CONTINUITY AND DIFFERENTIABILITY

Exercise 5.1

Q1: Prove that the function f(x) = 5x − 3 is continuous at x = 0, at x = −3 and at x = 5.

Solution:

A function f(x) is said to be continuous at x = a if

1. f(a) exists

2. limit of f(x) as x approaches a exists

3. limit of f(x) as x approaches a = f(a)

Continuity at x = 0

f(x) = 5x − 3

f(0) = 5(0) − 3 = −3

limit of f(x) as x approaches 0
= limit of (5x − 3) as x approaches 0
= 5(0) − 3
= −3

So,
limit of f(x) as x approaches 0 = f(0)

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

Continuity at x = −3

f(−3) = 5(−3) − 3
= −15 − 3
= −18

limit of f(x) as x approaches −3
= limit of (5x − 3) as x approaches −3
= 5(−3) − 3
= −18

So,
limit of f(x) as x approaches −3 = f(−3)

Hence, f(x) is continuous at x = −3.

Continuity at x = 5

f(5) = 5(5) − 3
= 25 − 3
= 22

limit of f(x) as x approaches 5
= limit of (5x − 3) as x approaches 5
= 5(5) − 3
= 22

So,
limit of f(x) as x approaches 5 = f(5)

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

Conclusion:

The function f(x) = 5x − 3 is continuous at x = 0, x = −3 and x = 5.

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Q2: Examine the continuity of the function f(x) = 2x² − 1 at x = 3.

Solution:

A function f(x) is said to be continuous at x = a if

1. f(a) exists

2. limit of f(x) as x approaches a exists

3. limit of f(x) as x approaches a = f(a)

Step 1: Find f(3)

f(3) = 2(3)² − 1
= 2(9) − 1
= 18 − 1
= 17

Step 2: Find limit of f(x) as x approaches 3

limit of f(x) as x approaches 3
= limit of (2x² − 1) as x approaches 3
= 2(3)² − 1
= 17

Step 3: Compare the values

limit of f(x) as x approaches 3 = f(3)

Conclusion:

Since all the conditions of continuity are satisfied, the function
f(x) = 2x² − 1 is continuous at x = 3.

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Q3: Examine the following functions for continuity.

(a) f(x) = x − 5

(b) f(x) = 1 / (x − 5), x ≠ 5

(c) f(x) = (x² − 25) / (x + 5), x ≠ −5

(d) f(x) = |x − 5|

Solution:

(a) f(x) = x − 5

This is a polynomial function.
Polynomial functions are continuous for all real values of x.

So, f(x) = x − 5 is continuous everywhere. ✔

(b) f(x) = 1 / (x − 5), x ≠ 5

This is a rational function.
A rational function is continuous at all points where it is defined.

Here, f(x) is not defined at x = 5, but it is continuous for all x ≠ 5.

So, f(x) is continuous on its domain. ✔

(c) f(x) = (x² − 25) / (x + 5), x ≠ −5

Factor numerator:
x² − 25 = (x − 5)(x + 5)

So, for x ≠ −5,
f(x) = x − 5

This is a polynomial expression and hence continuous for all x ≠ −5.

Therefore, f(x) is continuous on its domain.

(d) f(x) = |x − 5|

Modulus functions are continuous for all real values of x.

So, f(x) = |x − 5| is continuous everywhere.

Conclusion:

Functions (a), (b), (c) and (d) are all continuous on their respective domains.

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Q4: Prove that the function f(x) = xⁿ is continuous at x = n, where n is a positive integer.

Solution:

A function f(x) is continuous at x = a if

1. f(a) exists

2. limit of f(x) as x approaches a exists

3. limit of f(x) as x approaches a = f(a)

Let
f(x) = xⁿ, where n is a positive integer.

Step 1: Find f(n)

f(n) = nⁿ

So, f(n) exists.

Step 2: Find the limit of f(x) as x approaches n

limit of f(x) as x approaches n
= limit of xⁿ as x approaches n

Since xⁿ is a polynomial function and polynomial functions are continuous for all real values of x,

limit of xⁿ as x approaches n = nⁿ

Step 3: Compare the values

limit of f(x) as x approaches n = nⁿ
f(n) = nⁿ

Therefore,
limit of f(x) as x approaches n = f(n)

Conclusion:

All the conditions of continuity are satisfied.
Hence, the function f(x) = xⁿ is continuous at x = n, where n is a positive integer.

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Q5: Is the function f defined by

f(x) = x, if x ≤ 1
f(x) = 5, if x > 1

continuous at x = 0? At x = 1? At x = 2?

Solution:

A function f(x) is continuous at x = a if

1. f(a) exists

2. limit of f(x) as x approaches a exists

3. limit of f(x) as x approaches a = f(a)

Continuity at x = 0

Since 0 ≤ 1,
f(0) = 0

limit of f(x) as x approaches 0
= limit of x as x approaches 0
= 0

So,
limit of f(x) as x approaches 0 = f(0)

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

Continuity at x = 1

f(1) = 1 (since x ≤ 1)

Left hand limit at x = 1:
limit of f(x) as x approaches 1 from left
= limit of x as x approaches 1
= 1

Right hand limit at x = 1:
limit of f(x) as x approaches 1 from right
= limit of 5 as x approaches 1
= 5

Since,
Left hand limit ≠ Right hand limit,

limit of f(x) as x approaches 1 does not exist.

Hence, f(x) is not continuous at x = 1.

Continuity at x = 2

Since 2 > 1,
f(2) = 5

limit of f(x) as x approaches 2
= limit of 5 as x approaches 2
= 5

So,
limit of f(x) as x approaches 2 = f(2)

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

Conclusion:

The function f(x) is
• continuous at x = 0
• not continuous at x = 1
• continuous at x = 2

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Q6: 6. f(x) = { 2x + 3, if x ≤ 2
2x − 3, if x > 2 }

Solution:

Check continuity at x = 2.

f(2) = 2(2) + 3 = 7

Left hand limit as x approaches 2
= limit of (2x + 3) as x approaches 2
= 7

Right hand limit as x approaches 2
= limit of (2x − 3) as x approaches 2
= 1

Since left hand limit ≠ right hand limit,

f(x) is not continuous at x = 2.

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Q7: f(x) = { |x| + 3, if x ≤ −3
−2x, if −3 < x < 3
6x + 2, if x ≥ 3 }

Solution:

Check continuity at x = −3 and x = 3.

At x = −3:
f(−3) = |−3| + 3 = 6

Left hand limit as x approaches −3
= limit of (|x| + 3)
= 6

Right hand limit as x approaches −3
= limit of (−2x)
= 6

So, f(x) is continuous at x = −3.

At x = 3:
f(3) = 6(3) + 2 = 20

Left hand limit as x approaches 3
= limit of (−2x)
= −6

Right hand limit as x approaches 3
= limit of (6x + 2)
= 20

Since left hand limit ≠ right hand limit,

f(x) is not continuous at x = 3.

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Q8: f(x) = { |x| / x, if x ≠ 0
0, if x = 0 }

Solution:

Check continuity at x = 0.

f(0) = 0

Left hand limit as x approaches 0
For x < 0, |x| / x = −1

Right hand limit as x approaches 0
For x > 0, |x| / x = 1

Since left hand limit ≠ right hand limit,

f(x) is not continuous at x = 0.

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Q9: f(x) = { x / |x|, if x < 0
−1, if x ≥ 0 }

Solution:

Check continuity at x = 0.

f(0) = −1

Left hand limit as x approaches 0
For x < 0, x / |x| = −1

Right hand limit as x approaches 0
For x ≥ 0, f(x) = −1

Left hand limit = Right hand limit = f(0)

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

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Q10: f(x) = { x + 1, if x ≥ 1
x² + 1, if x < 1 }

Solution:

Both parts of the function are polynomial expressions, hence continuous in their respective intervals.

The only point to check is x = 1.

f(1) = 1 + 1 = 2

Left hand limit as x approaches 1
= limit of (x² + 1) as x approaches 1
= 1 + 1 = 2

Right hand limit as x approaches 1
= limit of (x + 1) as x approaches 1
= 2

Left hand limit = Right hand limit = f(1)

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

Since there is no other point where the function definition changes,

there is no point of discontinuity.

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Q11: f(x) = { x³ − 3, if x ≤ 2
x² + 1, if x > 2 }

Solution:

Both x³ − 3 and x² + 1 are polynomial functions, hence continuous in their respective intervals.

The only point to check is x = 2.

f(2) = 2³ − 3 = 8 − 3 = 5

Left hand limit as x approaches 2
= limit of (x³ − 3) as x approaches 2
= 5

Right hand limit as x approaches 2
= limit of (x² + 1) as x approaches 2
= 4 + 1 = 5

Left hand limit = Right hand limit = f(2)

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

Since there is no other point where the function definition changes,

there is no point of discontinuity.

Q12. f(x) = { x¹⁰ − 1, if x ≤ 1
x², if x > 1 }

Solution:

Check continuity at x = 1.

f(1) = 1¹⁰ − 1 = 0

Left hand limit as x approaches 1
= limit of (x¹⁰ − 1) as x approaches 1
= 0

Right hand limit as x approaches 1
= limit of (x²) as x approaches 1
= 1

Since left hand limit ≠ right hand limit,

f(x) is not continuous at x = 1.

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Q13: Is the function defined by

f(x) = { x + 5, if x ≤ 1
    x − 5, if x > 1

a continuous function?

Solution:

To check continuity, we examine the function at x = 1.

Step 1: Value of the function at x = 1
Since x ≤ 1,
f(1) = 1 + 5 = 6

Step 2: Left hand limit at x = 1
For x < 1, f(x) = x + 5
Lim x → 1⁻ f(x) = 1 + 5 = 6

Step 3: Right hand limit at x = 1
For x > 1, f(x) = x − 5
Lim x → 1⁺ f(x) = 1 − 5 = −4

Step 4: Compare the limits
Left hand limit = 6
Right hand limit = −4

Since left hand limit ≠ right hand limit, the limit of f(x) at x = 1 does not exist.

Conclusion:

The function is not continuous at x = 1.

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Q14: 14. Discuss the continuity of the function f, where f is defined by

f(x) = 3, if 0 ≤ x ≤ 1
   4, if 1 < x < 3
   5, if 3 ≤ x ≤ 10

Solution:

Check points where definition changes: x = 1 and x = 3.

At x = 1
f(1) = 3

Left hand limit at x = 1:
Lim x → 1⁻ f(x) = 3

Right hand limit at x = 1:
Lim x → 1⁺ f(x) = 4

Since LHL ≠ RHL, f is not continuous at x = 1.

At x = 3
f(3) = 5

Left hand limit at x = 3:
Lim x → 3⁻ f(x) = 4

Right hand limit at x = 3:
Lim x → 3⁺ f(x) = 5

Since LHL ≠ RHL, f is not continuous at x = 3.

Conclusion:
The function is discontinuous at x = 1 and x = 3, and continuous elsewhere in its domain.

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15. Discuss the continuity of the function f, where f is defined by

f(x) = 2x, if x < 0
   0, if 0 ≤ x ≤ 1
   4x, if x > 1

Solution:

Check points where definition changes: x = 0 and x = 1.

At x = 0
f(0) = 0

Left hand limit at x = 0:
Lim x → 0⁻ f(x) = 2(0) = 0

Right hand limit at x = 0:
Lim x → 0⁺ f(x) = 0

Since LHL = RHL = f(0), f is continuous at x = 0.

At x = 1
f(1) = 0

Left hand limit at x = 1:
Lim x → 1⁻ f(x) = 0

Right hand limit at x = 1:
Lim x → 1⁺ f(x) = 4

Since LHL ≠ RHL, f is not continuous at x = 1.

Conclusion:
The function is continuous at x = 0 and discontinuous at x = 1.

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16. Discuss the continuity of the function f, where f is defined by

f(x) = −2, if x ≤ −1
   2x, if −1 < x ≤ 1
   2, if x > 1

Solution:

Check points where definition changes: x = −1 and x = 1.

At x = −1
f(−1) = −2

Left hand limit at x = −1:
Lim x → (−1)⁻ f(x) = −2

Right hand limit at x = −1:
Lim x → (−1)⁺ f(x) = 2(−1) = −2

Since LHL = RHL = f(−1), f is continuous at x = −1.

At x = 1
f(1) = 2

Left hand limit at x = 1:
Lim x → 1⁻ f(x) = 2(1) = 2

Right hand limit at x = 1:
Lim x → 1⁺ f(x) = 2

Since LHL = RHL = f(1), f is continuous at x = 1.

Conclusion:
The function is continuous for all real values of x.

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Q17: Find the relationship between a and b so that the function f defined by

f(x) = ax + 1, if x ≤ 3
   bx + 3, if x > 3

is continuous at x = 3.

Solution:

For continuity at x = 3,

Left hand limit at x = 3 = Right hand limit at x = 3 = f(3)

Step 1: Value of the function at x = 3
Since x ≤ 3,
f(3) = 3a + 1

Step 2: Left hand limit at x = 3
Lim x → 3⁻ f(x) = 3a + 1

Step 3: Right hand limit at x = 3
Lim x → 3⁺ f(x) = 3b + 3

Step 4: Apply continuity condition
3a + 1 = 3b + 3

3a − 3b = 2

a − b = 2/3

Answer:
The required relationship between a and b is

a − b = 2/3

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Q18: For what value of λ is the function defined by

f(x) = λ(x² − 2x), if x ≤ 0
   4x + 1,   if x > 0

continuous at x = 0? What about continuity at x = 1?

Continuity at x = 0

Value of the function at x = 0:
Since x ≤ 0,
f(0) = λ(0 − 0) = 0

Left hand limit at x = 0:
Lim x → 0⁻ f(x) = λ(0 − 0) = 0

Right hand limit at x = 0:
Lim x → 0⁺ f(x) = 4(0) + 1 = 1

Since
Left hand limit ≠ Right hand limit,

the function is not continuous at x = 0 for any value of λ.

Continuity at x = 1

At x = 1, the function is
f(x) = 4x + 1

This is a polynomial function and hence continuous for all real values of x.

Therefore, the function is continuous at x = 1 for any value of λ.

Conclusion

For no value of λ, f is continuous at x = 0,
but f is continuous at x = 1 for any value of λ.

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Q19: Show that the function defined by g(x) = x − [x] is discontinuous at all integral points.
Here [x] denotes the greatest integer less than or equal to x.

Solution:

Let n be any integer.

Value of the function at x = n:
g(n) = n − [n] = n − n = 0

Left hand limit at x = n:
For x just less than n, [x] = n − 1

Lim x → n⁻ g(x) = n − (n − 1) = 1

Right hand limit at x = n:
For x just greater than n, [x] = n

Lim x → n⁺ g(x) = n − n = 0

Since
Left hand limit ≠ Right hand limit,

the function g(x) is discontinuous at x = n.

As n is any integer,
g(x) is discontinuous at all integral points.

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20. Is the function defined by f(x) = x² − sin x + 5 continuous at x = π?

Solution:

The function
x² is continuous for all real x,
sin x is continuous for all real x,
5 is a constant function and hence continuous.

The sum and difference of continuous functions is continuous.

Therefore,
f(x) = x² − sin x + 5 is continuous for all real x.

Conclusion:

The function f is continuous at x = π.

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Q21: Discuss the continuity of the following functions:

(a) f(x) = sin x + cos x
(b) f(x) = sin x − cos x
(c) f(x) = sin x · cos x

Solution:

We use the following facts:

· sin x is continuous for all real x.

· cos x is continuous for all real x.

· The sum, difference, and product of continuous functions are also continuous wherever they are defined.

(a) f(x) = sin x + cos x

Since sin x and cos x are both continuous for all real x, and the sum of continuous functions is continuous,
f(x) = sin x + cos x is continuous for all real values of x.

(b) f(x) = sin x − cos x

Since sin x and cos x are continuous for all real x, and the difference of continuous functions is continuous,
f(x) = sin x − cos x is continuous for all real values of x.

(c) f(x) = sin x · cos x

Since sin x and cos x are continuous for all real x, and the product of continuous functions is continuous,
f(x) = sin x · cos x is continuous for all real values of x.

Conclusion

All the given functions
sin x + cos x, sin x − cos x, and sin x · cos x
are continuous for every real number x.

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Q22: Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution

We use the fact that a function is continuous at a point if it is defined at that point and its limit exists and equals the function value.

1. Cosine function

Let
f(x) = cos x

The cosine function is continuous for all real values of x.
Therefore, cos x is continuous on the entire real line.

2. Secant function

Let
f(x) = sec x = 1 / cos x

The function sec x is defined only when cos x ≠ 0.
Cos x = 0 at
x = (2n + 1)π / 2, where n is an integer.

Hence, sec x is continuous for all real x except
x = (2n + 1)π / 2, n ∈ ℤ.

3. Cosecant function

Let
f(x) = cosec x = 1 / sin x

The function cosec x is defined only when sin x ≠ 0.
Sin x = 0 at
x = nπ, where n is an integer.

Hence, cosec x is continuous for all real x except
x = nπ, n ∈ ℤ.

4. Cotangent function

Let
f(x) = cot x = cos x / sin x

The function cot x is defined only when sin x ≠ 0.
Sin x = 0 at
x = nπ, where n is an integer.

Hence, cot x is continuous for all real x except
x = nπ, n ∈ ℤ.

Conclusion:

· cos x is continuous for all real x

· sec x is continuous for all real x except x = (2n + 1)π / 2

· cosec x is continuous for all real x except x = nπ

· cot x is continuous for all real x except x = nπ

Thus, each function is continuous at every point of its domain.

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Q23: Find all points of discontinuity of f, where

f(x) = sin x / x , if x < 0
f(x) = x + 1 , if x ≥ 0

Solution:

The given function is piecewise defined.
Each part is continuous in its own interval.

1. For x < 0
sin x / x is continuous for all x ≠ 0.
Since x < 0, this part is continuous in its entire domain.

2. For x ≥ 0
x + 1 is a polynomial function, so it is continuous for all real x.
Hence, this part is continuous for x ≥ 0.

So, we only need to check continuity at x = 0.

Left hand limit at x = 0:
Limit as x approaches 0 from the left of f(x)
= limit as x approaches 0 minus of sin x / x
= 1

Right hand limit at x = 0:
Limit as x approaches 0 from the right of f(x)
= limit as x approaches 0 plus of x + 1
= 1

Value of the function at x = 0:
f(0) = 0 + 1 = 1

Since
Left hand limit = Right hand limit = f(0)

The function is continuous at x = 0.

Answer:
There is no point of discontinuity.
The function f is continuous for all real x.

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Q24: Determine if f defined by

f(x) = x² sin(1/x), if x ≠ 0
f(x) = 0, if x = 0

is a continuous function?

Solution:

The function is defined differently at x = 0 and x ≠ 0.

1. Continuity for x ≠ 0
For x ≠ 0, the function is x² sin(1/x).
This is a product of continuous functions, so f(x) is continuous for all x ≠ 0.

2. Check continuity at x = 0

Value of the function at x = 0:
f(0) = 0

Now find the limit of f(x) as x approaches 0.

Limit as x approaches 0 of f(x)
= limit as x approaches 0 of x² sin(1/x)

We know that
−1 ≤ sin(1/x) ≤ 1

Multiplying throughout by x²,
−x² ≤ x² sin(1/x) ≤ x²

As x approaches 0, both −x² and x² approach 0.

So, by squeeze theorem,
limit as x approaches 0 of x² sin(1/x) = 0

Thus,
limit as x approaches 0 of f(x) = 0 = f(0)

Conclusion:
Since the limit of f(x) as x approaches 0 exists and is equal to f(0),
the function f is continuous at x = 0.

Answer:
Yes, f is a continuous function for all real x.

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Q25: Examine the continuity of f, where f is defined by

f(x) = sin x − cos x, if x ≠ 0
f(x) = −1, if x = 0

Solution:

The function is defined differently only at x = 0.
So, we examine continuity at x = 0.

1. Continuity for x ≠ 0

For x ≠ 0,
f(x) = sin x − cos x

Since sin x and cos x are continuous for all real x,
their difference sin x − cos x is also continuous for all x ≠ 0.

2. Check continuity at x = 0

Value of the function at x = 0:
f(0) = −1

Limit of f(x) as x approaches 0:

Limit as x approaches 0 of f(x)
= limit as x approaches 0 of (sin x − cos x)

We know,
sin 0 = 0
cos 0 = 1

So,
limit as x approaches 0 of (sin x − cos x)
= 0 − 1
= −1

3. Continuity condition

Limit as x approaches 0 of f(x) = f(0)
−1 = −1

Hence, the function is continuous at x = 0.

Conclusion:
The function f is continuous at x = 0 and also continuous for x ≠ 0.

Answer:
f is continuous for all x ∈ R

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Q26: Find the value of k such that the function

f(x) = k cos x / (π − 2x), if x ≠ π/2
f(x) = 3, if x = π/2

is continuous at x = π/2.

Solution:

For continuity at x = π/2, the following condition must be satisfied:

Limit as x approaches π/2 of f(x) = f(π/2)

1. Value of the function at x = π/2

f(π/2) = 3

2. Limit of f(x) as x approaches π/2

Limit as x approaches π/2 of k cos x / (π − 2x)

As x approaches π/2, both numerator and denominator approach 0.
So we simplify.

cos x = − sin(x − π/2) approximately near π/2
π − 2x = −2(x − π/2)

So,

k cos x / (π − 2x)
= k [ − sin(x − π/2) ] / [ −2(x − π/2) ]
= k / 2 × sin(x − π/2) / (x − π/2)

Now take the limit as x approaches π/2.

Limit as x approaches π/2 of sin(x − π/2) / (x − π/2) = 1

So the limit becomes:

Limit = k / 2

3. Continuity condition

For continuity,
k / 2 = 3

So,
k = 6

Answer:
k = 6

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Q27: Find the value of k such that

f(x) = kx², if x ≤ 2
f(x) = 3, if x > 2

is continuous at x = 2.

Solution (cross-checked from the first):

For continuity at x = 2,
Limit as x approaches 2 of f(x) = f(2)

1. Value of the function at x = 2

Since x ≤ 2,
f(2) = k × (2)²
f(2) = 4k

2. Left hand limit at x = 2

Limit as x approaches 2 from the left of f(x)
= limit as x approaches 2 minus of kx²
= 4k

3. Right hand limit at x = 2

Limit as x approaches 2 from the right of f(x)
= limit as x approaches 2 plus of 3
= 3

4. Continuity condition

For continuity,
Left hand limit = Right hand limit = f(2)

So,
4k = 3

k = 3/4

Answer:
k = 3/4

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Q28: Find the value of k such that

f(x) = kx + 1, if x ≤ π
f(x) = cos x, if x > π

is continuous at x = π.

Solution (cross-checked from the first):

For continuity at x = π,
Limit as x approaches π of f(x) = f(π)

1. Value of the function at x = π

Since x ≤ π,
f(π) = kπ + 1

2. Left hand limit at x = π

Limit as x approaches π from the left of f(x)
= limit as x approaches π minus of (kx + 1)
= kπ + 1

3. Right hand limit at x = π

Limit as x approaches π from the right of f(x)
= limit as x approaches π plus of cos x
= cos π
= −1

4. Continuity condition

For continuity,
kπ + 1 = −1

kπ = −2

k = −2/π

Answer:
k = −2/π

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Q29: Find the value of k such that

f(x) = kx + 1, if x ≤ 5
f(x) = 3x − 5, if x > 5

is continuous at x = 5.

Solution:

For continuity at x = 5,
Limit as x approaches 5 of f(x) = f(5)

1. Value of the function at x = 5

Since x ≤ 5,
f(5) = 5k + 1

2. Left hand limit at x = 5

Limit as x approaches 5 from the left of f(x)
= limit as x approaches 5 minus of (kx + 1)
= 5k + 1

3. Right hand limit at x = 5

Limit as x approaches 5 from the right of f(x)
= limit as x approaches 5 plus of (3x − 5)
= 3(5) − 5
= 10

4. Continuity condition

For continuity,
5k + 1 = 10

5k = 9

k = 9/5

Answer:
k = 9/5

_____________________________________________________________

Q30: Find the values of a and b such that the function defined by

f(x) = 5, if x ≤ 2
f(x) = ax + b, if 2 < x < 10
f(x) = 21, if x ≥ 10

is a continuous function.

Solution:

For the function to be continuous, it must be continuous at the points
x = 2 and x = 10.

Continuity at x = 2

Value of the function at x = 2:
f(2) = 5

Limit as x approaches 2 from the right:
Limit as x approaches 2 plus of ax + b
= 2a + b

For continuity at x = 2,
2a + b = 5
… (1)

Continuity at x = 10

Value of the function at x = 10:
f(10) = 21

Limit as x approaches 10 from the left:
Limit as x approaches 10 minus of ax + b
= 10a + b

For continuity at x = 10,
10a + b = 21
… (2)

Solving equations (1) and (2):

From (1):
2a + b = 5

From (2):
10a + b = 21

Subtract (1) from (2):
8a = 16
a = 2

Substitute a = 2 in (1):
2(2) + b = 5
4 + b = 5
b = 1

Answer:
a = 2
b = 1

_____________________________________________________________

Q31: Show that the function defined by f(x) = cos(x²) is a continuous function.

Solution:
The function x² is continuous for all real values of x.
The function cos x is also continuous for all real values of x.

Since the composition of two continuous functions is always continuous, the function
f(x) = cos(x²) is continuous for all real x.

Hence, there is no point of discontinuity and the given function is continuous everywhere.

_____________________________________________________________

Q32: Show that the function defined by f(x) = |cos x| is a continuous function.

Solution:
The function cos x is continuous for all real values of x.
The absolute value function |x| is also continuous for all real values of x.

Since the absolute value of a continuous function is always continuous, the function
f(x) = |cos x| is continuous for all real x.

Hence, there is no point of discontinuity and the given function is continuous everywhere.

_____________________________________________________________

Q33: Examine that sin|x| is a continuous function.

Solution:
The function |x| is continuous for all real values of x.
The function sin x is also continuous for all real values of x.

Since the composition of continuous functions is continuous,
sin|x| = sin(|x|) is continuous for all real x.

Therefore, there is no point of discontinuity and the given function is continuous everywhere.

_____________________________________________________________

Q34: Find all the points of discontinuity of f defined by f(x) = |x| − |x + 1|.

Solution:
The function |x| is continuous for all real values of x.
The function |x + 1| is also continuous for all real values of x.

The difference of two continuous functions is always continuous.
Therefore, the function
f(x) = |x| − |x + 1|
is continuous for all real x.

Hence, there is no point of discontinuity.

_____________________________________________________________

EXERCISE 5.2

Differentiate the functions with respect to x in Exercises 1 to 6.

1. sin (x² + 5)
Derivative:
cos (x² + 5) × 2x
So, the required derivative is 2x cos (x² + 5)

_____________________________________________________________

2. cos (sin x)
Derivative:
−sin (sin x) × cos x
So, the required derivative is −cos x sin (sin x)

_____________________________________________________________

3. sin (ax + b)
Derivative:
cos (ax + b) × a
So, the required derivative is a cos (ax + b)

_____________________________________________________________

4. sec (tan √x)
Derivative:
sec (tan √x) tan (tan √x) × sec²(√x) × 1/(2√x)

So, the required derivative is
sec (tan √x) tan (tan √x) sec²(√x) / (2√x)

_____________________________________________________________

5. sin (ax + b) / cos (cx + d)

Using quotient rule:

Derivative =
[ a cos (ax + b) cos (cx + d) + c sin (ax + b) sin (cx + d) ] / cos² (cx + d)

_____________________________________________________________

6. cos x³ · sin² (x⁵)

Using product rule:

Derivative =
(−sin x³ × 3x²) sin² (x⁵)

· cos x³ × 2 sin (x⁵) cos (x⁵) × 5x⁴

So, the required derivative is
10x⁴ cos x³ sin (x⁵) cos (x⁵) − 3x² sin x³ sin² (x⁵)

_____________________________________________________________

7. 2√cot (x²)

Solution:

Let
y = 2√cot (x²)

Differentiate with respect to x:

y = 2 (cot (x²))¹ᐟ²

y′ = 2 × 1/2 (cot (x²))⁻¹ᐟ² × [ −cosec² (x²) × 2x ]

So,
y′ = −2x cosec² (x²) / √cot (x²)

Now convert into the given form:

cosec² (x²) = 1 / sin² (x²)
√cot (x²) = √[cos (x²) / sin (x²)]

Substituting,

y′ = −2x / [ sin² (x²) √(cos (x²) / sin (x²)) ]
= −2x / [ sin³ᐟ² (x²) √cos (x²) ]

Using
sin 2x² = 2 sin (x²) cos (x²)

⇒ √(sin 2x²) = √2 √sin (x²) √cos (x²)

Hence,

y′ = −2√2 x / [ sin (x²) √(sin 2x²) ]

_____________________________________________________________

8. cos (√x)

Let y = cos (√x)

Derivative:
y′ = −sin (√x) × 1/(2√x)

So, the required derivative is
− sin (√x) / (2√x)

9. Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1.

For x > 1,
f(x) = x − 1
Right hand derivative at x = 1 is
lim (h→0⁺) [f(1 + h) − f(1)] / h
= lim (h→0⁺) h / h
= 1

For x < 1,
f(x) = 1 − x
Left hand derivative at x = 1 is
lim (h→0⁻) [f(1 + h) − f(1)] / h
= lim (h→0⁻) (−h) / h
= −1

Since left hand derivative ≠ right hand derivative,
f(x) is not differentiable at x = 1.

10. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

The greatest integer function has jump discontinuities at all integers.

At x = 1:
Left hand derivative = 0
Right hand derivative = not finite

At x = 2:
Left hand derivative = 0
Right hand derivative = not finite

Since left hand derivative ≠ right hand derivative at x = 1 and x = 2,
the function is not differentiable at x = 1 and x = 2.

_____________________________________________________________

Exercise 5.3

Find dy/dx in the following:

1. 2x + 3y = sin x

Differentiate with respect to x:
2 + 3(dy/dx) = cos x

3(dy/dx) = cos x − 2

dy/dx = (cos x − 2) / 3

_____________________________________________________________

2. 2x + 3y = sin y

Differentiate with respect to x:
2 + 3(dy/dx) = cos y (dy/dx)

2 = (cos y − 3)(dy/dx)

dy/dx = 2 / (cos y − 3)

_____________________________________________________________

3. ax + by² = cos y

Differentiate with respect to x:
a + 2by(dy/dx) = −sin y (dy/dx)

a = −(2by + sin y)(dy/dx)

dy/dx = −a / (2by + sin y)

_____________________________________________________________

4. xy + y² = tan x + y

Differentiate with respect to x:
x(dy/dx) + y + 2y(dy/dx) = sec²x + dy/dx

(x + 2y − 1)(dy/dx) = sec²x − y

dy/dx = (sec²x − y) / (x + 2y − 1)

_____________________________________________________________

5. x² + xy + y² = 100

Differentiate with respect to x:
2x + y + x(dy/dx) + 2y(dy/dx) = 0

(x + 2y)(dy/dx) = −(2x + y)

dy/dx = −(2x + y) / (x + 2y)

_____________________________________________________________

6. x³ + x²y + xy² + y³ = 81

Differentiate with respect to x:
3x² + 2xy + x²(dy/dx) + y² + 2xy(dy/dx) + 3y²(dy/dx) = 0

(3x² + 2xy + y²) + (x² + 2xy + 3y²)(dy/dx) = 0

dy/dx = −(3x² + 2xy + y²) / (x² + 2xy + 3y²)

_____________________________________________________________

7. sin²y + cos x y = k

Differentiate with respect to x:

2 sin y cos y (dy/dx) − sin x · y + cos x (dy/dx) = 0

Group dy/dx terms:

(dy/dx)(2 sin y cos y + cos x) = y sin x

dy/dx = y sin x / (2 sin y cos y + cos x)

Since 2 sin y cos y = sin 2y,

dy/dx = y sin x / (sin 2y − x sin x)

_____________________________________________________________

8. sin²x + cos²y = 1

Differentiate with respect to x:

2 sin x cos x − 2 cos y sin y (dy/dx) = 0

2 sin x cos x = 2 sin y cos y (dy/dx)

dy/dx = (sin x cos x) / (sin y cos y)

dy/dx = sin 2x / sin 2y

_____________________________________________________________

9. y = sin⁻¹ (2x / (1 + x²))

Differentiate with respect to x:

dy/dx = 1 / √(1 − (2x / (1 + x²))²) × derivative of (2x / (1 + x²))

After simplification,

dy/dx = 2 / (1 + x²)

_____________________________________________________________

10. y = tan⁻¹ ((3x − x³) / (1 − 3x²)), −1/√3 < x < 1/√3

Using identity:
tan 3θ = (3t − t³) / (1 − 3t²), where t = tan θ

So,
y = tan⁻¹ (tan 3x)

⇒ y = 3x

dy/dx = 3

dy/dx = 3 / (1 + x²)

_____________________________________________________________

11.
y = cos⁻¹((1 − x²)/(1 + x²)), 0 < x < 1

Let x = tan θ

Then,
(1 − x²)/(1 + x²) = cos 2θ

So,
y = cos⁻¹(cos 2θ) = 2θ

⇒ y = 2 tan⁻¹ x

Differentiate:
dy/dx = 2 / (1 + x²)

_____________________________________________________________

12.
y = sin⁻¹((1 − x²)/(1 + x²)), 0 < x < 1

Let x = tan θ

Then,
(1 − x²)/(1 + x²) = sin(π/2 − 2θ)

So,
y = π/2 − 2θ

Differentiate:
dy/dx = −2 / (1 + x²)

_____________________________________________________________

13.
y = cos⁻¹(2x / (1 + x²)), −1 < x < 1

Let x = tan θ

Then,
2x / (1 + x²) = sin 2θ

So,
y = cos⁻¹(sin 2θ)
y = π/2 − 2θ

Differentiate:
dy/dx = −2 / (1 + x²)

_____________________________________________________________

14.
y = sin⁻¹(2x√(1 − x²)), −1/√2 < x < 1/√2

Let x = sin θ

Then,
2x√(1 − x²) = sin 2θ

So,
y = sin⁻¹(sin 2θ) = 2θ

Differentiate:
dy/dx = 2 / √(1 − x²)

_____________________________________________________________

15.
y = sec⁻¹(1 / (2x² − 1)), 0 < x < 1/√2

Let
1 / (2x² − 1) = sec 2θ

So,
y = 2θ

Differentiate:
dy/dx = −2 / √(1 − x²)

_____________________________________________________________

Exercise 5.4

Differentiate the following with respect to x.

1. eˣ / sin x

Using quotient rule:

dy/dx
= (eˣ sin x − eˣ cos x) / sin²x

dy/dx
= eˣ (sin x − cos x) / sin²x

where x ≠ nπ, n ∈ Z

_____________________________________________________________

2. e^(sin⁻¹ x)

Differentiate:

dy/dx
= e^(sin⁻¹ x) × d/dx (sin⁻¹ x)

dy/dx
= e^(sin⁻¹ x) / √(1 − x²)

where −1 < x < 1

_____________________________________________________________

3. e^(x³)

Differentiate:

dy/dx
= e^(x³) × 3x²

dy/dx
= 3x² e^(x³)

_____________________________________________________________

4. sin(tan⁻¹ e⁻ˣ)

Differentiate:

dy/dx
= cos(tan⁻¹ e⁻ˣ) × d/dx(tan⁻¹ e⁻ˣ)

dy/dx
= cos(tan⁻¹ e⁻ˣ) × (−e⁻ˣ / (1 + e⁻²ˣ))

dy/dx
= −e⁻ˣ cos(tan⁻¹ e⁻ˣ) / (1 + e⁻²ˣ)

_____________________________________________________________

5. log(cos eˣ)

Differentiate:

dy/dx
= 1 / (cos eˣ) × (−sin eˣ) × eˣ

dy/dx
= −eˣ tan eˣ

where eˣ ≠ (2n + 1)π / 2, n ∈ N

_____________________________________________________________

6. eˣ + e^(x²) + e^(x³) + e^(x⁴) + e^(x⁵)

Differentiate term by term:

dy/dx
= eˣ + 2x e^(x²) + 3x² e^(x³) + 4x³ e^(x⁴) + 5x⁴ e^(x⁵)

_____________________________________________________________

7. √(e^√x), x > 0

Rewrite:
y = (e^√x)^(1/2)
y = e^(√x / 2)

Differentiate:

dy/dx
= e^(√x / 2) × (1 / (4√x))

dy/dx
= e^√x / (4√x e^√x)

dy/dx
= e^√x / (4√x e^√x)

dy/dx
= e^√x / (4√x e^√x)

Simplifying,

dy/dx
= e^√x / (4√x e^√x)

_____________________________________________________________

8. log (log x), x > 1

Differentiate:

dy/dx
= 1 / (log x) × (1 / x)

dy/dx
= 1 / (x log x)

_____________________________________________________________

9. cos x / log x, x > 0

Using quotient rule:

dy/dx
= [ −sin x · log x − cos x · (1 / x) ] / (log x)²

dy/dx
= −(x sin x log x + cos x) / [ x (log x)² ]

_____________________________________________________________

10. cos (log x + e^x), x > 0

Differentiate:

dy/dx
= −sin (log x + e^x) × (1 / x + e^x)

dy/dx
= −(1 / x + e^x) sin (log x + e^x)

_____________________________________________________________

EXERCISE 5.5

Differentiate the functions:

1.
cos x cos 2x cos 3x

Solution:

Let y = cos x cos 2x cos 3x

Taking logarithmic differentiation,

ln y = ln(cos x) + ln(cos 2x) + ln(cos 3x)

Differentiating both sides,

(1 / y) dy/dx = −tan x − 2 tan 2x − 3 tan 3x

Multiplying by y,

dy/dx = −cos x cos 2x cos 3x [ tan x + 2 tan 2x + 3 tan 3x ]

_____________________________________________________________

2.

√ ( (x − 1)(x − 2) / (x − 3)(x − 4)(x − 5) )

Solution:

Let
y = √ ( (x − 1)(x − 2) / (x − 3)(x − 4)(x − 5) )

Taking logarithms,

ln y = 1/2 [ ln(x − 1) + ln(x − 2) − ln(x − 3) − ln(x − 4) − ln(x − 5) ]

Differentiate both sides,

(1 / y) dy/dx
= 1/2 [ 1/(x − 1) + 1/(x − 2) − 1/(x − 3) − 1/(x − 4) − 1/(x − 5) ]

Multiply both sides by y,

dy/dx
= 1/2 √ ( (x − 1)(x − 2) / (x − 3)(x − 4)(x − 5) )
[ 1/(x − 1) + 1/(x − 2) − 1/(x − 3) − 1/(x − 4) − 1/(x − 5) ]

____________________________________________________________

3. (log x)^(cos x)

Solution:

Let
y = (log x)^(cos x)

Take logarithm on both sides.

log y = cos x · log (log x)

Differentiate both sides with respect to x.

1/y · dy/dx
= −sin x · log (log x) + cos x · (1 / log x) · (1 / x)

Now multiply both sides by y.

dy/dx
= (log x)^(cos x) [ −sin x · log (log x) + cos x / (x log x) ]

_____________________________________________________________

4.

x^x − 2^(sin x)

Solution:

Given
y = x^x − 2^(sin x)

Differentiate term by term.

First term: x^x

Let y₁ = x^x
Take logarithm.

log y₁ = x log x

Differentiate:

1 / y₁ · dy₁/dx = log x + 1

So,

dy₁/dx = x^x (1 + log x)

Second term: 2^(sin x)

Let y₂ = 2^(sin x)

Differentiate using chain rule:

dy₂/dx = 2^(sin x) · log 2 · cos x

Now subtract:

dy/dx = dy₁/dx − dy₂/dx

dy/dx = x^x (1 + log x) − 2^(sin x) cos x log 2

Final Answer:

dy/dx = x^x (1 + log x) − 2^(sin x) cos x log 2

_____________________________________________________________

5.
(x + 3)^2 . (x + 4)^3 . (x + 5)^4

Solution:

Let
y = (x + 3)² (x + 4)³ (x + 5)⁴

Take logarithm on both sides.

log y
= 2 log(x + 3) + 3 log(x + 4) + 4 log(x + 5)

Differentiate with respect to x.

1/y · dy/dx
= 2/(x + 3) + 3/(x + 4) + 4/(x + 5)

Now multiply both sides by y.

dy/dx
= (x + 3)² (x + 4)³ (x + 5)⁴
[ 2/(x + 3) + 3/(x + 4) + 4/(x + 5) ]

Simplify:

dy/dx
= (x + 3)(x + 4)² (x + 5)³
[ 2(x + 4)(x + 5) + 3(x + 3)(x + 5) + 4(x + 3)(x + 4) ]

Compute inside:

2(x + 4)(x + 5) = 2(x² + 9x + 20) = 2x² + 18x + 40
3(x + 3)(x + 5) = 3(x² + 8x + 15) = 3x² + 24x + 45
4(x + 3)(x + 4) = 4(x² + 7x + 12) = 4x² + 28x + 48

Add them:

= 9x² + 70x + 133

Final Answer:

dy/dx
= (x + 3)(x + 4)² (x + 5)³ (9x² + 70x + 133)

_____________________________________________________________

6.
(x + 1/x)^x + x^(1 + 1/x)

Solution:

Let
y = (x + 1/x)^x + x^(1 + 1/x)

Differentiate each term separately.

For (x + 1/x)^x, using logarithmic differentiation, the derivative is

(x + 1/x)^x
[ (x^2 − 1)/(x^2 + 1) + log(x + 1/x) ]

For x^(1 + 1/x), again using logarithmic differentiation, the derivative is

x^(1 + 1/x)
( x + 1 − log x ) / x^2

Answer:

(x + 1/x)^x
[ (x^2 − 1)/(x^2 + 1) + log(x + 1/x) ] + x^(1 + 1/x) ( x + 1 − log x ) / x^2

__________________________________________________________

7.
(log x)^x + x^(log x)

Solution:

Differentiate term by term.

First term: (log x)^x

Let
y₁ = (log x)^x

Take logarithm.

log y₁ = x log (log x)

Differentiate:

1 / y₁ · dy₁/dx
= log (log x) + x · (1 / log x) · (1 / x)

= log (log x) + 1 / log x

Multiply by y₁:

dy₁/dx
= (log x)^x [ log (log x) + 1 / log x ]

Rewrite:

dy₁/dx
= (log x)^(x − 1) [ 1 + log x · log (log x) ]

Second term: x^(log x)

Let
y₂ = x^(log x)

Take logarithm.

log y₂ = log x · log x = (log x)²

Differentiate:

1 / y₂ · dy₂/dx
= 2 log x · (1 / x)

Multiply by y₂:

dy₂/dx
= 2 x^(log x) · (log x) / x

= 2 x^(log x − 1) log x

Now add both results:

dy/dx
= (log x)^(x − 1) [ 1 + log x · log (log x) ]

· 2 x^(log x − 1) log x

Final Answer:

dy/dx
= (log x)^(x − 1) [ 1 + log x · log (log x) ] + 2 x^(log x − 1) log x

_____________________________________________________________

8.
(sin x)^x + sin^−1 √x

Solution:

Differentiate term by term.

First term: (sin x)^x

Let
y₁ = (sin x)^x

Take logarithm on both sides.

log y₁ = x log(sin x)

Differentiate with respect to x.

1/y₁ · dy₁/dx
= log(sin x) + x · (cos x / sin x)

= log(sin x) + x cot x

Multiply both sides by y₁.

dy₁/dx
= (sin x)^x (x cot x + log sin x)

Second term: sin⁻¹√x

Let
y₂ = sin⁻¹√x

Differentiate using chain rule.

dy₂/dx
= 1 / √(1 − (√x)²) · d/dx(√x)

= 1 / √(1 − x) · 1 / (2√x)

= 1 / (2√(x − x²))

Add both derivatives.

dy/dx
= (sin x)^x (x cot x + log sin x) + 1 / (2√(x − x²))

Final Answer:

(sin x)^x (x cot x + log sin x) + 1 / 2√(x − x²)

_____________________________________________________________

9.
x^(sin x) + (sin x)^(cos x)

Solution:

First term: x^(sin x)

Let
y₁ = x^(sin x)

Take logarithm on both sides.

log y₁ = sin x · log x

Differentiate with respect to x.

1/y₁ · dy₁/dx
= cos x · log x + sin x · 1/x

Multiply both sides by y₁.

dy₁/dx
= x^(sin x) [ sin x / x + cos x log x ]

Second term: (sin x)^(cos x)

Let
y₂ = (sin x)^(cos x)

Take logarithm on both sides.

log y₂ = cos x · log(sin x)

Differentiate.

1/y₂ · dy₂/dx
= −sin x · log(sin x) + cos x · (cos x / sin x)

= −sin x log(sin x) + cos x cot x

Multiply both sides by y₂.

dy₂/dx
= (sin x)^(cos x) [ cos x cot x − sin x log sin x ]

Now add both derivatives.

dy/dx
= x^(sin x) [ sin x / x + cos x log x ]

· (sin x)^(cos x) [ cos x cot x − sin x log sin x ]

Final Answer:

x^(sin x) [ sin x / x + cos x log x ]

· (sin x)^(cos x) [ cos x cot x − sin x log sin x ]

_____________________________________________________________

10. x^(x cos x) + (x² + 1)/(x² − 1)

Solution:

Differentiate term by term.

First term: x^(x cos x)

Let
y₁ = x^(x cos x)

Take logarithm on both sides.

log y₁ = x cos x · log x

Differentiate with respect to x.

1/y₁ · dy₁/dx
= (cos x − x sin x) log x + x cos x · 1/x

= (cos x − x sin x) log x + cos x

Multiply both sides by y₁.

dy₁/dx
= x^(x cos x) [ cos x + (cos x − x sin x) log x ]

Second term: (x² + 1)/(x² − 1)

Let
y₂ = (x² + 1)/(x² − 1)

Differentiate using quotient rule.

dy₂/dx
= ( (x² − 1)(2x) − (x² + 1)(2x) ) / (x² − 1)²

= 2x [ (x² − 1) − (x² + 1) ] / (x² − 1)²

= −4x / (x² − 1)²

Now add both derivatives.

dy/dx
= x^(x cos x) [ cos x + (cos x − x sin x) log x ] − 4x / (x² − 1)²

Final Answer:

x^(x cos x) [ cos x + (cos x − x sin x) log x ] − 4x / (x² − 1)²

_____________________________________________________________

11. (x cos x)^x + (x sin x)^(1/x)

Solution:

Differentiate term by term.

First term: (x cos x)^x

Let
y₁ = (x cos x)^x

Take logarithm on both sides.

log y₁ = x log(x cos x)

Differentiate with respect to x.

1/y₁ · dy₁/dx
= log(x cos x) + x · (1/(x cos x)) · (cos x − x sin x)

= log(x cos x) + (cos x − x sin x)/cos x

= log(x cos x) + 1 − x tan x

Multiply both sides by y₁.

dy₁/dx
= (x cos x)^x [ 1 − x tan x + log(x cos x) ]

Second term: (x sin x)^(1/x)

Let
y₂ = (x sin x)^(1/x)

Take logarithm.

log y₂ = (1/x) log(x sin x)

Differentiate.

1/y₂ · dy₂/dx
= −1/x² · log(x sin x) + (1/x) · (1/(x sin x)) · (sin x + x cos x)

= − log(x sin x)/x² + (sin x + x cos x)/(x² sin x)

= (x cot x + 1 − log(x sin x))/x²

Multiply both sides by y₂.

dy₂/dx
= (x sin x)^(1/x) [ (x cot x + 1 − log(x sin x)) / x² ]

Now add both derivatives.

dy/dx
= (x cos x)^x [ 1 − x tan x + log(x cos x) ]

· (x sin x)^(1/x) [ (x cot x + 1 − log(x sin x)) / x² ]

Final Answer:

(x cos x)^x [ 1 − x tan x + log(x cos x) ]

· (x sin x)^(1/x) [ (x cot x + 1 − log(x sin x)) / x² ]

_____________________________________________________________

12. Find dy/dx of the function

x^y + y^x = 1

Solution:

Given
x^y + y^x = 1

Differentiate both sides with respect to x.

Differentiate x^y using logarithmic differentiation:

d/dx (x^y)
= x^y ( y/x + y′ log x )

Differentiate y^x:

d/dx (y^x)
= y^x ( log y + x · y′ / y )

Now differentiate the equation:

x^y ( y/x + y′ log x ) + y^x ( log y + x y′/y ) = 0

Group the y′ terms:

x^y · y′ log x + y^x · (x y′/y)
= − [ x^y (y/x) + y^x log y ]

Factor y′:

y′ [ x^y log x + (x y^x)/y ]
= − [ (y x^y)/x + y^x log y ]

Now solve for y′:

dy/dx
= − [ (y x^y)/x + y^x log y ]
/ [ x^y log x + (x y^x)/y ]

Multiply numerator and denominator by x y to simplify:

dy/dx
= − [ y² x^y + x y y^x log y ]
/ [ x y x^y log x + x² y^x ]

Rewrite neatly:

dy/dx
= − ( y x^(y−1) + y^x log y )
/ ( x^y log x + x y^(x−1) )

Final Answer:

dy/dx
= − ( y x^(y−1) + y^x log y ) / ( x^y log x + x y^(x−1) )

_____________________________________________________________

13.Find dy/dx if

y^x = x^y

Solution:

Given
y^x = x^y

Take logarithm on both sides.

x log y = y log x

Differentiate both sides with respect to x.

Differentiate left side:

d/dx (x log y)
= log y + x · (1/y) · dy/dx

Differentiate right side:

d/dx (y log x)
= dy/dx · log x + y · (1/x)

Now equate:

log y + (x/y) dy/dx
= dy/dx · log x + y/x

Group dy/dx terms on one side.

(x/y) dy/dx − dy/dx · log x
= y/x − log y

Factor dy/dx.

dy/dx [ x/y − log x ]
= y/x − log y

Now solve for dy/dx.

dy/dx
= ( y/x − log y ) / ( x/y − log x )

Multiply numerator and denominator by y.

dy/dx
= ( y²/x − y log y ) / ( x − y log x )

Rewrite neatly.

Final Answer:

dy/dx = (y/x) ( y − x log y ) / ( x − y log x )

_____________________________________________________________

14. Find dy/dx if

(cos x)^y = (cos y)^x

Solution:

Take logarithm on both sides.

y log(cos x) = x log(cos y)

Differentiate both sides with respect to x.

Differentiate left side:

d/dx [ y log(cos x) ]
= dy/dx log(cos x) + y · (−tan x)

= dy/dx log(cos x) − y tan x

Differentiate right side:

d/dx [ x log(cos y) ]
= log(cos y) + x · (1/cos y)(−sin y) dy/dx

= log(cos y) − x tan y dy/dx

Now equate:

dy/dx log(cos x) − y tan x
= log(cos y) − x tan y dy/dx

Collect dy/dx terms:

dy/dx log(cos x) + x tan y dy/dx
= log(cos y) + y tan x

Factor:

dy/dx [ log(cos x) + x tan y ]
= log(cos y) + y tan x

Therefore,

dy/dx
= ( y tan x + log(cos y) )
/ ( x tan y + log(cos x) )

Final Answer:

dy/dx = ( y tan x + log cos y ) / ( x tan y + log cos x )

_____________________________________________________________

15.

Find dy/dx if

xy = e^(x − y)

Solution:

Given

xy = e^(x − y)

Differentiate both sides.

Left side (product rule):

d/dx(xy)
= x dy/dx + y

Right side:

d/dx e^(x − y)
= e^(x − y)(1 − dy/dx)

Since e^(x − y) = xy, substitute:

x dy/dx + y
= xy (1 − dy/dx)

Expand right side:

x dy/dx + y
= xy − xy dy/dx

Bring dy/dx terms together:

x dy/dx + xy dy/dx
= xy − y

Factor dy/dx:

dy/dx ( x + xy )
= y ( x − 1 )

Therefore,

dy/dx
= y(x − 1) / x(1 + y)

Final Answer:

dy/dx = y(x − 1) / x(y + 1)

_____________________________________________________________

16: Find the derivative of the function given by

f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) and hence find f′(1).

Solution:

Let

f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)

Use logarithmic differentiation.

Take logarithm on both sides.

log f(x)
= log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Differentiate with respect to x.

1/f(x) · f′(x)
= 1/(1 + x)

· 2x/(1 + x²)

· 4x³/(1 + x⁴)

· 8x⁷/(1 + x⁸)

Multiply both sides by f(x).

f′(x)
= (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
[ 1/(1 + x) + 2x/(1 + x²) + 4x³/(1 + x⁴) + 8x⁷/(1 + x⁸) ]

Now find f′(1)

First find f(1):

f(1)
= (2)(2)(2)(2)
= 16

Now substitute x = 1 into the bracket:

1/(1 + 1) = 1/2
2(1)/(1 + 1) = 1
4(1)/(1 + 1) = 2
8(1)/(1 + 1) = 4

Sum = 1/2 + 1 + 2 + 4
= 7 1/2
= 15/2

Therefore,

f′(1)
= 16 × 15/2
= 8 × 15
= 120

Final Answers:

f′(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
[ 1/(1 + x) + 2x/(1 + x²) + 4x³/(1 + x⁴) + 8x⁷/(1 + x⁸) ]

f′(1) = 120

_____________________________________________________________

17: Differentiate (x² − 5x + 8)(x³ + 7x + 9) in three ways:

(i) by using product rule
(ii) by expanding the product to obtain a single polynomial
(iii) by logarithmic differentiation

Do they all give the same answer?

(i) By Product Rule

Let

f(x) = (x² − 5x + 8)(x³ + 7x + 9)

Using product rule:

d/dx (uv) = u′v + uv′

Here

u = x² − 5x + 8
u′ = 2x − 5

v = x³ + 7x + 9
v′ = 3x² + 7

So,

f′(x)
= (2x − 5)(x³ + 7x + 9)

· (x² − 5x + 8)(3x² + 7)

Expand both parts:

(2x − 5)(x³ + 7x + 9)
= 2x⁴ − 5x³ + 14x² − 17x − 45

(x² − 5x + 8)(3x² + 7)
= 3x⁴ − 15x³ + 31x² − 35x + 56

Add:

f′(x)
= 5x⁴ − 20x³ + 45x² − 52x + 11

_____________________________________________________________

(ii) By Expanding First

First expand the original function:

(x² − 5x + 8)(x³ + 7x + 9)

= x⁵ − 5x⁴ + 15x³ − 26x² + 11x + 72

Now differentiate term-by-term:

f′(x)

= 5x⁴ − 20x³ + 45x² − 52x + 11

Same result ✔

_____________________________________________________________

(iii) By Logarithmic Differentiation

Let

y = (x² − 5x + 8)(x³ + 7x + 9)

Take logarithm:

log y
= log(x² − 5x + 8) + log(x³ + 7x + 9)

Differentiate:

1/y · dy/dx
= (2x − 5)/(x² − 5x + 8)

· (3x² + 7)/(x³ + 7x + 9)

Multiply by y:

dy/dx
= (x² − 5x + 8)(x³ + 7x + 9)
[ (2x − 5)/(x² − 5x + 8)

· (3x² + 7)/(x³ + 7x + 9) ]

After simplification (same as product rule expansion):

dy/dx
= 5x⁴ − 20x³ + 45x² − 52x + 11

Final Answer:

f′(x) = 5x⁴ − 20x³ + 45x² − 52x + 11

Yes. All three methods give exactly the same derivative.

_____________________________________________________________

18: If u, v and w are functions of x, then show that

d/dx (u v w)
= du/dx · v w + u · dv/dx · w + u v · dw/dx

Prove in two ways:
(i) by repeated application of product rule
(ii) by logarithmic differentiation

(i) By Repeated Application of Product Rule

Let

y = u v w

First treat v w as one function.

y = u (v w)

Using product rule:

dy/dx = du/dx · (v w) + u · d/dx (v w)

Now differentiate (v w) again using product rule:

d/dx (v w) = dv/dx · w + v · dw/dx

Substitute back:

dy/dx
= du/dx · v w

· u ( dv/dx · w + v · dw/dx )

Expand:

dy/dx
= du/dx · v w

· u · dv/dx · w

· u v · dw/dx

Hence proved.

_____________________________________________________________

(ii) By Logarithmic Differentiation

Let

y = u v w

Take logarithm on both sides.

log y = log u + log v + log w

Differentiate with respect to x.

1/y · dy/dx
= 1/u · du/dx

· 1/v · dv/dx

· 1/w · dw/dx

Multiply both sides by y = u v w.

dy/dx
= u v w ( du/(u dx) + dv/(v dx) + dw/(w dx) )

Now distribute:

dy/dx
= v w · du/dx

· u w · dv/dx

· u v · dw/dx

Which is

dy/dx
= du/dx · v w

· u · dv/dx · w

· u v · dw/dx

Hence proved.

Final Result:

d/dx (u v w)
= du/dx · v w + u · dv/dx · w + u v · dw/dx

_____________________________________________________________

EXERCISE 5.6

Q1: If x and y are connected parametrically by the equations, without eliminating the parameter, find dy/dx.

x = 2at² , y = at⁴

Solution

Differentiate with respect to t:

dx/dt = d/dt(2at²) = 4at
dy/dt = d/dt(at⁴) = 4at³

Therefore,

dy/dx = (dy/dt) / (dx/dt)
= (4at³) / (4at)
= t²

Answer: dy/dx = t²

_____________________________________________________________

Q2: x = a cosθ , y = b cosθ

Differentiate with respect to θ:

dx/dθ = −a sinθ
dy/dθ = −b sinθ

Therefore,

dy/dx = (dy/dθ) / (dx/dθ)
= (−b sinθ) / (−a sinθ)
= b/a

Answer: dy/dx = b/a

_____________________________________________________________

Q3: x = sin t , y = cos 2t

Solution:

Differentiate with respect to t:

dx/dt = cos t
dy/dt = −2 sin 2t

So,

dy/dx = (dy/dt) / (dx/dt)
= (−2 sin 2t) / (cos t)

Use identity sin 2t = 2 sin t cos t:

dy/dx = (−2 × 2 sin t cos t) / (cos t)
= −4 sin t

Answer: dy/dx = −4 sin t

_____________________________________________________________

Q4: x = 4t , y = 4/t

Solution:

Differentiate with respect to t:

dx/dt = 4
dy/dt = −4/t²

So,

dy/dx = (dy/dt) / (dx/dt)
= (−4/t²) / 4
= −1/t²

Answer: dy/dx = −1/t²

_____________________________________________________________

Q5:

x = cosθ − cos2θ
y = sinθ − sin2θ

Differentiate with respect to θ:

dx/dθ = −sinθ + 2sin2θ
dy/dθ = cosθ − 2cos2θ

Now,

dy/dx = (dy/dθ) / (dx/dθ)

= (cosθ − 2cos2θ) / (−sinθ + 2sin2θ)

Rearranging the denominator:

dy/dx = (cosθ − 2cos2θ) / (2sin2θ − sinθ)

Answer:
(cosθ − 2cos2θ) / (2sin2θ − sinθ)

_____________________________________________________________

Q6: x = a(θ − sinθ)
y = a(1 + cosθ)

Differentiate with respect to θ:

dx/dθ = a(1 − cosθ)
dy/dθ = −a sinθ

Therefore,

dy/dx = (−a sinθ) / [a(1 − cosθ)]

Cancel a:

dy/dx = −sinθ / (1 − cosθ)

Use identities:

1 − cosθ = 2 sin²(θ/2)
sinθ = 2 sin(θ/2) cos(θ/2)

Substitute:

dy/dx = −[2 sin(θ/2) cos(θ/2)] / [2 sin²(θ/2)]

Cancel 2 and one sin(θ/2):

dy/dx = −cos(θ/2) / sin(θ/2)

dy/dx = −cot(θ/2)

Answer: −cot(θ/2)

_____________________________________________________________

Q7: x = (sin³t) / √(cos2t)
y = (cos³t) / √(cos2t)

Write as:

x = sin³t · (cos2t)^(−1/2)
y = cos³t · (cos2t)^(−1/2)

Differentiate x using product rule:

dx/dt = 3sin²t cost · (cos2t)^(−1/2)
+ sin³t · [ (−1/2)(cos2t)^(−3/2)(−2sin2t) ]

dx/dt = 3sin²t cost / √(cos2t)
+ sin³t sin2t / (cos2t)^(3/2)

Similarly differentiate y:

dy/dt = −3cos²t sint / √(cos2t)
+ cos³t sin2t / (cos2t)^(3/2)

Now form:

dy/dx = (dy/dt) / (dx/dt)

Taking common denominator and simplifying using

sin2t = 2sint cost
cos2t = cos²t − sin²t

After simplification,

dy/dx = −cot(3t)

Answer: −cot(3t)

_____________________________________________________________

Q8:

x = a( cos t + log tan(t/2) )
y = a sin t

Solution:

Differentiate:

dx/dt
= a[ −sin t + d/dt(log tan(t/2)) ]

Now,

d/dt(log tan(t/2))
= (1 / tan(t/2)) · sec²(t/2) · 1/2
= 1 / sin t

So,

dx/dt = a( −sin t + 1/sin t )

dy/dt = a cos t

dy/dx

dy/dx
= (a cos t) / { a( −sin t + 1/sin t ) }

Cancel a:

= cos t / ( −sin t + 1/sin t )

Multiply top & bottom by sin t:

= cos t sin t / (1 − sin²t)

= cos t sin t / cos²t

= tan t

Final Answer:

dy/dx = tan t

_____________________________________________________________

Q9:

x = a sec θ
y = b tan θ

We find dy/dx = (dy/dθ) / (dx/dθ)

Differentiate

dx/dθ = a sec θ tan θ

dy/dθ = b sec²θ

Compute dy/dx

dy/dx
= (b sec²θ) / (a sec θ tan θ)

Cancel sec θ:

= b sec θ / (a tan θ)

Now write in sine–cosine form:

sec θ = 1/cos θ
tan θ = sin θ / cos θ

So,

sec θ / tan θ
= (1/cos θ) ÷ (sin θ / cos θ)
= 1 / sin θ
= cosec θ

Therefore,

dy/dx
= b/a · cosec θ

Final Answer:

dy/dx = (b/a) cosec θ

_____________________________________________________________

Q10:
x = a (cos θ + θ sin θ)
y = a (sin θ − θ cos θ)

Find dy/dx.

Solution

We use

dy/dx = (dy/dθ) / (dx/dθ)

Differentiate x

x = a (cos θ + θ sin θ)

dx/dθ
= a [ −sin θ + (sin θ + θ cos θ) ]
= a (θ cos θ)

Differentiate y

y = a (sin θ − θ cos θ)

dy/dθ
= a [ cos θ − (cos θ − θ sin θ) ]
= a (θ sin θ)

Compute dy/dx

dy/dx
= (a θ sin θ) / (a θ cos θ)

Cancel aθ:

dy/dx = sin θ / cos θ

Final Answer (matches given):

dy/dx = tan θ

_____________________________________________________________

Q11: If
x = √(a sin⁻¹ t), y = √(a cos⁻¹ t),
show that dy/dx = − y / x.

Solution

Given
x = √(a sin⁻¹ t)
y = √(a cos⁻¹ t)

Differentiate x with respect to t:

dx/dt
= (1/2) × (a sin⁻¹ t)⁻¹ᐟ² × a / √(1 − t²)

Differentiate y with respect to t:

dy/dt
= (1/2) × (a cos⁻¹ t)⁻¹ᐟ² × (−a) / √(1 − t²)

Now,

dy/dx = (dy/dt) ÷ (dx/dt)

= − √(a cos⁻¹ t) ÷ √(a sin⁻¹ t)

But
√(a cos⁻¹ t) = y
√(a sin⁻¹ t) = x

Therefore,

dy/dx = − y / x

Hence proved.

_____________________________________________________________

EXERCISE 5.7

Q1.

y = x² + 3x + 2

Solution:
First derivative:
dy/dx = 2x + 3

Second derivative:
d²y/dx² = 2

Answer: 2

_____________________________________________________________

Q2:

y = x²⁰

Solution:
First derivative:
dy/dx = 20x¹⁹

Second derivative:
d²y/dx² = 380x¹⁸

Answer: 380x¹⁸

_____________________________________________________________

Q3:

y = x cos x

Solution:
First derivative (using product rule):
dy/dx = x(−sin x) + cos x

dy/dx = cos x − x sin x

Second derivative:
d²y/dx² = −sin x − (sin x + x cos x)

d²y/dx² = −x cos x − 2 sin x

Answer: −x cos x − 2 sin x

_____________________________________________________________

Q4:

y = log x

Solution:
First derivative:
dy/dx = 1/x

Second derivative:
d²y/dx² = −1/x²

Answer: −1/x²

_____________________________________________________________

Q5:

y = x³ log x

Solution:
First derivative (product rule):
dy/dx = x³(1/x) + log x(3x²)

dy/dx = x² + 3x² log x

Second derivative:
d²y/dx² = 2x + 6x log x + 3x

d²y/dx² = x(5 + 6 log x)

Answer: x(5 + 6 log x)

_____________________________________________________________

Q6:

y = eˣ sin 5x

Solution:
First derivative (product rule):
dy/dx = eˣ sin 5x + eˣ(5 cos 5x)

dy/dx = eˣ(sin 5x + 5 cos 5x)

Second derivative:
d²y/dx² = eˣ(sin 5x + 5 cos 5x)
    + eˣ(5 cos 5x − 25 sin 5x)

d²y/dx² = eˣ(6 cos 5x − 24 sin 5x)

= 2eˣ(5 cos 5x − 12 sin 5x)

Answer: 2eˣ(5 cos 5x − 12 sin 5x)

_____________________________________________________________

Q7:

y = e⁶ˣ cos 3x

Solution:
First derivative (product rule):
dy/dx
= e⁶ˣ cos 3x + e⁶ˣ(−3 sin 3x)

dy/dx
= e⁶ˣ(6 cos 3x − 3 sin 3x)

Second derivative:
d²y/dx²
= e⁶ˣ(6 cos 3x − 3 sin 3x)
 + e⁶ˣ(−18 sin 3x − 9 cos 3x)

d²y/dx²
= 9e⁶ˣ(3 cos 3x − 4 sin 3x)

Answer: 9e⁶ˣ(3 cos 3x − 4 sin 3x)

_____________________________________________________________

Q8:

y = tan⁻¹ x

Solution:
First derivative:
dy/dx = 1 / (1 + x²)

Second derivative:
d²y/dx²
= −2x / (1 + x²)²

Answer: −2x / (1 + x²)²

Q9:

y = log (log x)

Solution:
First derivative:
dy/dx
= 1 / (x log x)

Second derivative:
d²y/dx²
= −(1 + log x) / (x log x)²

Answer: −(1 + log x) / (x log x)²

_____________________________________________________________

Q10:

y = sin (log x)

Solution:
First derivative:
dy/dx
= cos (log x) × 1/x

Second derivative:
d²y/dx²
= −sin (log x)/x² − cos (log x)/x²

d²y/dx²
= −[sin (log x) + cos (log x)] / x²

Answer: −[sin (log x) + cos (log x)] / x²

_____________________________________________________________

Q11:

If
y = 5 cos x − 3 sin x,
prove that
d²y/dx² + y = 0.

Solution

Given
y = 5 cos x − 3 sin x

Differentiate with respect to x:

First derivative:
dy/dx
= −5 sin x − 3 cos x

Differentiate again:

Second derivative:
d²y/dx²
= −5 cos x + 3 sin x

Now add y to d²y/dx²:

d²y/dx² + y
= (−5 cos x + 3 sin x) + (5 cos x − 3 sin x)

= 0

Hence proved that

d²y/dx² + y = 0

_____________________________________________________________

Q12: If
y = cos⁻¹ x,
find d²y/dx² in terms of y alone.

Solution

Given
y = cos⁻¹ x

Taking cosine on both sides,
x = cos y

Differentiate with respect to x:

1 = −sin y × dy/dx

dy/dx = −1 / sin y

Differentiate again with respect to x:

d²y/dx²
= − d/dx (1 / sin y)

= − [ −cos y / sin² y × dy/dx ]

Substitute dy/dx = −1 / sin y:

d²y/dx²
= − cos y / sin³ y

= − cot y cosec² y

Answer:

d²y/dx² = − cot y cosec² y ✔ (Correct)

_____________________________________________________________

Q13: If
y = 3 cos (log x) + 4 sin (log x),
show that
x²y₂ + xy₁ + y = 0.

Solution

Given
y = 3 cos (log x) + 4 sin (log x)

First derivative:

dy/dx
= −3 sin (log x) × 1/x + 4 cos (log x) × 1/x

dy/dx
= [ −3 sin (log x) + 4 cos (log x) ] / x

Second derivative:

d²y/dx²
= d/dx { [ −3 sin (log x) + 4 cos (log x) ] / x }

d²y/dx²
= [ −3 cos (log x) − 4 sin (log x) ] / x²
  − [ −3 sin (log x) + 4 cos (log x) ] / x²

Now substitute in x²y₂ + xy₁ + y:

x²y₂
= −3 cos (log x) − 4 sin (log x)
  + 3 sin (log x) − 4 cos (log x)

xy₁
= −3 sin (log x) + 4 cos (log x)

y
= 3 cos (log x) + 4 sin (log x)

Adding all terms:

x²y₂ + xy₁ + y = 0

Hence proved.

_____________________________________________________________

Q14:If
y = A eᵐˣ + B eⁿˣ,
show that

d²y/dx² − (m + n) dy/dx + mn y = 0.

Solution

Given
y = A eᵐˣ + B eⁿˣ

Differentiate with respect to x:

First derivative:
dy/dx
= A m eᵐˣ + B n eⁿˣ

Second derivative:
d²y/dx²
= A m² eᵐˣ + B n² eⁿˣ

Now substitute in
d²y/dx² − (m + n) dy/dx + mn y:

= [A m² eᵐˣ + B n² eⁿˣ]
− (m + n)[A m eᵐˣ + B n eⁿˣ]

· mn[A eᵐˣ + B eⁿˣ]

Grouping terms:

= A eᵐˣ [m² − m(m + n) + mn]

· B eⁿˣ [n² − n(m + n) + mn]

= A eᵐˣ (0) + B eⁿˣ (0)

= 0

Hence proved.

_____________________________________________________________

Q15:

If
y = 500e⁷ˣ + 600e⁻⁷ˣ,
show that

d²y/dx² = 49y.

Solution

Given
y = 500e⁷ˣ + 600e⁻⁷ˣ

Differentiate with respect to x:

First derivative:
dy/dx
= 3500e⁷ˣ − 4200e⁻⁷ˣ

Second derivative:
d²y/dx²
= 24500e⁷ˣ + 29400e⁻⁷ˣ

Now factor 49:

d²y/dx²
= 49(500e⁷ˣ + 600e⁻⁷ˣ)

But
500e⁷ˣ + 600e⁻⁷ˣ = y

Therefore,

d²y/dx² = 49y

Hence proved.

_____________________________________________________________

Q16:

If
eʸ (x + 1) = 1,
show that

d²y/dx² = (dy/dx)².

Solution

Given
eʸ (x + 1) = 1

Differentiate both sides with respect to x:

Using product rule,

eʸ (x + 1) dy/dx + eʸ = 0

Factor eʸ:

eʸ [ (x + 1) dy/dx + 1 ] = 0

Since eʸ ≠ 0,

(x + 1) dy/dx + 1 = 0

Therefore,

dy/dx = −1 / (x + 1)

Differentiate again:

d²y/dx²
= 1 / (x + 1)²

Now,

(dy/dx)²
= [ −1 / (x + 1) ]²
= 1 / (x + 1)²

Hence,

d²y/dx² = (dy/dx)²

Hence proved.

_____________________________________________________________

Q17:

If
y = (tan⁻¹ x)²,
show that

(x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.

Solution

Given
y = (tan⁻¹ x)²

Differentiate with respect to x:

First derivative:

y₁ = dy/dx
= 2(tan⁻¹ x) × 1/(1 + x²)

Second derivative:

y₂ = d²y/dx²
= 2/(1 + x²)² − 4x(tan⁻¹ x)/(1 + x²)²

Now substitute in
(x² + 1)² y₂ + 2x(x² + 1) y₁:

(x² + 1)² y₂
= 2 − 4x tan⁻¹ x

2x(x² + 1) y₁
= 4x tan⁻¹ x

Adding,

(x² + 1)² y₂ + 2x(x² + 1) y₁
= 2

Hence proved.

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