NCERT Solutions Class 12 Maths (Applications of Derivatives) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-6 (Applications of Derivatives) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Set 1

Question 1. Using differentials, find the approximate value of each of the following:

(i) (17/81)^1/4

(ii) 33^-1/5

Solution:

(i) (17/81)1/4

Let y = x1/4, x = 16/81 and △x = 1/81

△y = (x + △x)1/4 – x1/4

= (17/81)1/4 – (16/81)1/4

= (17/81)1/4 – (2/3)

So,

(17/81)1/4 = (2/3) + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{4x^{\frac{3}{4}}}(△x)$

= $\frac{1}{4(\frac{16}{81})^{\frac{3}{4}}}(\frac{1}{81})$

= 27/32 × 1/81

= 1/96

= 0.010

Hence, the approximate value of (17/81)1/4 = 2/3 + 0.010 = 0.677

(ii) 33-1/5

Let y = x-1/5, x = 32 and △x = 1

△y = (x + △x)-1/5 – x-1/5

= (33)-1/5 – (32)-1/5

= (33)-1/5 – 1/2

So,

(33)-1/5 = 1/2 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-1}{5x^{\frac{6}{5}}}(△x)$

= $\frac{-1}{5(32)^{\frac{6}{5}}}(1)$

= -1/320

= -0.003

Hence, the approximate value of (33)-1/5 = 1/2 – 0.003 = 0.497

Question 2. Show that the function given by f(x) = log x/x has maximum at x = e.

Solution:

The given function is f(x) = log x/x

$f'(x)=\frac{x(\frac{1}{x})-\log x}{x^2}=\frac{1-\log x}{x^2}$

Now, f'(x) = 0

1 – log x = 0

log x = 1

log x = log e

x = e

Now, $f''(x)=\frac{x^2(-\frac{1}{x})-(1-\log x)(2x)}{x^4}$

$=\frac{-x-2x(1-\log x}{x^3}$

$=\frac{-3+3\log x}{x^3}$

$f''(e)=\frac{-3+2\log e}{e^3}=\frac{-3+2}{e^3}=\frac{-1}{e^3}<0$

Therefore, by second derivatives test, f is the maximum at x = e.

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

Solution:

Given an isosceles triangle with fixed base b.

Let the other two sides be of length x.

Now its given that,

dx/dy = -3cm/s

Now semi-perimeter(s) = x + x + b/2

s = x + b/2

Area [by heron’s formula] = $\sqrt{s(s-x)(s-x)(s-b)}$

$A=(s-x).\sqrt{s(s-b)}$

$A=\frac{b}{2}\sqrt{(x+\frac{b}{2})(x-\frac{b}{2})}$

$A=\frac{b}{2}\sqrt{x^2}\frac{-b^2}{2}$

$A=\frac{b}{4}\sqrt{4x^2-b^2}$

To find: dA/dt = ?

DA/dt = ?

$\frac{dA}{dt}=\frac{b}{4}.\frac{1}{2\sqrt{4x^2-b^2}}.8x.\frac{dx}{dt}$

$\frac{dA}{dt}=\frac{b}{4}.\frac{1}{2b\sqrt{3}}.8b.(-3)$

dA/dt = -√3b cm2/s

Hence, the area is decreasing at the rate = √3b cm2/s

Question 4. Find the equation of the normal to curve x²= 4y which passes through the point (1, 2).

Solution:

Given area: x²= 4y

On Differentiating both sides with respect to y,

2x(dx/dy) = 4

dx/dy = 2/x

Slope = -1/m = -2/x

By point slope form equation of normal will be,

y – 2 = -1(x – 1)

x + y = 3 is the required equation of normal.

Question 5. Show that the normal at any point θ to the curve x = acosθ + aθsinθ, y = asinθ – aθ cosθ is at a constant distance from the origin.

Solution:

Given curve,

x = acosθ + aθsin θ

y = asinθ – aθcos θ

Now -dx/dy = slope of normal = $\frac{\frac{-dx}{dθ}}{\frac{dy}{dθ}}$ -(1)

$\frac{dx}{dθ}$ = -asinθ + asinθ + aθcosθ

$\frac{dx}{dθ}$ = aθcosθ -(2)

$\frac{dx}{dθ}$ = acosθ + aθsinθ – acosθ

$\frac{dx}{dθ}$ = aθsinθ -(3)

$\frac{-dx}{dy}=\frac{-aθ\cos θ}{aθ\sin θ}$ -(From 1, 2 & 3)

-dx/dy = -cotθ

Now using point slope from, equation of normal will be,

$y-a\sin θ+aθ\cos θ=\frac{-\cos θ}{\sin θ}(x-a\cos θ-aθ\sinθ)$

ysinθ – asin-2θ + aθcosθsinθ = -xcosθ + acos2θ + aθsinθcosθ

ysinθ + ysinθ − a = 0

$d=\frac{|0+0-a|}{\sqrt{\cos^2θ+\sin^2θ}}=a=$ constant.

Hence proved

Question 6. Find the intervals in which the function f given by

$f(x)=\frac{4\sin x-2x-x\cos x}{2t\cos x}$ is

(i) increasing (ii) decreasing

Solution:

$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$

$f'(x)=\frac{(2+\cos x)(4\cos x-2+x\sin x-\cos x)+(4\sin x-2x-x-\cos x)(\sin x)}{(2+\cos x)^2}$

$f'(x)=\frac{4\cos x-cos^2x}{(2+\cos x)^2}$

(i) For f(x) tp be increasing f'(x) ≥ 0

$\frac{4\cos x-cos^2x}{(2+\cos x)^2}>0$

$\cos x(4-\cos x)>0$

Now, 4 – cos x > 0 -(because 4 – cos x ≥ 3)

So, cos x > 0

Hence, f(x) is increasing for 0 < x < x/2 and 3π/2 < x < 2π

(ii) For f(x)to be decreasing,

f'(x) < 0

$\frac{4\cos x-\cos^2x}{(\cos x+2)^2}<0$

cosx(4 − cosx) < 0

cosx < 0

Hence, f(x) is decreasing for π/2 < x < 3π/2

Question 7. Find the intervals in which the function f given by f(x) $x^3+\frac{1}{x^3},x≠0$ is

(i)increasing (ii)decreasing

Solution:

f(x) = $x^3+\frac{1}{x^3},x≠ 0$

f'(x) $=3x^2-\frac{1}{3x^4}$

(i) For f(x) to be increasing,

f'(x) > 0

$3x^2-\frac{1}{3x^4}>0$

$\frac{9x^6-1}{3x^4}>0$

9x6 > 1

$x>(\frac{1}{9})^{\frac{1}{6}}$ or $x∈((\frac{1}{9}^{\frac{1}{6}}),∞)$

(ii) For f(x) to be decreasing,

f'(x) < 0 $3x^2-\frac{1}{3x^4}<0$

9x6 < 1

$x<(\frac{1}{9})^{\frac{1}{6}}$ or $x∈(∞,(\frac{1}{9}^{\frac{1}{6}}))$

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:

Given ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Its major axis is the x-axis

Using parametric form of ellipse, x = acosθ, y = bsinθ,

If coordinates of A are (acosθ, bsinθ)

Then B’S coordinates will be (acosθ, -bsin θ).

Now, OC = a, OD = acos θ, so CD = a(1 + cos θ)

AB = |AD| + |BD| = 2b sin θ

Area of △ABC = 1/2.AB.CD

= 1/2.2bsin θ.a(1 + cos θ)

△(θ) = ab.sinθ.(1 + sin θ)

For maxima/minima, put △'(θ) = 0

△'(θ) = ab[cosθ[1 + cosθ] + sinθ[-sinθ]]

△'(θ) = ab[2cos2θ + cosθ – 1] = 0

2cos2θ + cosθ – 1 = 0

2cos2θ + 2cosθ – cosθ – 1 = 0

2cosθ(cosθ + 1) – 1(cosθ + 1) = 0

(2cosθ – 1).(cosθ + 1) = 0

cosθ = 1/2 or cosθ = -1

If cosθ = -1, then sinθ = 0 & △(θ) = 0

But if cosθ = 1/2, sinθ = √3/2 & △(θ) = ab. $\frac{\sqrt{3}}{2}(1+\frac{1}{2})$

$△(θ)_{max}=\frac{2\sqrt{3}ab}{4}$

Question 9. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2m and volume is 8m3. If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square meter for sides. What is the cost of least expensive tank?

Solution:

Given:

Depth of tank = 2m

Volume = 8m3

Let the length be equal to x & width be to y

The base area will be equal to x.y.

Area of sides will be equal to; 2x, 2y, 2x, 2y

Now, volume = x.y.2 = 2xy = 8m3

so, xy = 4m2 -(1)

y = 4/x

Total cost = 70.base + 45.(sides)

c = 70xy + 45(2x + 2y + 2x + 2y)

c = 70.4 + 45.4(x + y) -(xy = 4)

c(x) = 180 – \ $\frac{180.4}{x^2}=0$

$1-\frac{4}{x^2}=0$

x2 = 4

x = ±2, x = 2, (Rejecting -ve value)

y = 4/x = 4/2 = 2

Now cost c(x) = 280 + 180(x + 4/x)

c = 280 + 180(2 + 2)

c = 1000 rupees

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 sides of square be x & radius of circle be r.

Perimeter of square = 4x

Circumference of circle = 2πr

Now given that, 4x + 2πr = k -(1)

x = $\frac{k-2πr}{4}$

Area of square = x2

Area of circle = πr2

Sum of areas = x2 + πr2

$s(r)=(\frac{k-2πr}{4})^2+πr^2$

Put s'(r) = 0

$s(r)=2(\frac{k-2πr}{4})(\frac{-π}{2})+2πr=0$ -(From eq(1))

$πr=\frac{π}{2}\frac{k-2πr}{4}$

8πr = kπ – 2π2r

8r = k – 2πr

8r = (4x + 2πr) – 2πr -(k = 4x + 2πr)

8r = 4x

x = 2r

Hence, proved that the sides of the square is double the radius of the circle.

Question 11. A window is in the form of a 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 the length of the rectangle = x

the breadth of the rectangle = y

and the radius of the semicircle = x/2

So given that total perimeter of the window = 10m

P = πx/2 + x + 2y = 10

x(1 + π/2) + 2y = 10

2y = 10 – x(1 + π/2)

y = 5 – x(1/2 – π/4) -(1)

Now, the area of the window

$A=\frac{πx^2}{2}+xy$ -(2)

From eq(1) put the value of y in eq(2), we get

$A=x.[5 - x(\frac{1}{2} - \frac{π}{4})]+\frac{πx^2}{2}$

= 5x – x2(1/2 + π/4) + πx2/8

On differentiating we get

A’ = 5 – 2x(1/2 + π/4) + 2xπ/8

= 5 – x(1 + π/2) + xπ/4

Put A’ = 0

5 – x(1 + π/2) + xπ/4 = 0

-x(1 + π/2) + xπ/4 = -5

x(-1 – π/2 + π/4) = -5

x(1 + π/4) = 5

x = 5/ (1 + π/4)

x = 20/ π + 4

Hence, the length of the rectangle = 20/ π + 4

Now put the value of x in eq(1)

y = 5 – (20/ π + 4)(1/2 – π/4)

y = 10/π + 4

Hence, breadth of the rectangle = 10/π + 4

and the radius of the semicircle = x/2 = $\frac{\frac{20}{π + 4}}{2}$ = 10/π + 4

Set 2

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 the hypotenuse is $(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$

Solution:

Given, a triangle ABC

Let, PE = a & PD = b

In the △ABC, ∠B = 90

Let ∠C = θ, so, ∠ DPA = θ

DP|| BC.

Now in △ADDP,

cosθ = DP/AP = b/AP

AP = b/cosθ

In △EPC,

sinθ = EP/CP = a/CP

CP = a/sin θ

Now AC = h = PA + PC

h = $\frac{b}{\cos θ}+\frac{a}{\sin θ}$

h(θ) = b sec θ + a cosec θ

Put h'(θ) = $\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}$

$\frac{b}{\cos θ}.\frac{\sin θ}{\cos θ}=\frac{a}{\sin θ}.\frac{\cos θ}{\sin θ}$

b sin3θ = a cos 3θ

tan3θ = a/b

tanθ = (a/b)1/3

secθ = $\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$

cosecθ = $\frac{\sqrt{a^{2/3}+b{2/3}}}{b^{1/3}}$

hmax = $b.\frac{\sqrt{b^{2/3}+a^{2/3}}}{b^{1/3}}+a.\frac{\sqrt{b^{2/3}+a^{2/3}}}{a^{1/3}}$

hmax = (b2/3+a2/3)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

On differentiating w.r.t x, we get

f'(x) = 4(x – 2)³(x + 1)³+ 3(x + 1)²(x – 2)⁴

Put f'(x) = 0

(x – 2)³(x + 1)² [4(x + 1) + 3(x – 2)] = 0

(x – 2)³(x + 1)²(7x – 2) = 0

Now,

Around x = -1, sign does not change, i.e

x = -1 is a point of inflation

Around x = 2/7, sign changes from +ve to -ve i.e.,

x = 2/7 is a point of local maxima.

Around x = 2, sign changes from -ve to +ve i.e.,

x = 2 is a point of local minima

Question 14. Find the absolute maximum and minimum values of the function f given by f(x) = cos² x + sin x, x ∈ [0, π]

Solution:

f(x) = cos2x + sin x; x ϵ [0, π]

On differentiating w.r.t x, we get

f'(x) = 2cos x(-sin x) + cos x = cos x – sin2x

Put f'(x) = 0

cos x(1 – 2sin x) = 0

cos x = 0; sin x = 1/2

In x ϵ[0, π] if cos x = 0, then x = π/2

and if sin x = 1/2, then x = π/6 & 5π/6

Now, f”(x) = -sin x – 2 cos2x

f”(π/2) = -1 + 2 = 1 > 0

x = π/2 is a point of local minima f(π/2) = 1

f”(π/6) = $\frac{-1}{2}-2.\frac{1}{2}=\frac{-3}{2}<0$

x = π/6 is a point of local maxima f(π/6) = 5/4

$f''(\frac{5π}{6})=\frac{-1}{2}-2.(\frac{-1}{2})>0$

x = 5π/6 is a point of local minima f(5π/6) = 5/4

Global/Absolute maxima = ma{f(0), f(π/6), f(π)}

= max{1, 5/4, 1}

= 5/4 = Absolute maxima value

Global/Absolute minima = min{f(0), f(π/2), f(π/6), f(π)}

= min{1, 1, 5/4, 1}

= 1 = Absolute minima value

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 4π/3

Solution:

Let ABC be the cone

and o is the centre of the sphere.

AO = BO = CO = R

AO = h = height of cone

BD = CD = r = radius of cone.

∠DOC = θ -(Properties of circle)

In △ DOC,

OD = R cosθ & CD = Rsinθ,

r = R sin θ

AD = AO + OD = R + Rcosθ

h = R(1 + cosθ)

Now, the volume of the cone is

V = $\frac{1}{3}πr^2h$

v(θ) = $\frac{1}{3}.πR^2\sin{-2}θ.R(1+\cosθ)$

$v'(θ)=\frac{}{}[\sin{-2}θ(-\sinθ)+(1+\cosθ)(2\sinθ\cosθ)]$

Put v(θ) = 0

sinθ[2cosθ + 2cos2θ − sin2θ] = 0

sinθ[2cosθ + 2cos2θ − 1] = 0

sinθ(3cosθ − 1)(1 + cosθ) = 0

sinθ = 0, cos = 1/3, cosθ = −1

If sinθ = 0, then volume will be 0.

If cosθ = -1, then sinθ = 0 & again volume will be 0.

But if cosθ = 1/3; sinθ = 2√2/3 and

Volume, v = 32/81πR3, which is maximum.

Height, h = R(1 + cosθ) = R( $1+\frac{1}{3}$ )

h = 4r/3

Hence proved

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:

Given that on [a, b] f'(x) > 0, for all x in interval I.

So let us considered x1, x2 belongs to I with x1 < x2

To prove: f(x) is increasing in (a, b)

According to the Lagrange’s Mean theorem

f(x2) – f(x1)/ x2 – x1 = f'(c)

f(x2) – f(x1) = f'(c)(x2 – x1)

Where x1 < c < x2

As we know that x1 < x2

so x1 < x2 > 0

It is given that f'(x) > 0

so, f'(c) > 0

Hence, f(x2) – f(x1) > 0

f(x2) < f(x1)

Therefore, for every pair of points x1, x2 belongs to I with x1 < x2

f(x2) < f(x1)

f(x) is strictly increasing in I

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:

In △ABC,

AC2 = BC2 + AB2

4R2 = 4r2 + h2

r2 = R2– $\frac{h^2}{4}$ ……….(1)

Now, volume of cylinder = πr2h

Put the value ov r2 from eq(1), we get

V = π( $R^2\frac{-h^2}{4}$ ).h

V(h) = $πR^2h-\frac{πh^3}{4}$

On differentiating both side we get

V ‘(h) = $πR^2h-\frac{3πh^3}{4}$

Now, put V'(h) = 0

πR2 = $\frac{3}{4}πh^2$

$h=\frac{2R}{\sqrt{3}}$

Now the maximum volume of cylinder = π[R2. 2R/√3 – 1/4.4R2/3.2R/√3]

= 4πR3/ 3√3

Question 18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α is one-third that of the cone and the greatest volume of the cylinder is 4/27πh³tan²α.

Solution:

Let,

XQ = r

XO = h’

AO = h

OC = r’

∠XAQ = α

In triangle AXQ and AOC = XQ/OC = AX/AO

So, r’/r = h-h’/h

hr’ = r(h-h’)

hr’ = rh – rh’

rh’ = rh – hr’

rh’ = h(r – r’)

h’ = h(r – r’)/r

The volume of cylinder = πr’²h’

v = πr’²(h(r – r’)/r)

= π(h(rr’² – r’³)/r)

On differentiating we get

v’ = πh/r(2rr’ – 3r’²)

Again on differentiating we get

v” = πh/r(2r – 6r’) ………(1)

Now put v’ = 0

πh/r(2rr’ – 3r’²) = 0

(2rr’ – 3r’²) = 0

2r’r = 3r’²

r’ = 2r/3

So, v is maximum at r’ = 2r/3

The maximum volume of cylinder = πh/r[r. 4r²/9 – 8r²/27]

= πhr²[4/27]

= 4/27πh(h tanα)²

= 4/27πh³ tan²α

Question 19. A cylindrical tank of a radius 10 m is being filled with wheat at the rate of 314 cubic meters per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h

Solution:

Given,

Radius of cylinder = 10m [radius is fixed]

Rate of increase of volume = 314m3/h

ie dv/dt = 314m3/h

Now, the volume of cylinder = πr2h

v = π.(10)2.h

v = 100πh

On differentiating w.r.t t, we get

dv/dt = 100π $\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{100π}.\frac{dv}{dt}=\frac{1}{314}.314$

$\frac{dh}{dt}=1m/h$

So 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) 22/7 (B) 6/7 (C) 7/6 (D) -6/7

Solution:

Given that the slope of the tangent to the curve x = t2 + 2t – 8 and y = 2t2 – 2t – 5

On differentiating we get

$\frac{dy}{dx}=2t+3 ;\frac{dy}{dt}=4t-2$

Now, when x = 2,

t2 + 3 – 8 = 2

t2 + 3 – 10 = 0

t2 – 2t + 5t – 10 = 0

(t – 2)(t + 5) = 0

Here, t = 2, t = -5 ……….(1)

When y = -1

2t2 – 2t – 5 = -1

2t2 – 2t – 4 = 0

t2 – t – 2 = 0

(t + 1)(t – 2) = 0

t = -1 or t = 2 ……….(2)

From eq(1) & eq(2) satisfies both,

Now, $\frac{dy}{dx}=slope=\frac{dy}{dt}$

$\frac{dy}{dx}=\frac{4t-2}{2t+3}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}$

So, option B is the correct.

Question 21.The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is

(A) 1 (B) 2 (C) 3 (D)1/2

Solution:

The curve if y2 = 4x …….(1)

On differentiating we get

$2y\frac{dy}{dx}=4$

$\frac{dy}{dx}=\frac{2}{y}$

The slope of the tangent to the given curve at point(x, y)

$\frac{dy}{dx}=\frac{2}{y}$

m = 2/y

y = 2/m

The equation of line is y = mx + 1

Now put the value of y, we get the value of x

2/m = mx + 1

x = 2 – m/m

Now put the value of y and x in eq(1), we get

(2/m)2 = 4(2 – m/m)

m = 1

Hence, the 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 = 1

Solution:

The equation of curve 2y + x2 = 3

On differentiating w.r.t x, we get

2 $\frac{dy}{dx}+2x=0$

dy/dx = -x

The slope of the tangent to the given curve at point(1, 1)

dy/dx = -x = -1

m = -1

And slope of normal = 1

Now the equation of normal

(y -1) = 1(x – 1)

x – y = 0

So, B option 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:

The equation of curve is x2 = 4y …….(1)

On differentiating w.r.t x, we get

2x = $4\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{x}{2}$

The slope of normal at (x, y)

-dx/dy = -2/x = m

The slope at given point(1, 2)

m = (y – 2)/(x – 1)

-2/x = (y – 2)/(x – 1)

y = 2/x

Now put the value of y in eq(1)

x2 = 4(2/x)

x = 2

and y = 1

So the point is (2, 1)

Now the slope of normal at point(2, 1) = -2/2 = -1

The equation of the normal is

(y – 1) = -1(x – 2)

x + y = 3

So option A is correct

Question 24. The points on the curve 9y²= x³, where the normal to the curve makes equal intercepts with the axes are

(A) $(4,±\frac{8}{3})$ (B) $(4,\frac{-8}{3})$
(C) $(4,±\frac{3}{8})$ (D) $(±4,\frac{3}{8})$

Solution:

Given equation 9y2 = x3

On differentiating w.r.t x, we get

18y dy/dx = 3x2

dy/dx = 3x2/18y

dy/dx = x2/6y

Now, the slope of the normal to the given curve at point (x1, y1) is

$-1=\frac{6y_1}{x^2_1}$

Hence, the equation of the normal to the curve at point (x1, y1) is

$y-y_1=\frac{-6y_1}{x^2_1}(x-x_1)$

$x^2_1y-x_1^2y=-6xy_1+6x_1y_1$

$\frac{6xy_1}{6x_1y_1+x_1^2y_1}+\frac{x_1^2y}{6x_1y_1+x^2_1y_1}$

$\frac{\frac{x}{x_1(6+x_1)}}{6}+\frac{\frac{y}{y_1(6+x_1)}}{x_1}=1$

According to the question it is given that the normal

make equal intercepts with the axes.

So,

$\frac{x_1(6+x_1)}{6}+\frac{y_1(6+x_1)}{x_1}$

$x_1^2=6y_1$ …………(1)

The point (x1, y1)lie on the curve,

$9_1^2=x_1^22$ …………(2)

From eq(1) and (2), we get

$9(\frac{x_1^2}{6})^2=x_1^3=\frac{x_1^4}{4}=x^3_1=x_1=4$

From eq(2), we get

$9y_1^2=(4)^3=64$

$y_1^2=\frac{64}{9}$

$y_1=±\frac{8}{3}$

Hence, the required points are $(4,±\frac{8}{3})$

So, option A is correct.

NCERT Solutions Class 12 Maths (Applications of Derivatives) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Applications of Derivatives) Miscellaneous Exercise

Set 1

Set 2

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NCERT Solutions Class 12 Maths (Applications of Derivatives) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Applications of Derivatives) Miscellaneous Exercise

Set 1

Set 2

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