Application Of Derivatives(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

APPLICATION OF DERIVATIVES MCQs

Total Questions : 58 | Page 6 of 6 pages

Question 51. The minimum value of the function defined by f(x) = minimum {x, x+1,2 - x } is

Discuss Question

Answer: Option D. -> 32
:
D

∵

f(x) = maximum{x, x + 1, 2-x}

The Minimum Value Of The Function Defined By F(x) = Minimum ...

∴

Minimum value of function

= \frac{3}{2}

Question 52. The value of m for which the area of the triangle included between the axes and any tangent to the curve

x^{m} y = b^{m}

is constant is

Discuss Question

Answer: Option B. -> 1
:
B

∴ x^{m} y = b^{m}

Taking logarithm

m l o g_{e} x + l o g_{e} y = m l o g_{e} b ∴ \frac{m}{x} + \frac{1}{y} \frac{d y}{d x} = 0 ∴ \frac{d y}{d x} = - \frac{m y}{x}

Equation of tangent at (x,y) is

Y - y = - \frac{m y}{x} (X - x) x Y - x y = - m y X + m x y \Rightarrow m y X = x Y = x y (1 + m) \Rightarrow \frac{\frac{X}{x (1 + m)}}{m} + \frac{x}{x (1 + m)} = 1

∴

Area of triangle OAB

= \frac{1}{2} . O A . O B

The Value Of M For Which The Area Of The Triangle Included ...

= \frac{1}{2} ∣ ∣ \frac{x (1 + m)}{m} ∣ ∣ | y (1 + m) | = \frac{| x y | (1 + m)^{2}}{2 | m |} F o r m = 1, = \frac{| x y | (4)}{2} = 2 | x y | (∵ x y = b) = 2 | b | =

constant

Question 53. If a + b +c = 0, then the equation

3 a x^{2} + 2 b x + c = 0

has, in the interval (0, 1)

Atleast one root
Atmost one root
No root
exactly one root

Discuss Question

Answer: Option A. -> Atleast one root
:
A
Let

f (x) = a_{n} X^{n} + a_{n - 1} X^{n - 1} + . . . + a_{2} x^{2} + a_{1} x + a_{0}

Which is a polynomial function in x of degree n. Hence f(x) is continuos and differentiable for all x.
Let

\infty < β

. We given, f

(\infty) = 0 = f (β)

.
By Rolle's theorem, f' (c) = 0 for some value c,

\infty < c < β

.
Hence the equation

F^{'} (x) = n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + . . . + a_{1} = 0

has atleast one root between

\infty a n d β

.
Hence (c) is the correct answer.

Question 54.

f (x) = \frac{x}{l o g_{e} x}, x \neq 1

, is decreasing in interval

(0, e)
(1, e)
(e, ∞)
R

Discuss Question

Answer: Option A. -> (0, e)
:
A
f'(x) =

\frac{l o g_{e} x . 1 - x . \frac{1}{x}}{(l o g_{e} x)^{2}}

\frac{(l o g_{e} x - 1)}{(l o g_{e} x)^{2}} < 0

It is decreasing at (0, e) – {1}

F(x) = xloge x, x ≠ 1, Is Decreasing In Interval

Question 55. The value of

(127)^{1 / 3}

to four decimal places is

5.0267
5.4267
5.5267
5.001

Discuss Question

Answer: Option A. -> 5.0267
:
A
Let y =

x^{1 / 3}, x = 125

and

x + Δ x = 127.

Then,

\frac{d y}{d x} = \frac{1}{3 x^{2 / 3}} a n d Δ x = 0

When, x = 125, we have

y = 5 a n d \frac{d y}{d x} = \frac{1}{75}

∴ y = \frac{d y}{d x} Δ x \Rightarrow Δ y \frac{1}{75} \times 2 = \frac{2}{75}

∴ (127)^{1 / 3} = y + Δ y = 5 + \frac{2}{75} = 5 + \frac{8}{3} \times \frac{1}{100}

\Rightarrow (127)^{1 / 3} = 5 + \frac{(2.6667)}{100} = 5.02667 = 5.0267

Hence (a) is the correct answer.

Question 56.

f (x) = c o s (\frac{π}{x}), x \neq 0

increases in the interval

(12k+1, 12k)
(12k+2, 12k+1)
(12k+1, 12k+3)
(12k+1, 12k+3)

Discuss Question

Answer: Option A. -> (12k+1, 12k)
:
A
f(x) = cos

(\frac{π}{x})

Thus
f(x) = cos

(\frac{π}{x})

increases in

\frac{1}{3} \to \frac{1}{2}

generalizing it

(\frac{1}{2 k + 1}, \frac{1}{2 k})

F(x) = cos (πx), x ≠ 0 Increases In The Interval

Question 57. A ladder20 ft long has one end on the ground and the other end in contact with a vertical wall. The lower end slips along the ground. If the lower end of the ladder is 16 ft away from the wall, upper end is moving

λ

times as fast as the lower end, then

λ

Discuss Question

Answer: Option C. -> 43
:
C
Let OC be the wall. Let AB be the position of the ladder at any time t such that OA =x and OB=y. Length of the ladder AB =20 ft.
In