Free Objective Test 02 Practice Test - 11th and 12th

Question 1

A man of height 2m walks directly away from a lamp of height 5m, on a level road at 3 m/s. The rate at which the length of his shadow is increasing is

1m/s

2m/s

3m/s

4m/s

SOLUTION

Solution : B

Let be the lamp and PQ be the man and OQ=x metre be his shadow and let MQ =y metre.

$∴ \frac{d y}{d t}$ =speed of the man
=3 m/s (given)
$∵ Δ O P Q$ and $Δ O L M$ are similar
$∴ \frac{O M}{O Q} = \frac{L M}{P Q} \Rightarrow \frac{x + y}{x} = \frac{5}{2} \Rightarrow y = \frac{3}{2} x ∴ \frac{d y}{d t} = \frac{3}{2} \frac{d x}{d t} \Rightarrow 3 = \frac{3}{2} \frac{d x}{d t} \Rightarrow \frac{d x}{d t} = 2 m / s$

Question 2

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 $λ$ is

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{5}{3}$

SOLUTION

Solution : 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 $Δ A O B$ ,

$x^{2} + y^{2} = (20)^{2}$
$\Rightarrow 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 ∴ \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = \frac{- x}{\sqrt{400} - x^{2}} . \frac{d x}{d t} = - \frac{16}{\sqrt{400} - (16)^{2}} . \frac{d x}{d t} = - \frac{4}{3} \frac{d x}{d t}$
-ve sign indicates, that when X increases with time, y decreases. Hence, the upper end is moving $\frac{4}{3}$ times as fast as the lower end.

Question 3

The angle of intersection of the curves $y = 2 s i n^{2} x a n d y = c o s 2 x a t x = \frac{π}{6}$ is

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{2 π}{3}$

SOLUTION

Solution : B and D

We have,
$y = s i n^{2} x . . . (1) y = c o s 2 x . . . (2)$ And
On differentiating equation (1) w.r.t x, we get
$\frac{d y}{d x} = 4 s i n x c o s x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = 4 (\frac{1}{2}) \frac{\sqrt{3}}{2} = \sqrt{3} = m_{1} (s a y)$
On differentiating equation (2) w.r.t x, we get
$\frac{d y}{d x} = - 2 s i n 2 x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = - 2 s i n \frac{π}{3} = - \sqrt{3} = m_{2} (s a y)$
Hence, angle between the two curves is
$θ = \pm t a n^{- 1} (\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}) = \pm t a n^{- 1} \sqrt{3} = \frac{π}{3} o r \frac{2 π}{3}$
Hence (b) is the correct answer.

Question 4

The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point t = 2 is

\frac{7}{6}

$\frac{5}{6}$

$\frac{6}{7}$

$1$

SOLUTION

Solution : C

We have,
$\frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 t - 2}{2 t + 3}$
Thus, slope of the tangent to the curve at the point t = 2 is
${[\frac{d y}{d x}]}_{t - 2} = \frac{4 (2) - 2}{2 (2) + 3} = \frac{6}{7}$
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

Question 5

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

$\frac{1}{2}$

$\frac{3}{2}$

SOLUTION

Solution : 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
$= \frac{1}{2} . O A . O B$

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

Question 6

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

\frac{1}{2}

\frac{3}{2}

SOLUTION

Solution : D

$∵$ f(x) = maximum {x, x + 1, 2-x}

$∴$ Minimum value of function $= \frac{3}{2}$

Question 7

The value of $(127)^{1 / 3}$ to four decimal places is

5.0267

5.4267

5.5267

5.001

SOLUTION

Solution : 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 8

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

SOLUTION

Solution : 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 9

If $f (x) = ∣ ∣ ∣ ∣ \begin{matrix} s i n x & s i n a & s i n b c o s x & c o s a & c o s b t a n x & t a n a & t a n b \end{matrix} ∣ ∣ ∣ ∣,$

where $0 < a < b < \frac{π}{2}$
then the equation
$f^{'} (x) = 0$ has in the interval $(a, b)$

Atleast one root

Atmost one root

No root

exactly one root

SOLUTION

Solution : A

Here $f (a) = ∣ ∣ ∣ ∣ \begin{matrix} s i n a & s i n a & s i n b c o s a & c o s a & c o s b t a n a & t a n a & t a n b \end{matrix} ∣ ∣ ∣ ∣ = 0. Also f (b) = 0.$
Moreover, as sin x, cos x and tan x are continuos and differentiable in (a, b) for 0 < a < b < $\frac{π}{2}$ , therefore f(x) is also continuos and differentiable in [a, b]. Hence, by Rolle's theorem, there exists atleast one real number c in (a, b) such that f ' (c) = 0.
Hence (a) is the correct answer.

Question 10

$f (x) = c o s (\frac{π}{x}), x \neq 0$ increases in the interval

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

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

(\frac{1}{2 k} + 1, \frac{1}{2 k} + 3)

(\frac{1}{2 k + 1}, \frac{1}{2 k} + 3)

SOLUTION

Solution : 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
$(\frac{1}{2 k + 1}, \frac{1}{2 k})$

Question 11

$f (x) = \frac{x}{l o g_{e} x}, x \neq 1$ , is decreasing in interval

(0, e)

(1, e)

(e, \infty)

D. R

SOLUTION

Solution : 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}

Question 12

If $f " (x) > 0 \forall x ϵ R$ then for any two real numbers $x_{1} a n d x_{2}, (x_{1} \neq x_{2})$

f (\frac{x_{1} + x_{2}}{2}) > \frac{f (x_{1}) + f (x_{2})}{2}

f (\frac{x_{1} + x_{2}}{2}) < \frac{f (x_{1}) + f (x_{2})}{2}

f^{'} (\frac{x_{1} + x_{2}}{2}) > \frac{f^{'} (x_{1}) + f^{'} (x_{2})}{2}

f^{'} (\frac{x_{1} + x_{2}}{2}) < \frac{f^{'} (x_{1}) + f^{'} (x_{2})}{2}

SOLUTION

Solution : B

Let A = $(x_{1}, f (x_{1}))$ and B = $(x_{2}, f (x_{2}))$ be any two points on the graph of y = f(x).
Since f"(x) $>$ 0, in the graph of the function tangent will always lie below the curve. Hence chord AB will lie completely above the graph of y = f(x).
Hence $\frac{f (x_{1}) + f (x_{2})}{2} > f (\frac{x_{1} + x_{2}}{2})$

Question 13

$f (x) = x^{2} - 4 | x | a n d g (x) = {\begin{matrix} m i n {f (t) : - 6 \leq t \leq x}, x ϵ [- 6, 0] m a x {f (t) : 0 \leq t \leq x}, x ϵ [0, 6] \end{matrix}, t h a n g (x)$

A. Exactly one point of local minima

B. Exactly one point of local maxima

C. No point of local maxima but exactly one point of local minima

D. Neither a point of local maxima nor minima

SOLUTION

Solution : D

Bold line represents the graph of y = g(x) clearly g(x) has neither a point of local maxima nor a point of local minima.

Question 14

Let S be the set of real values of parameter $λ$ for which the equation $f (x) = 2 x^{3} - 3 (2 + λ) x^{2} + 12 λ x$ has exactly one local maximum and exactly one local minimum. Then S is a subset of

(- 4, \infty)

(- 3, 3)

(3, \infty)

D. R

SOLUTION

Solution : C

$f (x) = 2 x^{3} - 3 (2 + λ) x^{2} + 12 λ x \Rightarrow f^{'} (x) = 6 x^{2} - 6 (2 + λ) x + 12 λ f^{'} (x) = 0 \Rightarrow x = 2, λ$
If f(x) has exactly one local maximum and exactly one local minimum, then $λ \neq 2.$

Question 15

The tangent to the curve x = a $\sqrt{c o s 2 θ} c o s θ, y = a \sqrt{c o s 2 θ} s i n θ$ at the point corresponding to $θ = \frac{π}{6}$ is

A. parallel to the x-axis

B. parallel to the y-axis

C. parallel to line y = x

D. 3X-4Y+2=0

SOLUTION

Solution : A

$\frac{d x}{d θ} = - a \sqrt{c o s 2 θ} s i n θ + \frac{- a c o s θ s i n θ}{\sqrt{c o s 2 θ}} = - a \frac{(c o s 2 θ s i n θ + c o s θ s i n 2 θ)}{\sqrt{c o s 2 θ}} = \frac{- a s i n 3 θ}{\sqrt{c o s 2 θ}} = \frac{d y}{d θ} = a \sqrt{c o s 2 θ} c o s θ - a \frac{s i n θ s i n 2 θ}{\sqrt{c o s 2 θ}} = \frac{a c o s 3 θ}{\sqrt{c o s 2 θ}}$
Hence $\frac{d y}{d x} = - c o t 3 θ \Rightarrow \frac{d y}{d x} |_{θ = \frac{π}{6}}$ = 0
So the tangent to the curve at $θ = \frac{π}{6}$ is parallel to the x-axis.

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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