>
Exams
>
Mathematics
>
Derivatives of Functions in Parametric Forms
>
if u sec 1 sec 2 theta and v cos theta then frac d
Question:
If $u=\sec^{-1}(-\sec 2\theta)$ and $v=\cos \theta$, then $\frac{du}{dv}$ at $\theta=\frac{\pi}{4}$ is equal to:
Show Hint
Simplify the inverse trigonometric expression using identities before differentiating to save time.
KEAM - 2025
KEAM
Updated On:
May 10, 2026
$\sqrt{2}$
$2\sqrt{2}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2\sqrt{2}}$
$-\sqrt{2}$
Show Solution
Verified By Collegedunia
The Correct Option is
B
Solution and Explanation
Step 1: Concept
Use the derivative of a function with respect to another function: $\frac{du}{dv} = \frac{du/d\theta}{dv/d\theta}$.
Step 2: Analysis
$u = \sec^{-1}(-\sec 2\theta) = \pi - 2\theta$ (using property $\sec^{-1}(-x) = \pi - \sec^{-1}x$). $du/d\theta = -2$. $v = \cos \theta \implies dv/d\theta = -\sin \theta$.
Step 3: Calculation
$\frac{du}{dv} = \frac{-2}{-\sin \theta} = \frac{2}{\sin \theta}$. At $\theta = \pi/4$: $\sin(\pi/4) = 1/\sqrt{2}$. $\frac{du}{dv} = \frac{2}{1/\sqrt{2}} = 2\sqrt{2}$.
Final Answer:
(B)
Download Solution in PDF
Was this answer helpful?
0
0
Top KEAM Mathematics Questions
If
$\int e^{2x}f' \left(x\right)dx =g \left(x\right)$
, then
$ \int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$
KEAM - 2017
Mathematics
Methods of Integration
View Solution
The value of
$ \cos [{{\tan }^{-1}}\{\sin ({{\cot }^{-1}}x)\}] $
is
KEAM - 2009
Mathematics
Inverse Trigonometric Functions
View Solution
The solutions set of inequation
$\cos^{-1}x < \,\sin^{-1}x$
is
KEAM - 2011
Mathematics
Inverse Trigonometric Functions
View Solution
Let
$\Delta= \begin{vmatrix}1&1&1\\ 1&-1-w^{2}&w^{2}\\ 1&w&w^{4}\end{vmatrix}$
, where
$w \neq 1$
is a complex number such that
$w^3 = 1$
. Then
$\Delta$
equals
KEAM
Mathematics
Determinants
View Solution
Let
$p : 57$
is an odd prime number,
$\quad \, q : 4$
is a divisor of
$12$
$\quad$
$r : 15$
is the
$LCM$
of
$3$
and
$5$
Be three simple logical statements. Which one of the following is true?
KEAM
Mathematics
mathematical reasoning
View Solution
View More Questions
Top KEAM Derivatives of Functions in Parametric Forms Questions
If \( x = 2\cos t - \cos 2t \) and \( y = 2\sin t - \sin 2t \), then \( \frac{dy}{dx} \) at \( t = \frac{\pi}{2} \) is
KEAM - 2019
Mathematics
Derivatives of Functions in Parametric Forms
View Solution
If $x = \frac{3t}{1+t^3}$ and $y = \frac{3t^2}{1+t^3}$, then $\frac{dy}{dx}$ at $t=1$ equals
KEAM - 2019
Mathematics
Derivatives of Functions in Parametric Forms
View Solution
If $s = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)$ and $t = \sqrt{1 - x^2}$, then $\frac{ds}{dt}$ at $x = \frac{1}{2}$ is:
KEAM - 2016
Mathematics
Derivatives of Functions in Parametric Forms
View Solution
If \( x = \sin t \) and \( y = \tan t \), then \( \frac{dy}{dx} = \)
KEAM - 2014
Mathematics
Derivatives of Functions in Parametric Forms
View Solution
If \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \), then \( 1 + \left( \frac{dy}{dx} \right)^2 \) is:
KEAM - 2014
Mathematics
Derivatives of Functions in Parametric Forms
View Solution
View More Questions
Top KEAM Questions
i.
$\quad$
They help in respiration ii.
$\quad$
They help in cell wall formation iii.
$\quad$
They help in DNA replication iv.
$\quad$
They increase surface area of plasma membrane Which of the following prokaryotic structures has all the above roles?
KEAM - 2015
Prokaryotic Cells
View Solution
A body oscillates with SHM according to the equation (in SI units),
$x = 5 cos \left(2\pi t +\frac{\pi}{4}\right) .$
Its instantaneous displacement at
$t = 1$
second is
KEAM - 2014
Energy in simple harmonic motion
View Solution
The pH of a solution obtained by mixing 60 mL of 0.1 M BaOH solution at 40m of 0.15m HCI solution is
KEAM - 2016
Acids and Bases
View Solution
Kepler's second law (law of areas) of planetary motion leads to law of conservation of
KEAM - 2016
Keplers Laws
View Solution
If
$\int e^{2x}f' \left(x\right)dx =g \left(x\right)$
, then
$ \int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$
KEAM - 2017
Methods of Integration
View Solution
View More Questions