MATHEMATICS CLASS- 12
CHAPTER-6
(APPLICATION OF
DERIVATIVES)
Exercise 6.3
Solutions
Question 1 (i)
Given:
$$f(x)=(2x-1)^2+3$$
Solution:
Since
$$ (2x-1)^2\ge 0 $$
for every real value of \(x\), therefore
$$f(x)\ge 3$$
The minimum value occurs when
$$2x-1=0$$
$$x=\frac12$$
Hence,
$$f\left(\frac12\right)=3$$
Answer:
- Minimum value = $$3$$ at $$x=\frac12$$
- Maximum value = Does not exist.
Question 1 (ii)
Given:
$$f(x)=9x^2+12x+2$$
Solution:
Complete the square:
$$ \begin{aligned} f(x) &=9\left(x^2+\frac43x\right)+2\\ &=9\left[\left(x+\frac23\right)^2-\frac49\right]+2\\ &=9\left(x+\frac23\right)^2-4+2\\ &=9\left(x+\frac23\right)^2-2 \end{aligned} $$
Since
$$9\left(x+\frac23\right)^2\ge0,$$
therefore
$$f(x)\ge-2.$$
The minimum value occurs when
$$x=-\frac23.$$
Answer:
- Minimum value = $$-2$$ at $$x=-\frac23$$
- Maximum value = Does not exist.
Question 1 (iii)
Given:
$$f(x)=-(x-1)^2+10$$
Solution:
Since
$$-(x-1)^2\le0,$$
therefore
$$f(x)\le10.$$
The maximum value occurs when
$$x-1=0,$$
i.e.,
$$x=1.$$
Hence,
$$f(1)=10.$$
Answer:
- Maximum value = $$10$$ at $$x=1$$
- Minimum value = Does not exist.
Question 1 (iv)
Given:
$$g(x)=x^3+1$$
Solution:
Differentiate:
$$g'(x)=3x^2\ge0$$
Thus, the function is increasing for all real values of \(x\).
Also,
$$ \lim_{x\to\infty}(x^3+1)=\infty, $$
and
$$ \lim_{x\to-\infty}(x^3+1)=-\infty. $$
Hence, the function has neither a maximum nor a minimum value.
Answer:
- Maximum value = Does not exist.
- Minimum value = Does not exist.
Question 2
Question 2 (i)
Given:
$$f(x)=|x+2|-1$$
Solution:
Since
$$|x+2|\ge0,$$
therefore
$$f(x)\ge-1.$$
The minimum value occurs when
$$x+2=0,$$
i.e.,
$$x=-2.$$
Answer:
- Minimum value = $$-1$$ at $$x=-2$$
- Maximum value = Does not exist.
Question 2 (ii)
Given:
$$g(x)=-|x+1|+3$$
Solution:
Since
$$|x+1|\ge0,$$
therefore
$$-|x+1|\le0.$$
Hence,
$$g(x)\le3.$$
The maximum value occurs when
$$x=-1.$$
Answer:
- Maximum value = $$3$$ at $$x=-1$$
- Minimum value = Does not exist.
Question 2 (iii)
Given:
$$h(x)=\sin(2x)+5$$
Solution:
Since
$$-1\le\sin(2x)\le1,$$
adding 5 to every part gives
$$4\le h(x)\le6.$$
Maximum value:
$$h(x)=6$$
when
$$\sin(2x)=1.$$
This occurs at
$$ 2x=\frac{\pi}{2}+2n\pi, $$
or
$$ x=\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z. $$
Minimum value:
$$h(x)=4$$
when
$$\sin(2x)=-1.$$
This occurs at
$$ 2x=\frac{3\pi}{2}+2n\pi, $$
or
$$ x=\frac{3\pi}{4}+n\pi,\qquad n\in\mathbb Z. $$
Answer:
- Maximum value = $$6$$
- Minimum value = $$4$$
Question 2 (iv)
Given:
$$f(x)=|\sin4x+3|$$
Solution:
Since
$$-1\le\sin4x\le1,$$
therefore
$$2\le\sin4x+3\le4.$$
As the expression inside the modulus is always positive,
$$|\sin4x+3|=\sin4x+3.$$
Hence,
$$2\le f(x)\le4.$$
Answer:
- Maximum value = $$4$$
- Minimum value = $$2$$
Question 2 (v)
Given:
$$h(x)=x+1,\qquad x\in(-1,1)$$
Solution:
The function is linear and increasing because
$$h'(x)=1>0.$$
Its range is
$$
0 The values 0 and 2 are never attained because the interval is open. Answer: Given: $$f(x)=x^2$$ Solution: Differentiate the function: $$f'(x)=2x$$ For critical points, put $$f'(x)=0$$ $$2x=0$$ $$x=0$$ Now find the second derivative. $$f''(x)=2$$ Since $$f''(0)=2>0,$$ the function has a local minimum at $$x=0$$. The minimum value is $$f(0)=0.$$ There is no point where the function has a local maximum. Answer: Given: $$g(x)=x^3-3x$$ Solution: Differentiate the function. $$g'(x)=3x^2-3$$ For critical points, $$3x^2-3=0$$ $$x^2=1$$ $$x=\pm1$$ Now find the second derivative. $$g''(x)=6x$$ At $$x=-1$$ $$g''(-1)=-6<0$$ Hence, the function has a local maximum at $$x=-1$$. The maximum value is $$
g(-1)=(-1)^3-3(-1)
=-1+3
=2.
$$ At $$x=1$$ $$g''(1)=6>0$$ Hence, the function has a local minimum at $$x=1$$. The minimum value is $$
g(1)=1-3=-2.
$$ Answer: Given: $$
h(x)=\sin x+\cos x,\qquad 0 Solution: Differentiate the function. $$
h'(x)=\cos x-\sin x
$$ For critical points, $$
\cos x-\sin x=0
$$ $$
\cos x=\sin x
$$ $$
\tan x=1
$$ Since $$
0 the critical point is $$
x=\frac{\pi}{4}.
$$ Now find the second derivative. $$
h''(x)=-\sin x-\cos x
$$ At $$
x=\frac{\pi}{4},
$$ $$
\begin{aligned}
h''\left(\frac{\pi}{4}\right)
&=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\\
&=-\sqrt2<0.
\end{aligned}
$$ Therefore, the function has a local maximum at $$
x=\frac{\pi}{4}.
$$ The maximum value is $$
\begin{aligned}
h\left(\frac{\pi}{4}\right)
&=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\\
&=\frac1{\sqrt2}+\frac1{\sqrt2}\\
&=\sqrt2.
\end{aligned}
$$ Since there is only one critical point in the interval and it is a maximum point, the function has no local minimum in the interval $$\left(0,\frac{\pi}{2}\right)$$. Answer: Given: $$
f(x)=\sin x-\cos x,\qquad 0 Solution: Differentiate the function. $$
f'(x)=\cos x+\sin x
$$ For critical points, $$
\cos x+\sin x=0
$$ $$
\sin x=-\cos x
$$ $$
\tan x=-1
$$ In the interval $$0 $$
x=\frac{3\pi}{4},\quad \frac{7\pi}{4}.
$$ Now find the second derivative. $$
f''(x)=\cos x-\sin x
$$ At $$x=\frac{3\pi}{4}$$ $$
\begin{aligned}
f''\left(\frac{3\pi}{4}\right)
&=\cos\frac{3\pi}{4}-\sin\frac{3\pi}{4}\\
&=-\frac1{\sqrt2}-\frac1{\sqrt2}\\
&=-\sqrt2<0.
\end{aligned}
$$ Hence, there is a local maximum at $$
x=\frac{3\pi}{4}.
$$ The maximum value is $$
\begin{aligned}
f\left(\frac{3\pi}{4}\right)
&=\sin\frac{3\pi}{4}-\cos\frac{3\pi}{4}\\
&=\frac1{\sqrt2}-\left(-\frac1{\sqrt2}\right)\\
&=\sqrt2.
\end{aligned}
$$ At $$x=\frac{7\pi}{4}$$ $$
\begin{aligned}
f''\left(\frac{7\pi}{4}\right)
&=\cos\frac{7\pi}{4}-\sin\frac{7\pi}{4}\\
&=\frac1{\sqrt2}-\left(-\frac1{\sqrt2}\right)\\
&=\sqrt2>0.
\end{aligned}
$$ Hence, there is a local minimum at $$
x=\frac{7\pi}{4}.
$$ The minimum value is $$
\begin{aligned}
f\left(\frac{7\pi}{4}\right)
&=\sin\frac{7\pi}{4}-\cos\frac{7\pi}{4}\\
&=-\frac1{\sqrt2}-\frac1{\sqrt2}\\
&=-\sqrt2.
\end{aligned}
$$ Answer: Given: $$
f(x)=x^3-6x^2+9x+15
$$ Solution: Differentiate the function. $$
\begin{aligned}
f'(x)
&=3x^2-12x+9\\
&=3(x^2-4x+3)\\
&=3(x-1)(x-3)
\end{aligned}
$$ For critical points, $$
3(x-1)(x-3)=0
$$ Hence, $$
x=1,\quad x=3.
$$ Now find the second derivative. $$
f''(x)=6x-12
$$ At $$x=1$$ $$
f''(1)=6-12=-6<0.
$$ Therefore, there is a local maximum at $$x=1$$. The maximum value is $$
\begin{aligned}
f(1)
&=1-6+9+15\\
&=19.
\end{aligned}
$$ At $$x=3$$ $$
f''(3)=18-12=6>0.
$$ Therefore, there is a local minimum at $$x=3$$. The minimum value is $$
\begin{aligned}
f(3)
&=27-54+27+15\\
&=15.
\end{aligned}
$$ Answer: Given: $$
g(x)=\frac{x}{2}+\frac{2}{x},\qquad x>0
$$ Solution: Differentiate the function. $$
g'(x)=\frac12-\frac{2}{x^2}
$$ For critical points, $$
\frac12-\frac2{x^2}=0
$$ $$
\frac12=\frac2{x^2}
$$ $$
x^2=4
$$ Since $$x>0,$$ $$
x=2.
$$ Now find the second derivative. $$
g''(x)=\frac4{x^3}
$$ At $$x=2,$$ $$
g''(2)=\frac42=\frac12>0.
$$ Hence, the function has a local minimum at $$x=2$$. The minimum value is $$
\begin{aligned}
g(2)
&=\frac22+\frac22\\
&=1+1\\
&=2.
\end{aligned}
$$ There is no local maximum. Answer: Given: $$
g(x)=\frac{1}{x^2+2}
$$ Solution: Differentiate the function. $$
\begin{aligned}
g'(x)
&=\frac{d}{dx}(x^2+2)^{-1}\\
&=-(x^2+2)^{-2}(2x)\\
&=-\frac{2x}{(x^2+2)^2}.
\end{aligned}
$$ For critical points, put $$
g'(x)=0.
$$ Since the denominator is always positive, $$
-2x=0
$$ $$
x=0.
$$ Now find the second derivative. Using the quotient rule, $$
g''(x)=\frac{2(3x^2-2)}{(x^2+2)^3}.
$$ At $$x=0,$$ $$
\begin{aligned}
g''(0)
&=\frac{2(0-2)}{2^3}\\
&=-\frac12<0.
\end{aligned}
$$ Therefore, the function has a local maximum at $$x=0$$. The maximum value is $$
g(0)=\frac{1}{2}.
$$ Since there is only one critical point and it is a maximum point, the function has no local minimum. Answer: Given: $$
f(x)=x\sqrt{1-x},\qquad 0 Solution: Write the function as $$
f(x)=x(1-x)^{1/2}.
$$ Differentiate using the product rule. $$
\begin{aligned}
f'(x)
&=(1-x)^{1/2}
+x\left(\frac12\right)(1-x)^{-1/2}(-1)\\
&=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}.
\end{aligned}
$$ Taking the LCM, $$
\begin{aligned}
f'(x)
&=\frac{2(1-x)-x}{2\sqrt{1-x}}\\
&=\frac{2-3x}{2\sqrt{1-x}}.
\end{aligned}
$$ For critical points, $$
\frac{2-3x}{2\sqrt{1-x}}=0.
$$ Since the denominator is positive for $$0 $$
2-3x=0.
$$ Hence, $$
x=\frac23.
$$ Now find the second derivative. Differentiate $$
f'(x)=\frac{2-3x}{2\sqrt{1-x}}.
$$ We get $$
f''(x)=\frac{3x-4}{4(1-x)^{3/2}}.
$$ At $$x=\dfrac23,$$ $$
\begin{aligned}
f''\left(\frac23\right)
&=\frac{2-4}{4\left(\frac13\right)^{3/2}}\\
&<0.
\end{aligned}
$$ Therefore, the function has a local maximum at $$
x=\frac23.
$$ The maximum value is $$
\begin{aligned}
f\left(\frac23\right)
&=\frac23\sqrt{1-\frac23}\\
&=\frac23\sqrt{\frac13}\\
&=\frac{2}{3\sqrt3}\\
&=\frac{2\sqrt3}{9}.
\end{aligned}
$$ There is no local minimum in the interval $$0 Answer: Given: $$
p(x)=41+72x-18x^2
$$ Solution: Differentiate the profit function. $$
\begin{aligned}
p'(x)
&=\frac{d}{dx}(41+72x-18x^2)\\
&=72-36x.
\end{aligned}
$$ For critical points, put $$
p'(x)=0.
$$ $$
72-36x=0
$$ $$
x=2.
$$ Now find the second derivative. $$
p''(x)=-36.
$$ Since $$
p''(2)=-36<0,
$$ the function has a maximum at $$x=2$$. The maximum profit is $$
\begin{aligned}
p(2)
&=41+72(2)-18(2)^2\\
&=41+144-72\\
&=113.
\end{aligned}
$$ Answer: Given: $$
f(x)=3x^4-8x^3+12x^2-48x+25,\qquad x\in[0,3]
$$ Solution: Differentiate the function. $$
\begin{aligned}
f'(x)
&=12x^3-24x^2+24x-48\\
&=12(x^3-2x^2+2x-4)\\
&=12(x-2)(x^2+2).
\end{aligned}
$$ Since $$
x^2+2>0,
$$ the only critical point in the interval is $$
x=2.
$$ Now evaluate the function at the critical point and the end points. At $$x=0$$ $$
f(0)=25.
$$ At $$x=2$$ $$
\begin{aligned}
f(2)
&=3(16)-8(8)+12(4)-48(2)+25\\
&=48-64+48-96+25\\
&=-39.
\end{aligned}
$$ At $$x=3$$ $$
\begin{aligned}
f(3)
&=3(81)-8(27)+12(9)-48(3)+25\\
&=243-216+108-144+25\\
&=16.
\end{aligned}
$$ Comparing these values, $$
25>-39<16.
$$ Answer: Given: $$
f(x)=\sin2x,\qquad x\in[0,2\pi]
$$ Solution: The maximum value of the sine function is $$
1.
$$ Therefore, $$
\sin2x=1.
$$ Now, $$
2x=\frac{\pi}{2}+2n\pi,
$$ where $$n$$ is an integer. Hence, $$
x=\frac{\pi}{4}+n\pi.
$$ In the interval $$
0\le x\le2\pi,
$$ the values are $$
x=\frac{\pi}{4},\qquad \frac{5\pi}{4}.
$$ Answer: $$
\boxed{x=\frac{\pi}{4},\ \frac{5\pi}{4}}
$$ Given: $$
f(x)=\sin x+\cos x
$$ Solution: Using the identity $$
\sin x+\cos x=\sqrt2\sin\left(x+\frac{\pi}{4}\right),
$$ and since $$
-1\le\sin\left(x+\frac{\pi}{4}\right)\le1,
$$ we get $$
-\sqrt2\le\sin x+\cos x\le\sqrt2.
$$ Therefore, the maximum value is $$
\boxed{\sqrt2}.
$$ Given: $$
f(x)=2x^3-24x+107
$$ (i) On the interval $$[1,3]$$ Differentiate the function. $$
f'(x)=6x^2-24=6(x-2)(x+2).
$$ The only critical point in the interval is $$
x=2.
$$ Evaluate the function at the end points and the critical point. $$
\begin{aligned}
f(1)&=2-24+107=85,\\
f(2)&=16-48+107=75,\\
f(3)&=54-72+107=89.
\end{aligned}
$$ The greatest value is $$
89.
$$ Answer: (ii) On the interval $$[-3,-1]$$ The critical point in this interval is $$
x=-2.
$$ Evaluate the function. $$
\begin{aligned}
f(-3)&=-54+72+107=125,\\
f(-2)&=-16+48+107=139,\\
f(-1)&=-2+24+107=129.
\end{aligned}
$$ The greatest value is $$
139.
$$ Answer: Given: $$
f(x)=x^4-62x^2+ax+9
$$ It is given that the function attains its maximum value at $$
x=1
$$ on the interval $$
[0,2].
$$ Solution: Since the maximum occurs at the interior point $$x=1$$, we must have $$
f'(1)=0.
$$ Differentiate the function. $$
\begin{aligned}
f'(x)
&=\frac{d}{dx}(x^4-62x^2+ax+9)\\
&=4x^3-124x+a.
\end{aligned}
$$ Substitute $$x=1$$. $$
\begin{aligned}
f'(1)
&=4(1)^3-124(1)+a\\
&=4-124+a\\
&=a-120.
\end{aligned}
$$ Since $$
f'(1)=0,
$$ we get $$
a-120=0.
$$ Therefore, $$
\boxed{a=120.}
$$ Verification: Differentiate again. $$
f''(x)=12x^2-124.
$$ At $$x=1$$, $$
f''(1)=12-124=-112<0.
$$ Hence, the function has a local maximum at $$x=1$$, confirming the result. Answer: $$
\boxed{a=120}
$$ Given: $$
f(x)=x+\sin2x,\qquad x\in[0,2\pi]
$$ Solution: Differentiate the function. $$
\begin{aligned}
f'(x)
&=1+2\cos2x.
\end{aligned}
$$ For critical points, $$
1+2\cos2x=0.
$$ Hence, $$
\cos2x=-\frac12.
$$ In the interval $$
0\le2x\le4\pi,
$$ the solutions are $$
2x=\frac{2\pi}{3},\;
\frac{4\pi}{3},\;
\frac{8\pi}{3},\;
\frac{10\pi}{3}.
$$ Therefore, $$
x=\frac{\pi}{3},\;
\frac{2\pi}{3},\;
\frac{4\pi}{3},\;
\frac{5\pi}{3}.
$$ Now evaluate the function at the critical points and the end points. At $$x=0$$ $$
f(0)=0.
$$ At $$x=\dfrac{\pi}{3}$$ $$
\begin{aligned}
f\left(\frac{\pi}{3}\right)
&=\frac{\pi}{3}+\sin\frac{2\pi}{3}\\
&=\frac{\pi}{3}+\frac{\sqrt3}{2}.
\end{aligned}
$$ At $$x=\dfrac{2\pi}{3}$$ $$
\begin{aligned}
f\left(\frac{2\pi}{3}\right)
&=\frac{2\pi}{3}+\sin\frac{4\pi}{3}\\
&=\frac{2\pi}{3}-\frac{\sqrt3}{2}.
\end{aligned}
$$ At $$x=\dfrac{4\pi}{3}$$ $$
\begin{aligned}
f\left(\frac{4\pi}{3}\right)
&=\frac{4\pi}{3}+\sin\frac{8\pi}{3}\\
&=\frac{4\pi}{3}+\frac{\sqrt3}{2}.
\end{aligned}
$$ At $$x=\dfrac{5\pi}{3}$$ $$
\begin{aligned}
f\left(\frac{5\pi}{3}\right)
&=\frac{5\pi}{3}+\sin\frac{10\pi}{3}\\
&=\frac{5\pi}{3}-\frac{\sqrt3}{2}.
\end{aligned}
$$ At $$x=2\pi$$ $$
f(2\pi)=2\pi.
$$ Comparing all the values, $$
\frac{4\pi}{3}+\frac{\sqrt3}{2}
$$ is the greatest, and $$
0
$$ is the smallest. Answer: Given: Two numbers have sum $$
24.
$$ Their product is to be as large as possible. Solution: Let one number be $$
x.
$$ Then the other number is $$
24-x.
$$ The product is $$
\begin{aligned}
P(x)
&=x(24-x)\\
&=24x-x^2.
\end{aligned}
$$ Differentiate. $$
P'(x)=24-2x.
$$ For maximum product, $$
24-2x=0.
$$ $$
x=12.
$$ Now find the second derivative. $$
P''(x)=-2.
$$ Since $$
P''(12)=-2<0,
$$ the product is maximum when $$
x=12.
$$ The second number is $$
24-12=12.
$$ The maximum product is $$
12\times12=144.
$$ Answer: Given: Find two positive numbers \(x\) and \(y\) such that $$
x+y=60
$$ and $$
xy^3
$$ is maximum. Solution: Since $$
x+y=60,
$$ we have $$
x=60-y.
$$ The function to be maximized is $$
\begin{aligned}
P(y)
&=(60-y)y^3\\
&=60y^3-y^4.
\end{aligned}
$$ Differentiate with respect to \(y\). $$
\begin{aligned}
P'(y)
&=180y^2-4y^3\\
&=4y^2(45-y).
\end{aligned}
$$ For critical points, $$
4y^2(45-y)=0.
$$ Since both numbers are positive, $$
y=45.
$$ Then $$
x=60-45=15.
$$ Now find the second derivative. $$
\begin{aligned}
P''(y)
&=360y-12y^2\\
&=12y(30-y).
\end{aligned}
$$ At $$
y=45,
$$ $$
P''(45)=12(45)(30-45)=-8100<0.
$$ Hence, the product is maximum. Answer: Given: Find two positive numbers \(x\) and \(y\) such that $$
x+y=35
$$ and the product $$
x^2y^5
$$ is maximum. Solution: Since $$
x+y=35,
$$ we have $$
x=35-y.
$$ Therefore, $$
P(y)=(35-y)^2y^5.
$$ Taking logarithm on both sides, $$
\log P=2\log(35-y)+5\log y.
$$ Differentiate. $$
\frac{P'}{P}
=
-\frac{2}{35-y}
+\frac{5}{y}.
$$ For maximum value, $$
-\frac2{35-y}+\frac5y=0.
$$ Multiply by \(y(35-y)\). $$
5(35-y)=2y.
$$ $$
175-5y=2y.
$$ $$
7y=175.
$$ $$
y=25.
$$ Hence, $$
x=35-25=10.
$$ To verify, differentiate again. $$
\frac{d}{dy}\left(\frac{P'}P\right)
=
-\frac2{(35-y)^2}
-\frac5{y^2}<0.
$$ Therefore, the product is maximum. Answer: Given: Find two positive numbers whose sum is $$
16
$$ and the sum of their cubes is minimum. Solution: Let the numbers be $$
x
$$ and $$
16-x.
$$ The required function is $$
\begin{aligned}
S(x)
&=x^3+(16-x)^3.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
S'(x)
&=3x^2-3(16-x)^2.
\end{aligned}
$$ For critical points, $$
3x^2-3(16-x)^2=0.
$$ $$
x^2=(16-x)^2.
$$ Since both numbers are positive, $$
x=16-x.
$$ $$
2x=16.
$$ $$
x=8.
$$ The other number is also $$
8.
$$ Now find the second derivative. $$
\begin{aligned}
S''(x)
&=6x+6(16-x)\\
&=96.
\end{aligned}
$$ Since $$
96>0,
$$ the function has a minimum value at $$
x=8.
$$ The minimum sum of cubes is $$
8^3+8^3
=512+512
=1024.
$$ Answer: Given: A square piece of tin of side $$
18\text{ cm}
$$ is to be made into an open box by cutting equal squares of side \(x\) cm from each corner and folding the flaps. Find the side of the square to be cut off so that the volume of the box is maximum. Solution: After cutting squares of side \(x\) cm, Therefore, the volume is $$
\begin{aligned}
V(x)
&=x(18-2x)^2.
\end{aligned}
$$ Expanding, $$
\begin{aligned}
V(x)
&=x(324-72x+4x^2)\\
&=324x-72x^2+4x^3.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
V'(x)
&=324-144x+12x^2.
\end{aligned}
$$ For critical points, $$
324-144x+12x^2=0.
$$ Divide by 12. $$
27-12x+x^2=0.
$$ $$
x^2-12x+27=0.
$$ $$
(x-3)(x-9)=0.
$$ Thus, $$
x=3,\quad 9.
$$ Since $$
18-2x>0,
$$ we must have $$
0 Hence, only $$
x=3
$$ is admissible. Now find the second derivative. $$
\begin{aligned}
V''(x)
&=-144+24x.
\end{aligned}
$$ At $$
x=3,
$$ $$
V''(3)=-144+72=-72<0.
$$ Therefore, the volume is maximum when $$
x=3\text{ cm}.
$$ The maximum volume is $$
\begin{aligned}
V(3)
&=3(18-6)^2\\
&=3(12)^2\\
&=432\text{ cm}^3.
\end{aligned}
$$ Answer: Given: A rectangular sheet of tin measuring $$
45\text{ cm}\times24\text{ cm}
$$ is to be made into an open box by cutting equal squares of side \(x\) cm from each corner. Find the side of the square to be cut off so that the volume of the box is maximum. Solution: After cutting squares of side \(x\) cm, Hence, the volume is $$
\begin{aligned}
V(x)
&=x(45-2x)(24-2x).
\end{aligned}
$$ Expand the expression. $$
\begin{aligned}
V(x)
&=x(1080-138x+4x^2)\\
&=1080x-138x^2+4x^3.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
V'(x)
&=1080-276x+12x^2.
\end{aligned}
$$ For critical points, $$
1080-276x+12x^2=0.
$$ Divide by 12. $$
90-23x+x^2=0.
$$ $$
x^2-23x+90=0.
$$ Factorising, $$
(x-5)(x-18)=0.
$$ Thus, $$
x=5,\quad18.
$$ Since $$
24-2x>0,
$$ we must have $$
0 Therefore, $$
x=5
$$ is the only feasible value. Now find the second derivative. $$
\begin{aligned}
V''(x)
&=-276+24x.
\end{aligned}
$$ At $$
x=5,
$$ $$
V''(5)=-276+120=-156<0.
$$ Hence, the volume is maximum. The maximum volume is $$
\begin{aligned}
V(5)
&=5(45-10)(24-10)\\
&=5(35)(14)\\
&=2450\text{ cm}^3.
\end{aligned}
$$ Answer: Given: Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. Solution: Let the radius of the circle be $$
r.
$$ Let the sides of the inscribed rectangle be $$
2x \quad \text{and} \quad 2y.
$$ Since the rectangle is inscribed in the circle, its diagonal is equal to the diameter of the circle. Hence, by the Pythagoras theorem, $$
(2x)^2+(2y)^2=(2r)^2.
$$ Therefore, $$
x^2+y^2=r^2.
$$ The area of the rectangle is $$
A=(2x)(2y)=4xy.
$$ From $$
x^2+y^2=r^2,
$$ we get $$
y=\sqrt{r^2-x^2}.
$$ Hence, $$
A(x)=4x\sqrt{r^2-x^2}.
$$ Differentiate using the product rule. $$
\begin{aligned}
A'(x)
&=4\left[\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}\right]\\
&=\frac{4(r^2-2x^2)}{\sqrt{r^2-x^2}}.
\end{aligned}
$$ For maximum area, $$
A'(x)=0.
$$ Since the denominator is positive, $$
r^2-2x^2=0.
$$ Thus, $$
x=\frac{r}{\sqrt2}.
$$ Using $$
x^2+y^2=r^2,
$$ we obtain $$
y=\frac{r}{\sqrt2}.
$$ Hence, $$
x=y.
$$ Therefore, both sides of the rectangle are equal. Thus, the rectangle is a square. Now find the second derivative. At $$
x=\frac{r}{\sqrt2},
$$ the second derivative is negative. Hence, the area is maximum. Answer: The rectangle of maximum area inscribed in a given circle is a $$
\boxed{\text{Square}.}
$$ Given: Show that the right circular cylinder of given surface area and maximum volume has its height equal to the diameter of the base. Solution: Let The total surface area is constant. $$
2\pi r(r+h)=S.
$$ Hence, $$
h=\frac{S}{2\pi r}-r.
$$ The volume of the cylinder is $$
V=\pi r^2h.
$$ Substituting the value of \(h\), $$
\begin{aligned}
V
&=\pi r^2\left(\frac{S}{2\pi r}-r\right)\\
&=\frac{Sr}{2}-\pi r^3.
\end{aligned}
$$ Differentiate with respect to \(r\). $$
\begin{aligned}
V'(r)
&=\frac{S}{2}-3\pi r^2.
\end{aligned}
$$ For maximum volume, $$
\frac{S}{2}-3\pi r^2=0.
$$ Therefore, $$
S=6\pi r^2.
$$ Using $$
S=2\pi r(r+h),
$$ we get $$
2\pi r(r+h)=6\pi r^2.
$$ Dividing both sides by \(2\pi r\), $$
r+h=3r.
$$ Hence, $$
h=2r.
$$ Also, $$
V''(r)=-6\pi r<0.
$$ Therefore, the volume is maximum. Answer: $$
\boxed{h=2r}
$$ Thus, the height is equal to the diameter of the base. Given: A closed cylindrical can has volume $$
100\text{ cm}^3.
$$ Find its dimensions so that the total surface area is minimum. Solution: Let The volume is $$
\pi r^2h=100.
$$ Hence, $$
h=\frac{100}{\pi r^2}.
$$ The total surface area of the closed cylinder is $$
S=2\pi rh+2\pi r^2.
$$ Substitute the value of \(h\). $$
\begin{aligned}
S(r)
&=2\pi r\left(\frac{100}{\pi r^2}\right)+2\pi r^2\\
&=\frac{200}{r}+2\pi r^2.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
S'(r)
&=-\frac{200}{r^2}+4\pi r.
\end{aligned}
$$ For minimum surface area, $$
-\frac{200}{r^2}+4\pi r=0.
$$ $$
4\pi r^3=200.
$$ $$
r^3=\frac{50}{\pi}.
$$ Therefore, $$
\boxed{r=\sqrt[3]{\frac{50}{\pi}}\text{ cm}.}
$$ Now find the height. $$
\begin{aligned}
h
&=\frac{100}{\pi r^2}\\
&=\frac{100}{\pi\left(\sqrt[3]{\frac{50}{\pi}}\right)^2}\\
&=2\sqrt[3]{\frac{50}{\pi}}\text{ cm}.
\end{aligned}
$$ Hence, $$
h=2r.
$$ Also, $$
S''(r)=\frac{400}{r^3}+4\pi>0,
$$ which confirms that the surface area is minimum. Answer: Given: A wire of length $$
28\text{ m}
$$ is cut into two pieces. One piece is made into a square and the other into a circle. Find the lengths of the two pieces so that the combined area is minimum. Solution: Let $$
x\text{ m}
$$ be the length of the wire used for making the square. Then the length used for making the circle is $$
28-x\text{ m}.
$$ Area of the square Perimeter of the square is $$
x.
$$ Hence, side of the square is $$
\frac{x}{4}.
$$ Therefore, its area is $$
\begin{aligned}
A_1
&=\left(\frac{x}{4}\right)^2\\
&=\frac{x^2}{16}.
\end{aligned}
$$ Area of the circle Circumference of the circle is $$
28-x.
$$ Therefore, $$
2\pi r=28-x.
$$ Hence, $$
r=\frac{28-x}{2\pi}.
$$ The area of the circle is $$
\begin{aligned}
A_2
&=\pi r^2\\
&=\pi\left(\frac{28-x}{2\pi}\right)^2\\
&=\frac{(28-x)^2}{4\pi}.
\end{aligned}
$$ Total area $$
\begin{aligned}
A(x)
&=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
A'(x)
&=\frac{x}{8}-\frac{28-x}{2\pi}.
\end{aligned}
$$ For minimum area, $$
A'(x)=0.
$$ Therefore, $$
\frac{x}{8}=\frac{28-x}{2\pi}.
$$ Cross-multiplying, $$
2\pi x=8(28-x).
$$ $$
2\pi x=224-8x.
$$ $$
(2\pi+8)x=224.
$$ Hence, $$
\boxed{
x=\frac{112}{4+\pi}\text{ m}.
}
$$ The remaining wire is $$
\begin{aligned}
28-x
&=28-\frac{112}{4+\pi}\\
&=\frac{28\pi}{4+\pi}.
\end{aligned}
$$ Now differentiate again. $$
A''(x)=\frac18+\frac1{2\pi}.
$$ Since $$
A''(x)>0,
$$ the total area is minimum. Answer: Given: Prove that the volume of the largest cone that can be inscribed in a sphere of radius $$
R
$$ is $$
\frac{8}{27}
$$ of the volume of the sphere. Solution: Let the sphere have radius $$
R.
$$ Let the height of the cone be $$
h.
$$ Let the radius of its base be $$
r.
$$ From the geometry of the sphere, $$
r^2=R^2-(R-h)^2.
$$ Simplifying, $$
\begin{aligned}
r^2
&=R^2-\left(R^2-2Rh+h^2\right)\\
&=2Rh-h^2.
\end{aligned}
$$ The volume of the cone is $$
\begin{aligned}
V
&=\frac13\pi r^2h\\
&=\frac13\pi h(2Rh-h^2)\\
&=\frac13\pi(2Rh^2-h^3).
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
V'(h)
&=\frac13\pi(4Rh-3h^2).
\end{aligned}
$$ For maximum volume, $$
4Rh-3h^2=0.
$$ $$
h(4R-3h)=0.
$$ Ignoring the trivial solution, $$
\boxed{
h=\frac{4R}{3}.
}
$$ Now find the second derivative. $$
\begin{aligned}
V''(h)
&=\frac13\pi(4R-6h).
\end{aligned}
$$ At $$
h=\frac{4R}{3},
$$ $$
\begin{aligned}
V''\left(\frac{4R}{3}\right)
&=\frac13\pi\left(4R-8R\right)\\
&=-\frac{4\pi R}{3}<0.
\end{aligned}
$$ Hence, the volume is maximum. Now find the corresponding base radius. $$
\begin{aligned}
r^2
&=2R\left(\frac{4R}{3}\right)-\left(\frac{4R}{3}\right)^2\\
&=\frac{8R^2}{3}-\frac{16R^2}{9}\\
&=\frac{8R^2}{9}.
\end{aligned}
$$ Therefore, $$
r=\frac{2\sqrt2R}{3}.
$$ The maximum volume of the cone is $$
\begin{aligned}
V_{\max}
&=\frac13\pi\left(\frac{8R^2}{9}\right)\left(\frac{4R}{3}\right)\\
&=\frac{32\pi R^3}{81}.
\end{aligned}
$$ The volume of the sphere is $$
V_s=\frac43\pi R^3.
$$ Hence, $$
\begin{aligned}
\frac{V_{\max}}{V_s}
&=\frac{\frac{32\pi R^3}{81}}
{\frac43\pi R^3}\\
&=\frac{32}{81}\times\frac34\\
&=\frac{8}{27}.
\end{aligned}
$$ Therefore, $$
\boxed{
V_{\max}
=\frac{8}{27}\times
(\text{Volume of the sphere}).
}
$$ Given: Show that the right circular cone of least curved surface area and given volume has an altitude equal to $$
\sqrt2
$$ times the radius of its base. Solution: Let The volume of the cone is constant. $$
\frac13\pi r^2h=V.
$$ Hence, $$
h=\frac{3V}{\pi r^2}.
$$ The curved surface area of the cone is $$
S=\pi rl.
$$ Substituting $$
l=\sqrt{r^2+h^2},
$$ we get $$
S=\pi r\sqrt{r^2+h^2}.
$$ To simplify differentiation, minimize $$
S^2.
$$ Since $$
S^2=\pi^2r^2(r^2+h^2),
$$ substituting $$
h=\frac{3V}{\pi r^2},
$$ gives $$
\begin{aligned}
S^2
&=\pi^2r^4+\frac{9V^2}{r^2}.
\end{aligned}
$$ Differentiate with respect to \(r\). $$
\begin{aligned}
\frac{d(S^2)}{dr}
&=4\pi^2r^3-\frac{18V^2}{r^3}.
\end{aligned}
$$ For minimum curved surface area, $$
4\pi^2r^3-\frac{18V^2}{r^3}=0.
$$ Therefore, $$
4\pi^2r^6=18V^2.
$$ Using $$
V=\frac13\pi r^2h,
$$ we obtain $$
V^2=\frac19\pi^2r^4h^2.
$$ Substitute this value. $$
\begin{aligned}
4\pi^2r^6
&=18\left(\frac19\pi^2r^4h^2\right)\\
&=2\pi^2r^4h^2.
\end{aligned}
$$ Dividing both sides by $$
2\pi^2r^4,
$$ we get $$
2r^2=h^2.
$$ Hence, $$
\boxed{
h=\sqrt2\,r.
}
$$ Also, $$
\frac{d^2(S^2)}{dr^2}
=12\pi^2r^2+\frac{54V^2}{r^4}>0,
$$ therefore the curved surface area is minimum. Hence proved. Given: Show that the semi-vertical angle of the cone of maximum volume and given slant height is $$
\tan^{-1}\sqrt2.
$$ Solution: Let Since the slant height is fixed, $$
r^2+h^2=l^2.
$$ Hence, $$
h=\sqrt{l^2-r^2}.
$$ The volume of the cone is $$
V=\frac13\pi r^2h.
$$ Substituting \(h\), $$
V=\frac13\pi r^2\sqrt{l^2-r^2}.
$$ To simplify differentiation, maximize $$
V^2.
$$ $$
\begin{aligned}
V^2
&=\frac{\pi^2}{9}r^4(l^2-r^2).
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
\frac{d(V^2)}{dr}
&=\frac{\pi^2}{9}
\left(4r^3(l^2-r^2)-2r^5\right).
\end{aligned}
$$ For maximum volume, $$
4r^3(l^2-r^2)-2r^5=0.
$$ Factorising, $$
2r^3(2l^2-3r^2)=0.
$$ Hence, $$
3r^2=2l^2.
$$ Now, $$
h^2=l^2-r^2.
$$ Therefore, $$
h^2=l^2-\frac23l^2=\frac13l^2.
$$ Thus, $$
\frac{r^2}{h^2}
=\frac{\frac23l^2}{\frac13l^2}
=2.
$$ Hence, $$
\frac{r}{h}=\sqrt2.
$$ Since $$
\tan\theta=\frac{r}{h},
$$ we obtain $$
\boxed{
\theta=\tan^{-1}\sqrt2.
}
$$ Hence proved. Given: Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is $$
\sin^{-1}\left(\frac13\right).
$$ Solution: Let The total surface area is constant. $$
S=\pi r(r+l).
$$ Hence, $$
l=\frac{S}{\pi r}-r.
$$ Since $$
l^2=r^2+h^2,
$$ the volume is $$
V=\frac13\pi r^2h.
$$ Using the condition of maximum volume for a cone of fixed surface area (obtained by differentiation), we get $$
l=3r.
$$ Now, $$
\sin\theta=\frac{r}{l}.
$$ Substituting $$
l=3r,
$$ we have $$
\sin\theta=\frac13.
$$ Therefore, $$
\boxed{
\theta=\sin^{-1}\left(\frac13\right).
}
$$ Hence proved. Given: Find the point on the curve $$
x^2=2y
$$ which is nearest to the point $$
(0,5).
$$ Solution: The equation of the curve is $$
y=\frac{x^2}{2}.
$$ Let $$
P\left(x,\frac{x^2}{2}\right)
$$ be any point on the curve. The square of the distance from $$
(0,5)
$$ to \(P\) is $$
\begin{aligned}
D^2
&=x^2+\left(\frac{x^2}{2}-5\right)^2.
\end{aligned}
$$ Expanding, $$
\begin{aligned}
D^2
&=x^2+\frac{x^4}{4}-5x^2+25\\
&=\frac{x^4}{4}-4x^2+25.
\end{aligned}
$$ Differentiate. $$
\begin{aligned}
\frac{d(D^2)}{dx}
&=x^3-8x\\
&=x(x^2-8).
\end{aligned}
$$ For critical points, $$
x=0,\qquad x=\pm2\sqrt2.
$$ Now evaluate \(D^2\). At $$x=0$$ $$
D^2=25.
$$ At $$x=\pm2\sqrt2$$ $$
y=\frac{(2\sqrt2)^2}{2}=4.
$$ Therefore, $$
D^2=8+(4-5)^2=9.
$$ Since $$
9<25,
$$ the nearest point is $$
(2\sqrt2,4)
$$ or $$
(-2\sqrt2,4).
$$ Among the given options, $$
\boxed{(2\sqrt2,4)}
$$ is the correct answer. Answer: (A) $$\mathbf{(2\sqrt2,4)}$$ Given: Find the minimum value of $$
f(x)=\frac{1-x+x^2}{1+x+x^2}.
$$ Solution: Differentiate using the quotient rule. $$
\begin{aligned}
f'(x)
&=\frac{(2x-1)(1+x+x^2)-(1-x+x^2)(2x+1)}
{(1+x+x^2)^2}.
\end{aligned}
$$ Simplifying, $$
f'(x)=\frac{2(x^2-1)}
{(1+x+x^2)^2}.
$$ For critical points, $$
x^2-1=0.
$$ Hence, $$
x=\pm1.
$$ Now evaluate the function. At $$x=1$$ $$
\begin{aligned}
f(1)
&=\frac{1-1+1}{1+1+1}\\
&=\frac13.
\end{aligned}
$$ At $$x=-1$$ $$
\begin{aligned}
f(-1)
&=\frac{1+1+1}{1-1+1}\\
&=3.
\end{aligned}
$$ Hence, the minimum value is $$
\boxed{\frac13.}
$$ Answer: (D) $$\mathbf{\dfrac13}$$ Given: Find the maximum value of $$
\left[x(x-1)+1\right]^{\frac13},
\qquad 0\le x\le1.
$$ Solution: Since the cube root function is increasing, it is sufficient to maximize $$
g(x)=x(x-1)+1.
$$ Simplify. $$
g(x)=x^2-x+1.
$$ Differentiate. $$
g'(x)=2x-1.
$$ For critical point, $$
2x-1=0.
$$ $$
x=\frac12.
$$ Now evaluate the function at the critical point and the end points. At $$x=0$$ $$
g(0)=1.
$$ At $$x=1$$ $$
g(1)=1.
$$ At $$x=\dfrac12$$ $$
\begin{aligned}
g\left(\frac12\right)
&=\frac14-\frac12+1\\
&=\frac34.
\end{aligned}
$$ The maximum value of \(g(x)\) is $$
1.
$$ Therefore, the maximum value of the given function is $$
\sqrt[3]{1}=1.
$$ Answer: (C) $$\mathbf{1}$$
Solutions
Question 3 (i)
Question 3 (ii)
Question 3 (iii)
Question 3 (iv)
Question 3 (v)
Question 3 (vi)
Question 3 (vii)
Question 3 (viii)
Solutions
Question 6
Question 7
Question 8
Question 9
Question 10
Solutions
Question 11
Question 12
Question 13
Question 14
Question 15
Question 16
Question 17
Question 18
Question 19
Question 20
Question 21
Solutions
Question 22
Question 23
Question 24
Question 25
Question 26
Solutions
Question 27
Question 28
Question 29