NCERT Solutions for Class 12 Mathemetics Chapter 6 - Application of Derivatives

Answer:

Let ‘r’ be the radius of the circle.
Then A, the area of the circle = $\mathrm{\pi }$r².

Hence, the rate of change of the area of the circle = 2$\mathrm{\pi }$r.
(a) When r = 3 cm, then the rate of change of the area of the circle = 2$\mathrm{\pi }$ (3) = 6$\mathrm{\pi }$ cm²/s.
(b) When r = 4 cm, then the rate of change of the area of the circle = 2$\mathrm{\pi }$ (4) = 8$\mathrm{\pi }$ cm²/s.

Answer:

Answer:

Answer:

Let ‘x’ be the edge of the cube.

Answer:

Let ‘r’ be the radius of the circular wave.
Then A= $\mathrm{\pi }$r²,
where A is the enclosed area at time t.
Differentiating w.r.t. t, we get :

Hence, the enclosed area is increasing at the rate of 80$\mathrm{\pi }$ cm²/s, where r = 8 cm.

Answer:

Let ‘r’ be the radius of the circle.

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Answer:

V, the volume of the balloon

Answer:

Let ‘r’ be the variable radius of the balloon.

Answer:

Let MP be the wall and AB = (5 m) be the ladder.
Let BM = x metres.

Answer:

We have :

Answer:

Let ‘r’ be the radius of the air bubble.

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Radius (say r) of sphere

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Let ‘r’ and ‘h’ be the radius and height of the sand cone respectively at time t.

Answer:

Marginal cost is the rate of change of total cost with respect to output.
Now C (x) = 0·007 x³ – 0·003 x² + 15x + 4000.

= (0·007) (3x²) – 0·003 (2x) + 15
When x = 17, MC = (0·007) (3(17)²) – 0·003 (2 (17)) + 15
= 6·069 – 0·102 + 15 = 20·967.
Hence, the reqd. marginal cost is ₹ 21 (nearly).

Answer:

Marginal revenue is the rate of change of total revenue with respect to number of units sold.
Now R(x) = 13x² + 26x + 15.

= 26x + 26.
When x = 7, MR = 26 (7) + 26
= 182 + 26 = 208.
Hence, the reqd. marignal revenue is ₹ 208.

Answer:

12$\mathrm{\pi }$

Answer:

126

Answer:

Answer:

Answer:

We have :
f (x) = sin x.
$\therefore$ f′ (x) = cos x.

$\Rightarrow$ f′ (x) does not have same sign in (0, $\mathrm{\pi }$).
Hence, ‘f’ is neither increasing nor decreasing in (0, $\mathrm{\pi }$).

Answer:

We have :
f (x) = 2x² – 3x.
$\therefore$ f′ (x) = 4x – 3.
(a) For f (x) to be strictly increasing function of x,
f′ (x) > 0
$\Rightarrow$ 4x – 3> 0

Answer:

Answer:

We have : f (x) = 2x³ – 3x² – 36x + 7.
$\therefore$ f′ (x) = 6x² – 6x – 36.
For f (x) to be strictly increasing function of x :
f′ (x) > 0 i.e. 6x² – 6x – 36 > 0
$\Rightarrow$ x² – x – 6 > 0
$\Rightarrow$ (x – 3) (x + 2) > 0
$\Rightarrow$ (x – (– 2)) (x – 3) > 0
$\Rightarrow$ x $\in$ (– $\infty$, – 2) ∪ (3, $\infty$).
Hence, ‘f’ is strictly increasing in (– $\infty$, – 2) ∪ (3, $\infty$).

Answer:

For f (x) to be strictly decreasing function of x,
f′ (x) < 0 i.e. 6x² – 6x – 36 < 0
$\Rightarrow$ x² – x – 6 < 0
$\Rightarrow$ (x – 3) (x + 2) < 0
$\Rightarrow$ (x – (– 2)) (x – 3) < 0
$\Rightarrow$ x $\in$ (– 2, 3).
Hence, ‘f’ is strictly decreasing in (– 2, 3).

Answer:

We have : f (x) = x² + 2x – 5.
$\therefore$ f′ (x) = 2x + 2.
(I) For f (x) to be strictly increasing function of x,
f ′ (x) > 0
i.e., if 2x + 2 > 0 i.e., if x > – 1.
Hence, f is strictly increasing in (–1, $\infty$).
(II) For f (x) to be strictly decreasing function of x,
f ′ (x) < 0
i.e.if 2x + 2 < 0 i.e., if x < – 1.
Hence, ‘f ’ is strictly decreasing in (– $\infty$,–1).

Answer:

We have : f (x) = 10 – 6x – 2x².
$\therefore$ f ′ (x) = – 6 – 4x.
(I) For f (x) to be strictly increasing function of x,
f ′ (x) > 0
i.e. if – 6 – 4x > 0 i.e. if 3 + 2x < 0

Answer:

We have :
f (x) = – 2x³ – 9x² – 12x + 1.
$\therefore$ f′ (x) = – 6x² – 18x – 12.
(I) For f (x) to be strictly increasing function of x,
f ′ (x) > 0
i.e. if – 6x² – 18x – 12 > 0
i.e. if x² + 3x + 2 < 0
i.e. if (x + 2) (x + 1) < 0
i.e. (x – (– 2)) (x – (– 1) < 0
i.e. if – 2 < x < – 1.
Hence, ‘f’ is srictly increasing in (– 2, – 1).
(II) For f (x) to be strictly decreasing function of x,
f′ (x) < 0
i.e. if – 6x² – 18x – 12 < 0
i.e. if x² + 3x + 2 > 0
i.e. if (x + 2) (x + 1) > 0
i.e. if (x – (– 2)) (x – (– 1) > 0
i.e. if x < – 2 or x > – 1.
Hence, ‘f’ is strictly decreasing in (– $\infty$, – 2) ∪ (– 1, $\infty$).

Answer:

We have : f (x) = 6 – 9x – x².
$\therefore$ f′ (x) = – 9 – 2x.
(I) For f (x) to be strictly increasing function of x,
f′ (x) > 0
i.e. if – 9 – 2x > 0 i.e. if 2x < – 9

Answer:

We have : f (x) = (x + 1)³ (x – 3)³.
$\therefore$ f′ (x) = (x + 1)³ . 3(x – 3)2 + 3 (x + 1)2 (x – 3)³
= 3 (x + 1)² (x – 3)² [(x + 1) + (x – 3)]
= 3 (x + 1)² (x – 3)² (2x – 2)
= 6 (x + 1)² (x – 3)² (x – 1).
(I) For f (x) to be strictly increasing function of x,
f ′ (x) > 0
i.e.if 6 (x + 1)² (x – 3)² (x – 1) > 0
i.e.if x – 1 > 0 [$\because$ 6 (x + 1)² (x – 3)² > 0]
i.e.if x > 1.
But f′ (x) = 0 at x = 3.
Hence, ‘f ’ is strictly increasing in (1, 3) ∪ (3, $\infty$).
(II) For f (x) to be strictly decreasing function of x,
f ′ (x) > 0
i.e.if 6 (x + 1)² (x – 3)² (x – 1) > 0
i.e.if x – 1 < 0 [$\because$ 6 (x + 1)² (x – 3)² > 0]
i.e. if x < 1.
But f′ (x) = 0 at x = – 1.
Hence, ‘f’ is strictly decreasing in (– $\infty$, – 1) ∪ (– 1, 1).

Answer:

Answer:

We have :
y = [x (x – 2)]² = x² (x² – 4x + 4)
= x⁴ – 4x³ + 4x².

Answer:

Answer:

Let f (x) = log x for x > 0.

Hence, ‘f’ is strictly increasing on (0, −$\infty$).

Answer:

We have : f (x) = x² – x + 1.

Hence, f (x) is neither increasing nor decreasing on (– 1, 1).

Answer:

We have : f (x) = cos x.

Answer:

We have: f (x) = cos 2x.
$\therefore$ f′ (x) = – 2 sin 2x.

Answer:

We have :
f (x) = cos 3x.
$\therefore$ f′ (x) = – 3 cos 3x.

Answer:

We have : f (x) = tan x.

Answer:

For – 1 < x < 1
$\Rightarrow$ – 1 < x⁹⁹ < 1
$\Rightarrow$ – 100 < 100 x⁹⁹ < 100.
Also 0 < cos x < 1
$\therefore$ f′ (x) is either + ve or – ve as (– 1, 1).
$\Rightarrow$ f (x) is neither increasing nor decreasing on (– 1, 1).

Answer:

For 0 < x < 1, x⁹⁹ and cos x are both + ve
$\Rightarrow$ f′ (x) > 0
$\Rightarrow$ f (x) is increasing on (0, 1).

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Answer:

We have :
f (x) = x² + ax + 1.
$\therefore$ f′ (x) = 2x + a.
Now 1 < x < 2
$\Rightarrow$ 2 < 2x < 4
$\Rightarrow$ 2 + a < 2x + a < 4 + a
$\Rightarrow$ 2 + a < f′ (x) < 4 + a.
For f (x) to be increasing, f′ (x) $\ge$ 0
$\Rightarrow$ 2 + a $\ge$ 0
$\Rightarrow$ a $\ge$ – 2.
Hence, the least value of a is – 2.

Answer:

Answer:

We have : f (x) = log sin x.

Answer:

Answer:

We have :
f (x) = x³ – 3x² + 3x – 100.
$\therefore$ f′ (x) = 3x² – 6x + 3 = 3 (x – 1)²,
which is $\ge$ 0 for all x $\in$ R.
Hence, f (x) is increasing function on R.

Answer:

(0, 2)

Answer:

The given curve is y = 3x⁴ – 4x.

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Answer:

We have : y = x³ – x + 1.

Hence, the reqd. slope of the tangent = 11.

Answer:

The given curve is y = x³ – 3x + 2

Answer:

The given curve is :
x = a cos³ $\mathrm{\theta }$, y = a sin³ $\mathrm{\theta }$.

Answer:

We have :
x = 1 – a sin $\mathrm{\theta }$ and y = b cos² $\mathrm{\theta }$.

Answer:

The given curve is :
y = x³ – 3x² – 9x + 7 ...(1)

Answer:

The given curve is :
y = f(x) = (x – 2)² ...(1)
$\therefore$ f′ (x) = 2 (x – 2).

$\Rightarrow$ 2x – 4 = 2
$\Rightarrow$ 2x = 6
$\Rightarrow$ x = 3.
When x = 3, then from (1), y = (3 – 2)² = 1.
Hence, the reqd. point is (3, 1).

Answer:

The given curve is :
y = x³ – 11x + 5 ...(1)
The given line is y = x – 11 ...(2)
Slope of the line = 1.

Thus 3x² – 11 = 1 $\Rightarrow$ 3x² = 12
$\Rightarrow$ x² = 4 $\Rightarrow$ x = ± 2.
When x = 2, then from (1),
y = 8 – 22 + 5 = – 9.
When x = – 2, then from (1),
y = – 8 + 22 + 5 = 19.
Equation of tangent at (2, – 9) is :
y + 9 = 1. (x – 2)
$\Rightarrow$ y = x – 11.
Hence, (2, – 9) is the reqd. point.

Answer:

The given curve is :

$\therefore$ The equation of reqd. lines are :
y + 1 = (– 1) (x – 0)
$\Rightarrow$ y + 1 = – x
$\Rightarrow$ y + x + 1 = 0
and y – 1 = (– 1) (x – 2)
$\Rightarrow$ y – 1 = – x + 2 $\Rightarrow$ y + x – 3 = 0.

Answer:

The given curve is :

Answer:

The given curve is :

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The given curve is :

Answer:

The given curve is :
y = x⁴ – 6x³ + 13x² – 10x + 5.

Answer:

Answer:

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Answer:

The given curve is :
y = x² – 2x + 7 ...(1)

Slope of the tangent = 2 (x – 1).
Slope of the line 2x – y + 9 = 0 is 2.
Since the tangent is parallel to the given line,
$\therefore$ 2 (x – 1) = 2 [$\because$ m₁ = m₂ ]
$\Rightarrow$ 2x = 4 $\Rightarrow$ x = 2.
Putting in (1),y = 4 – 4 + 7 = 7.
$\therefore$The equation of the tangent at (2, 7) parallel to
2x – y + 9 = 0 is :
y – 7 = 2 (x – 2) $\Rightarrow$ 2x – y + 3 = 0
$\Rightarrow$y – 2x – 3 = 0.

Answer:

Slope of the tangent = 2 (x – 1).
Slope of the line 5y – 15x = 13 is 3.
Since the tangent is perp. to the given line,
$\therefore$ (2(x – 1)) (3)= – 1 [$\because$m₁ m₂ =−1]

Answer:

The given curve is y = 7x³ + 11.

Hence, the tangents at the points where x = 2 and x = – 2 are parallel.

Answer:

Let P (x₁, y₁) be any point on the curve y = x³ ...(1)

$\Rightarrow$ x₁ = 0 or 3.
When x₁ = 0, then y₁ = (0)³ = 0.
When x₁ = 3, then y₁ = 3³ = 27.
Hence, the reqd. points are (0, 0) and (3, 27).

Answer:

Let (x₁, y₁) be the reqd. point on the given curve :
y = 4x³ – 2x⁵ ...(1)

$\Rightarrow$ x₁ = 0, x₁ = ± 1.
When x₁ = 0, then from (2),
y₁ = 0.
When x₁ = 1, then from (2),
y₁ = (1) – 2 (1) = 2.
When x₁ = – 1, then from (2),
y₁ = 4 (– 1)³ – 2 (– 1)⁵
= – 4 + 2 = – 2.
Hence, the reqd. points are (0, 0), (1, 2) and (– 1, – 2).

Answer:

The given curve is x² + y² – 2x – 3 = 0 ...(1)

$\Rightarrow$ 1 – x = 0 $\Rightarrow$ x = 1.
Putting in (1), 1 + y² – 2 (1) – 3 = 0
$\Rightarrow$ y² = 4 $\Rightarrow$ y = ± 2.
Hence, the reqd. points are (1, ± 2).

Answer:

The given curve is ay² = x³

Answer:

The given curve is :
y = x3 + 2x + 6 ...(1)

When x = 2, then from (1),
y = 8 + 4 + 6 = 18.
When x = – 2, then from (1),
y = – 8 – 4 + 6 = – 6.
Thus the co-ordinates of the points are (2, 18) and (– 2, – 6).
$\therefore$ The equations of the normals are :

$\Rightarrow$ 14y – 252 = – x + 2 and 14y + 8y = – x – 2
$\Rightarrow$ x + 14y – 254 = 0 and x + 14y + 86 = 0.

Answer:

The given parabola is y2 = 4ax.

Answer:

The given curves are :
x = y² ...(1)
and xy = k ...(2)
From (2), using (1), we get :
y²y = k $\Rightarrow$ y³ = k $\Rightarrow$ y = k^1/3.
Putting in (1), x = k^2/3.
Thus the given curves intersect at P (k^2/3, k^1/3).
Diff. (1) w.r.t. x,

Since the given curves cut at right angles at P,
$\therefore$ m^{1m2 = – 1}

Answer:

Answer:

The given curve is :

Answer:

Answer:

(1, 2)

Answer:

Let y = x^1/2, x = 25 and dx = 0·3.

Answer:

Let y = x^1/2, x = 49 and dx = 0·5.

Answer:

Let y = x^1/2, x = 0·64 and dx = – 0·04.

Answer:

Let y = x^1/3, x = 0·008
and dx = 0·001 so that x + dx = 0·009.
We require (x + dx)^1/3.
Now $∆$y = (x + $∆$x)^1/3 – x^1/3
= (0·008 + 0·001)^1/3 – (0·008)^1/3
= (0·009)^1/3 – 0·2
$\Rightarrow$(0·009)^1/3 = 0·2 + $∆$y.
Now $∆$y is approximately equal to :

Answer:

Answer:

Let y = x^1/4 , x = 16,
dx = – 1 so that x + dx = 15.
We require (x + dx)^1/4.
Now $∆$y = (x +$∆$ ^)1/4 – x ^1/4
= (16 − 1)^1/4 – (16)^1/4
= (15 )^−1/4 – 2
$\Rightarrow$(15)^1/4 = 2 + $∆$y.
Now $∆$y is approximately equal to :

Answer:

y = x^1/3, x = 27
and dx = – 1 so that x + dx = 26.
We require (x + dx)^1/3.
Now $∆$y = (x + $∆$y)^1/3 – x^1/3
= (27 – 1)^1/3 – 3 = (26)^1/3 – 3
$\Rightarrow$ (26)^1/3 = 3 + $∆$y.
Now $∆$y is approximately equal to :

Answer:

y = x^1/4, x = 256, dx = – 1 so that x + dx = 255.
We require (x + dx)^1/4.
Now $∆$y = (x +$∆$x)^1/4 – x^1/4
= (256 − 1)^1/4 – 4 = (255)^1/4 −4
$\Rightarrow$(255)^1/4 = 4 + $∆$y.
Now $∆$y, approximately equal to dy and

Answer:

y = x^1/4, x = 81, dx = 1

Answer:

y = x^1/2, x = 400 and dx = 1.

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Answer:

Answer:

Let y = x^1/4, x = 81, dx = ·5
so that x + dx = 81·5.
We require (x + dx)^1/4
Now $∆$y = (x + $∆$x)^1/4 − x^1/4

Answer:

Answer:

Answer:

Let x = 2 and $∆$x = 0·01 ...(1)
Then f (2·01) = f (x + $∆$x)
= 4 (x + $∆$x)² + 5 (x + $∆$x) + 2.
Now $∆$y = f (x + $∆$x) – f (x)
$\Rightarrow$ f (x + $∆$x) = f (x) + $∆$y
= f (x) + f′ (x) $∆$x [$\because$ dx = $∆$x]
$\therefore$ f (2·01) = (4x2 + 5x + 2) + (8x + 5) $∆$x
= (4 (2)2 + 5 (2) + 2) + (8 (2) + 5) (0·01)
[Using (1)]
= (16 + 10 + 2) + (16 + 5) (0·01)
= 28 + 0·21 = 28·21.
Hence, approximate value of f (2·01) = 28·21.

Answer:

We have : f (x) = x³ – 7x² + 15.
$\therefore$ f′ (x) = 3x² – 14x.
Now f (x + $∆$x) = f(x) + $∆$x f′ (x)
= (x³ – 7x² + 15) + $∆$x (3x² – 14x).
Taking x = 5 and $∆$x = 0·001, we get :
f (5·001) = (5)³ – 7 (5)² + 15) + (0·001) (3 (5)² – 14 (5))
= (125 – 175 + 15) + (0·001) (75 –70)
= – 35 + (0·001) (5)
= – 35 + 0·005 = – 34·995.

Answer:

V, the volume of the cube = x3.

= (3x²) $∆$x
= (3x²) [0·01x] [$\because$ 1% 0f x = 0⋅01x]
= 0·03 x³ m³.
Hence, the approximate change in the volume
= 0·03 x³ m³.

Answer:

Answer:

Answer:

Let ‘r’ be the radius of the sphere and $∆$r be the error in measuring the radius.
Then r = 9 m and $∆$r = 0·03 m.
Now S, the surface area of the sphere is given by :
S = 4$\mathrm{\pi }$r²

= (8$\mathrm{\pi }$ (9)) (0·03) = 2·16$\mathrm{\pi }$ m².
Hence, the approximate error in calculating the surface area is 2·16$\mathrm{\pi }$ m².

Answer:

77·66

Answer:

0·09 x³ m³

Answer:

We have : f (x) = (2x – 1)² + 3.
Since (2x – 1)² $\ge$0,
$\therefore$ Min. Value of (2x – 1)² = 0.
$\therefore$ Min. Value of f (x) = 3.
Max. value of f (x) does not exist.

Answer:

We have : f (x) = 9x² + 12x + 2
= (3x + 2)² – 2.
Min. value of (3x + 2)² = 0
$\therefore$ Min. value of f (x) = – 2.
Max. value of f (x) does not exist.

Answer:

We have : f (x) = – (x – 1)² + 10.
Max. value of – (x – 1)² = 0.
$\therefore$ Max. value of f (x) = 10.
Min. value of f (x) does not exist.

Answer:

Answer:

Answer:

Answer:

We have : f (x) = sin 2x + 5.
Max. value of sin 2x =1
$\Rightarrow$ Max. value of sin 2x + 5 =1 + 5 = 6.
Min. value of sin 2x =– 1
$\Rightarrow$ Min. value of sin 2x + 5 = – 1 + 5 = 4.

Answer:

We have : f (x) = | sin 4x + 3 |.
Max. value of sin 4x= 1
$\Rightarrow$ Max. value of | sin 4x + 3 | = | 1 + 3 | = | 4 | = 4.
Min. value of sin 4x= – 1
$\Rightarrow$ Min. value of | sin 4x + 3 | = | – 1 + 3 | = | 2 | = 2.

Answer:

We have :h (x) = x + 1.
As x $\to$ $\infty$, h (x) $\to$ $\infty$.
As x $\to$ – $\infty$, h (x) $\to$ – $\infty$.
Thus there is no max. and min. value.
However, when x = 1, h (x) = 2
when x = – 1, h (x) = 0.
Hence, max. value is 2 and min. value is 0.

Answer:

We have : f (x) = x².
$\therefore$ f′ (x) = 2x, f ″ (x) = 2.
Now f′ (x) = 0 $\Rightarrow$2x = 0 $\Rightarrow$ x = 0.
And f″ (0) = 2 (+ ve)
Hence, local minimum value at x = 0,
local min. value = f(0) = 0.

Answer:

We have : g (x) = x³ – 3x.
$\therefore$ g′ (x) = 3x² – 3, g″ (x) = 6x.
Now g′ (x)= 0 $\Rightarrow$ 3x2 – 3 = 0 $\Rightarrow$ x2 = 1
$\Rightarrow$ x = ± 1.
And g″ (1) = 6 (1) = 6 ( + ve) and
g″ (– 1) = 6 (– 1) = – 6 ( – ve)
Hence, local minimum value is, at x = 1, local min. value =
g (1) = 1 – 3 (1) = – 2.
Local maximum value is, at x = – 1, local max. value
= g (– 1) = (– 1)3 – 3 (– 1) = – 1 + 3 = 2.

Answer:

We have h (x) = sin x + cos x.
$\therefore$ h′ (x) = cos x – sin x and
h′′ (x) = – sin x – cos x.
Now h′(x) = 0 $\Rightarrow$ cos x – sin x = 0
$\Rightarrow$ sin x = cos x $\Rightarrow$ tan x = 1

Answer:

We have : f (x) = sin x – cos x.
$\therefore$ f′ (x) = cos x + sin x
and f″ (x) = – sin x + cos x.
Now f′ (x) = 0
$\Rightarrow$ cos x + sin x= 0 $\Rightarrow$ sin x = – cos x

Answer:

We have : f(x) = x³ – 6x² + 9x + 15.
$\therefore$ f′ (x) = 3x² – 12x + 9
and f ″ (x) = 6x – 12.
Now f ′ (x) = 0 $\Rightarrow$ 3x² – 12x + 9 = 0
$\Rightarrow$ x² – 4x + 3 = 0
$\Rightarrow$ (x – 1) (x – 3) = 0 $\Rightarrow$ x = 1, 3.
And f ″ (1) = 6 (1) – 12 = – 6 (– ve),
f ″ (3) = 18 – 12 = 6 (+ ve)
Hence, local max. is at x = 1, local max. value
= 1 – 6 + 9 + 15 = 19.
Local min is at x = 3, local min. value
= 27 – 6 (9) + 9 (3) + 15
= 27 – 54 + 27 + 15 = 15.

Answer:

Answer:

Answer:

Answer:

We have : f (x) = ex.
$\therefore$ f ′ (x) = ex.
Now f ′ (x) = 0
$\Rightarrow$ ex= 0,
which gives no real value of x.
Hence, f (x) does not have maximum or minimum values.

Answer:

We have f (x) = log x.

which gives no real value of x.
Hence, f (x) does not have maximum or minimum values.

Answer:

We have : f (x) = x³ + x² + x + 1.
$\therefore$ f′ (x) = 3x² + 2x + 1.
Now f′ (x) = 0
$\Rightarrow$ 3x² + 2x + 1 = 0

which are non-real.
Hence, f (x) does not have maximum and minimum values.

Answer:

We have : f (x) = x3, x $\in$ [– 2, 2].
$\therefore$ f′ (x) = 3x².
For extreme values,
f′ (x) = 0 $\Rightarrow$ 3x² = 0
$\Rightarrow$ x = 0 $\in$ [– 2, 2].
Now f (– 2) = (– 2)³ = – 8
f (0) = (0)³ = 0
and f (2) = 2³ = 8.
Hence, absolute max. value is 8,which is attained at x = 2
and absolute min. value is – 8, which is attained at
x = – 2.

Answer:

We have : f (x) = sin x + cos x, x $\in$ [0, $\mathrm{\pi }$].
$\therefore$ f′ (x) = cos x – sin x.
For extreme values,
f′ (x) = 0
$\Rightarrow$ cos x – sin x= 1
$\Rightarrow$ sin x = cos x

Answer:

We have :

Answer:

We have :
f (x) = (x – 1)² + 3, x $\in$ [– 3, 1].
$\therefore$ f′ (x) = 2 (x – 1).
For extreme values, f ′ (x) = 0
$\Rightarrow$ 2 (x – 1) = 0
$\Rightarrow$ x – 1 = 0 $\Rightarrow$ x = 1 $\in$ [– 3, 1].
Now f (– 3) = (– 4)² + 3 = 16 + 3 = 19
and f (1) = (0)² + 3 = 0 + 3 = 3.
Hence, absolute max. value is 19, which is attained at x = – 3.
and absolute min. value is 3, which is attained at x = 1.

Answer:

p (x) = 41 – 24x – 18x².
$\therefore$ p′ (x) = – 24 – 36x and p″ (x) = – 36.
Now p′ (x) = 0
$\Rightarrow$ – 24 – 36x = 0 $\Rightarrow$ 3x = – 2

Answer:

We have :
f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25.
$\therefore$ f′ (x) = 12x³ – 24x² + 24x – 48.
Now f′ (x) = 0
$\Rightarrow$ 12x³ – 24x² + 24x – 48 = 0
$\Rightarrow$ x³ – 2x² + 2x – 4 = 0
$\Rightarrow$ x² (x – 2) + 2 (x – 2) = 0
$\Rightarrow$ (x² + 2) (x – 2) = 0
$\Rightarrow$ x = 2.
$\therefore$ f (0) = 25
f (2) = 3 (16) – 8 (8) + 12 (4) – 48 (2) + 25
= 48 – 64 + 48 – 96 + 25
= 121 – 160 = – 39.
f (3) = 3 (81) – 8 (27) + 12 (9) – 48 (3) + 25
= 243 – 216 + 108 – 144 + 25
= 376 – 360 = 16.
Hence, max. value is 25 and min value is – 39.

Answer:

We have :
f (x) = sin 2x, 0 ≤ x ≤ 2$\mathrm{\pi }$
so that f′ (x) = 2 cos 2x, 0 < x < 2$\mathrm{\pi }$.
Now f′ (x)= 0
$\Rightarrow$ 2 cos 2x= 0, 0 < x < 2$\mathrm{\pi }$
$\Rightarrow$ cos 2x = 0, 0 < x < 2$\mathrm{\pi }$

Answer:

We have :
f (x) = sin x + cos x ; x $\in$ [0, 2$\mathrm{\pi }$] (say)
$\therefore$ f ′ (x) = cos x – sin x.
For extreme values, f ′ (x) = 0
$\Rightarrow$ cos x – sin x= 0
$\Rightarrow$ sin x = cos x $\Rightarrow$ tan x = 1

Answer:

We have :
f (x) = 2x³ – 24x + 107, x $\in$ [1, 3].
$\therefore$ f′ (x) = 6x² – 24.
(i) For extreme values, f′ (x) = 0
$\Rightarrow$ 6x² – 24 = 0
$\Rightarrow$ x² = 4 $\Rightarrow$ x = ± 2
$\Rightarrow$ x = 2 $\in$ [1, 3].
Now f (1) = 2 (1) – 24 (1) + 107
= 2 – 24 + 107 = 85
f (2) = 2 (8) – 24 (2) + 107
= 16 – 48 + 107 = 75
and f (3) = 2 (27) – 24 (3) + 107
= 54 – 72 + 107 = 89.
Hence, max. value = 89, which is attained at x = 3.
(ii) We have :
f (x) = 2x³ – 24x + 107,x $\in$ [– 3, – 1].
$\therefore$ f′ (x) = 6x² – 24.
For extreme values,
f ′ (x) = 0
$\Rightarrow$ 6x² – 24 = 0 $\Rightarrow$ x² – 4 = 0 $\Rightarrow$ x² = 4
$\Rightarrow$ x = ± 2 $\Rightarrow$ x = – 2 $\in$ [– 3, – 1].
Now f (– 3) = 2 (– 27) – 24 (– 3) + 107
= – 54 + 72 + 107 = 125
f (– 2) = 2 (– 8) – 24 (– 2) + 107
= – 16 + 48 + 107 = 139
f (– 1) = 2 (– 1) – 24 (– 1) + 107
= – 2 + 24 + 107 = 129.
Hence, max. value = 139,
which is attained at x = – 2.

Answer:

We have : f (x) = x⁴ – 62x² + ax + 9.
$\therefore$ f′ (x) = 4x³ – 124x + a
and f″ (x) = 12x² – 124.
Now f′ (1) = 0 [Given]
$\Rightarrow$ 4 (1) – 124 (1) + a = 0
$\Rightarrow$ a = 120.
Now f ″(1) = 12 (1) – 124 = – 112 < 0.
Thus f (x) attains its maximum value at x = 1, where a = 120.
Hence, a = 120.

Answer:

We have : f (x)= x + sin 2x.
$\therefore$ f′ (x) = 1 + 2 cos 2x.
Now f′ (x) = 0
$\Rightarrow$ 1 + 2 cos 2x = 0

Answer:

Let x and y be two numbers.
By the question, x + y = 24 ...(1)
Let the product, P = xy
$\Rightarrow$ P = x (24 – x)
$\Rightarrow$ P = 24 – x².

Answer:

We have :
x + y = 60 ...(1)
Let P = xy³
= x (60 – x)³ [Using (1)]

Answer:

We have : x + y = 35 ...(1)
Let P = x²y⁵
= x² (35 – x)⁵.

Answer:

Let ‘x’, and ‘y’ be the two positive numbers.
By the question, x + y = 16
$\Rightarrow$ y = 16 – x ...(1)
Let S = x³
+ y³ = x³ + (16 – x)³.

Answer:

Let ‘x’ cm, be the side of the square cut from each corner.

Length of the resulting box = 18 – 2x ;
Breadth of the resulting box = 18 – 2x ;
Height of the resulting box = x.
$\therefore$ Volume = (18 – 2x)². x
Let V (x) = x (18 – 2x)².
$\therefore$ V′ (x) = x.2 (18 – 2x) (– 2) + (18 – 2x)² (1)
= (18 – 2x) [– 4x + 18 – 2x]
= (18 – 2x) (18 – 6x).
For max./min. volume, V′ (x) = 0
$\Rightarrow$ (18 – 2x) (18 – 6x) = 0
$\Rightarrow$ x = 9 or x = 3.
But x = 9 is impossible.
[$\because$ The whole of thin sheet will be cut off]
Thus x = 3.
Now V″ (x) = (18 – 2x) (– 6) + (18 – 6x) (– 2)
= – 108 + 12x – 36 + 12x
= 24x – 144.
$\therefore$V″ (x = 3) = 24 (3) – 144
= 72 – 144 = – 72,
which is – ve.
$\therefore$V (x) is max. when x = 3.
Hence, the length of the side of the square cut off from each corner = 3 cm.

Answer:

Let ‘x’ cm. be the side of the square cut off from each corner.
Length of the resulting box = 45 – 2x.

Breadth of the resulting box = 24 – 2x.
Height of the resulting box = x.
$\therefore$Volume, V = (45 – 2x) (24 – 2x) x
= 2x (45 – 2x) (12 – x).
$\therefore$ V (x) = 2 (2x³ – 69x² + 540x).
$\therefore$ V′ (x) = 2 (6x² – 138x + 540)
= 12 (x² – 23x + 90).
For max. / min., V′= 0
$\Rightarrow$ 12 (x² – 23x + 90) = 0
$\Rightarrow$ (x – 5) (x – 18) = 0
$\Rightarrow$ x = 5, 18.
But x =18 is impossible.
$\therefore$ x = 5.
Also V″ (x) = 12 (2x – 23).
$\therefore$ V″ (5) = 12 (10 – 23)
= (12) (– 13) = – 156 < 0
Thus x = 5 for max. value.
Hence, length of the side of the square cut off for each corner is 5 cm.

Answer:

Let ‘x’ and ‘y’ be the length and breadth of the rectangle, which is inscribed in a fixed circle of radius ‘a’.
Then x² + y² = (2a)²
$\Rightarrow$ x² + y² = 4a² ...(1)
Now A, the area of the rectangle = xy

Answer:

Let ‘S’ and ‘V’ be the surface area and volume
of the cylinder, whose radius is ‘r’ and height ‘h’.
$\therefore$Surface area, S = 2$\mathrm{\pi }$r² + 2$\mathrm{\pi }$rh

Answer:

Let ‘r’ be the radius and ‘h’, the height of cylindrical can.
$\therefore$Volume = $\mathrm{\pi }$r²h = 100

Answer:

Let one piece be x m.
$\therefore$ Other piece is (28 – x) m.
Let ‘x’ metres be made into a circle and (28 – x) metres be made into a square.
$\therefore$ 2$\mathrm{\pi }$r = x

Answer:

Let the cone ABC be inscribed in a sphere of radius R.
Let ∠BOL = $\mathrm{\theta }$.
$\therefore$BL, radius of the base of the cone = R sin $\mathrm{\theta }$
and AL height of the cone = AO + OL
= R + R cos $\mathrm{\theta }$.

Answer:

Let ‘r’ and ‘h’ be the radius and height respectively of the cone ABC.

Answer:

Let ‘l’ be the given slant height and ‘$\mathrm{\theta }$’ be the semi-vertical angle.
$\therefore$r, radius of the base = l sin $\mathrm{\theta }$
and h, height of the cone = l cos $\mathrm{\theta }$.

Answer:

Let ‘S’ be the given surface of the cone whose radius is ‘r’, ‘h’ the height and ‘l’, the slant height.
$\therefore$ S = $\mathrm{\pi }$r² + $\mathrm{\pi }$rl ...(1)

Answer:

Answer:

Answer:

1

Answer:

12 $\mathrm{\pi }$

Answer:

126

Answer:

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is :

1. 2
2. –3

3. 

4. 

Answer:

(1, 2)

Answer:

77.66.

Answer:

0.09 x³ m³

Answer:

Answer:

Answer:

1

Answer:

1 m³/minute

Answer:

Answer:

1

Answer:

x – y = 0

Answer:

x + y = 3

Answer: