strictly decreasing on R (D) strictly decreasing on (−∞,0)
›Reveal solutionSolution
The derivative f′(x)=3x2−6x+12 is always positive (its discriminant is negative and leading coefficient positive), so f(x) is strictly increasing on R. The correct option is (B).
The core question here is about monotonicity — whether a function is always increasing, always decreasing, or neither. For a polynomial, the sign of its derivative tells us everything. If the derivative is positive everywhere, the function is strictly increasing; if negative everywhere, strictly decreasing; if it changes sign, the function is neither.
Let’s see what f′(x) looks like.
- Find the derivative. f(x)=x3−3x2+12x−18 Differentiating term by term:
f′(x)=3x2−6x+12
- Analyze the sign of f′(x). This is a quadratic: 3x2−6x+12. To check if it ever becomes negative or zero, compute its discriminant:
D=(−6)2−4⋅3⋅12=36−144=−108
Since D<0, the quadratic has no real roots — it never touches or crosses the x-axis.
- What does a negative discriminant mean for sign? The leading coefficient 3>0, so the parabola opens upward. A quadratic that opens upward and has no real roots is always positive. Therefore, f′(x)>0 for every real x. …
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