ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.6

Find the maximum and the minimum values (if any) of the following (1 to 7) functions:

Question 1.
(i) f(x) = x²
(ii) f(x) = (2x – 1)² + 3. (NCERT)
Solution:
(i) Given f(x) = x²
Since x² ≥ 0 ∀ x ∈ R
∴ f(x) ≥ 0 ∀ x ∈ R
Thus, 0 be the minimum value of f(x) which attains at x = 0
and f(x) can be made as large as we please.
Therefore the maximum value of f(x) does not exists which can be observed from the below graph.

(ii) f(x) = (2x – 1)² + 3 ≥ 3
[∵ (2x – 1)² ≥ 0 ∀ x ∈ R]
Here, D_f = R and f(x) ≥ 3
Hence the minimum value is 3.
However, maximum value does not exists.
[∵ f(x) can be made as large as possible]

Question 2.
(i) f (x) = – (x – 1)² + 10 (NCERT)
(ii) f(x) = 9x² + 12x + 2 (NCERT)
Solution:
(1) Given f(x) = – (x – 1)² + 10 ≤ 10,
D_f = R
∴ f(x) ≤ 10
[∵ (x – 1)² ≥ 0 ∀ x ∈ R]
∴ the maximum value of f(x) = 10.
However minimum value does not exists.
[∵ f(x) can be made as small as we please]

(ii) Given f(x) = 9x² + 12x + 6
= 9 \(\left(x^2+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{6}{9}\right)\)
= 9 \(\left(x+\frac{2}{3}\right)^2\) + 2 ≥ 2
∴ Minimum value 2 and Max. value does not exists.
[Since f(x) can be made as large as possible].

Question 3.
(i) f(x) = |x|
(ii) f(x) = x³ + 1 (NCERT)
Solution:
(i) f(x) = |x| ≥ 0 ∀ x ∈ R
Thus 0 be the minimum value of f(x) which it attains at x = 0.
Further F (x) can be made as large as we please.
Therefore, the maximum value of f(x) does not exists and clearly it can be observed from given below graph.

(ii) Given g(x) = x³ + 1
∴ D_f = R
∴ g(x) = (x + 1) (x² – x + 1)
= (x + 1) \(\left[\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\right]\)
= (x + 1) \(\left[\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]\)
Since \(\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\) > 0 ∀ x ∈ R
∴ g (x) > 0 or g (x) < 0 according as x + 1 > 0 or < 0
∴ g (x) has no maximum and minimum value.

Question 4.
(i) f(x) = | x + 2 | – 1 (NCERT)
(ii) f(x) = 3 + | x | (NCERT)
Solution:
(i) Given f(x) = | x + 2 | – 1
∴ f(x) ≥ – 1
[∵ |x + 2| ≥ 0 ∀ x ∈ R]
∴ min. value = – 1 and Max. value does not exists.
[Since f (x) can be made as small as we please].

(ii) f(x) = 3 + | x |
Since | x | ≥ 0 ∀ x ∈ R
⇒ 3 + | x | ≥ 3
⇒ f{x) ≥ 3 ∀ x ∈ R
Hence 3 be the minimum value of f (x) which is attains at x = 0 and f (x) can be made as large as we please.
Thus the maximum value of/(x) does not exists.

Question 5.
(i) f(x) = – | x + 1 | + 3 (NCERT)
(ii) f (x) = sin 2x
Solution:
(i) Given, f(x) = – | x + 1 | + 3 on R
since |x + 1| ≥ 0 ∀ x ∈ R
⇒ – |x + 1 |≤ 0 ∀ x ∈ R
⇒ – | x + 1| + 3 ≤ 3 ∀ x ∈ R
⇒ f (x) ≤ 3 ∀ x ∈ R
⇒ f (x) ≤ f (- 1) ∀ x ∈ R
Thus, 1 be the maximum value of f(x) and it attains at x = – 1.
Since, f(x) can be made as small as we please thus minimum value of f(x) does not exists.

(ii) Given f(x) = sin 2x
since – 1 ≤ sin 2x ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ f(x) ≤ 1 ∀ x ∈ R
Thus maximum value of f(x) = 1
and minimum value of f(x) = – 1.

Question 6.
(i) f(x) = sin 2x + 5 (NCERT)
(ii) f(x) = | sin 4x + 3 | (NCERT)
Solution:
(i) Given f(x) = sin 2x + 5
Since, -1 ≤ sin 2x ≤ 1 ∀ x ∈ R
⇒ – 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ; ∀ x ∈ R
⇒ 4 ≤ sin 2x + 5 ≤ 6,∀ x ∈ R
⇒ 4 ≤ f(x) ≤ 6, ∀ x ∈ R
∴ Minimum value of f(x) = 4
and maximum value of f(x) = 6.

(ii) Given f(x) = | sin 4x + 3 | = sin 4x + 3
[∵ | sin x | ≤ 1]
Since – 1 ≤ sin 4x ≤ 1 ∀ x ∈ R
– 1 + 3 ≤ sin 4x + 3 ≤ 1 + 3, ∀ x ∈ R
⇒ 2 ≤ sin 4x + 3 ≤ 4 ∀ x ∈ R
2 ≤ f(x) ≤ 4 ∀ x ∈ R
∴ Min. value of f(x) = 2
and max. value of f(x) = 4.

Question 7.
(i) f(x) = | sin 3x | – 3
(ii) f(x) = 4 – | cos 3x |
Solution:
(i) Given f(x) = | sin 3x | – 3
Since 0 ≤ | sin 3x | ≤ 1
[∵ | sin x | ≤ 1 ∀ x ∈ R]
⇒ 0 – 3 ≤ | sin 3x | – 3 ≤ 1 – 3
⇒ – 3 ≤ f(x) ≤ – 2
Maximum value of f(x) = – 2
and Minimum value of f(x) = – 3

(ii) Given f(x) = 4 – | cos 3x |
since 0 ≤ | cos 3x | ≤ 1
⇒ 0 ≥ – | cos 3x | ≥ -1
⇒ 4 – 1 ≤ 4 – | cos 3x | ≤ 4 + 0
⇒ 3 ≤ 4 – | cos 3x | ≤ 4
⇒ 3 ≤ f(x) ≤ 4
∴ Maximum value of f(x) = 5
and Minimum value of f(x) = 3

Question 8.
(i) Is the minimum value of cos x zero ?
(ii) What are the minimum and maximum values of 2 – 3 cos x ?
Solution:
(i) Since, | cos x | ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ cos x ≤ 1
Thus, minimum value of f(x) = cos x be – 1
∴ minimum value of cos x is not 0.

(ii) Since, | cos x | ≤ 1 ∀ x ∈ R
⇒ – 1 ≤ cos x ≤ 1, ∀ x ∈ R
⇒ – 3 ≤ 3 cos x ≤ 3
⇒ – 3 ≤ – 3 cos x ≤ 3
⇒ 2 – 3 ≤ 2 – 3 cos x <≤ 2 + 3
⇒ – 1 ≤ 2 – 3 cos x ≤ 5 ∀ x ∈ R
⇒ – 1 ≤ f(x) ≤ 5, ∀ x ∈ R
∴ Maximum value of f(x) = 5
Minimum value of f(x) = – 1.

Question 9.
What are the maximum and minimum values of 3 sin x + 4 cos x ?
Solution:
Let f(x) = 3 sin x + 4 cos x
put 3 = r cos α ; 4 = r sin α
On squaring and adding ; we have
r = \(\sqrt{16+9}\) = 5 ;
tan α = \(\frac{4}{3}\)
⇒ α = tan^-1 (\(\frac{4}{3}\) )
∴ f(x) = r sin (x + α) = 5 sin (x + α)
since – 1 ≤ sin (x + α) < 1 – 5 < 5 sin (x + α) < 5
⇒ – 5 ≤ f(x) ≤ 5
Thus maximum value of f(x) = 5
and Minimum value of f(x) = – 5.

Question 10.
Find the maximum and minimum values, if any, of the following functions :
(i) x in (0, 1) (NCERT)
(ii) x in [0, 1].
Solution:
(i) Let f(x) = x, x ∈ (0, 1)
So it is differentiable ∀ x ∈ (0, 1)
Diff. w.r.t. x, we have
f'(x) = 1 ≠ 0
Thus, f'(x) ≠ 0 for any x ∈ (0, 1)
Hence given function f(x) has no turning point.
Thus, the given function has neither maximum nor minimum values.

(ii) Let f(x) = x, x ∈ [0, 1] …………..(1)
Clearly it is differentiable ∀ x ∈ [0, 1]
DifF. eqn. (1) w.r.t. x, we have
f'(x) = 1 ≠ 0 ∀ x ∈ (- 1, 1)
and hence f(x) has no critical / turning points.
Also, f(0) = 0 ;
f(1) = 1
Maximum value of f(x) = 1
and Minimum value = 0
The point of maxima is 1 and point of minima be 0.

Question 11.
Prove that the function f(x) = x³ + x² + x + 1 does not have maxima or minima.
Solution:
Given f(x) = x³ + x² + x + 1
∴ f (x) = 3x² + 2x + 1
Now f'(x) = 0
⇒ 3x² + 2x + 1 = 0 does not gives any real values of x.
∴ f(x) has no point of maxima or minima.

Question 12.
Find the (absolute) maximum and the (absolute) minimum values of the following functions in the indicated intervals. Also find the points of (absolute) maxima and minima :
(i) f(x) = (x – 1)² + 3 in [- 3, 1] (NCERT)
(ii) f(x) = 2x³ – 15x² + 36x + 1 on the interval [1, 5]
(iii) f (x) – 3x⁴ – 8x³ + 12x² – 48x + 25 in [0, 3] (NCERT)
(iv) f(x) = x + sin x in [0, 2π]
(v) f(x) = sin x + √3 cos x in [0, π]
(vi) f(x) = 3 + | x + 1 | in [- 2, 3].
Solution:
(i) Given f(x) = (x – 1)² + 3 in [- 3, 1]
∴ f(x) = 2(x – 1)
∴ f(x) = 0
⇒ 2x – 2 = 0
⇒ x = 1 ∈ [- 3, 1]
Now y]_{x = 1} = 3 ;
y]_{x = – 3} = 16 + 3 = 19
Absolute maximum value = 19
and Absolute minimum value = 3.

(ii) Given f(x) = 2x³ – 15x² + 36x + l in [1, 5]
∴ f'(x) = 6x² – 30x + 36
∴ f (x) = 0
⇒ 6(x² – 5x + 6) = 0
⇒ x = 2, 3
Now x = 2, 3 ∈ [1, 5]
Now y]_{x = 1} = 2 – 15 + 36 + 1 = 24 ;
y]_{x = 2} = 16 – 60 + 72 + 1 = 29;
y]_{x = 3} = 54 – 135 + 108 + 1 = 23;
y]_{x = 5} = 250 – 375 + 180 + 1 = 56
∴ y has absolute maximum value is 56
and absolute min value = 24.
Thus point of maxima be at x = 5
and point of minima be x = 1

(iii) Given f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
f'(x) = 12x³ – 24x² + 24x – 48
= 12 (x³ – 2x² + 2x – 4)
For local maxima/minima, we have f'(x) = 0
⇒ 12 [x³ – 2x² + 2x – 4] = 0
⇒ x² (x – 2) + 2 (x – 2) = 0
⇒ (x – 2) (x² + 2) = 0
⇒x-2 = 0 or x² + 2 = 0 does not gives real values of x
∴ x = 2
Now let us compute the values of f(x) at stationary points and also at the end points of given interval [0, 3].
f(2) = 3 × 16 – 8 × 8 + 12 × 4 – 96 + 25
= 48 – 64 + 48 – 96 + 25 = – 39
f(0) = 0 – 0 + 0 – 0 + 25 = 25
f(3) = 3 × 81 -8 × 27 + 12 × 9 – 48 × 3 + 25
= 243 – 216 + 108 – 144 + 25 = 16
Out of these values, maximum value be f(0) = 25
and least value be f(2) = – 39
Hence absolute maximum value = 25 at x = 0
and absolute minimum value = – 39 at x = 2.

(iv) Given f(x) = x + sin x in [0, 2π]
Diff. both sides w.r.t. x, we have
f'(x) = 1 + cos x
Now, f'(x) = 0
⇒ 1 + cos x = 0
⇒ cos x = – 1 = cos π
But x ∈ [0, 2π]
∴ x = π
Now, f(π) = π + sin π
= π + 0 = π
f(0) = 0 + sin 0
= 0 + 0 = 0
f(2π) = 2π + sin 2π
= 2π + 0 = 2π
Thus, absolute maximum value = 2π
and point of maxima be 2π.
Also, absolute minimum value of f(x) be 0
and it attains at x = 0.

(v) Given f(x) = sin x + √3 cos x
∴ f'(x) = cos x – √3 sin x
Now critical points are given by putting f'(x) = 0
⇒ cos x – √3 sin x = 0
⇒ cos x = √3 sin x
⇒ tan x = \(\frac{1}{\sqrt{3}}\) = tan \(\frac{\pi}{6}\)
⇒ x = nπ + \(\frac{\pi}{6}\), n ∈ Z
But x ∈ [0, π]
∴ x = \(\frac{\pi}{6}\)
Now we computed f(x0 at critical points and end points
i.e. x = 0, \(\frac{\pi}{6}\), π
∴ f(0) = 0 + √3 × 1 = √3;
f(\(\frac{\pi}{6}\)) = \(\frac{1}{2}+\sqrt{3} \times \frac{\sqrt{3}}{2}\) = 2
f(π) = 0 + √3 cos π = – √3
Out pf these values, maximum value of f(x) be 2 at x = \(\frac{\pi}{6}\)
and minimum value of f(x) = – √3 at x = π
∴ points of maxima and minima are x = \(\frac{\pi}{6}\) and π.

Question 12 (old).
(i) f(x) = x³ in [- 2, 2] (NCERT)
(ii) f(x) = 4x – x² in [- 2, \(\frac{9}{2}\)]
(iii) f(x) = (\(\frac{1}{2}\) – x)² + x³ in [- 2, 2-5]
(iv) f(x) = sin x + cos x in [0, π] (NCERT)
Solution:
(i) Given f(x) = x³ in [- 2, 2]
∴ f'(x) = 3x² i.e. f(x) = 0
⇒ 3x² = 0
⇒ x = 0 ∈ [- 2, 2]
Now y]_{x = 0} = 0;
y]_{x = 2} = 8;
y]_{x = – 2} = – 8
absolute max. value = 8 ;
absolute min value = – 8.

(ii) Given f(x) = 4x – \(\frac{1}{2}\) x² ; x ∈ [- 2, \(\frac{9}{2}\)]
∴ f'(x) = 4 – x
Now f'(x) = 0
⇒ x = 4 ∈ [- 2, \(\frac{9}{2}\)]
∴ y]_{x = – 2} = – 8 – \(\frac{1}{2}\) × 4 = – 10;
y]_{x = \(\frac{9}{2}\)} = 18 – \(\frac{1}{2} \times \frac{81}{4}\)
= \(\frac{144-81}{8}=\frac{63}{8}\)
y]_{x = 4} = 4 × 4 – \(\frac{1}{2}\) (4)² = 8
∴ y is absolute max. at x = 4
and absolute maximum value = 8
and y is absolute minimum at x = -2
and absolute minimum value = -10.

(iii) f(x) = (\(\frac{1}{2}\) – x)² + x³ in [- 2, 2.5]
∴ f'(x) = 2 (\(\frac{1}{2}\) – x) (- 1) + 3x²
∴ f'(x) = 0
⇒ 3x² – 1 + 2x = 0
⇒ (x + 1) (3x – 1) = 0
∴ x = – 1, \(\frac{1}{3}\) ∈ [- 2, 2.5]
Now y]_{x = – 1} = \(\frac{9}{4}\) + (- 1) = \(\frac{5}{4}\) ;
y]_{x = \(\frac{1}{3}\)} = \(\frac{1}{36}+\frac{1}{27}=\frac{7}{108}\)
y]_{x = – 2} = \(\frac{5}{4}\) – 8 = \(\frac{-7}{4}\)
y]_{x = \(\frac{5}{2}\)} = 4 + \(\frac{125}{8}\) = \(\frac{157}{8}\)
∴ absolute max. value = \(\frac{157}{8}\)
and absolute min. value = \(\frac{-7}{4}\)
Thus point of maxima be at x = \(\frac{5}{2}\) and
point of minima be x = – 2.

(iv) Given f(x) = sin x + cos x, [0, π]
∴ f'(x) = cos x – sin x,
f'(x) = 0
⇒ cos x – sin x = 0
⇒ tan x = 1
∴ x = \(\frac{\pi}{4}\) ∈ [0, π]
Now y]_{x = 0} = 1 ;
y]_{x = π} = – 1 ;
y]_{x = \(\frac{\pi}{4}\)} = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2
Absolute max. value = √2 and attains at x = \(\frac{\pi}{4}\)
and Absolute min value = – 1 and attains at x = π

(vi) Given f(x) = 3 + | x + 1 | ∀ x ∈ [- 2, 3]
Clearly f(x) is differentiable ∀ x ∈ [- 2, 3] except at x = – 1.
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{x+1}{|x+1|}\)
Now f'(x) = 0
⇒ \(\frac{x+1}{|x+1|}\) = 0, x ≠ – 1
which gives no values of x i.e. has no solution.
So f(x) has no critical points.
So, f(- 1) = 3 + |- 1 + 1| = 3
f(- 2) = 3 + |- 2 + 1|
= 3 + |- 1| = 4
and f(3) = 3 + |3 + 1|
= 3+ 4 = 7
Thus absolute max. value of f(x) = 7 and it attains at x = 3.
and Absolute minimum value of f(x) = 3 and it attains at x = – 1.

Question 13.
Find the maximum and minimum values of x + sin 2x on [0, 2π]. Also find points of maxima and minima. (NCERT)
Solution:
Let y = x + sin 2x on [0, 2π]
∴ \(\frac{d y}{d x}\) = 1 + 2 cos 2x,
Now \(\frac{d y}{d x}\) = 0
⇒ 1 + 2 cos 2x = 0
⇒ cos 2x = – \(\frac{1}{2}\)
∴ cos 2x = – cos (\(\frac{\pi}{3}\))
= cos (π – \(\frac{\pi}{3}\))
⇒ cos x = cos \(\frac{2 \pi}{3}\)
⇒ 2x = 2nπ ± \(\frac{2 \pi}{3}\)
[∵ cos θ = cos α
⇒ 0 = 2nπ ± α]
⇒ x = nπ ± \(\frac{\pi}{3}\)
∴ x = \(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\)
∴ y]_{x = 0} = 0 ;
y]_{x = 2π} = 2π ;
y]_{x = \(\frac{\pi}{3}\)} = \(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{2\pi}{3}\)} = \(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{4 \pi}{3}\)} = \(\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}\) ;
y]_{x = \(\frac{5 \pi}{3}\)} = \(\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}\)
∴ y is absolute maximum at x = 2π
and absolute Max. value = 2π
and y is absolute minimum at x = 0
and absolute minimum value = 0.

Question 14.
Find the absolute maximum and absolute minimum values of the function f(x) = cos² x + sin x, x ∈ [0, π].
Solution:
Given f(x) = cos² x + sin x, x ∈ [0, π]
∴ f’(x) = – 2 cos x sin x + cos x
= cos x (1 – 2 sin x)
The critical points are given by putting f’(x) = 0
⇒ cos x (1 – 2 sin x) = 0
⇒ cos x = 0 or sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\), \(\frac{5 \pi}{6}\)
Thus, we compute f(x) at x = 0, \(\frac{\pi}{6}\), \(\frac{\pi}{2}\), \(\frac{5 \pi}{6}\), π
Here, f(0) = 1;
f(\(\frac{\pi}{6}\)) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}\)
= \(\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
f(\(\frac{\pi}{2}\)) = 1 ;
f(\(\frac{5 \pi}{6}\)) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}\) ;
f(π) = 1
Thus, maximum value of f(x) = \(\frac{5}{4}\)
and minimum value of f(x) = 1.