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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.6

### Question 1. x = 2at2, y = at4

Solution:

Here, x = 2at2, y = at

= 2a

= 2a (2t)

= 4at

And, now

= a

= a (4t3)

= 4at3

Now, as

= t2

### Question 2. x = a cos(θ), y = b cos(θ)

Solution:

Here, x = a cos(θ), y = b cos(θ)

= a

= a (-sin(θ))

= – a sin(θ)

And, now

= b

= b (-sin(θ))

= – b sin(θ)

Now, as

### Question 3. x = sin(t), y = cos(2t)

Solution:

Here, x = sin(t), y = cos(2t)

= cos(t)

And, now

= -sin(2t)

= – 2sin(2t)

Now, as

(Using the identity: sin(2θ) = 2 sinθ cosθ)

= – 4 sin(t)

### Question 4. x = 4t, y =

Solution:

Here, x = 4t, y = 4/t

= 4

= 4

And, now

= 4

= 4

= 4

= 4

Now, as

### Question 5. x = cos(θ) – cos(2θ), y = sin(θ) – sin(2θ)

Solution:

Here, x = cos(θ) – cos(2θ), y = sin(θ) – sin(2θ)

= – sin(θ) – (-sin(2θ))

= – sin(θ) + 2sin(2θ)

And, now

= cos(θ) – (cos(2θ))

= cos(θ) – (2 cos(2θ)

Now, as

### Question 6. x = a (θ – sin(θ)), y = a (1 + cos(θ))

Solution:

Here, x = a (θ – sin(θ)), y = a (1 + cos(θ))

= a ()

= a (1 – cos(θ))

And, now

= a ()

= a (0 + (- sin (θ)))

= – a sin (θ)

Now, as

=

(Using identity: sin(2θ) = 2 sinθ cosθ and 1- cos(2θ) = 2 sin2θ)

= – cot(θ/2)

Solution:

Here, x =

=

And, now

=

Now, as

= – cot 3(t)

### Question 8. x = a (cos(t) + log tan), y = a sin(t)

Solution:

Here, x = a (cos(t) + log tan ), y = a sin(t)

= a ()

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )              (Using identity: 2 sinθ cosθ = sin(2θ))

= a ( – sin(t))

= a ()

= a ()

=

And, now

= a

= a cos(t)

Now, as

= tan(t)

### Question 9. x = a sec(θ), y = b tan(θ)

Solution:

Here, x = a sec(θ), y = b tan(θ)

= a ()

= a (sec(θ) tan(θ))

= a sec(θ) tan(θ)

And, now

= b ()

= b (sec2(θ))

Now, as

### Question 10. x = a (cos(θ) + θ sin(θ)), y = a (sin(θ) – θ cos(θ))

Solution:

Here, x = a (cos(θ) + θ sin(θ)), y = a (sin(θ) – θ cos(θ))

= a ()

= a (- sin(θ) + (θ.) + sin(θ).)

= a (- sin(θ) + (θ.(cos(θ) + sin(θ).1))

= a (- sin(θ) + θ cos(θ) + sin(θ))

= aθ cos(θ)

And, now

= a ()

= a (cos (θ) – (θ.) + cos(θ).)

= a (cos(θ) – (θ.(-sin (θ) + cos(θ).1))

= a (cos(θ) + θ sin(θ) – cos(θ))

= aθ sin(θ)

Now, as

= tan(θ)

### Question 11. If x = , y = , show that

Solution:

Here, Let multiply x and y.

xy = (

= ()

= ()                           (Using identity: sin-1θ + cos-1θ = )

Let’s differentiate w.r.t x,

x.+ y. = 0

x. + y = 0

Hence, Proved !!!

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