## Normal to x^2/3 + y^2/3 = a^2/3 in the form y cosθ – x sinθ = a cos2θ ….

Q: Normal to x2/3 + y2/3 = a2/3 in the form y cosθ – x sinθ = a cos2θ ; Where θ is the angle which the normal makes with the x – axis .

Solution : x2/3 + y2/3 = a2/3

Differentiating with respect to x

$\displaystyle \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = – \frac{x^{-1/3}}{y^{-1/3}} = – \frac{y^{1/3}}{x^{1/3}}$

Hence Slope of normal is

$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = tan\theta$ ; Since it is given that normal makes an angle θ with x -axis .

$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = \frac{sin\theta}{cos\theta}$

$\displaystyle = \frac{x^{1/3}}{sin\theta} = \frac{y^{1/3}}{cos\theta} = [\frac{x^{2/3} + y^{2/3}}{sin^2 \theta + cos^2 \theta}]^{1/2} = a^{1/3}$

Hence , x = a sin3θ , y = a cos3θ

Therefore Equation of normal whose slope is tanθ will be

$\displaystyle y – a cos^3 \theta = \frac{sin\theta}{cos\theta} (x – a sin^3 \theta)$

$\displaystyle y cos\theta – a cos^4 \theta = x sin\theta – a sin^4 \theta)$

$\displaystyle y cos\theta – x sin\theta = a ( cos^4 \theta = – sin^4 \theta )$

$\displaystyle y cos\theta – x sin\theta = a cos 2 \theta$

## Find the equation of tangent and Normal to the curve at any point (x,y) x = a sin^3θ , y = a cos^3θ

Q: Find the equation of tangent and Normal to the curve at any point (x,y) x = a sin3θ , y = a cos3θ

Solution : x = a sin3θ , y = a cos3θ

$\displaystyle \frac{dx}{d\theta} = 3 a sin^2 \theta \; cos\theta$

$\displaystyle \frac{dy}{d\theta} = – 3 a cos^2 \theta \; sin\theta$

$\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta}/\frac{dx}{d\theta}$

$\displaystyle \frac{dy}{dx} = – \frac{- 3 a cos^2 \theta \; sin\theta}{3 a sin^2 \theta \; cos\theta}$

$\displaystyle \frac{dy}{dx} = – \frac{cos\theta}{sin\theta}$

= – cotθ

Equation of tangent is

$\displaystyle y – a cos^3 \theta = – \frac{cos\theta}{sin\theta} (x – a sin^3 \theta )$

$\displaystyle x cos\theta + y sin\theta = (a/2) sin2\theta$

Equation of Normal :

As Slope of tangent is – cotθ ; therefore Slope of normal is + tanθ .

Equation of Normal is

$\displaystyle y – a cos^3 \theta = \frac{sin\theta}{cos\theta} (x – a sin^3 \theta )$

$\displaystyle y cos\theta – x sin\theta = a (cos^4 \theta – sin^4 \theta )$

$\displaystyle y cos\theta – x sin\theta = a cos2 \theta$

## Find the equation of the tangent to the curve at any point (x,y) $\frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Q: Find the equation of the tangent to the curve at any point (x,y) $\displaystyle \frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Solution : $\displaystyle \frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Differentiating with respect to x we get ,

$\displaystyle m . \frac{x^{m-1}}{a^m} + m . \frac{y^{m-1}}{b^m} . \frac{dy}{dx} = 0$

$\displaystyle m . \frac{x^{m-1}}{a^m} = – m . \frac{y^{m-1}}{b^m} . \frac{dy}{dx}$

$\displaystyle \frac{x^{m-1}}{a^m} = – \frac{y^{m-1}}{b^m} . \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = -\frac{1}{a} (\frac{x}{a})^{m-1}.b (\frac{b}{y})^{m-1}$

Equation of tangent is

$\displaystyle Y – y = \frac{dy}{dx} (X-x)$

$\displaystyle Y- y = -\frac{1}{a} (\frac{x}{a})^{m-1}.b (\frac{b}{y})^{m-1} (X-x)$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} – (\frac{x}{a})^m = -\frac{Y}{b}(\frac{y}{b})^{m-1} + (\frac{y}{b})^m$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} + \frac{Y}{b}(\frac{y}{b})^{m-1} = (\frac{x}{a})^m + (\frac{y}{b})^m$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} + \frac{Y}{b}(\frac{y}{b})^{m-1} = 1$

## Find the domain and range of the function $y = \frac{1}{\sqrt{x-[x]}}$

Q: Find the domain and range of the function $\displaystyle y = \frac{1}{\sqrt{x-[x]}}$

Solution : The function is defined for all x ∈ R except when x – [x] = 0

⇒ x is an integer i.e. x ∈ I

Domain = R – I , where I is set of integers .

Also x – [x] > 0

Again $\displaystyle x – [x] = \frac{1}{y^2}$ from given relation

By the definition of x-[x] we know that

0 < x-[x] < 1

$\displaystyle 0 < \frac{1}{y^2} < 1$

y2 > 1

⇒ y2 – 1 > 0

(y+1)(y-1) > 0

y < -1 , y > 1 , but y is +ve

Hence y > 1

Hence Range = (1 , ∞)

## Find the domain and Range of $y = \frac{1}{\sqrt{4 + 3 cosx}}$

Q:  Find the domain and Range of $\displaystyle y = \frac{1}{\sqrt{4 + 3 cosx}}$

Solution: Domain = R , Since  – 1 ≤ cosx ≤ 1

$\displaystyle y \in [\frac{1}{\sqrt{7}} , 1 ]$

The function is many-one but not one-one as for so many values of x we will have same image as

cos(2nπ+x) = cos x

## Find the domain and Range of the function y = (x+2)/(x^2 – 8x -4)

Q: Find the domain and Range of the function $\displaystyle y = \frac{x+2}{x^2 – 8x -4}$

Solution : As we know that denominator should not be zero .

x2 – 8x -4 = 0

$x = 4 \pm 2 \sqrt{5}$

Domain $R – ( 4 \pm 2 \sqrt{5})$

Let $\displaystyle y = \frac{x+2}{x^2 – 8x -4}$

$\displaystyle x^2 y – (8y+1 )x – (4y+2) = 0$

Since x is real Δ = B2 – 4 A C ≥ 0

80 y2 + 24 y + 1 ≥ 0

(20 y + 1)(4y+1) = +ve

Hence , y ≤-1/4 or ≥ -1/20

Also when y = 0 , x = -2 ∈ R

Y = 0 also belongs to Range

Range = (-∞ , -1/4) U (-1/20 , ∞)

## What is the domain and Range of the function y = f(x) = (x^2 – 3x + 2)/(x^2+ x -6) .Find the limit of f(x) as x approaches 2 .

Q: What is the domain and Range of the function $\displaystyle y = f(x) = \frac{x^2 – 3x + 2}{x^2+ x -6}$ .Find the limit of f(x) as x approaches 2 .

Solution : $\displaystyle f(x) = \frac{x^2 – 3x + 2}{x^2 + x -6}$

$\displaystyle y = f(x) = \frac{(x-1)(x-2)}{(x+3)(x-2)}$

Therefore function is not defined at x = 2 and x= -3

Hence domain of f = {x: x ∈R , x ≠-3 , x ≠2}

$\displaystyle lim_{x \rightarrow 2} \frac{x^2 – 3x + 2}{x^2+ x -6}$

$\displaystyle = lim_{\rightarrow 2}\frac{(x-1)(x-2)}{(x+3)(x-2)}$

$\displaystyle = lim_{\rightarrow 2}\frac{(x-1)}{(x+3)} = \frac{1}{5}$

To find the range of f we first observe that it can not take the value of 1/5 since it is not defined at x = 2 . For x ≠ 2 we have

$\displaystyle y = f(x) = \frac{x-1}{x+3}$

x y + 3 y = x -1

$\displaystyle x = -\frac{3y+1}{y-1}$

Hence y ≠1 in the domain of x . Also at x = -3 , y = ∞

Thus f(x) takes all real values in the domain of x except y =1/5 and y = 1

Hence range of f = {y : y∈R , y ≠1/5 , y ≠ 1}

## If f(x) = 1 + x^2 and f[g(x)] = 1 + x^2 – 2x^3 + x^4 , then determine the function g(x) along with its domain and range .

Q: If f(x) = 1 + x2 and f[g(x)] = 1 + x2 – 2x3 + x4 , then determine the function g(x) along with its domain and range .

Solution : f[g(x)] = 1 + [g(x)]2

1 + x2 – 2x3 + x4 = 1 + [g(x)]2

[g(x)]2 = x2 – 2x3 + x4

= x2 (x2 – 2 x + 1 )

= x2 (x-1)2

g(x) = ± x (x-1)

g(x) = ± (x2 – x)

$\displaystyle = \pm (x^2 – x + \frac{1}{4} – \frac{1}{4})$

$\displaystyle = \pm [(x-\frac{1}{2})^2 – \frac{1}{4}]$

$\displaystyle g(x) = (x-\frac{1}{2})^2 – \frac{1}{4}$ ….(i)

or , $\displaystyle g(x) = \frac{1}{4} – (x-\frac{1}{2})^2$ …(ii)

In either case domain of g(x) is R .

In (i) of g(x) it is always ≥ -1/4

∴ Range is [-1/4 , ∞)

In the form (ii) of g(x) it is always ≤ 1/4

∴ Range is (-∞ , 1/4]

## Find the range of the function y = 3 sinx + 4 cos(x + π/3) + 7

Q: Find the range of the function y = 3 sinx + 4 cos(x + π/3) + 7

Solution : $\displaystyle cos (x + \frac{\pi}{3}) = cos x cos\frac{\pi}{3} – sinx sin\frac{\pi}{3}$

$\displaystyle cos (x + \frac{\pi}{3}) = \frac{1}{2} cos x – \frac{\sqrt{3}}{2} sinx$

$\displaystyle y = 3 sinx + 4 (\frac{1}{2} cos x – \frac{\sqrt{3}}{2} sinx) + 7$

$\displaystyle y = 3 sinx + 2 cos x – 2 \sqrt{3} sinx + 7$

$\displaystyle y = ( 3 – 2 \sqrt{3} ) sinx + 2 cos x + 7$

Put $\displaystyle 3 – 2\sqrt{3} = r cos \alpha$ …(i)

and $2 = r sin\alpha$  …(ii)

$\displaystyle y = r sinx . cos \alpha + r cos x . sin\alpha + 7$

$\displaystyle y = r sin(x + \alpha) + 7$

Range of sin(x+α) is -1 to 1

Range of y = -r + 7 , r + 7

Now by squaring & adding equation (i) & (ii) we can get the value of r .

## Find the range of the function $y = sin^{-1}[\frac{1}{2} + x^2]$ ; Where [] denotes the greatest integer function

Q: Find the range of the function $\displaystyle y = sin^{-1}[\frac{1}{2} + x^2]$ ; Where [] denotes the greatest integer function .

Solution : $\displaystyle siny = [\frac{1}{2} + x^2] \le 1$

Since $\frac{1}{2} + x^2$ is +ve for all x .

$\displaystyle [\frac{1}{2} + x^2]$ is a +ve integer ≤ 1 i.e. 0 and 1 only .

Therefore siny = 0 or 1

y = 0 , π/2 ;

Range of y = {0 , π/2} only .