Differentiation — Page 7 | JEE Mathematics

Q61

\text{ If } y(x)=\left|\begin{array}{ccc} \sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{array}\right|, x \in \mathbb{R} \text{, then } \frac{d^2 y}{d x^2}+y \text{ is equal to }

A 28

B 27

C -1

D 1

Correct Answer

Option C

Solution

To solve for $y(x)$ , we start with the determinant of the given matrix: $\begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$ Using the formula for a $3 \times 3$ determinant, we expand along the first row: $f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix}$ Calculating the smaller $2 \times 2$ determinants: $\begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1$ $\begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0$ $\begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1$ Substituting these into the determinant calculation gives: $f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1)$ $f(x) = \sin x - (\sin x + \cos x + 1)$ $f(x) = \sin x - \sin x - \cos x - 1$ $f(x) = -\cos x - 1$ Now, calculate the derivatives: First derivative $f'(x)$ : $f'(x) = \dfrac{d}{dx}(-\cos x - 1) = \sin x$ Second derivative $\dfrac{d^2 f}{dx^2}$ : $\dfrac{d^2 f}{dx^2} = \dfrac{d}{dx}(\sin x) = \cos x$ Finally, compute $\dfrac{d^2 y}{dx^2} + y$ : $\dfrac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1$ Thus, $\dfrac{d^2 y}{dx^2} + y$ is equal to $-1$ .

Q62

If f(x) = sin-1

\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),

then f'

\left( { - {1 \over 2}} \right)

equals :

A

- \sqrt 3 {\log _e}\sqrt 3

B

\sqrt 3 {\log _e}\sqrt 3

C

- \sqrt 3 {\log _e}\, 3

D

\sqrt 3 {\log _e}\, 3

Correct Answer

Option B

Solution

Since f(x) = sin

\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)

Suppose 3x = tan t $\Rightarrow$ f(x) = sin $-$ 1

\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)

= sin $-$ 1 (sin2t) = 2t = 2tan $-$ 1 (3x) So, f'(x) =

{2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3

$\therefore$ f '

\left( { - {1 \over 2}} \right)

=

{2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}

. loge 3 =

{1 \over 2} \times \sqrt 3 \times {\log _e}3

=

\sqrt 3 \times {\log _e}\sqrt 3

Q63

If x2 + y2 + sin y = 4, then the value of

{{{d^2}y} \over {d{x^2}}}

at the point (

-

2,0) is :

A

-

34

B

-

32

C 4

D

-

2

Correct Answer

Option A

Solution

Given, x2 + y2 + sin y = 4 After differentiating the above equation w.r.t.x we get 2x + 2y

{{dy} \over {dx}}

+ cos y

{{dy} \over {dx}}

= 0 . . . . (1) $\Rightarrow$ 2x + (2y + cos y)

{{dy} \over {dx}}

= 0 $\Rightarrow$

{{dy} \over {dx}}

=

{{ - 2x} \over {2y + \cos y}}

At ( $-$ 2, 0),

{\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}

=

{{ - 2x - 2} \over {2 \times 0 + \cos 0}}

$\Rightarrow$

{\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}

=

{4 \over {0 + 1}}

$\Rightarrow$

{\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}

= 4 . . . . .(2) Again differentiating equation (1) w.r.t to x, we get 2 + 2

{\left( {{{dy} \over {dx}}} \right)^2}

+ 2y

{{{d^2}y} \over {d{x^2}}}

$-$ sin y

{\left( {{{dy} \over {dx}}} \right)^2}

+ cos y

{{{d^2}y} \over {d{x^2}}}

= 0 $\Rightarrow$ 2 + (2 $-$ sin y)

{\left( {{{dy} \over {dx}}} \right)^2}

+ (2y + cos y)

{{{d^2}y} \over {d{x^2}}}

= 0 $\Rightarrow$ (2y + cos y)

{{{d^2}y} \over {d{x^2}}}

= $-$ 2 $-$ (2 $-$ sin y)

{\left( {{{dy} \over {dx}}} \right)^2}

$\Rightarrow$

{{{d^2}y} \over {d{x^2}}}

=

{{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}

So, at ( $-$ 2, 0),

{{{d^2}y} \over {d{x^2}}}

=

{{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}

$\Rightarrow$

{{{d^2}y} \over {d{x^2}}}

=

{{ - 2 - 2 \times 16} \over 1}

$\Rightarrow$

\,\,\,

{{{d^2}y} \over {d{x^2}}}

= $-$ 34

Q64

Let f : R

\to

R be a function such that f(x) = x3 + x2f'(1) + xf''(2) + f'''(3), x

\in

R. Then f(2) equals -

A 30

B

-

2

C

-

4

D 8

Correct Answer

Option B

Solution

f(x) = x3 + x2f '(1) + xf ''(2) + f '''(3) $\Rightarrow$ f '(x) = 3x2 + 2xf '(1) + f ''(x) . . . . . (1) $\Rightarrow$ f ''(x) = 6x + 2f '(1) . . . . . . (2) $\Rightarrow$ f '''(x) = 6 . . . . . .(3) put x = 1 in equation (1) : f '(1) = 3 + 2f '(1) + f ''(2) . . . . .(4) put x = 2 in equation (2) : f ''(2) = 12 + 2f '(1) . . . . .(5) from equation (4) & (5) : $-$ 3 $-$ f '(1) = 12 + 2f'(1) $\Rightarrow$ 3f '(1) = $-$ 15 $\Rightarrow$ f '(1) = $-$ 5 $\Rightarrow$ f ''(2) = 2 . . . . .(2) put x = 3 in equation (3) : f ''' (3) = 6 $\therefore$ f(x) = x3 $-$ 5x2 + 2x + 6 f(2) = 8 $-$ 20 + 4 + 6 = $-$ 2

Q65

If x

=

3 tan t and y

=

3 sec t, then the value of

{{{d^2}y} \over {d{x^2}}}

at t

= {\pi \over 4},

is :

A

{1 \over {3\sqrt 2 }}

B

{1 \over {6\sqrt 2 }}

C

{3 \over {2\sqrt 2 }}

D

{1 \over 6}

Correct Answer

Option B

Solution

x = 3 tan t and y = 3 sec t So that

{{dx} \over {dt}}

= 3sec2t and

{{dy} \over {dt}}

= 3 sec t tan t

{{dy} \over {dx}}

=

{{dy/dt} \over {dx/dt}}

= sin t

{{{d^2}y} \over {d{x^2}}}

= (cos t)

.{{dt} \over {dx}}

{{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}

{{{d^2}y} \over {d{x^2}}}

=

{1 \over 3}

(cos3 t)

{\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}

=

{1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}

Q66

If xloge(logex)

-

x2 + y2 = 4(y > 0), then

{{dy} \over {dx}}

at x = e is equal to :

A

{{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}

B

{{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}

C

{{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}

D

{e \over {\sqrt {4 + {e^2}} }}

Correct Answer

Option C

Solution

Differentiating with respect to x,

x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0

at

x = e

we get

1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}

\Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,

as

y(e) = \sqrt {4 + {e^2}}