Answer :

The given function is defined at all points of real line. So, let d be any point on the number line.

Case I: When d < -3

For this case, we can define function as,

f(x) = -x + 3, x ≤ -3

So,

At d < -3, f(x) = -x + 3

And f(d) = -d + 3

That is,

Therefore, the function is continuous at all points x < -3.

Case II: When d = -3

We have the function f(x) = -x + 3, x = -3

So,

At d = -3, f(x) = -x + 3

And f(-3) = -(-3) + 3

⇒ f(-3) = 3 + 3

⇒ f(-3) = 6

For the function to be continuous at x = -3,

Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(-3)

That is,

For LHL,

We have the function f(x) = -x + 3, x < -3

So,

For RHL,

We have the function f(x) = -2x, x > -3

So,

Since,

Therefore, the function is continuous at x = -3.

Case III: When -3 < d < 3

For this case, we have the function f(x) = -2x, -3 < x < 3

At -3 < d < 3, f(x) = -2x

And f(d) = -2d

That is,

Therefore, the function is continuous at all points -3 < x < 3.

Case IV: When d = 3

We have the function f(x) = 6x + 2, x = 3

So,

At d = 3, f(x) = 6x + 2

And f(3) = 6(3) + 2

⇒ f(3) = 18 + 2

⇒ f(3) = 20

For the function to be continuous at x = 3,

Left Hand Limit (LHL) = Right Hand Limit (RHL) = f(3)

That is,

For LHL,

We have the function f(x) = -2x, x < 3

So,

For RHL,

We have the function f(x) = 6x + 2, x > 3

So,

Since,

Therefore, the function is discontinuous at x = 3.

Case V: When d > 3

For this case, we have the function f(x) = 6x + 2, x > 3

So,

At d > 3, f(x) = 6x + 2

And f(d) = 6d + 2

That is,

Therefore, the function is continuous at all points x > 3.

**Thus, the function is discontinuous at x = 3.**

**OR**

We need to find the differentiation of y with respect to x.

Let u = (cos x)^{x} …(i)

Let v = (sin x)^{1/x} …(ii)

For u = (cos x)^{x}

Taking log on both sides, we get

log u = log (cos x)^{x}

⇒ log u = x log (cos x)

Differentiate u with respect to x,

Substitute value of u from eq (i),

…(iii)

For v = (sin x)^{1/x}

Taking log on both sides, we get

log v = log (sin x)^{1/x}

Differentiate v with respect to x,

Substitute value of v from eq (ii),

…(iv)

We know,

y = u + v

Substituting the value of and from eq (iii) and (iv),

Rearranging it,

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