Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Question 1.
(3x² – 9x + 5)⁹
Solution:
Let y = (3x² – 9x + 5)⁹ = ⁹, where u = 3x² – 9x+ 5
∴ (frac { dy }{ dx }) = (frac { dy }{ du }) x (frac { du }{ dx }) = 9u^9-1 x (6x – 9)
= 9(3x² – 9x + 5)⁸ . 3(2x – 3).

Question 2.
sin³ x + cos⁶x
Solution:

Question 3.
(5x)^{3 cos 2x}
Solution:
Let y = (5x)^{3 cos 2x}
Taking log both sides, we get
log y = 3 cos 2x log 5x
Differentiating w.r.t. x,

Question 4.
sin^-1(x(sqrt{x})), 0 ≤ x ≤ 1
Solution:

Question 5.
(frac{cos ^{-1} frac{x}{2}}{sqrt{2 x+7}}), – 2 < x < 2
Solution:
Let y = (frac{cos ^{-1} frac{x}{2}}{sqrt{2 x+7}}), – 2 < x < 2

Question 6.
cot^-1(left[frac{sqrt{1+sin x}+sqrt{1-sin x}}{sqrt{1+sin x}-sqrt{1-sin x}}right]), 0 < x < (frac { π }{ 2 })
Solution:

Question 7.
(log x)^{log x}, x > 1
Solution:
Let (log x)^{log x}, x > 1
Taking log of both sides, we get
log y = (log x) log (log x)
Here u = log x and v = log log x.
So, = (frac { du }{ dx }). Also, v = log t, where t = log x.

Question 8.
cos(a cos x + b sin x), for some constant a and b.
Solution:
Let cos(a cos x + b sin x).
Put a cos x + b sin x = t.

Question 9.
(sin x – cos x)^{(sin x – cos x)}, (frac { π }{ 4 }) < x < (frac { 3π }{ 4 })
Solution:
Let y = (sin x – cos x)^{sin x – cos x}
Taking log of both sides, we get
log y = (sin x – cosx)log(sin x – cos x)
Differentiating both sides w.r.t. x
(frac { 1 }{ y })(frac { dy }{ dx }) = (cos x + sin x) log (sin x – cos x) + (sin x – cos x) x (frac { d }{ dx }) log(sin x – cos x)
= (cos x + sin x) log (sin x – cos x) + (sin x – cos x) x (frac { 1 }{ sin x – cos x }) x (frac { d }{ dx })(sin x – cos x)
= (cos x + sin x) log (sin x – cos x) + (cos x + sin x)
= (sin x + cos x) [1 + log (sin x – cos x)]
∴ (frac { dy }{ dx }) = y (sin x + cos x) [1 + log (sin x – cos x)]
= (sin x + cos x) (sin x – cos x) [1 + log (sin x – cos x)^{sin x – cos x} × [1 + log (sin x – cos x)].

Question 10.
x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0.
Solution:
Let y = x^x + x^a + a^x + a^a
Differentiating w.r.t. x,

Question 11.
x^x²-3 + (x – 3)^x² for x > 3
Solution:
Let y = x^x²-3 + (x – 3)^x² = u + v.
∴ u = x^x²-3
Taking log of both sides, we get
log u = log x^x²-3 = (x² – 3) log x.
Differentiating w.r.t. x,

Question 12.
Find (frac { dy }{ dx }), if y = 12(1 – cos t), x = 10(t – sin t), – (frac { π }{ 2 }) < t < (frac { π }{ 2}).
Solution:

Question 13.
Find (frac { dy }{ dx }), if y = sin^-1x + sin^-1(sqrt{1-x^{2}}).
Solution:
y = sin^-1x + sin^-1(sqrt{1-x^{2}}).
Put x = sin θ
∴ y = θ + sin^-1(sqrt{1-sin ^{2} theta})
= θ + sin^-1 (cos θ)
= θ + (frac { π }{ 2 }) – θ = (frac { π }{ 2 }) .
∴ (frac { dy }{ dx }) = 0.

Question 14.
If x(sqrt{1+y}) + y(sqrt{1+x}) = 0 for – 1 < x < 1, prove that (frac { dy }{ dx }) = – (frac{1}{(1+x)^{2}})
Solution:
The given equation may be written as
x(sqrt{1+y}) = – y(sqrt{1+x})
Squaring both sides, we get,
x²(1 + y) = y²(1 + x) ⇒ x² – y² = y² x – x²y
⇒ (x + y) (x – y) = – xy(x – y)
⇒ x + y = – xy [∵ x – y ≠ 0 ]
⇒ x = – y – xy
⇒ y(1 + x) = – x
⇒ y = – (frac { x }{ 1+x })
⇒ (frac { dy }{ dx }) = – (left{frac{(1+x) cdot 1-x(0+1)}{(1+x)^{2}}right})
⇒ (frac { dy }{ dx }) = – (frac{1}{(1+x)^{2}})

Question 15.
If (x – a)² + (y – b)² = c², for some c > 0, prove that (frac{left{1+left(frac{d y}{d x}right)^{2}right}^{3 / 2}}{frac{d^{2} y}{d x^{2}}}) is a constant, independent of a and b.
Solution:
Given (x – a)² + (y – b)² = c² … (1)
Differentiating (1) w.r.t. x, we get
2(x – a) + 2(y – b)(frac { dy }{ dx }) = 0
⇒ (x – a) + (y – b)(frac { dy }{ dx }) = 0 … (2)
Differentiating (2) w.r.t. x, we get
1 + (y – b) . (frac{d^{2} y}{d x^{2}}) + ((frac { dy }{ dx }))² = 0

Putting these values of (y-b) and (x – a) from (3) and (4) in (1), we get

which is a constant, independent of a and b.

Question 16.
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that (frac { dy }{ dx }) = (frac{cos ^{2}(a+y)}{sin a})
Solution:
Given: cos y = x cos (a + y).

Question 17.
If x = a (cos t + t sin t) and y = a (sin t – 1 cos t), find (frac{d^{2} y}{d x^{2}}). Mention the domain in which it is valid.
Solution:
We have : x = a (cos t + t sin t)
∴ (frac { dx }{ dt }) = a (- sin t + sin t + t cos t) = at cos t.
Also, we have : y = a (sin t – t cos t)
∴ (frac { dy }{ dx }) = a(cos t – cos t + t sin t) = at sin t.
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d t}}{frac{d x}{d t}}) = (frac{a t sin t}{a t cos t}) = tan t.
It is valid when t ≠ 0, t ≠ (2n + 1)(frac { π }{ 2 }).

Here, also t ≠ 0, t ≠ (2n + 1)(frac { π }{ 2 }),
So, domain is t ∈R – [ (2n + 1)(frac { π }{ 2 }) ].

Question 18.
If f(x) = | x |³, show that f”(x) exists for all real x and find it.
Solution:
When x ≥ 0, then f(x) = | x |³ = x³.
∴ f’ (x) = – 3x² and f”(x) = 6x, which exists for all real values of x.
When x < 0, then
f(x) = | x |³ = (- x)³ = – x³.
∴ f'(x) = – 3x²
and f”(x) = – 6x,
which exists for all real values of x.
Hence, f’”(x) = (left{begin{array}{r} 6 x, text { if } x geq 0 \ -6 x, text { if } x<0 end{array}right.)

Question 19.
Using Mathematical Induction, prove that (frac { d }{ dx })(xⁿ) = nx^n-1, ∀ n ∈ N, for all positive integers n.
Solution:
Let P(n) be the given statement in the problem.
∴ P(n) : (frac { d }{ dx })(xⁿ) = nx^n-1 … (1)
To verify the statement for n = 1,
put n = 1 in (1). We get
P(1) : (frac { d }{ dx })(x¹) = (1)x^1-1 = (1)(x)⁰ = (1)(1) = 1,
which is true as (frac { d }{ dx }) (x) = 1.
We suppose P(x) is true for n = m.
∴ P(m) : (frac { d }{ dx })(x^m) = m x^m-1 … (2)
To establish the truth of P(m + 1), we prove
P(m + 1) : (frac { d }{ dx })(x^m+1) = (m + 1) x^m.
Since x^m+1 = x¹. x^m
∴ (frac { d }{ dx })(x^m+1) = (frac { d }{ dx })(x . x^m) = x (frac { d }{ dx })(x^m) + x^m (frac { d }{ dx }) (x)
= x . mx^m-1 + x^m . 1 [Using (2)]
= mx^m + x^m = (m + 1)x^m
= (m + 1) x^(m+1)-1.
∴ p(m + 1) is also true, if P(m) is true. But P(1) is true.
∴ By principle of mathematical induction, P(n) is true for all n ∈ N.

Question 20.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution:
sin (A + B) = sin A cos B + cos A sin B … (1)
Consider A and B as functions of t and differentiating both sides of (1) w.r.t. t, we have:

Question 21.
Does there exist a function which is continuous every-where but not differentiable at exactly two points? Justify your answer.
Solution:
Consider the function
f(x) = | x | + |x-1|
f is continuous everywhere. But it is not differentiable at x – 0 and x = 1.

Question 22.
If y = (left|begin{array}{ccc}
f(x) & g(x) & h(x) \
l & m & n \
a & b & c
end{array}right|), prove that (left|begin{array}{ccc}
f^{prime}(x) & g^{prime}(x) & h^{prime}(x) \
l & m & n \
a & b & c
end{array}right|)
Solution:
To differentiate a determinant, we differentiate one row (or one column) at a time, keeping other unchanged.

Question 23.
If y = (e^{a cos ^{-1} x}), show that (1 – x²)(frac{d^{2} y}{d x^{2}}) – x(frac { dy }{ dx }) – a²y = 0
Solution:
We have :