1. Do concentric circles have an envelope?
a) Yes
b) No
Explanation: Concentric circles are circles having the same center but different radii. No curve can be a tangent to all the circles. Hence, there is no envelope for concentric circles.
2. What is the relation between evolutes and envelopes?
a) Evolutes and envelopes are same
b) Evolute is the envelope of normals to a curve
c) Evolute is the envelope of tangents to a curve
d) Envelope is the evolute of normal to a curve
Explanation: Evolute is the locus of the all the centres of curvature of the curve whereas envelope is the curve which touches all the members of the family of the curve i.e Envelope is tangent to all the curves in a family of curves. Hence, evolute is the envelope of normal to a curve.
3. Find the envelope of the family of lines \(\frac{x}{t} \) + yt = 2c, t being the parameter.
a) xy = c
b) xy = 2c
c) xy = 2
d) xy = c2
Explanation: Given equation can be written as yt2 – 2ct + x = 0 ——–> eq(1)
The envelope of At2 + Bt + C = 0 is B2 – 4AC = 0 ———> eq(2)
From eq(1), A = y, B= -2c, C = x
Putting the values in eq(2),
(-2c)2 – 4(y)(x) = 0
4c2 – 4xy = 0
xy = c2.
4. What is the envelope of the family of straight lines y = mx +\(\frac{a}{m} \), m is the parameter?
a) y = 4ax
b) y2 = 4ax
c) x = 4ay
d) x2 = 4ay
Explanation: The given equation can be written as m2x – ym + a = 0 —-> eq(1) is in the form At2 + Bt + C =0
The envelope is given by B2 – 4AC = 0 ——> eq(2)
From eq(1), A = x , B = -y , C = a
Putting the values in eq(2),
(-y)2 – 4(x)(a) = 0
y2 = 4ax.
5. What is the envelope of straight lines given by x cos b + y sin b = a sec b, where b is the parameter?
a) y2 + 4a(a-x) = 0
b) y + 4ax = 0
c) y + 4a(a-x) = 0
d) y2 = 4a(a-x)
Explanation: Dividing the given equation by cos b,
x + y tan b = a \(\frac{secb}{cosb}\) = a sec2 b = a(1 + tan2 b)
The above equation can be written as a tan2 b – y tan b + (a-x) = 0 which is a quadratic equation in tan b
Hence, A = a, B = -y, C = a-x
The envelope is B2 – 4AC = 0
(-y)2 – 4(a)(a-x) = 0
y2 = 4a(a-x).
6. Find \(lt_{(x,y)\rightarrow(0,0)}\frac{121.x^{-5}.y^{\frac{13}{3}}}{y+(x)\frac{3}{2}}\)
a) ∞
b) 0
c) Does Not Exist
d) 121
Explanation: Put x = t : y = a.t3⁄2 we have
=\(lt_{(x,y)\rightarrow(0,0)}\frac{121.t^{-5}.(at^{\frac{3}{2}})^{\frac{13}{3}}}{t^{\frac{3}{2}}+t^{\frac{3}{2}}}\)
=\(lt_{(x,y)\rightarrow(0,0)}\frac{121.at^{\frac{13}{3}}.t^{\frac{3}{2}}}{2.t^{\frac{3}{2}}}\)
=\(lt_{(x,y)\rightarrow(0,0)}\frac{121.at^{\frac{13}{3}}}{2}\)
By varying a we get different limits along different paths
Hence, Does Not Exist is the right answer.
7. Find \(lt_{(x,y)\rightarrow(0,0)}\frac{y^6}{x^{10}y^2+x^{15}}\)
a) 0
b) 1
c) Does Not exist
d) ∞
Explanation: Put Put x = t : y = a.t5⁄2 we have
=\(lt_{(x,y)\rightarrow(0,0)}\frac{(a.t^{\frac{5}{2}})^6}{t^{10}.(a.t^{\frac{5}{2}})^2+t^{15}}\)
=\(lt_{(x,y)\rightarrow(0,0)}\frac{a^6}{a^2+1}\)
By varying a we get different limits along different paths
Hence, Does Not exist is the right answer.
8. Find \(lt_{(x,y)\rightarrow(0,0)}\frac{sec(y).sin(x)}{x}\)
a) ∞
b) 1⁄2
c) 1
d) 1⁄3
Explanation: Treating limits separately we have
lt(x, y)→(0, 0) sin(x)⁄x * lt(x, y)→(0,0) sec(y)
= 1 * 1
= 1.
9. Find \(lt_{(x,y)\rightarrow(0,0)}\frac{x^3-y^3}{(x-y)}\)
a) -1⁄2
b) 0
c) ∞
d) -90
Explanation: Simplifying the expression we have
\(lt_{(x,y)\rightarrow(0,0)}\frac{(x-y)(x^2+xy+y^2)}{(x-y)}\)
\(lt_{(x,y)\rightarrow(0,0)}\frac{(x^2+xy+y^2)}{1}\)=(02+0.0+02)
= 0.
10. Find \(lt_{(x,y)\rightarrow(0,1)}\frac{x+y-1}{\sqrt{x+y}-1}\)
a) 9
b) 0
c) 6
d) 2
Explanation: Simplifying the expression we have
=\(lt_{(x,y)\rightarrow(0,1)}\frac{(\sqrt{x+y}^2)-(1)^2}{\sqrt{x+y}-1}=lt_{(x,y)\rightarrow(0,1)}\frac{(\sqrt{x+y}+1).(\sqrt{x+y}-1)}{\sqrt{x+y}-1}\)
=\(lt_{(x,y)\rightarrow(0,1)}(\sqrt{x+y}+1)=\sqrt{1}+1\)
=2