12th NCERT Application of Derivatives Miscellaneous Exercise
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Miscellaneous 6

Question (1)

Using differentials, find the approximate value of each of the following:
$\text{(a)} \quad {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}}$ $\text{(b)} \quad {\left( {33} \right)^{\frac{1}{5}}}$

Solution

$\text{(a)} \quad {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}}$ $f\left( {x \pm \delta x} \right) = f\left( x \right) \pm \delta xf'\left( x \right)$ $f\left( x \right) = {x^{\frac{1}{4}}}$ $f'\left( x \right) = \frac{1}{4}{x^{\frac{{ - 3}}{4}}} = \frac{1}{{4{x^{\frac{3}{4}}}}}$ $f\left( {\frac{{16}}{{81}}} \right) = {\left( {\frac{{16}}{{81}}} \right)^{\frac{1}{4}}} = \frac{2}{3}$ $f\left( {\frac{{16}}{{81}}} \right) = \frac{1}{{4\cdot8}} \times 27 = \frac{{27}}{{32}}$ $f\left( {\frac{{17}}{{81}}} \right) = f\left( {\frac{{16}}{{81}} + \frac{1}{{81}}} \right)$ $f\left( {\frac{{17}}{{81}}} \right) = f\left( {\frac{{16}}{{81}}} \right) + \frac{1}{{81}}f'\left( {\frac{{16}}{{81}}} \right)$ $f\left( {\frac{{17}}{{81}}} \right) = \frac{2}{3} + \frac{1}{{\require{cancel} \cancel{81}_3}} \times \frac{{\cancel{27}}}{{32}}$ $f\left( {\frac{{17}}{{81}}} \right) = \frac{2}{3} + \frac{1}{{96}}$ $f\left( {\frac{{17}}{{81}}} \right) = 0.667 + 0.010 = 0.677$
$\text{(b)} \quad {\left( {33} \right)^{\frac{1}{5}}}$ Let f(x) = x1/5
$\text{Let} \quad f\left( x \right) = {x^{\frac{1}{5}}}$ $f\left( {32} \right) = {\left( {32} \right)^{\frac{1}{5}}} = {\left( {{2^5}} \right)^{\frac{1}{5}}} = 2$ $f'\left( x \right) = \frac{1}{5}{x^{\frac{{ - 4}}{5}}} = \frac{1}{{5{x^{\frac{4}{5}}}}}$ $f'\left( {32} \right) = \frac{1}{{5{{\left( {32} \right)}^{\frac{4}{5}}}}} = \frac{1}{{5{{\left( {{2^5}} \right)}^{\frac{4}{5}}}}}$ $f'\left( {32} \right) = \frac{1}{{5\left( {16} \right)}} = \frac{1}{{80}} = 0.0125$ $f\left( {33} \right) = f\left( {32 + 1} \right)$ $f\left( {33} \right) = f\left( {32} \right) + 1f'\left( {32} \right)$ $f\left( {33} \right) = 2 + 0.0125 = 2.0125$

Question (2)

Show that the function given by $f\left( x \right) = \frac{{\log x}}{x}$ has maximum at x =e

Solution

$f\left( x \right) = \frac{{\log x}}{x}$ $f'\left( x \right) = \frac{{x\frac{d}{{dx}}\log x - \log x\frac{d}{{dx}}x}}{{{x^2}}}$ $f'\left( x \right) = \frac{{x \cdot \frac{1}{x} - \log x\left( 1 \right)}}{{{x^2}}}$ $f'\left( x \right) = \frac{{1 - \log x}}{{{x^2}}}$ $f"\left( x \right) = \frac{{{x^2}\left( {\frac{{ - 1}}{x}} \right) - \left( {1 - \log x} \right)2x}}{{{x^4}}}$ $f"\left( x \right) = \frac{{ - x - 2x + 2x\log x}}{{{x^4}}}$ $f"\left( x \right) = \frac{{x\left( { - 3 + 2\log x} \right)}}{{{x^4}}}$ $f"\left( x \right) = \frac{{ - 3 + 2\log x}}{{{x^3}}}$ Now f'(x) = 0 $\Rightarrow \frac{{1 - \log x}}{{{x^2}}} = 0$ $1 - \log x = 0$ $\log _e^x = 1 = \log _e^e \quad \text{as} \quad \log _e^e = 1$ ∴ x = e
$f"\left( e \right) = \frac{{ - 3 + 2\log e}}{{{e^3}}} = \frac{{ - 3 + 2}}{{{e^3}}}$ $f"\left( e \right) = \frac{{ - 1}}{{{e^3}}} < 0$ ⇒ f is maximum at x = e

Question (3)

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate 3cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

Solution

Let two equal sides of isosceles triangle be x
$\frac{{dx}}{{dt}} = - 3\frac{{cm}}{{\sec }}$ In Δ ABC, AM ⊥BC
M is mid point of BC
BM = ½BC = b/2
Let length of altitude be 'h' ${h^2} + {\left( {\frac{b}{2}} \right)^2} = {x^2}$ ${h^2} = {x^2} - {\left( {\frac{b}{2}} \right)^2} = \frac{{4{x^2} - {b^2}}}{4}$ Area of Δ = ½ b h
$A = \frac{b}{2}\sqrt {\frac{{4{x^2} - {b^2}}}{4}}$ $A = \frac{b}{4}\sqrt {4{x^2} - {b^2}}$ Differentiate with respect to "t" $\frac{{dA}}{{dt}} = \frac{b}{\cancel{4}} \cdot \frac{1}{{\cancel{2}\sqrt {4{x^2} - {b^2}} }} \cdot \cancel{8}x\frac{{dx}}{{dt}}$ $\frac{{dA}}{{dt}} = \frac{{bx\frac{{dx}}{{dt}}}}{{\sqrt {4{x^2} - {b^2}} }}\quad \text{at} \quad x = b$ $\frac{{dA}}{{dt}} = \frac{{b^2\left( { - 3} \right)}}{{\sqrt {4{x^2} - {b^2}} }} = \frac{{ - 3b^2}}{{\sqrt 3 b}} = - \sqrt 3b$ Area decreases at the rate √3 cm2/ sec

Question (4)

Find the equation of the normal to curve y2 = 4x at the point (1, 2)

Solution

normal y2 = 4x     (1, 2) y2 = 4x
Differentiate with respect to "x" $2y\frac{{dy}}{{dx}} = 4$ $\frac{{dy}}{{dx}} = \frac{2}{y}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = \frac{2}{2} = 1$ $\text{ slope of normal} = \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {1,2} \right)}}}} = \frac{{ - 1}}{1} = - 1$ Equation of normal with slope -1 and passing through (1, 2) is
y -2 = -1( x - 1)
y - 2 = -x + 1
x + y - 3 = 0

Question (5)

Show that the normal at any point θ to the curve
x = acosθ + a θsinθ and y - asinθ - aθcosθ
is at a constant distance from the origin

Solution

x = acosθ + aθsinθ
Differentiate with respect to "θ" $\frac{{dx}}{{d\theta }} = - a\sin \theta + a\theta \cos \theta + a\sin \theta$ $\frac{{dx}}{{d\theta }} = a\theta \cos \theta$ y = asinθ - aθcosθ
Differentiate with respect to "θ" $\frac{{dy}}{{d\theta }} = a\cos \theta - a\left[ {\theta \left( { - \sin \theta } \right) + \cos \theta \left( 1 \right)} \right]$ $\frac{{dy}}{{d\theta }} = a\cos \theta + a\theta \sin \theta - a\cos \theta$ $\frac{{dy}}{{d\theta }} = a\theta \sin \theta$ $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$ $\frac{{dy}}{{dx}} = \frac{{a\theta \sin \theta }}{{a\theta \cos \theta }} = \tan \theta$ ${\left( {\frac{{dy}}{{dx}}} \right)_\theta } = \tan \theta$ $\text{slope of normal}= \frac{{ - 1}}{{\left( {\frac{{dy}}{{dx}}} \right)}} = \frac{{ - 1}}{{\tan \theta }}$ slope of normal = -cotθ Equation of normal is
$y - {y_1} = m\left( {x - {x_1}} \right)$ $y - \left( {a\sin \theta - a\theta \cos \theta } \right) = \frac{{ - \cos \theta }}{{\sin \theta }}\left( {x - a\cos \theta - a\theta \sin \theta } \right)$ $\sin \theta y - a{\sin ^2}\theta + \cancel{ a\theta \cos \theta \sin \theta} = - x\cos \theta + a{\cos ^2}\theta + \cancel{a\theta \sin \theta \cos \theta}$ ∴ cosθx + sinθy = a cos2θ + asin2θ
cosθx + sinθy = a (cos2θ + sin2θ)
cosθ x + sinθ y = a
The perpendicular distance from the origin
$p = \frac{{\left| c \right|}}{{\sqrt {{a^2} + {b^2}} }}$ $p = \frac{{\left| a \right|}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }}$ p = |a| = constant
so the distance of normal from origin is constant for any θ point

Question (6)

Find the intervals in which the function f given by $f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}$ is (i) increasing (ii) decreasing

Solution

$f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}$ $f\left( x \right) = \frac{{4\sin x}}{{2 + \cos x}} - \frac{{x\left( {2 + \cos x} \right)}}{{2 + \cos x}}$ $f\left( x \right) = \frac{{4\sin x}}{{2 + \cos x}} - x$ $f'\left( x \right) = \frac{{\left( {2 + \cos x} \right)\left( {4\cos x} \right) - 4\sin x\left( { - \sin x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}} - 1$ $f'\left( x \right) = \frac{{8\cos x + 4{{\cos }^2}x + 4{{\sin }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}} - 1$ $f'\left( x \right) = \frac{{8\cos x + 4 - {{\left( {2 + \cos x} \right)}^2}}}{{{{\left( {2 + \cos x} \right)}^2}}}$ $f'\left( x \right) = \frac{{8\cos x + \cancel{4} - \cancel{4} - 4\cos x - {{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}}$ $f'\left( x \right) = \frac{{4\cos x - {{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}}$ $f'\left( x \right) = \frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}}$ ∀ x, ( 4 - cosx) > 0, (2+cosx)2 > 0 The value of f'(x) depends upon the sign of cosx
(i) for increasing function
f'(x) > 0
⇒ cos x > 0; cos is +ve in 1st and 4th quadrant
f(x) is increasing 0 < x < π/2 and 3π/2 < x < 2π
(ii) f(x) is decreasing in 2nd and 3rd quadrant
i.e π/2 < x < 3π/2
x ∈ ( π/2, 3π/2)

Question (7)

Find the intervals in which the function f given by $f\left( x \right) = {x^3} + \frac{1}{{{x^3}}},x \ne 0$ is (i) increasing (ii) decreasing

Solution

$f\left( x \right) = {x^3} + \frac{1}{{{x^3}}};x \ne 0$ $f'\left( x \right) = 3{x^2} + \frac{{ - 3}}{{{x^4}}}$ $f'\left( x \right)= \frac{{3{x^6} - 3}}{{{x^4}}}$ $f'\left( x \right) = \frac{{3\left( {{x^6} - 1} \right)}}{{{x^4}}}$ $f\left( x \right) = \frac{{3\left( {{x^3} + 1} \right)\left( {{x^3} - 1} \right)}}{{{x^4}}}$ So at x = 1 and x = -1, f'(x) = 0
Intervals breaks as (-∞ , -1) (-1, 1) (1, ∞)
 intervals x3+1 x3-1 f'(x) (-∞ -1) - ve - ve +ve ↑ (-1, 1) + ve - ve - ve ↓ (1, ∞ + ve + ve + ve ↑
f(x) is increasing in the interval ( -∞, -1) and ( 1, -∞)
i.e. x < -1 or x > 1
f(x) is decreasing in the interval (-1, 1)

Question (8)

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with its vertex at one end of the major axis

Solution

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \text{ is equation of ellipse}$ Δ APQ is isosceles triangle whose vertex A is at (a, 0)
Let OM = x, OA = A
∴ AM = a - x
as OM = x therefore x-coordinate P and Q is x $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ $\frac{{{y^2}}}{{{b^2}}} = 1 - \frac{{{x^2}}}{{{a^2}}}$ $\frac{{{y^2}}}{{{b^2}}} = \frac{{{a^2} - {x^2}}}{{{a^2}}}$ $y = \pm \frac{b}{a}\sqrt {{a^2} - {x^2}}$ $\text{coordinate of P}\left( {x,\frac{b}{a}\sqrt {{a^2} - {x^2}} } \right)$ $\text{coordinate of Q}\left( {x,\frac{{ - b}}{a}\sqrt {{a^2} - {x^2}} } \right)$ $PQ = \left| {{y_1} - {y_2}} \right| = \frac{{2b}}{a}\sqrt {{a^2} - {x^2}}$ Area of Δ APQ= A = ½ PQ × AM $A = \frac{1}{2}\left( {\frac{{2b}}{a}\sqrt {{a^2} - {x^2}} } \right)\left( {a - x} \right)$ $A = \frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} } \right)\left( {a - x} \right)$ As the area is maximum square of area is also maximum
${A^2} = \frac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right){\left( {a - x} \right)^2}$ $f\left( x \right) = \frac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right){\left( {a - x} \right)^2}$ Now $f'\left( x \right) = \frac{{{b^2}}}{{{a^2}}}\left[ {\left( {{a^2} - {x^2}} \right)2\left( {a - x} \right)\left( { - 1} \right) + {{\left( {a - x} \right)}^2}\left( { - 2x} \right)} \right]$ $f'\left( x \right) = \frac{{{b^2}}}{{{a^2}}} \times \left[ { - 2\left( {a - x} \right)} \right]\left[ {{a^2} - {x^2} + x\left( {a - x} \right)} \right]$ $f'\left( x \right) = \frac{{ - 2{b^2}}}{{{a^2}}}\left( {a - x} \right)\left[ {{a^2} - 2{x^2} + ax} \right]$ $f"\left( x \right) = \frac{{ - 2{b^2}}}{{{a^2}}}\left[ {\left( {a - x} \right)\left( { - 4x + a} \right) + \left( {{a^2} - 2{x^2} + ax} \right)\left( { - 1} \right)} \right]$ $f"\left( x \right) = \frac{{ - 2{b^2}}}{{{a^2}}}\left[ { - 4ax + \cancel{{a^2}} + 4{x^2} - ax - \cancel{{a^2}} + 2{x^2} - ax} \right]$ $f"\left( x \right) = \frac{{ - 2{b^2}}}{{{a^2}}}\left[ {6{x^2} - 6ax} \right]$ $f"\left( x \right) = \frac{{ - 2{b^2}}}{{{a^2}}}6x\left( {x - a} \right)$ $f"\left( x \right) = \frac{{ - 12{b^2}x}}{{{a^2}}}\left( {x - a} \right)$ Now f'(x) = 0 $\frac{{ - 2{b^2}}}{{{a^2}}}\left( {a - x} \right)\left( {{a^2} - 2{x^2} + 4x} \right) = 0$ a - x = 0 OR a2 - 2x2+ax = 0
a =x OR 2x2 - ax -a2 = 0
x = a OR 2x2 - 2ax + ax - a2=0
2x(x-a) + a(x-a) = 0
(x-a) (2ax+a)= 0
x = a or x = -a/2
f"(a) = 0
⇒ No maxima or minima at x = a
$f"\left( {\frac{{ - a}}{2}} \right) = \frac{{-12{b^2}}}{{{a^2}}}x\left( {x - a} \right)$ $f"\left( {\frac{{ - a}}{2}} \right) = \frac{{ - 12{b^2}}}{{{a^2}}}\left( {\frac{{ - a}}{2}} \right)\left( {\frac{{ - a}}{2} - a} \right)$ $f"\left( {\frac{{ - a}}{2}} \right) = \frac{{ - \cancel{12}^3{b^2}}}{{\cancel{{a^2}}}}\left( {\frac{{ - \cancel{a}}}{\cancel{2}}} \right)\left( {\frac{{ - 3\cancel{a}}}{\cancel{2}}} \right)$ $f\left( {\frac{{ - a}}{2}} \right) = - 9{b^2} < 0$ $\text{since} \quad {b^2} > 0$ ⇒ f is maximum at x = -a/2 $\text{maximum area} = \frac{b}{a}\left( {\sqrt {{a^2} - {x^2}} } \right)\left( {a - x} \right)$ $\text{maximum area}= \frac{b}{a}\left( {\sqrt {{a^2} - {{\frac{a}{4}}^2}} } \right)\left( {a + \frac{a}{2}} \right)$ $\text{maximum area}= \frac{b}{a}\left( {\sqrt {{{\frac{{3a}}{4}}^2}} } \right)\left( {\frac{{3a}}{2}} \right)$ $\text{maximum area}= \frac{{3\sqrt 3 }}{4}ab$

Question (9)

A tank with rectangular base and rectangular side, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tanks costs Rs 70 per sq. meters for the bases and Rs 45 per square meter for sides. What is the cost of least expensive tank?

Solution

Let the length of box be "l", breadth be "b" given height h = 2m
Volume of the box = lbh
8 lb(2)
⇒ lb = 4
⇒ b = 4/l
cost of box = cost of base + cost of all sides of the box
= Rate × base + rate × area of 4 sides
= 70(lb) + 45[ 2h(l+b)
=70 (4) + 45[2(2) (l+b)] = 280 + 180( l +b) $\text{cost of box} = 280 + 180\left( {l + \frac{4}{l}} \right)$ $\text{cost of box} = 280 + 180l + \frac{{720}}{l}$ so function is $f\left( l \right) = 280 + 180l + \frac{{720}}{l}$ $f'\left( l \right) = 180 - \frac{{720}}{{{l^2}}}$ $f"\left( l \right) = - 720 \times \frac{{ - 2}}{{{l^3}}}$ $f"\left( l \right) = \frac{{1440}}{{{l^3}}}$ Now f'(l) = 0 $180 - \frac{{720}}{{{l^2}}} = 0$ $180 = \frac{{720}}{{{l^2}}}$ ${l^2} = \frac{{720}}{{180}} = 4$ l = 2 $f"\left( l \right) = \frac{{1440}}{8} > 0$ ⇒ f(l) is minimum at l = 2
$\text{cost} = 280 + 180l + \frac{{720}}{l}$ $\text{cost}= 280 + 180 \times 2 + \frac{{720}}{2}$ cost = 280 +360+360
cost = 1000
minimum cost is Rs 1000/-

Question (10)

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle

Solution

Let the radius of circle be 'r' and side of square be x
∴ Perimeter of circle + Perimeter of square = k
2πr + 4x = k
4x = k - 2π r
$x = \frac{{k - 2\pi r}}{4}$ f: Area of circle + Area of square
f = π r2 + x2 $f = \pi {r^2} + {\left( {\frac{{k - 2\pi r}}{4}} \right)^2}$ $f = \pi {r^2} + \frac{1}{{16}}{\left( {k - 2\pi r} \right)^2}$ $f'\left( r \right) = 2\pi r + \frac{1}{{16}}2{\left( {k - 2\pi r} \right)}\left( { - 2\pi } \right)$ $f'\left( r \right) = 2\pi r - \frac{\pi }{4}\left( {k - 2\pi r} \right)$ $f"\left( r \right) = 2\pi - \frac{\pi }{4}\left( { - 2\pi } \right)$ $f"\left( r \right) = 2\pi + \frac{{{\pi ^2}}}{2} > 0$ Now f'(r) = 0 $2\pi r - \frac{\pi }{4}\left( {k - 2\pi r} \right) = 0$ $2\pi r = \frac{\pi }{4}\left( {k - 2\pi r} \right)$ $8\cancel{\pi} r = \cancel{\pi} \left( {k - 2\pi r} \right)$ 8r = 2πr + 4 x - 2πr
8r = 4x
x = 2r
r = x/2
f"(x) > 0
f is maximum at r = x/2
x = 2r
sides of square is double, radius of circle

Question (11)

A window is in the form of rectangle surmounted by a semicircular opening. the total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening

Solution

Let radius of semicircle be ' r'
∴ length of window = 2r
Perimeter of window = 10
⇒ l + 2b + length of semicircle = 10
2r +2x + πr = 10
$x = \frac{{10 - 2r - \pi r}}{2}$ To get maximum height, Area of opening must be maximum
f = area of window
f = area of rectangle + Area of semicircle
$f = lb + \frac{1}{2}\pi {r^2}$ $f = 2rx + \frac{1}{2}\pi {r^2}$ $f = \cancel{2}r\left( {\frac{{10 - 2r - \pi r}}{\cancel{2}}} \right) + \frac{1}{2}\pi {r^2}$ $f = 10r - 2{r^2} - \pi {r^2} + \frac{{\pi {r^2}}}{2}$ $f\left( r \right) = 10r - 2{r^2} - \frac{{\pi {r^2}}}{2}$ $f'\left( r \right) = 10 - 4r - \frac{\pi }{\cancel{2}}\cancel{2}r$ f"(r) = -4 - π
f"(r) = 0
10 - 4r - πr = 0
10 = 4r + π r
$r = \frac{{10}}{{\pi + 4}}$ f"(r) = -4-π < 0
⇒ f is maximum at $\Rightarrow \text{f is maximum at }\quad r = \frac{{10}}{{\pi + 4}}$ $\text{ length l} \quad = 2r = \frac{{20}}{{\pi + 4}}$ $\text{ breadth x} = \frac{{10 - 2r - \pi r}}{2}$ $x = \frac{1}{2}\left( {10 - \frac{{20}}{{\pi + 4}} - \frac{{10\pi }}{{\pi + 4}}} \right)$ $x = \frac{1}{2}\left( {\frac{{\cancel{10\pi} + 40 - 20 - \cancel{10\pi} }}{{\pi + 4}}} \right)$ $x = \frac{{10}}{{\pi + 4}}$

Question (12)

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.
Show that the minimum length of hypotenuse is ${\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}$

Solution

Let P be the point of hypotenuse AC, such that its distance from AB is 'a' and from BC be 'b' i.e. PN = a, PM = b
Let ∠C = θ
$\overline {PN} ||\overline {BC} \ \text {as ∠N = ∠B = 90)}$ ∴ ∠P = θ
In ΔPMC $\sin \theta = \frac{{PM}}{{PC}}$ $\sin \theta = \frac{b}{{PC}}$ $PC = \frac{b}{{\sin \theta }} = b cosec\theta$ In ΔAPN $\cos \theta = \frac{{PN}}{{AP}}$ $\cos \theta = \frac{a}{{AP}}$ $AP = \frac{a}{{\cos \theta }} = a\sec \theta$ Length of hypotenuse AC = AP + PC
AC = asecθ + b cosecθ It is a function of θ
f(θ) = a secθ + b cosecθ
f'(θ) =a secθtanθ + b (-cosec θ) cot θ
= secθtanθ - b cosec θ cot θ
f"(θ) = a [ secθ sec2θ + tanθ secθtanθ] - b[cosecθ ( -cosec2θ) + cotθ (-cosec θ cot θ)]
= a sec3 θ + asecθ tan2 θ + b cosec3 θ + b cosec θ cot2 θ
Now f'(θ) = 0
⇒ a secθ tanθ - b cosecθ cotθ = 0
$a\frac{1}{{\cos \theta }}\frac{{\sin \theta }}{{\cos \theta }} = b\frac{1}{{\sin \theta }}\frac{{\cos \theta }}{{\sin \theta }}$ $a{\sin ^3}\theta = b{\cos ^3}\theta$ ${\tan ^3}\theta = \frac{b}{a}$ $\tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}$ $\text{for}\quad \tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}$ $f"\left( \theta \right) > 0$ $\therefore \text{ f is minimum at} \quad \tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}} = \frac{{{b^{\frac{1}{3}}}}}{{{a^{\frac{1}{3}}}}}$ Now AC = a secθ + b cosecθ $AC = \frac{{a\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{a^{\frac{1}{3}}}}} + \frac{{b\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{b^{\frac{1}{3}}}}}$ $AC = {a^{\frac{2}{3}}}\left( {\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} } \right) + {b^{\frac{2}{3}}}\left( {\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} } \right)$ $AC = \left( {\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} } \right)\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)$ $AC = {\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{1}{2}}}\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)$ $AC = {\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}$ $\text{ minimum length of hypotenuse is} {\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}$

Question (13)

Find the points at which the function f given by f(x) = (x-2)4 (x + 1)3 has
(i) local maxima (ii) local minima (iii) point of inflexion

Solution

f(x) = (x-2)4 (x + 1)3
$f'\left( x \right) = {\left( {x - 2} \right)^4}\frac{d}{{dx}}{\left( {x + 1} \right)^3} + {\left( {x + 1} \right)^3}\frac{d}{{dx}}{\left( {x - 2} \right)^4}$ f'(x) = (x - 2)4 . 3(x+1)2 + (x+1)3. 4(x-2)3
= (x-2)3 (x+1)2 [ 3( x - 2) + 4 (x +1)]
= (x-2)3(x+1)2 (3x - 6 +4x +4 )
= (x-2)3(x+1)2 (7x - 2)
$f''(x) = {\left( {x + 1} \right)^2}{\left( {x - 2} \right)^3}(7) + {\left( {x + 1} \right)^2}(7x - 2)3{\left( {x - 2} \right)^2} + {\left( {x - 2} \right)^3}(7x - 2)2\left( {x + 1} \right)$ $= \left( {x + 1} \right){\left( {x - 2} \right)^2}\left[ {7(x + 1)(x - 2) + 3(x + 1)(7x - 2) + 2(x - 2)(7x - 2)} \right]$ Now f'(x) = 0
⇒ x = 2 OR x = -1, x = 2/7
f"(-1) = 0, f"(2) = 0 .
So we have to use first order test to find local maxima, minima and point of inflection.
We will check one by one for all values of x for which f'(x) = 0
Now for x = 2/7 we will find value of f'(2/7) when x tends left of 2/7.
$As\;x \to {\left( {\frac{2}{7}} \right)^ - } \Rightarrow 7x - 2 < 0,{(x - 2)^3} < 0$ ∴ f'(x) > 0
$\text{ as} \quad x \to \frac{{{2^ + }}}{7},7x - 2 > 0,{\left( {x - 2} \right)^3} < 0$ $f'\left( x \right) < 0$ $\Rightarrow x = \frac{2}{7}$ is point of local maxima
Now let check for x = 2
$for\;x \to {2^ - } \Rightarrow x - 2{\rm{ < }}0, \Rightarrow {(x - 2)^3} < 0,7x - 2 > 0$ f'(x) < 0
as x → 2+ ⇒ x-2 > 0 , (x-2)3 > 0 ⇒ f'(x) >0
∴ x = 2 is point of local minima
For x = -1
As x⇒- 1- , x+1 < 0 but (x+1)2 > 0 (x-2)3 < 0, (7x-2) < 0
∴ f'(x) > 0
As x⇒- 1+ , x+1 > 0
but (x+1)2 > 0 (x-2)3 < 0, (7x-2) < 0
∴ f'(x) > 0
So there is no change in f'(x) at x = -1
Thus x = -1 is inflection point.

Question (14)

Find the absolute maximum and minimum value of the function f given by
f(x) = cos2x + sin x , x ∈ [0, π]

Solution

f(x) = cos2x + sinx ; x ∈ [0, π]
f'(x) = 2cosx(-sinx) + cos x
f'(x) = -2sinxcosx + cos x
f"(x) = -2[(sinx)(-sinx) + cosx (cosx)] + (-sinx)
f"(x) = -2(cos2x - sin2x) - sinx
Now f'(x) = 0
⇒ -2sinxcosx + cosx = 0
cosx ( 1-2sinx) = 0
cos x = 0 OR 1 -2sinx = 0
cos x = 0 OR sinx = ½
$x = \frac{\pi }{2},\frac{{3\pi }}{2}\quad OR \quad x = \frac{\pi }{6},\frac{{5\pi }}{6}$ $as\quad x \in \left( {0,\pi } \right);x \ne \frac{{3\pi }}{2}$ $\therefore x = \frac{\pi }{2},\frac{\pi }{6},\frac{{5\pi }}{6}$ $f"\left( {\frac{\pi }{2}} \right) = - 2\left( {{{\cos }^2}\frac{\pi }{2} - {{\sin }^2}\frac{\pi }{2}} \right) - \sin \frac{\pi }{2}$ $f"\left( {\frac{\pi }{2}} \right) = - 2\left( {0 - 1} \right) - 1$ $f"\left( {\frac{\pi }{2}} \right) = 2 - 1 = 1 > 0$ ⇒ f is minimum at x = π/2
$f\left( {\frac{\pi }{2}} \right) = {\cos ^2}\frac{\pi }{2} + \sin \frac{\pi }{2}$ $f\left( {\frac{\pi }{2}} \right) = 0 + 1 = 1$ $f"\left( {\frac{\pi }{6}} \right) = - 2\left( {{{\cos }^2}\frac{\pi }{6} - {{\sin }^2}\frac{\pi }{6}} \right) - \sin \frac{\pi }{6}$ $f"\left( {\frac{\pi }{6}} \right) = - 2\left( {\frac{3}{4} - \frac{1}{4}} \right) - \frac{1}{2}$ $f"\left( {\frac{\pi }{6}} \right) = - 2\left( {\frac{2}{4}} \right) - \frac{1}{2}$ $f"\left( {\frac{\pi }{6}} \right) = \frac{{ - 3}}{2} < 0$ ⇒ f is maximum at x = π/6
$f\left( {\frac{\pi }{6}} \right) = {\cos ^2}\frac{\pi }{6} + sin\frac{\pi }{6}$ $f\left( {\frac{\pi }{6}} \right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ $f"\left( {\frac{{5\pi }}{6}} \right) = - 2\left[ {{{\cos }^2}\frac{{5\pi }}{6} - {{\sin }^2}\frac{{5\pi }}{6}} \right] - \sin \frac{{5\pi }}{6}$ $f"\left( {\frac{{5\pi }}{6}} \right) = - 2\left[ {\frac{3}{4} - \frac{1}{4}} \right] - \frac{1}{2}$ $f"\left( {\frac{{5\pi }}{6}} \right) = - 2\left( {\frac{2}{4}} \right) - \frac{1}{2} = \frac{{ - 3}}{2} < 0$ f is maximum at x = 5π/6 $f\left( {\frac{{5\pi }}{6}} \right) = {\cos ^2}\frac{{5\pi }}{6} + \sin \frac{{5\pi }}{6}$ $f\left( {\frac{{5\pi }}{6}} \right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ $f\left( 0 \right) = {\cos ^2}0 + \sin 0 = 1 + 0 = 1$ $f\left( \pi \right) = {\cos ^2}\pi + \sin \pi = 1 + 0 = 1$ $\text{Absolute max} = \max \left\{ {1,1,\frac{5}{4},\frac{5}{4},1} \right\} = \frac{5}{4}$ $\text{Absolute min} = \min \left\{ {1,1,\frac{5}{4},\frac{5}{4},1} \right\} = 1$

Question (15)

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

Solution

Let cone of radius 'R' and height of 'H' is inscribed in sphere of given radius 'r'
Now from diagram CM = H, OC = r
AM = R, OM = H - r
In Δ OAM, ∠M = 90
By Pythagoras' theorem
OM2 + AM2 = OA2
(H-r)2 + R2 = r2
R2 = r2 - (H-r)2
R2 = r2 - H2 + 2rH - r2
R2 = 2rH - H2
Now function f : Volume of cone $f = \frac{1}{3}\pi {R^2}H$ $f = \frac{1}{3}\pi H\left( {2rH - {H^2}} \right)$ $f\left( H \right) = \frac{\pi }{3}\left( {2r{H^2} - {H^3}} \right)$ $f'\left( H \right) = \frac{\pi }{3}\left( {4r{H} - 3{H^2}} \right)$ $f"\left( H \right) = \frac{\pi }{3}\left( {4r - 6H} \right)$ Now f'(H) = 0 $\frac{\pi }{3}\left( {4rH - 3{H^2}} \right) = 0$ $4rH = 3{H^2}$ $\frac{4}{3}r = H$ $f"\left( H \right) = \frac{\pi }{3}\left[ {4r - \cancel{6}^2\left( {\frac{{4r}}{\cancel{3}}} \right)} \right]$ $f"\left( H \right) = \frac{\pi }{3}\left( { - 4r} \right) < 0$ ⇒ f is maximum at H = 4r/3
So, altitude of cone of maximum volume inscribed in sphere of radius 'r' is 4r/3

Question (16)

Let f be a function defined on [a, b] such that f'(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b)

Solution

If f(x), [a, b] f'(x) > 0
It mean f(x) is differentiable as f(x) is differentiable it is continuous
f'(x) > 0 ⇒ f(a) ≠ f(b)
so mean value theorem is applicable $\therefore f'\left( x \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ But f'(x) > 0
$\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} > 0$ $f\left( b \right) - f\left( a \right) > 0$ $f\left( b \right) > f\left( a \right)$ a < b and f(a) < f(b)
It implies f(x) is increasing function

Question (17)

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume.

Solution

Let 'r' be radius of cylinder and height be 'h' which is inscribed in sphere of radius 'R'
In diagram OM = h/2, OA = R, AM=r
ΔOAM, ∠M = 90
Pythagoras' theorem
AM2 + OM2 = OA2
${r^2} + {\left( {\frac{h}{2}} \right)^2} = {R^2}$ ${r^2} = {R^2} - {\left( {\frac{h}{2}} \right)^2}$ ${r^2} = {R^2} - \frac{{{h^2}}}{4}$ $\text{Volume of cylinder (V)} = \pi {r^2}h$ $V = \pi h\left( {{R^2} - \frac{{{h^2}}}{4}} \right)$ $f\left( h \right) = \pi {R^2}h - \frac{{\pi {h^3}}}{4}$ $f'\left( h \right) = \pi {R^2} - \frac{{3\pi {h^2}}}{4}$ $f"\left( h \right) = - \frac{{3\pi }}{4} \cdot 2h = - \frac{{3\pi h}}{2}$ Now f'(h) = 0 $\pi {R^2} - \frac{{3\pi {h^2}}}{4} = 0$ $4\cancel{\pi} {R^2} = 3\cancel{\pi} {h^2}$ $\frac{4}{3}{R^2} = {h^2}$ $\Rightarrow h = \frac{{2R}}{{\sqrt 3 }}$ $f"\left( {\frac{{2R}}{{\sqrt 3 }}} \right) = \frac{{ - 3\pi }}{2} \cdot \frac{{2R}}{{\sqrt 3 }} < 0$ $\Rightarrow \text{f is minimum at} h = \frac{{2R}}{{\sqrt 3 }}$ ${r^2} = {R^2} - \frac{{{h^2}}}{4}$ ${r^2} = {R^2} - \frac{1}{\cancel{4}}\left( {\frac{{\cancel{4}{R^3}}}{3}} \right)$ ${r^2} = \frac{{3{R^2} - {R^2}}}{3} = \frac{2}{3}{R^2}$ $\text{volume (V)} = \pi {r^2}h$ $V = \pi \frac{{2{R^2}}}{3} \times \frac{{2R}}{{\sqrt 3 }}$ $V = \frac{{4\pi {R^3}}}{{3\sqrt 3 }}$

Question (18)

Show that height of the cylinder of greatest volume which can be inscribed in right circular cone of height h and semi vertical angle α is one -third that of the cone and the greatest volume of cylinder is $\frac{4}{{27}}\pi {h^3}{\tan ^2}\alpha$

Solution

Cylinder of radius R and Heigt H be inscribed in the cone
In the diagram AC = h, BC = H, BE = R
∴ AB = h - H
Δ ABE ∼ Δ ACD $\therefore \frac{R}{r} = \frac{{h - H}}{h}$ $Rh = hr - Hr$ $H = \frac{{hr - Rh}}{r}$ Now volume of cylinder V $V = \pi {R^2}H$ $V = \pi {R^2}\left( {\frac{{hr - Rh}}{r}} \right)$ $f\left( R \right) = \pi {R^2}h - \frac{{\pi {R^3}h}}{r}$ $f'\left( R \right) = \pi h\left( {2R} \right) - \frac{{\pi h}}{r}3{R^2}$ $f'\left( R \right) = 2\pi Rh - \frac{{3\pi h{R^2}}}{r}$ $f"\left( R \right) = 2\pi h - \frac{{6\pi hR}}{r}$ Now f'(R) = 0 $2\pi hR - \frac{{3\pi h{R^2}}}{r} = 0$ $2\cancel{\pi}\cancel{h}\cancel{R} = \frac{{3\cancel{\pi}\cancel{h} {R^\cancel{2}}}}{r}$ $\frac{2}{3}r = R$ $f"\left( R \right) = 2\pi h - \frac{{\cancel{6}^2\pi h}}{\cancel{r}} \cdot \frac{{2r}}{\cancel{3}}$ $f"\left( R \right) = 2\pi h - 4\pi h$ $f"\left( R \right) = - 2\pi h < 0$ ⇒ f is maximum at R= 2r/3
$H = \frac{{hr - Rh}}{r}$ $H = \frac{{h\cancel{r}}}{\cancel{r}} - \frac{R}{r}h$ $H = h - \frac{2}{3}\frac{\cancel{r}}{\cancel{r}}h$ $H = \frac{h}{3}$ ∴ Height of cylinder having maximum volume is h/3.
i.e. one third of height of cone
$\text{volume of cylinder(V}= \pi {R^2}H$ $V = \pi \frac{{4{r^2}}}{9} \cdot \frac{h}{3}$ $V = \frac{4}{{27}}\pi {r^2}h$ From diagram r = htanα $V = \frac{4}{{27}}\pi {h^2}{\tan ^2}\alpha \cdot h$ $V = \frac{4}{{27}}\pi {h^3}{\tan ^2}\alpha$

Choose the correct answer in the Exercises from 19 to 24

Question (19)

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m3 / h    (B) 0.1 m3 / h    (C) 1.1 m3 / h    (D) 0.5 m3 / h

Solution

Given radius of cylinder r = 10 m and dV/dt = 314 m3/hr
$\text{volume of cylinder (V)} = \pi {r^2}h$ $V = \pi {\left( {10} \right)^2}h$ $V = 100\pi h$ Differentiating w.r.t t $\frac{{dV}}{{dt}} = 100\pi \frac{{dh}}{{dt}}$ $314 = 100 \times \frac{{314}}{{100}} \times \frac{{dh}}{{dt}}$ $\frac{{dh}}{{dt}} = 1\frac{m}{{hr}}$ Option (A) is correct

Question (20)

The slope of the tangent to the curve x = t2 + 3t - 8, y = 2t2 - 2t -5 at the point (2, -1) is
( A) $\frac{{22}}{7}$     (B) $\frac{{6}}{7}$     (C) $\frac{{7}}{6}$     (D) $\frac{{-6}}{7}$

Solution

x= t2 + 3t - 8
when x = 2
2 = t2 +3t - 8
t2 + 3t - 10 = 0
(t+5) (t - 2) = 0
t = -5 OR t = 2
y = 2t2 - 2t - 5
y = -1
-1 = 2t2 - 2t - 5
2t2 - 2t - 4 = 0
2( t2 - t - 2) = 0
2( t- 2) (t+1) = 0
t = 2, t = -1
Common value of t = 2
t2 + 3t - 8
Differentiating w.r.t 't' $\frac{{dx}}{{dt}} = 2t + 3$ y = 2t2 - 2t - 5
Differentiating w.r.t 't' $\frac{{dy}}{{dt}} = 4t - 2$ $\text{Now} \quad \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}$ $\frac{{dy}}{{dx}} = \frac{{4t - 2}}{{2t + 3}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{t = 2}} = \frac{{4\left( 2 \right) - 2}}{{2\left( 2 \right) + 3}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{t = 2}} = \frac{{8 - 2}}{{4 + 3}} = \frac{6}{7}$ $\text{slope of tangent =}{\left( {\frac{{dy}}{{dx}}} \right)_{t = 2}}$ ${\left( {\frac{{dy}}{{dx}}} \right)_{t = 2}} = \frac{6}{7}$ Option (B) is correct

Question (21)

The line y = mx + 1 is tangent to the curve y2 = 4x if the value of m is
(A) 1    (B) 2    (C) 3    (D) $\frac{{1}}{2}$

Solution

y = mx +1 is tangent
y2 = 4x ---(1)
Replacing the value of y = mx + 1 in (1) we get
(mx+1)2 = 4x
m2x2 + 2mx + 1 - 4x = 0
m2x2 + (2m-4)x + 1 =0
D = b2 - 4ac
D = (2m -4)2 -4m2(1)
D= 4m2 - 16m + 16 - 4m2
D = 16 - 16m
The curve has only one tangent therefore D = 0
16 - 16m = 0
m = 1
Option (A) is correct

Question (22)

The normal at the point (1,1) on the curve 2y+x2 = 3 is
(A) x + y = 0    (B) x - y = 0    (C) x + y + 1 = 0    (D) x - y = 0

Solution

2y + x2 = 3
Differentiating w.r.t 'x' $2\frac{{dy}}{{dx}} + 2x = 0$ ${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = - 1$ ${\text{slope of normal}} = \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {1,1} \right)}}}} = \frac{{ - 1}}{{ - 1}} = 1$ Slope of normal = 1
Equation of normal y - 1 = 1 (x - 1)
x = y
x - y = 0
Option (B) is correct

Question (23)

The normal to the curve x2 = 4y passing (1, 2) is
(A) x+y = 3    (B) x - y = 3    (C) x + y = 1    (D) x - y = 1

Solution

x2 = 4y
Let point (x1, y1) be on curve at which normal is drawn
x12 = 4y1
Differentiating w.r.t 'x' $2x = 4\frac{{dy}}{{dx}}$ $\frac{{dy}}{{dx}} = \frac{x_1}{2}$ $\text{slope of normal} = \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},{y_1}} \right)}}}}$ $\text{slope of normal}= \frac{{ - 1}}{{\frac{{{x_1}}}{2}}}$ $\text{slope of normal} = \frac{{ - 2}}{{{x_1}}}$ Equation of normal is
$y - {y_1} = \frac{{ - 2}}{{{x_1}}}\left( {x - {x_1}} \right)$ It passes through (1, 2) $2 - {y_1} = \frac{{ - 2}}{{{x_1}}}\left( {1 - {x_1}} \right)$ $\cancel{2{x_1}} - {y_1}{x_1} = - 2 + \cancel{2{x_1}}$ $- {y_1}{x_1} = - 2$ ${x_1}\left( {\frac{{x_1^2}}{4}} \right) = 2$ $x_1^3 = 8$ ${x_1} = 2$ If x1 = 2, y1 = 1
$\text{slope of normal}= \frac{{ - 2}}{{{x_1}}} = \frac{{ - 2}}{2} = - 1$ Equation of normal is
y - 1 = -1(x-2)
y-1 = -x + 2
x + y = 3
Option (A) is correct

Question (24)

The points on the curve 9y2 = x3, where the normal the curve makes equal intercepts with the axes are
$\left( A \right)\left( {4, \pm \frac{8}{3}} \right)$     $\left( B \right)\left( {4,\frac{{ - 8}}{3}} \right)$     $\left( C \right)\left( {4, \pm \frac{3}{8}} \right)$     $\left( D \right)\left( { \pm 4,\frac{3}{8}} \right)$

Solution

9y2 = x3
Normal make equal intercept on axis therefore m = -1 ---(1)
Differentiating w.r.t 'x' $18y\frac{{dy}}{{dx}} = 3{x^2}$ $\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{18y}} = \frac{{{x^2}}}{{6y}}$ $\text{slope of normal normal m} = \frac{{ - 1}}{{\left( {\frac{{dy}}{{dx}}} \right)}}$ $- 1 = \frac{{ - 1}}{{\frac{{{x^2}}}{{6y}}}}$ ${x^2} = 6y$ $y = \frac{{{x^2}}}{6}$ $9{y^2} = {x^3}$ Replacing the value of $y = \frac{{{x^2}}}{6}$ we get $9{\left( {\frac{{{x^2}}}{6}} \right)^2} = {x^3}$ $\cancel{9}\frac{{{x^2}}}{{\cancel{36}_4}} = {x^3}$ $x = 4$ If x = 4
$9{y^2} = {4^3}$ $9{y^2} = 64$ ${y^2} = \frac{{64}}{9}$ $y = \pm \frac{8}{3}$ $\text{Point is} \quad \left( {4, \pm \frac{8}{3}} \right)$ Therefore Option (A) is correct