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**Q.1. Examine the continuity of the function ****f(x) = x ^{3} + 2x^{2} â€“ 1 at x = 1**

We know that y = f(x) will be continuous at x = a if

Given: f(x) = x

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

Since

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

As

âˆ´ f(x) is discontinuous at x = 0.

Hence, f(x) is continuous at x = 2.**Q.5. at x = 4****Ans.**

âˆ´

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 Ã— [a number oscillating between â€“ 1 and 1]

= 0

= 0 Ã— [a number oscillating between â€“ 1 and 1]

As

Hence, f(x) is continuous at x = a.**Q.8. at x = 0****Ans.**

[âˆµ e^{âˆž} = 0]

[e^{â€“ âˆž}= 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 = â€“ 1

Hence, the value of k is â€“ 1.**Q.14. at x = 0****Ans.**

â‡’ k^{2} = 1 â‡’ k = Â± 1

Hence, the value of k is Â± 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.

âˆ´

â€“ 1 + a = a + b = 1 + b

âˆ´ â€“ 1 + a=a + b â‡’ b = â€“ 1

1 + b =a + b â‡’ a = 1

Hence, the value of a = 1 and b = â€“ 1.**Q.17. Given the function f (x) =Find the points of discontinuity of the composite function y = f (f (x)).****Ans.**

âˆ´

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

â‡’

So, if f(t) is discontinuous, then 2 â€“ x = 0

âˆ´ x = 2 and 2x â€“ 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 = Ï€.****Ans.**

Given that f(x) =

Put g(x) = sin x + cos x and h(x) =

âˆ´ 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 = Ï€.

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

Hence, f(x) =is continuous function at x = Ï€.**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 â€˜aâ€™ in its domain if

Lf â€²(c) = Rf â€²(c)

where Lf â€²(c) =

Lf â€²(2) â‰ Rf â€²(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:â€²

Lf â€²(c) = Rf â€²(c)

So, Lf â€²(0) = Rf â€²(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 â€²(2) = Rf â€²(2)

So, Lf â€²(2) â‰ Rf â€²(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) =

â‡’

For continuity at x = 5

L.H.L. = R.H.L.

So, f(x) is continuous at x = 5.

Now, for differentiability

âˆµ Lf â€²(5) â‰ Rf â€²(5)

Hence, f(x) is not differentiable at x = 5.**Q.24.**** A function f : R â†’ R satisfies the equation f( x + y) = f(x) f(y) for all x, y âˆˆ R, f(x) â‰ 0. Suppose that the function is differentiable at x = 0 and f â€² (0) = 2. ****Prove that f â€²(x) = 2 f (x).****Ans.**

Given that: f : R â†’ R satisfies the equation f(x + y) = f(x).f(y) âˆ€ x, y âˆˆ R, f(x) â‰ 0.

Let us take any point x = 0 at which the function f(x) is differentiable.

âˆ´

â‡’

Now

Hence, fâ€²(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 =

â‡’ log y = log 8^{x} â€“ log x^{8} â‡’ log y = x log 8 â€“ 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 x ^{5} )]**

Let y = log [log (log x

Differentiating both sides w.r.t. x

Hence,

Let y = sin âˆšx + cos

Differentiating both sides w.r.t. x

Hence,

Let y = sin

Differentiating both sides w.r.t. x

= n.sin

Hence,

Let y =

Differentiating both sides w.r.t. x

Hence,

Let y = sinx

Differentiating both sides w.r.t. x,

Hence,

Q.33.

Let

Differentiating both sides w.r.t. x

Hence,

Let y = (sin x)

Taking log on both sides,

log y = log (sin x)

â‡’ log y = cos x.log (sin x) [âˆµ log x

Differentiating both sides w.r.t. x,

Hence,

Let y = sin

Differentiating both sides w.r.t. x

= n.sin

= â€“ n.sin

Hence,

Let y = (x + 1)

Taking log on both sides,

log y = log [( x + 1)

â‡’ log y = log (x + 1)

[âˆµ log xy = log x + log y]

â‡’ log y = 2 log (x + 1) + 3 log (x + 2) + 4 log (x + 3)

[âˆµ log x

Differentiating both sides w.r.t. x,

= (x + 1)(x + 2)

= (x + 1)(x + 2)

Hence,

Let

Differentiating both sides w.r.t. x

Let

âˆµ

Differentiating both sides w.r.t. x

Hence,

Let y = tan

Differentiating both sides w.r.t. x

Hence,

Alternate solution

Let y =

[Dividing the Nr. and Den. by cos x/2]â‡’

âˆ´

Differentiating both sides w.r.t. x

Hence,

Let

â‡’

â‡’

â‡’

â‡’

Differentiating both sides with respect to x

Hence,

Let

Put x = cos Î¸ âˆ´ Î¸ = cos

â‡’

â‡’ y = sec

y = 3 cos

Differentiating both sides w.r.t. x

Hence,

Let

Put x = a tan Î¸

â‡’

â‡’

â‡’ y = tan

â‡’

Differentiating both sides w.r.t. x

Hence,

Let

Putting x

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

Differentiating both sides w.r.t. x

Hence,

Given that:

Differentiating both the given parametric functions w.r.t. t

âˆ´

Hence,

Given that:

Differentiating both the parametric functions w.r.t. Î¸.

âˆ´

Hence,

Given that: x = 3cosÎ¸ â€“ 2cos

Differentiating both the parametric functions w.r.t. Î¸

= â€“ 3 sin Î¸ â€“ 6 cos

= â€“ 3 sin Î¸ + 6 cos

= 3 cos Î¸ â€“ 6 sin

âˆ´

â‡’

Hence,

Given that sin x =

âˆ´ Taking sin x =

Differentiating both sides w.r.t t, we get

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

Now taking, tan y =

Differentiating both sides w.r.t, t, we get

â‡’

â‡’

â‡’

â‡’

Given that:

Differentiating both the parametric functions w.r.t. t

Given that: x = e

â‡’ cos 2t = log x and sin 2t = log y.

Differentiating both the parametric functions w.r.t. t

Now y = e

Hence,

Given that: x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 â€“ 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

= a[2(cos

= a [2 cos 4t + 2 cos 2t] [âˆµ cos 2x = cos

= 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] [âˆµ sin 2x = 2 sin x cos x]

= b [2 sin 4t - 2 sin 2t ] = 2b (sin 4t â€“ sin 2t)

Put

Hence,

Given that: x = 3 sin t â€“ sin 3t, y = 3 cos t â€“ cos 3t.

Differentiating both parametric functions w.r.t. t

= 3 cos t â€“ cos 3t.3 = 3(cos t â€“ cos 3t)

= â€“ 3 sin t + sin 3t.3 = 3(â€“ sin t + sin 3t)

âˆ´

Put

Hence,

Letand z = sin x.

Differentiating both the parametric functions w.r.t. x,

âˆ´

Hence,

Let

Put x = tan Î¸.

âˆ´

â‡’

â‡’

â‡’

â‡’

Differentiating both parametric functions w.r.t. Î¸

âˆ´

Given that:

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

â‡’

Hence,

Given that: sec (x + y) = xy

Differentiating both sides w.r.t. x

Given that: tan

â‡’ x

Differentiating both sides w.r.t. x.

â‡’

â‡’

Hence,

Given that: (x

â‡’ x

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

Hence,

Given that: ax

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

â‡’

â‡’

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

â‡’

â‡’

â‡’

â‡’

â‡’

Hence, Hence, proved.

Q.59. Ifprove that

Given that: x = e

Taking log on both the sides,

log x = log

â‡’

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

â‡’

Hence,

Given that: y

Taking log on both sides log y

â‡’x log y = (y â€“ x) log e

â‡’ x log y = y â€“ x [âˆµ log e = 1]

â‡’ x log y + x = y

â‡’ x (log y + 1) = y

â‡’

Differentiating both sides w.r.t. y

We know that

Hence,

Given that

â‡’ y = (cos x)

Taking log on both sides log y = y.log (cos x)

Differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

Hence,Hence, proved.

Given that: x sin (a + y) + sin a cos (a + y) = 0

â‡’ x sin (a + y) = â€“ sin a cos (a + y)

â‡’

Differentiating both sides w.r.t. y

â‡’

â‡’

â‡’

âˆ´

Hence, Hence proved.

Given that:

Put x = sin Î¸ and y = sin Ï•.

âˆ´ Î¸ = sin

â‡’

â‡’ cos Î¸ + cos Ï• = a(sin Î¸ â€“ sin Ï•)

â‡’

â‡’

â‡’

â‡’ sin

Differentiating both sides w.r.t. x

â‡’

âˆ´

Hence,

Given that: y = tan

Differentiating both sides w.r.t. y

Again differentiating both sides w.r.t. x

â‡’

â‡’

â‡’

âˆ´

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