NCERT Exemplar: Continuity and Differentiability- 1 | Mathematics (Maths) for JEE Main & Advanced PDF Download

Q.1. Examine the continuity of the function 
f(x) = x³ + 2x² – 1 at x = 1
Ans.
We know that y = f(x) will be continuous at x = a if

Given: f(x) = x³ + 2x² – 1


Hence, f(x) is continuous at x = 1.

Find which of  the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:
Q.2. at x = 2
Ans.



Since
Hence f (x) is discontinuous at x = 2.

Q.3. at x = 0
Ans.






As
∴ f(x) is discontinuous at x = 0.

Q.4. at x = 2
Ans.

Hence, f(x) is continuous at x = 2.

Q.5. at x = 4
Ans.



∴
Hence, f(x) is discontinuous at x = 4.

Q.6. at x = 0
Ans.





Hence, f(x) is continuous at x = 0.

Q.7. at x = a
Ans.


= 0 × [a number oscillating between – 1 and 1]
= 0


= 0 × [a number oscillating between – 1 and 1]

As
Hence, f(x) is continuous at x = a.

Q.8. at x = 0
Ans.


[∵ e^∞ = 0]


[e^{– ∞}= 0]

As
Hence, f(x) is discontinuous at x = 0.

Q.9. at x = 1
Ans.



Hence, f(x) is continuous at x = 1.

Q.10.
Ans.




Hence, f(x) is continuous at x = 1.

Find the value of k in each of the Exercises 11 to 14 so that the function f is continuous at the indicated point:
Q.11. at x = 5
Ans.


As the function is continuous at x = 5

Hence, the value of k is

Q.12. at x = 2
Ans.



As the function is continuous at x = 2.

Hence, value of k is

Q.13. at x = 0
Ans.






As the function is continuous at x = 0.

k = – 1
Hence, the value of k is – 1.

Q.14. at x = 0
Ans.








⇒ k² = 1 ⇒ k = ± 1
Hence, the value of k is ± 1.

Q.15. Prove that the function f defined by

remains discontinuous at x = 0, regardless the choice of k.
Ans.





Hence, f(x) is discontinuous at x = 0 regardless the choice of k.

Q.16. Find the values of a and b such that the function f  defined by

is a continuous function at x = 4.
Ans.




As the function is continuous at x = 4.
∴
– 1 + a = a + b = 1 + b
∴ – 1 + a=a + b ⇒ b = – 1
1 + b =a + b ⇒ a = 1
Hence, the value of a = 1 and b = – 1.

Q.17. Given the function f (x) =Find the points of discontinuity of the composite function y = f (f (x)).
Ans.


∴
This function will not be defined and continuous where
    
Hence,is the point of discontinuity.

Q.18 Find all points of discontinuity of the function
where
Ans.
We have
⇒


So, if f(t) is discontinuous, then 2 – x = 0 
∴ x = 2 and 2x – 1 = 0
Hence, the required points of discontinuity are 2 and 1/2.

Q.19. Show that the function f(x) = sin x + cos x is continuous at x = π.
Ans.
Given that f(x) =
Put g(x) = sin x + cos x and h(x) =
∴ h[g(x)] = h(sin x + cos x) =

Now, g(x) = sin x + cos x is a continuous function since sin x and cos x are two continuous functions at x = π.

We know that every modulus function is a continuous function everywhere.

Hence, f(x) =is continuous function at x = π.

Examine the differentiability of f, where f is defined by

Q.20. at x = 2.
Ans.
We know that a function f is differentiable at a point ‘a’ in its domain if
Lf ′(c) = Rf ′(c)
where Lf ′(c) =






Lf ′(2) ≠ Rf ′(2)
Hence, f(x) is not differentiable at x = 2.

Q.21. Examine the differentiability of f, where f is defined by
at x = 0.
Ans.
Given that:

For differentiability we know that:′
Lf ′(c) = Rf ′(c)




So, Lf ′(0) = Rf ′(0)  = 0
Hence, f(x) is differentiable at x = 0.

Q.22. at x = 2.
Ans.
 f(x) is differentiable at x = 2 if
Lf ′(2) = Rf ′(2)




So, Lf ′(2) ≠ Rf ′(2)
Hence, f(x) is not differentiable at x = 2.

Q.23. Show that f(x) =  is continuous but not differentiable at x = 5.
Ans.
We have f(x) =
⇒
For continuity at x = 5


L.H.L. = R.H.L.
So, f(x) is continuous at x = 5.
Now, for differentiability


∵ Lf ′(5) ≠ Rf ′(5)
Hence, f(x) is not differentiable at x = 5.

Q.24. A function f : R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f ′ (0) = 2. Prove that f ′(x) = 2 f (x).
Ans.
Given that: f : R → R satisfies the equation f(x + y) = f(x).f(y) ∀ x, y ∈ R, f(x) ≠ 0.
Let us take any point x = 0 at which the function f(x) is differentiable.
∴

⇒
Now

Hence, f′(x) = 2f(x).

Differentiate each of the following w.r.t. x (Exercises 25 to 43) :
Q.25.
Ans.
Let
Taking log on both sides, we get

Differentiating both sides w.r.t. x


Hence,

Q.26.
Ans.
Let
Taking log on both sides, we get,  log y =
⇒ log y = log 8^x – log x⁸ ⇒ log y = x log 8 – 8 log x
Differentiating both sides w.r.t. x

Hence,

Q.27.
Ans.
Let
Differentiating both sides w.r.t. x



Hence,

Q.28. log [log (log x⁵ )]
Ans.
Let y = log [log (log x⁵ )]
Differentiating both sides w.r.t. x



Hence,

Q.29. sin √x + cos² √x
Ans.
Let y = sin √x + cos² √x
Differentiating both sides w.r.t. x



Hence,

Q.30. sinⁿ (ax² + bx + c)
Ans.
Let y = sinⁿ (ax² + bx + c)
Differentiating both sides w.r.t. x


= n.sin^{n – 1}(ax² + bx + c).cos(ax² + bx + c).(2ax + b)
Hence,

Q.31.  
Ans.
Let y = 
Differentiating both sides w.r.t. x


Hence,

Q.32. sinx² + sin²x + sin²(x²)
Ans.
Let y = sinx² + sin²x + sin²(x²)
Differentiating both sides w.r.t. x,


Hence,

Q.33. 
Ans.
Let 
Differentiating both sides w.r.t. x



Hence,

Q.34. ( sin x )^{cos x}
Ans.
Let y = (sin x)^{cos x}
Taking log on both sides,
log y = log (sin x)^{cos x}
⇒ log y = cos x.log (sin x) [∵ log x^y = y log x]
Differentiating both sides w.r.t. x,



Hence,

Q.35. sin^mx . cosⁿx
Ans.
 Let y = sin^m x . cosⁿ x
Differentiating both sides w.r.t. x


= n.sin^m x.cos^{n – 1} x.(– sin x) + m.cosⁿ x.sin^{m –1} x.cos x
= – n.sin^m + 1 x.cos^{n – 1} x + m cos^n + 1 x. sin^{m – 1}x

Hence,

Q.36. (x + 1)² (x + 2)³ (x + 3)⁴
Ans.
Let y = (x + 1)²(x + 2)³(x + 3)⁴ 
Taking log on both sides,
log y = log [( x + 1)² .(x+ 2)³ .( x + 3)⁴ ]
⇒ log y = log (x + 1)² + log (x + 2)³ + log (x + 3)⁴
[∵ log xy = log x + log y]
⇒ log y = 2 log (x + 1) + 3 log (x + 2) + 4 log (x + 3)
[∵ log x^y = y log x]
Differentiating both sides w.r.t. x,



= (x + 1)(x + 2)²(x + 3)³(2x² + 10x + 12 + 3x² + 12x + 9 + 4 x² + 12x + 8)
= (x + 1)(x + 2)²(x + 3)³(9x² + 34x + 29)
Hence,

Q.37.
Ans.
Let


Differentiating both sides w.r.t. x


Q.38.
Ans.
Let

∵
Differentiating both sides w.r.t. x

Hence,

Q.39.
Ans.
Let y = tan^{– 1}(sec x + tan x)
Differentiating both sides w.r.t. x



Hence,
Alternate solution
Let y = 


[Dividing the Nr. and Den. by cos x/2]⇒

∴
Differentiating both sides w.r.t. x

Hence,

Q.40.
Ans.
Let 
⇒
⇒
⇒

⇒
Differentiating both sides with respect to x

Hence,

Q.41.
Ans.
Let
Put x = cos θ ∴ θ = cos^{– 1} x

⇒
⇒ y = sec^{– 1} (sec 3θ) ⇒ y = 3θ
y = 3 cos^{– 1} x
Differentiating both sides w.r.t. x

Hence,

Q.42.
Ans.
Let
Put x = a tan θ

⇒
⇒
⇒ y = tan ^{- 1}[tan 3θ]
⇒
Differentiating both sides w.r.t. x

Hence,

Q.43.
Ans.
Let
Putting x² = cos 2θ

⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Differentiating both sides w.r.t. x


Hence,   

Find of each of the functions expressed in parametric form in Exercises from 44 to 48.
Q.44.
Ans.
Given that:
 
Differentiating both the given parametric functions w.r.t. t

∴
Hence,

Q.45.
Ans.
Given that:
 
Differentiating both the parametric functions w.r.t. θ.




∴

Hence,

Q.46. x = 3cosθ – 2cos³θ, y = 3sinθ – 2sin³θ.
Ans.
Given that: x = 3cosθ – 2cos³θ, y = 3sinθ – 2sin³θ.
Differentiating both the parametric functions w.r.t. θ

= – 3 sin θ – 6 cos² θ . (– sin θ) 
= – 3 sin θ + 6 cos² θ . sin θ

= 3 cos θ – 6 sin² θ . cos θ
∴
⇒

Hence,

Q.47.
Ans.
Given that sin x =
∴ Taking sin x =
Differentiating both sides w.r.t t, we get


⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Now taking, tan y =
Differentiating both sides w.r.t, t, we get



⇒

⇒
⇒
⇒


Q.48.
Ans.
Given that:
Differentiating both the parametric functions w.r.t. t





Q.49. If x = e^cos2t and y = e^sin2t, prove that
Ans.
Given that: x = e^cos2t and y = e^sin2t 
⇒ cos 2t = log x and sin 2t = log y.
Differentiating both the parametric functions w.r.t. t

Now y = e^sin2t


Hence,

Q.50. If x = a sin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that

Ans.
Given that: x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t).
Differentiating both the parametric functions w.r.t. t

= a [sin 2t . ( - sin 2t).2 + (1+ cos 2t) (cos 2t).2]
= a[- 2 sin ² 2t + 2 cos 2t+ 2 cos² 2t]
= a[2(cos² 2t - sin² 2t)+ 2 cos 2t]
= a [2 cos 4t + 2 cos 2t] [∵ cos 2x = cos² x - sin² x]
= 2a [cos 4t + cos 2t]
y = b cos 2t (1 - cos 2t)

= b [cos 2t. sin 2t.2 + (1- cos 2t ).( - sin 2t ).2]
= b [2 sin 2t. cos 2t - 2 sin 2t+ 2 sin 2t cos 2t]
= b [sin 4t - 2 sin 2t+ sin 4t] [∵ sin 2x = 2 sin x cos x]
= b [2 sin 4t - 2 sin 2t ] = 2b (sin 4t – sin 2t)

Put

Hence,

Q.51. If x = 3sint – sin 3t, y = 3cost – cos 3t, find
Ans.
Given that: x = 3 sin t – sin 3t, y = 3 cos t – cos 3t.
Differentiating both parametric functions w.r.t. t
= 3 cos t – cos 3t.3 = 3(cos t – cos 3t)
= – 3 sin t + sin 3t.3 = 3(– sin t + sin 3t)
∴
Put


Hence,

Q.52. Differentiatew.r.t. sinx.
Ans.
Letand z = sin x.
Differentiating both the parametric functions w.r.t. x,

∴

Hence,

Q.53. Differentiate tan^–1w.r.t. tan^–1 x when x ≠ 0.
Ans.
Let
Put x = tan θ.
∴
⇒
⇒
⇒
⇒
Differentiating both parametric functions w.r.t. θ

∴

Find  when x and y are connected by the relation given in each of the Exercises 54 to 57.
Q.54.
Ans.
Given that:
Differentiating both sides w.r.t. x


⇒
⇒
⇒
⇒

Hence,

Q.55. sec (x + y) = xy
Ans.
Given that: sec (x + y) = xy
Differentiating both sides w.r.t. x





Q.56. tan^–1 (x² + y²) = a
Ans.
Given that: tan^{– 1}(x² + y²) = a
⇒ x² + y² = tan a.
Differentiating both sides w.r.t. x.

⇒
⇒
Hence,

Q.57. (x² + y²)² = xy
Ans.
Given that: (x² + y²)² = xy
⇒ x⁴ + y⁴ + 2x²y² = xy
Differentiating both sides w.r.t. x


⇒
⇒
⇒
Hence,

Q.58. If ax² + 2hxy + by² + 2gx + 2fy + c = 0, then show that
Ans.
Given that: ax² + 2hxy + by² + 2gx + 2fy + c = 0.
Differentiating both sides w.r.t. x


⇒
⇒
⇒
⇒
⇒
Now, differentiating the given equation w.r.t. y.

⇒
⇒
⇒
⇒
⇒

Hence, Hence, proved.

Q.59. Ifprove that
Ans.
Given that: x = e^x/y 
Taking log on both the sides,
log x = log ^ex/y
⇒
Differentiating both sides w.r.t. x

⇒
⇒
⇒
⇒
Hence,

Q.60. If y^x = e^{y − x} , prove that
Ans.
Given that: y^x = e^{y – x}
Taking log on both sides  log y^x = log e^{y – x}
⇒x log y = (y – x) log e
⇒ x log y = y – x [∵ log e = 1]
⇒ x log y + x = y
⇒ x (log y + 1) = y
⇒
Differentiating both sides w.r.t. y


We know that

Hence,

Q.61. If  , show that
Ans.
Given that 
⇒ y = (cos x)^y
Taking log on both sides log y = y.log (cos x)
Differentiating both sides w.r.t. x

⇒
⇒

⇒
Hence,Hence, proved.

Q.62. If x sin (a + y) + sin a cos (a + y) = 0, prove that
Ans.
Given that:  x sin (a + y) + sin a cos (a + y) = 0
⇒ x sin (a + y) = – sin a cos (a + y)
⇒
Differentiating both sides w.r.t. y
⇒
⇒
⇒
∴
Hence, Hence proved.

Q.63. If = a (x – y), prove that
Ans.
Given that: = a (x – y)
Put x = sin θ and y = sin ϕ.
∴ θ = sin^{– 1} x and ϕ = sin^{– 1} y

⇒
⇒ cos θ + cos ϕ = a(sin θ – sin ϕ)
⇒

⇒
⇒
⇒ sin^{– 1} x – sin^{– 1} y = 2 cot^{– 1} a
Differentiating both sides w.r.t. x

⇒
∴
Hence,

Q.64. If y = tan^–1x, findin terms of y alone.
Ans.
Given that: y = tan^{– 1} x ⇒ x = tan y
Differentiating both sides w.r.t. y

Again differentiating both sides w.r.t. x

⇒
⇒
⇒
∴