CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Differentiation under Integral Sign

Let p x =3 ...

Question

Let $p (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 3 x & 3 x^{2} + 2 3 x & 3 x^{2} + 2 & 3 x^{3} + 6 x 3 x^{2} + 2 & 3 x^{3} + 6 x & 3 x^{4} + 12 x^{2} + 2 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$q (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 2 & 2 2 + x & 3 + x & 4 + x (2 + x)^{2} & (3 + x)^{2} & (4 + x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$r (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + π / 4) & sin (x + π / 4) & 1 cos (x + π / 2) & sin (x + π / 2) & \sqrt{2} cos (x + 3 π / 4) & sin (x + 3 π / 4) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$s (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x + 1 4 x & 2 x (2 x - 1) & 2 x (2 x + 1) 6 x (2 x - 1) & 4 x (2 x - 1) (x - 1) & 2 x (4 x^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$
Match the expressions on the left with their properties or values on the right.

Open in App

Solution

A) $p (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 3 x & 3 x^{2} + 2 3 x & 3 x^{2} + 2 & 3 x^{3} + 6 x 3 x^{2} + 2 & 3 x^{3} + 6 x & 3 x^{4} + 12 x^{2} + 2 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $C_{3} \to C_{3} - x C_{2}, C_{2} \to C_{2} - x C_{1}$ , we get
$p (x) = ∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 2 3 x & 2 & 4 x 3 x^{2} + 2 & 4 x & 6 x^{2} + 2 \end{matrix} ∣ ∣ ∣ ∣$
Now applying $R_{3} \to R_{3} - x R_{2}, R_{2} \to R_{2} - x R_{1}$
$p (x) = ∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 2 0 & 2 & 2 x 2 & 2 x & 2 x^{2} + 2 \end{matrix} ∣ ∣ ∣ ∣ = 3 (4 x^{2} + 4 - 4 x^{2}) + 2 (- 4) = 4$
B) $q (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 2 & 2 2 + x & 3 + x & 4 + x (2 + x)^{2} & (3 + x)^{2} & (4 + x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $C_{3} \to C_{3} - C_{2}, C_{2} \to C_{2} - C_{1}$ , we get
$q (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 0 & 0 2 + x & 1 & 1 (2 + x)^{2} & 2 x + 5 & 2 x + 7 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 (2 x + 7 - 2 x - 5) = 4$
C) $r (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + π / 4) & sin (x + π / 4) & 1 cos (x + π / 2) & sin (x + π / 2) & \sqrt{2} cos (x + 3 π / 4) & sin (x + 3 π / 4) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Differentiating w.r.t, we get
$r^{'} (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - sin (x + π / 4) & sin (x + π / 4) & 1 - sin (x + π / 2) & sin (x + π / 2) & \sqrt{2} - sin (x + 3 π / 4) & sin (x + 3 π / 4) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + π / 4) & cos (x + π / 4) & 1 cos (x + π / 2) & cos (x + π / 2) & \sqrt{2} cos (x + 3 π / 4) & cos (x + 3 π / 4) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + π / 4) & sin (x + π / 4) & 0 cos (x + π / 2) & sin (x + π / 2) & 0 cos (x + 3 π / 4) & sin (x + 3 π / 4) & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 + 0 + 0 = 0$
$\Rightarrow r (x)$ is constant
D) $s (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x + 1 4 x & 2 x (2 x - 1) & 2 x (2 x + 1) 6 x (2 x - 1) & 4 x (2 x - 1) (x - 1) & 2 x (4 x^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 x (2 x) (2 x - 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 2 x + 1 2 & (2 x - 1) & 2 x + 1 3 & 2 (x - 1) & (2 x + 1) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 x^{2} (2 x + 1) (2 x - 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x & 1 2 & (2 x - 1) & 1 3 & 2 (x - 1) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $C_{2} \to C_{2} + C_{1}$ , we get
$s (x) = 4 x^{2} (4 x^{2} - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 x + 1 & 1 2 & 2 x + 1 & 1 3 & 2 x + 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 4 x^{2} (4 x^{2} - 1) (2 x + 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 & 1 & 1 3 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Evaluate the integral

\int_{- 3 π / 4}^{5 π / 4} (\frac{s i n x + c o s x}{e^{x - π / 4} + 1}) d x

\int \frac{(cos x + \sqrt{3}) d x}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} =

\int \frac{cos x + \sqrt{3}}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} d x

c

is constant of integration

Q. Assertion :Let

f : X \to Y

be a function defined by

f (x) = 2 sin (x + \frac{π}{4}) - \sqrt{2} cos x + c

Statement I: For set

X, x \in [0, \frac{π}{2}] \cup [π, \frac{3 π}{2}], f (x)

is one-one function

Reason: Statement II:

f^{'} (x) \geq 0, x \in [0, \frac{π}{2}]

Q. Minimum value of the expression

4 sin (x - \frac{π}{6}) - 6 \sqrt{2} cos (x + \frac{π}{4})

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Differentiation under Integral Sign

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Differentiation under Integral Sign

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon