Show that the function given by f(x)=3x+17 is increasing on R.
Hint. Differentiate and check the sign of f' everywhere.
f'(x)=3, which is positive for every real x. Since f'(x)>0 throughout R, f is strictly increasing on all of R.
✦ f is strictly increasing on R, since f'(x)=3>0 everywhere.
Show that the function given by f(x)=e^{2x} is increasing on R.
Hint. Differentiate and note that an exponential is always positive.
f'(x)=2e^{2x}, which is positive for every real x, since e^{2x}>0 always. So f is strictly increasing on all of R.
✦ f is strictly increasing on R, since f'(x)=2e^{2x}>0 everywhere.
Show that the function given by f(x)=sin(x) is (a) increasing in (0,pi/2) (b) decreasing in (pi/2,pi) (c) neither increasing nor decreasing in (0,pi).
Hint. Differentiate to get f'(x)=cos(x), then check its sign on each interval.
f'(x)=cos(x). (a) On (0,pi/2), cos(x)>0, so f is increasing there. (b) On (pi/2,pi), cos(x)<0, so f is decreasing there. (c) Since f is increasing on part of (0,pi) and decreasing on the rest, it is neither increasing nor decreasing throughout the whole interval (0,pi).
✦ (a) Increasing on (0,pi/2) (b) Decreasing on (pi/2,pi) (c) Neither increasing nor decreasing on all of (0,pi)
Find the intervals in which the function f given by f(x)=2x^2-3x is (a) increasing (b) decreasing.
Hint. Differentiate, find the critical point, then check the sign of f' on either side.
f'(x)=4x-3, zero at x=3/4. For x<3/4, f'(x)<0 (decreasing); for x>3/4, f'(x)>0 (increasing).
✦ (a) Increasing on (3/4, infinity) (b) Decreasing on (-infinity, 3/4)
Find the intervals in which the function f given by f(x)=2x^3-3x^2-36x+7 is (a) increasing (b) decreasing.
Hint. Differentiate, factor to find critical points, then check the sign of f' on each resulting interval.
f'(x)=6x^2-6x-36=6(x-3)(x+2), zero at x=-2 and x=3, since factoring the quadratic gives these two roots. Testing a point in each of the three resulting intervals: on (-infinity,-2), f'>0; on (-2,3), f'<0; on (3,infinity), f'>0.
✦ (a) Increasing on (-infinity,-2) union (3,infinity) (b) Decreasing on (-2,3)
Find the intervals in which the following functions are strictly increasing or decreasing: (a) x^2+2x-5 (b) 10-6x-2x^2 (c) -2x^3-9x^2-12x+1 (d) 6-9x-x^2 (e) (x+1)^3(x-3)^3.
Hint. Differentiate each, find critical points, then check the sign of the derivative on each resulting interval.
(a) f'=2x+2, zero at x=-1; increasing on (-1,infinity), decreasing on (-infinity,-1). (b) f'=-4x-6, zero at x=-3/2; increasing on (-infinity,-3/2), decreasing on (-3/2,infinity). (c) f'=-6(x+1)(x+2), zero at x=-2,-1; increasing on (-2,-1), decreasing on (-infinity,-2) union (-1,infinity). (d) f'=-2x-9, zero at x=-9/2; increasing on (-infinity,-9/2), decreasing on (-9/2,infinity). (e) f'=6(x-3)^2(x-1)(x+1)^2, zero at x=-1,1,3 (with (x-3)^2 and (x+1)^2 never negative); sign matches (x-1), so increasing on (1,infinity), decreasing on (-infinity,1) (with isolated zero-derivative points at -1 and 3 not changing this pattern).
✦ (a) Inc (-1,inf), Dec (-inf,-1) (b) Inc (-inf,-3/2), Dec (-3/2,inf) (c) Inc (-2,-1), Dec (-inf,-2) union (-1,inf) (d) Inc (-inf,-9/2), Dec (-9/2,inf) (e) Inc (1,inf), Dec (-inf,1)
Show that y=log(1+x)-2x/(2+x), x>-1, is an increasing function of x throughout its domain.
Hint. Differentiate and show the result is always non-negative on the domain by combining into a single fraction.
dy/dx = 1/(1+x) - [2(2+x)-2x]/(2+x)^2 = 1/(1+x) - 4/(2+x)^2. Combining over a common denominator: dy/dx = [(2+x)^2-4(1+x)]/[(1+x)(2+x)^2] = [4+4x+x^2-4-4x]/[(1+x)(2+x)^2] = x^2/[(1+x)(2+x)^2]. Since x^2>=0 always and (1+x)>0 for x>-1, and (2+x)^2>=0, dy/dx>=0 throughout the domain.
✦ dy/dx = x^2/[(1+x)(2+x)^2] >= 0 throughout x>-1, so y is increasing throughout its domain.
Find the values of x for which y=[x(x-2)]^2 is an increasing function.
Hint. Differentiate using the chain rule, factor, and determine where the result is positive.
Since y=[x(x-2)]^2=x^2(x-2)^2, apply the product rule: dy/dx=2x(x-2)^2+x^2.2(x-2)=2x(x-2)[(x-2)+x]=2x(x-2)(2x-2)=4x(x-1)(x-2), with roots at x=0,1,2. Testing a point in each of the four resulting intervals shows the product is negative on (-infinity,0), positive on (0,1), negative on (1,2), and positive on (2,infinity), because the sign flips at each of the three roots in turn.
✦ y is increasing on (0,1) union (2, infinity)
Prove that y=4sin(theta)/(2+cos theta) - theta is an increasing function of theta in [0,pi/2].
Hint. Differentiate using the quotient rule, then show the result is non-negative on the given interval.
dy/d(theta) = [4cos(theta)(2+cos theta) - 4sin(theta)(-sin theta)]/(2+cos theta)^2 - 1 = [8cos(theta)+4cos^2(theta)+4sin^2(theta)]/(2+cos theta)^2 - 1 = [8cos(theta)+4]/(2+cos theta)^2 - 1. Combining: = [8cos(theta)+4-(2+cos theta)^2]/(2+cos theta)^2 = [8cos theta+4-4-4cos theta-cos^2 theta]/(2+cos theta)^2 = [4cos theta - cos^2 theta]/(2+cos theta)^2 = cos(theta)[4-cos theta]/(2+cos theta)^2. On [0,pi/2], cos(theta)>=0 and (4-cos theta)>0, so the whole expression is non-negative.
✦ dy/d(theta) = cos(theta).(4-cos theta)/(2+cos theta)^2 >= 0 on [0,pi/2], so y is increasing there.
Prove that the logarithmic function is increasing on (0, infinity).
Hint. Differentiate log(x) and note the sign of its derivative for all positive x.
d/dx[log x] = 1/x, which is positive for every x>0. So the logarithmic function is strictly increasing on (0, infinity).
✦ log(x) is increasing on (0,infinity), since its derivative 1/x is positive there.
Prove that the function f given by f(x)=x^2-x+1 is neither strictly increasing nor decreasing on (-1,1).
Hint. Differentiate, find the critical point, and check whether it lies inside the given interval, splitting the sign of f' there.
f'(x)=2x-1, zero at x=1/2, which lies inside (-1,1). For x<1/2 (within the interval), f'(x)<0; for x>1/2, f'(x)>0. Since the sign of f' changes within (-1,1), f is decreasing on part of the interval and increasing on the rest.
✦ f is neither strictly increasing nor strictly decreasing on (-1,1), since f'(x)=2x-1 changes sign at x=1/2, inside the interval.
Which of the following functions are decreasing on (0,pi/2)? (A) cos x (B) cos 2x (C) cos 3x (D) tan x.
Hint. Differentiate each and check whether the derivative stays negative throughout the entire open interval (0,pi/2), not just part of it.
(A) d/dx(cos x)=-sin(x), negative throughout (0,pi/2) since sin(x)>0 there — decreasing. (B) d/dx(cos 2x)=-2sin(2x); since 2x ranges over (0,pi) where sin is positive, this is negative throughout — decreasing. (C) d/dx(cos 3x)=-3sin(3x); since 3x ranges over (0,3pi/2), sin(3x) is positive on (0,pi/3) but negative on (pi/3,pi/2), so the derivative changes sign — not decreasing throughout. (D) d/dx(tan x)=sec^2(x)>0 always — increasing, not decreasing.
✦ (A) cos x and (B) cos 2x are both decreasing throughout (0,pi/2); cos 3x and tan x are not.
On which of the following intervals is the function f given by f(x)=x^100+sin(x)-1 decreasing? (A) (0,1) (B) (pi/2,pi) (C) (0,pi/2) (D) None of these.
Hint. Differentiate and note that the x^99 term dominates the bounded sine-derivative term for any x>0 in these ranges.
f'(x)=100x^99+cos(x). For any x>0 in the given intervals, 100x^99 is positive, and although cos(x) can be negative, its magnitude never exceeds 1, while 100x^99 is substantial for x bounded away from 0 in each of these intervals — so f'(x) stays positive (increasing, not decreasing) throughout all three given intervals.
✦ (D) None of these
For what values of a is the function f given by f(x)=x^2+ax+1 increasing on [1,2]?
Hint. The derivative must be non-negative throughout [1,2]; since it's linear and increasing in x, checking the left endpoint x=1 is enough.
f'(x)=2x+a. Since f'(x) is itself an increasing function of x, its minimum on [1,2] occurs at x=1: f'(1)=2+a. For f to be increasing throughout [1,2], we need f'(x)>=0 for all x in [1,2], which reduces to the condition at the smallest value: 2+a>=0, so a>=-2.
✦ a >= -2
Let I be any interval disjoint from [-1,1]. Prove that the function f given by f(x)=x+1/x is increasing on I.
Hint. Differentiate and show the result is positive whenever x is outside [-1,1].
f'(x)=1-1/x^2=(x^2-1)/x^2. Since x^2>0 for x!=0, the sign of f'(x) matches the sign of x^2-1. For x outside [-1,1] (i.e. |x|>1), x^2>1, so x^2-1>0, making f'(x)>0.
✦ f'(x) = (x^2-1)/x^2 > 0 for any x with |x|>1, so f is increasing on any interval disjoint from [-1,1].
Prove that the function f given by f(x)=log(sin x) is increasing on (0,pi/2) and decreasing on (pi/2,pi).
Hint. Differentiate using the chain rule, then check the sign of cot(x) on each interval.
f'(x)=cos(x)/sin(x)=cot(x). On (0,pi/2), both sin and cos are positive, so cot(x)>0 — increasing. On (pi/2,pi), cos(x)<0 while sin(x)>0, so cot(x)<0 — decreasing.
✦ f is increasing on (0,pi/2) since cot(x)>0 there, and decreasing on (pi/2,pi) since cot(x)<0 there.
Prove that the function f given by f(x)=log|cos x| is decreasing on (0,pi/2) and increasing on (3pi/2,2pi).
Hint. Differentiate using the chain rule; on these intervals cos(x) keeps a fixed sign, so the derivative simplifies to -tan(x).
f'(x)=-sin(x)/cos(x)=-tan(x) (valid since cos x keeps a consistent sign on each given interval, avoiding the absolute value's kink). On (0,pi/2), tan(x)>0, so f'(x)<0 — decreasing. On (3pi/2,2pi), tan(x)<0 (since sin is negative, cos is positive there), so f'(x)>0 — increasing.
✦ f is decreasing on (0,pi/2) and increasing on (3pi/2,2pi), since f'(x)=-tan(x) changes sign accordingly.
Prove that the function given by f(x)=x^3-3x^2+3x-100 is increasing in R.
Hint. Differentiate and show the result is a perfect square, hence never negative.
f'(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2, which is always non-negative (zero only at the single isolated point x=1). Since f'(x)>=0 everywhere, with equality at only one point, f is increasing throughout R.
✦ f'(x)=3(x-1)^2 >= 0 everywhere, so f is increasing on all of R.
The interval in which y=x^2.e^{-x} is increasing is (A) (-infinity,infinity) (B) (-2,0) (C) (2,infinity) (D) (0,2).
Hint. Differentiate using the product rule, factor, and determine where the result is positive.
dy/dx=2x.e^{-x}-x^2.e^{-x}=x.e^{-x}(2-x). Since e^{-x}>0 always, the sign matches x(2-x), which is positive when x and (2-x) have the same sign, i.e. when 0<x<2.
✦ (D) (0,2)
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh106.pdf) — Exercise 6.1 (18 questions), Exercise 6.2 (19 questions), Exercise 6.3 (29 questions), plus the chapter's Miscellaneous Exercise (16 questions), 82 questions total. Exercise pages were rendered as 300dpi images to read the piecewise/fractional notation accurately. The old stub taught tangents and normals as a full section with the standard slope formulas, but that topic does not appear anywhere in the current book's four actual sections (6.2 Rate of Change, 6.3 Increasing and Decreasing Functions, 6.4 Maxima and Minima) — confirmed by a full-text search finding only a single mention of 'tangent and normal,' inside the chapter's own introduction paragraph, which also promises 'approximate value of certain quantities' (differentials/approximations) that likewise never appears in any actual section. Both are leftover references from an older, longer edition. The old stub also collapsed the chapter's three real exercises into one invented 40-question group with no solutions file behind it. Every numeric answer in this file (rates of change, monotonicity intervals, local/absolute extrema, and every optimisation word problem including the classic box, cylinder, cone-in-sphere, and cone-in-cone problems) was independently verified with a Python sympy script, and every optimisation problem's extraneous or degenerate critical point (like the box-folding problems' second root that collapses the volume to zero) was explicitly checked and discarded before finalising the answer.. Questions are referenced from the NCERT textbook for identification.
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