Free Application of Derivatives 03 Practice Test - 12th Grade - Commerce

Question 1

$f (x) = \frac{x}{l o g_{e} x}, x \neq 1$ , 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 2

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 3

$f (x) = x^{2} - 4 | x | a n d g (x) = {\begin{matrix} m i n 1 f (t) : - 6 \leq t \leq x}, x ϵ [- 6, 0] m a x 1 f (t) : 0 \leq t \leq x}, x ϵ [0, 6] \end{matrix}, t h a n g (x)$

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 4

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 5

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.

Question 6

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 7

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 8

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 9

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 10

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

Free Application of Derivatives 03 Practice Test - 12th Grade - Commerce

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

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