Free Objective Test 01 Practice Test - 11th and 12th

Question 1

The distance moved by the particle in time t is given by $x = t^{3} - 12 t^{2} + 6 t + 8$ . At the instant when its acceleration is zero, then the velocity is

-42

-48

SOLUTION

Solution : B

We have,
$x = t^{3} - 12 t^{2} + 6 t + 8$
$\Rightarrow \frac{d x}{d t} = 3 t^{2} - 24 t + 6 a n d \frac{d^{2} x}{d t^{2}} = 6 t - 24$
Now, Acceleration =0
$\Rightarrow \frac{d^{2} x}{d t^{2}} = 0 \Rightarrow 6 t - 24 = 0 \Rightarrow t = 4$
Att=4, we have
Velocity = ${(\frac{d x}{d t})}_{r - 4} = 3 \times 4^{2} - 24 \times 4 + 6 = - 42$ .
Hence (b) is the correct answer.

Question 2

For what values of x is the rate of increase of $x^{3} - 5 x^{2} + 5 x + 8$ is twice rate of increase of x ?

$- 3, - \frac{1}{3}$

$- 3, \frac{1}{3}$

$3, - \frac{1}{3}$

3, \frac{1}{3}

SOLUTION

Solution : D

Let $y = x^{3} - 5 x^{2} + 5 x + 8$ . Then,
$\frac{d y}{d x} = (3 x^{2} - 10 x + 5) \frac{d x}{d t} W h e n \frac{d y}{d t} = 2 \frac{d x}{d t}, w e h a v e (3 x^{2} - 10 x + 5) \frac{d x}{d t} = 2 \frac{d x}{d t} \Rightarrow 3 x^{2} - 10 x + 3 = 0 \Rightarrow (3 x - 1) (x - 3) = 0 \Rightarrow x = 3, \frac{1}{3}$ .
Hence (d) is the correct answer.

Question 3

The two curves $x^{3} - 3 x y^{2} + 2 = 0 a n d 3 x^{2} y - y^{3} - 2 = 0$

Cut at right angles

Touch each other

Cut at an angle π/3

Cut at an angle π/4

SOLUTION

Solution : A

$x^{3} - 3 x y^{2} + 2 = 0... (1) 3 x^{2} y - y^{3} - 2 = 0... (2)$
On differentiating equations (1) and (2) w.r.t x, we obtain
${(\frac{d y}{d x})}_{c 1} = \frac{x^{2} - y^{2}}{2 x y} a n d {(\frac{d y}{d x})}_{c 2} = \frac{- 2 x y}{x^{2} - y^{2}}$
Since $m_{1} . m_{2} = - 1$ .Therefore the two curves cut at right angles.
Hence (a) is the correct answer.

Question 4

Number of possible tangents to the curve $y = c o s (x + y), - 3 π \leq x \leq 3 π$ that are parallel to the line x+2y = 0, is

SOLUTION

Solution : C

We have, y = cos (x + y)
$\frac{d y}{d x} = s i n (x + y) (1 + \frac{d y}{d x})$
Since, the tangents are parallel to the line x + 2y = 0
$- \frac{1}{2} = - s i n (x + y) (1 - \frac{1}{2}) \Rightarrow s i n (x + y) = 1 \Rightarrow x + y = \frac{π}{2}, \frac{5 π}{2}, \frac{3 π}{2} - 1 \leq y \leq 1.$
Hence (c) is the correct answer.

Question 5

If the tangent to the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ at any point on it cuts the axes OX and OY at P and Q respectively, then OP +OQ is

$\frac{a}{2}$

SOLUTION

Solution : B

$\sqrt{x} + \sqrt{y} = \sqrt{a} . . . . . (i)$
$\Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0$

$∴ \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}}$
Equation of tangent at $(x_{1} y_{1}) i s y - y_{1} = - \frac{\sqrt{y_{1}}}{\sqrt{x_{1}}} (x - x_{1})$
$\Rightarrow \frac{x}{\sqrt{x_{1}}} + \frac{y}{\sqrt{y_{1}}} = \sqrt{a}; \Rightarrow o p = \sqrt{a} \sqrt{x_{1}}, O Q = \sqrt{a} \sqrt{y_{1}} ∴ O P + O Q = \sqrt{a}$

Question 6

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1,$ where a > 0, attains its maximum and minimum at p and q respectively such that $p^{2} = q$ , then a equals

\frac{1}{2}

SOLUTION

Solution : C

We have, $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1 ∴ f (x) = 6 x^{2} - 18 a x + 12 a^{2} = 0 \Rightarrow 6 [x^{2} - 3 a x + 2 a^{2}] = 0 \Rightarrow x^{2} - 3 a x + 2 a^{2} = 0 \Rightarrow x^{2} - 2 a x - a x + 2 a^{2} = 0 \Rightarrow x (x - 2 a) - a (x - 2 a) = 0 \Rightarrow (x - a) (x - 2 a) = 0 \Rightarrow x = a, x = 2 a$
Now, $f^{'} (x) = 12 x - 18 a$
$∴ f^{'} (a) = 12 a - 18 a = - 6 a < 0 ∴ f (x)$ will be maximum at x = a
i.e. p = a
Also, $f^{'} (2 a) = 24 a - 18 a = 6 a ∴ f (x)$ will be minimum at x = 2a
i.e.q = 2a
Given, $p^{2} = q$
$\Rightarrow a^{2} = 2 a \Rightarrow a = 2.$
Hence (c) is the correct answer.

Question 7

A function f such that $f (a) = f^{''} (a) = . . . . . . f^{2 n} (a) = 0$ and f has a local maximum value b at x = a, if f (x) is

$(x - a)^{2 n + 2}$

$b - 1 - (x + 1 - a)^{2 n + 1}$

$b - (x - a)^{2 n + 2}$

$(x - a)^{2 n + 2} - b .$

SOLUTION

Solution : C

For local maximum or local minimum odd derivative must be equal to zero.
For local maxima, even derivative must be negative.
Since maximum value at x = a is b.
$∴ f (x) = b - (x - a)^{2 n + 2} (∵ f^{2 n + 2} (a) = - v e)$
Hence (c) is the correct answer.

Question 8

The number of values of x where the function f(x) = 2 (cos 3x + cos $\sqrt{3 x}$ attains its maximum, is

Infinite

SOLUTION

Solution : A

We have,
$f (x) = 2 (c o s 3 x + c o s \sqrt{3} x) = 4 c o s (\frac{3 + \sqrt{3}}{2}) x c o s (\frac{3 - \sqrt{3}}{2}) x ⩽ 4$
and it is equal to 4 when both cos $(\frac{3 + \sqrt{3}}{2}) x a n d c o s (\frac{3 - \sqrt{3}}{2})$
Are equal to 1 for a value of x. This is possible only when x = 0.
Hence (a) is the correct answer.

Question 9

The point in the interval $[0, 2 π]$ where $f (x) = e^{x} s i n x$ has maximum slope, is

\frac{π}{4}

\frac{π}{2}

$π$

3 \frac{π}{2}

SOLUTION

Solution : B

We have, $f^{'} (x) = e^{x} + c o s x + s i n x e^{x} A n d f^{'} (x) = - s i n x e^{x} + c o s x e^{x} + c o s x e^{x} + s i n x c o s x e^{x} . N o w, f^{'} (x) = 2 c o s x c o s x e^{x} = 0 \Rightarrow c o s x = 0 \Rightarrow x = \frac{π}{2} . A l s o, f^{'} (x) = - 2 s i n x e^{x} + 2 c o s x e^{x} = - v e$
$∴$ Slope is maximum at $x = \frac{π}{2} .$
Hence (b) is the correct answer.

Question 10

The approximate value of square root of 25.2 is

5.01

5.02

5.03

5.04

SOLUTION

Solution : B

Let f (x) = $\sqrt{x}$
Now, $f (x + δ x) - f (x) = f^{'} (x) . δ x = \frac{δ x}{2 \sqrt{x}}$
We may write, 25.2 = 25 + 0.2
Taking x = 25 and $δ x = 0.2$ We have
$f (25.2) - f (25) = \frac{0.2}{2 \sqrt{25}} = 0.02 ∴ f (25.2) = f (25) + 0.02 = \sqrt{25} + 0.02 = 5.02 \Rightarrow \sqrt{(25.2)} = 5.02$

Question 11

If f (x) is differentiable in the interval [2, 5], where $f (2) = \frac{1}{5}$ and $f (5) = \frac{1}{2}$ , then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

$\frac{1}{2}$

$\frac{1}{5}$

$\frac{1}{10}$

$7$

SOLUTION

Solution : C

As f (x) is differentiable in [2 , 5], therefore, it is also continuos in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that
$f^{'} (c) = \frac{f (5) - f (2)}{5 - 2} \Rightarrow f^{'} (c) = \frac{\frac{1}{2} - \frac{1}{5}}{3} = \frac{1}{10} .$
Hence (c) is the correct answer.

Question 12

The equation x log x = 3 - x has, in the interval (1, 3),

Exactly one root

Atmost one root

Atleast one root

No root

SOLUTION

Solution : C

Let f (x) = (x - 3) log x
Then, f (1) = - 2 log 1 = 0 and f (3) = (3-3) log 3 = 0. As, (x-3) and log x are continuos and differentiable in [1, 3], therefore (x-3) log x = f (x) is also continuos and differentiable in [1, 3]. Hence, by Rolle's theorem, there exists a value of x in (1, 3) such that
f ' (x) = 0 $\Rightarrow$ log x+(x-3) $\frac{1}{x}$ = 0
$\Rightarrow$ x log x = 3 - x.
Hence (c) is the correct answer.

Question 13

Between any two real roots of the equation $e^{x}$ sin x = 1, the equation $e^{x}$ cos x = - 1 has

Atleast one root

Exactly one root

Atmost one root

No root

SOLUTION

Solution : A

Let $\propto, β (\propto< β)$ be any two real roots of
f(x) = e - x - sin x
Then, f $(\propto) = 0 = f (β)$
Moreover, f(x) is continuos and differentiable for $x ε [\propto, β] .$
Hence, from Rolle's thereom, thereom, there exists atleast one x in $\propto, β$ such t
$f^{'} (x) = 0 \Rightarrow - e^{- x} - c o s x = 0 \Rightarrow - e^{- x} (1 + e^{x} c o s x) = 0 \Rightarrow e^{x} c o s x = - 1.$
Hence (a) is the correct answer.

Question 14

Let f (x) = sinx + ax + b. Then f(x) = 0 has

A. only one real root which is positive if

a > 1, b < 0

B. only one real root which is negative if

a > 1, b < 0

C. only one real root which is negative if

a < - 1, b > 0

D. CAN'T SAY ANYTHING

SOLUTION

Solution : A

f'(x) = - cosx + a, if a $>$ 1,then f(x) entirely increasing. So f(x) =0 has only one real root, which is positive if f(0) $<$ 0 and negative if f(0) $>$ 0.

Similarly when a $<$ -1. Then f(x) entirely decreasing. So f(x) has only one real root which is negative if f(0) $<$ 0 and positive if f(0) $>$ 0

Question 15

Let f(x) = ${\begin{matrix} 1 + s i n x, x < 0 x^{2} - x + 1, x \geq 0 \end{matrix}$ . Then

A. f has a local maximum at x = 0

B. f has a local minimum at x = 0

C. f is increasing every where

D. f is decreasing everywhere

SOLUTION

Solution : A

f is continuous at ‘0’ and f' (0-) $>$ 0 and f' ( 0 +) $<$ 0 . Thus f has a local maximum at ‘0’.

Free Objective Test 01 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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