Q.A function π(π₯) = 10 β π₯ β 2π₯2 is increasing on the interval
(A) (ββ, β 1/4]
(B) (ββ, 1/4)
(C) [β 1/4, β)
(D) [β 1/4, 1/4]
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π Start your 14-day free trial to unlock the full solution βConcept understanding β Increasing Function Test
The Intuition: What Does "Increasing" Really Mean?
Imagine walking along the graph of a function from left to right. If the function is increasing, then as you step right (increasing x), you always move upward β your height f(x) never drops. You might stay flat briefly, but you never go down.
That's the visual idea. But we need a precise way to check it without drawing the entire graph β that's the Increasing Function Test, which uses the derivative to tell you where a function is rising.
A function f is increasing on an interval if, for any two points x1β<x2β in it, f(x1β)β€f(x2β). With strict inequality (<), it's strictly increasing.
The Core Idea: Derivative as a Slope Detector
The derivative fβ²(x) gives the slope of the tangent line β the instantaneous rate of change. Positive slope means the function is rising at that instant; negative means falling. So the natural question: if the derivative is positive everywhere on an interval, does that guarantee the function is increasing on that whole interval? The answer is yes β and that's the Increasing Function Test.
The Precise Statement
Increasing Function Test
Let f be continuous on [a,b] and differentiable on (a,b).
- If fβ²(x)>0 for every x in (a,b), then f is strictly increasing on [a,b].
- If fβ²(x)β₯0 for every x in (a,b), then f is increasing (non-decreasing) on [a,b].
The conditions "continuous on the closed interval" and "differentiable on the open interval" ensure there are no jumps or corners that could break the logic.
Why Does This Work? (A Quick Proof Sketch)
The proof relies on the Mean Value Theorem. For x1β<x2β in [a,b], there exists some c between them such that:
f(x2β)βf(x1β)=fβ²(c)(x2ββx1β)
Since x2ββx1β>0, if fβ²(c)>0 the right-hand side is positive, so f(x2β)>f(x1β). This holds for any pair x1β<x2β β exactly the definition of strictly increasing.
The converse is not true. A function can be strictly increasing even if its derivative is zero at some isolated points. Example: f(x)=x3 is strictly increasing everywhere, but fβ²(0)=0. The test gives a sufficient condition, not a necessary one.
How to Use It in Practice
- Compute fβ²(x).
- Solve fβ²(x)>0 β the solution intervals tell you where f is strictly increasing.
- Check endpoints if needed. β¦
The key idea is the Increasing Function Test: a differentiable function is increasing where its derivative is non-negative (fβ²(x)β₯0).
Step 1: Differentiate f(x)=10βxβ2x2.
fβ²(x)=β1β4x
Step 2: Set fβ²(x)β₯0 for increasing behaviour.
β1β4xβ₯0ββ4xβ₯1βxβ€β41β β¦
A function is increasing where its derivative is non-negative. For f(x)=10βxβ2x2, the derivative fβ²(x)=β1β4x is β₯0 when xβ€β41β, so the function increases on (ββ,β41β].
The key idea is the Increasing Function Test: a differentiable function f is increasing on an interval if its derivative fβ²(x)β₯0 for all x in that interval. This is not a trick β itβs the direct definition of what βincreasingβ means in calculus: the slope of the tangent must be non-negative.
For a quadratic like this, the derivative is linear, so the inequality is simple to solve. Letβs work through it.
-
Find the derivative.
f(x)=10βxβ2x2
Differentiate term by term:
fβ²(x)=0β1β4x=β1β4x
-
Set up the increasing condition.
We need fβ²(x)β₯0:
β1β4xβ₯0
-
Solve the inequality.
Add 1 to both sides: β4xβ₯1
Divide by β4 (remember: dividing by a negative flips the inequality sign):
xβ€β41β
So f is increasing for all x less than or equal to β41β.
A common mistake is forgetting to flip the inequality when dividing by a negative number. If you wrote xβ₯β41β, youβd get the decreasing interval instead.
- Interpret the result. β¦
Method: Determining Where a Function Is Increasing (or Decreasing)
This is the standard approach for any question that asks you to find the interval(s) on which a function is increasing or decreasing, or to identify which of several given intervals is correct.
Steps
Step 1: Differentiate the function
Find fβ²(x) using the standard differentiation rules (power rule, etc.). This derivative tells you the slope of the tangent at every point.
Step 2: Set up the correct inequality
- For increasing (non-decreasing): solve fβ²(x)β₯0.
- For strictly increasing: solve fβ²(x)>0.
- For decreasing: solve fβ²(x)β€0.
This follows directly from the Increasing/Decreasing Function Test: the sign of the derivative tells you the direction the function is moving.
Step 3: Solve the inequality for x
Since fβ²(x) is usually linear or quadratic here, solving the inequality is routine algebra. Remember: multiplying or dividing an inequality by a negative number flips its direction.
fβ²(x)β·0βΉsolveΒ forΒ theΒ intervalΒ ofΒ x β¦
Common Mistakes
Mistake 1: Forgetting to flip the inequality sign when dividing by a negative number
Solving β1β4xβ₯0 gives β4xβ₯1; dividing both sides by β4 must flip the inequality to xβ€β41β. A student who forgets this rule gets xβ₯β41β, which is exactly the decreasing interval, not the increasing one β and would wrongly match a different option.
Mistake 2: Excluding the point where the derivative is zero β¦
- KEAM 2026Set eng-2026-04214 marksMCQQ.The function f(x)=x4β2x2 is strictly increasing on (A) (β2,0) and [1,β) (B) [β1,0] and [2,β) (C) [β1,0] and [1,β) (D) (β2,0] and [0,β) (E) [β2,0] and (1,β)
βΊReveal solutionSolution
fβ²(x)=4x(xβ1)(x+1)>0 on (β1,0) and (1,β); hence strictly increasing on [β1,0] and [1,β).
f(x)=x4β2x2βfβ²(x)=4x3β4x=4x(x2β1)=4x(xβ1)(x+1).
Sign chart with roots β1,0,1:
- x<β1: negative (decreasing)
- β1<x<0: positive (increasing)
- 0<x<1: negative (decreasing) β¦
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.The derivative of a function f is given by fβ²(x)=x2+4βxβ5β. Then the interval in which f is increasing, is (A) (5,β) (B) (0,β) (C) (β4,β) (D) (ββ,β4) (E) (ββ,5)
βΊReveal solutionSolution
f is increasing on (5,β).
Concept and Intuition
A function increases where its derivative is positive. Here the positive denominator means the sign of fβ² is controlled entirely by the numerator xβ5.
Step-by-Step Solution
- x2+4β>0 for all x.
- So fβ²(x)>0βΊxβ5>0βΊx>5.
- Therefore f is increasing on (5,β).
Common Mistakes β¦
- KEAM 2025Set eng-2025-04264 marksMCQQ.The function f(x)=exβx is increasing in the interval (A) (0,4) (B) (ββ,0) (C) (β1,1) (D) (β1,0) (E) (0,β)
βΊReveal solutionSolution
fβ²(x)=exβ1>0 exactly when x>0.
For f(x)=exβx, fβ²(x)=exβ1. This is positive precisely when ex>1, i.e. x>0. Therefore f is incre β¦
- KEAM 2025Set eng-2025-04274 marksMCQQ.If g(x)=x2βx, xβR, then g(x) is increasing in (A) (ββ,β) (B) (ββ,0) (C) (0,ββ) (D) (β5,5) (E) [21β,β)
βΊReveal solutionSolution
gβ²(x)=2xβ1β₯0 for xβ₯21β, so g increases on [21β,β).
For g(x)=x2βx, the derivative is
gβ²(x)=2xβ1.
The function is increasing where gβ²(x)β₯0:
2xβ1β₯0βΉxβ₯21β. β¦
- KEAM 2024Set eng-2024-06074 marksMCQQ.The function f(x)=6x4β3x2β5 is increasing in the set (A) (ββ,2β1β)βͺ(21β,1) (B) (2β1β,0)βͺ(21β,β) (C) (2β1β,21β) (D) (ββ,21β) (E) (ββ,2β1β)βͺ(21β,β)
βΊReveal solutionSolution
Factor fβ² and take a sign chart across its roots 0,Β±21β.
fβ²(x)=24x3β6x=6x(4x2β1)=6x(2xβ1)(2x+1),
with roots at x=β21β,0,21β. Sign of fβ²:
- x<β21β: negative,
- β21β<x<0: positive,
- 0<x<21β: negative,
- x>21β: positive. β¦
- KEAM 2025Set eng-2025-04234 marksMCQQ.The function f(x)=2x3β3x2β36x+28 is increasing in (A) (ββ,β1]βͺ[3,β) (B) (ββ,β2]βͺ[3,β) (C) (ββ,β2]βͺ[5,β) (D) (ββ,β5]βͺ[3,β) (E) (ββ,β2]βͺ[8,β)
βΊReveal solutionSolution
f'(x)=6(x-3)(x+2) >= 0 for x <= -2 or x >= 3.
Concept and Intuition
A function is increasing where its derivative is non-negative. Factor f' and read off the intervals outside its roots.
Step-by-Step Solution
- f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2).
- Roots at x = -2 and x = 3; the upward parabola f' is >= 0 outside the roots. β¦
- KEAM 2026Set eng-2026-04214 marksMCQQ.If the function f(x)=x2+ax+1 is increasing on [1,2], then a is greater than or equal to (A) β2 (B) β5 (C) β4 (D) β7 (E) β3
βΊReveal solutionSolution
Require fβ²(x)=2x+aβ₯0 throughout [1,2]; the minimum of 2x there is at x=1, giving aβ₯β2.
f(x)=x2+ax+1βfβ²(x)=2x+a. β¦
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.The function f(x)=x5eβx is increasing in the interval (A) (5,β) (B) (4,β) (C) (β4,β) (D) (ββ,5) (E) (β5,β)
βΊReveal solutionSolution
f is increasing exactly where 5βx>0, i.e. on (ββ,5).
Concept and Intuition
A function increases where its derivative is positive. Differentiate the product x5eβx and factor to read off the sign.
Step-by-Step Solution
- f(x)=x5eβx.
- fβ²(x)=5x4eβxβx5eβx=x4eβx(5βx).
- x4β₯0 and eβx>0 for all x, so the sign of fβ² equals the sign of (5βx).
- fβ²(x)>0βΊ5βx>0βΊx<5.
- Hence f is increasing on (ββ,5).
Common Mistakes β¦
- KEAM 2026Set eng-2026-04194 marksMCQQ.Let f(x)=1+xlog(x+x2+1β)βx2+1β,xβ₯0. Then (A) f(x) is increasing on (0,β) (B) f(x) is increasing only on (10,β) (C) f(x) is increasing only on (0,e) (D) f(x) is decreasing on (0,β) (E) f(x) is decreasing only on (100,β)
βΊReveal solutionSolution
Differentiate; the x/x2+1β terms cancel, leaving fβ²=log(x+x2+1β)β₯0.
f(x)=1+xlog(x+x2+1β)βx2+1β.
Using dxdβlog(x+x2+1β)=x2+1β1β and dxdβx2+1β=x2+1βxβ: β¦
- KEAM 2024Set eng-2024-06064 marksMCQQ.The function f(x)=x3/5(5xβ12) is increasing in the set (A) (125β,β) (B) (ββ,0)βͺ(109β,β) (C) (ββ,0)βͺ(125β,β) (D) (0,109β) (E) (109β,β)
βΊReveal solutionSolution
The factor xβ2/5 is always positive, so the sign of fβ² follows 8xβ536β; increasing on (109β,β).
Write f(x)=5x8/5β12x3/5. Then
fβ²(x)=8x3/5β536βxβ2/5=xβ2/5(8xβ536β).
Since xβ2/5=(x2)β1/5>0 for all xξ =0, the sign of fβ² equals the sign of 8xβ536β. β¦
- KEAM 2026Set eng-2026-04204 marksMCQQ.Let f(x)=log(Ο+x)log(e+x)β, β2<x<β. Then f is (A) decreasing on (β2,β) (B) decreasing only on (0,β) (C) increasing only on (0,e) (D) increasing on (β2,β) (E) increasing only on (0,Ο)
βΊReveal solutionSolution
The derivative's sign reduces to comparing (Ο+x)log(Ο+x) with (e+x)log(e+x); tlogt is increasing here, so fβ²>0 everywhere.
f(x)=log(Ο+x)log(e+x)β. Its numerator (of fβ²) has the sign of
N=e+xlog(Ο+x)ββΟ+xlog(e+x)β.
Multiplying by (e+x)(Ο+x)>0, the sign of N equals the sign of (Ο+x)log(Ο+x)β(e+x)log(e+x). β¦
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