Let f be an odd function defined on the set of real numbers such that for x≥0,f(x)=3sinx+4cosx . Then f(x) at x=−11π6 is equal to

f(x)=3sinx+4cosxf(x) is an odd function f(−x)=−f(x) f(−11π6)=−[3sin11π6+4cos11π6] =−3sin(2π−π6)−4cos(2π−π6) =3sinπ6−4cosπ6 =32−432=32−23

Let f be an odd function defined on the set of real numbers such that for x≥0,f(x)=3sinx+4cosx . Then f(x) at x=−11π6 is equal to

f(x)=3sinx+4cosxf(x) is an odd function f(−x)=−f(x) f(−11π6)=−[3sin11π6+4cos11π6] =−3sin(2π−π6)−4cos(2π−π6) =3sinπ6−4cosπ6 =32−432=32−23

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $f$ be an odd function defined on the set of real numbers such that for $x \geq 0, f (x) = 3 \sin x + 4 \cos x$ . Then $f (x)$ at $x = \frac{- 11 π}{6}$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- \frac{3}{2} + 2 \sqrt{3}$

$- \frac{3}{2} - 2 \sqrt{3}$

$\frac{3}{2} - 2 \sqrt{3}$

$\frac{3}{2} + 2 \sqrt{3}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = 3 \sin x + 4 \cos x$
$f (x)$ is an odd function
$f (- x) = - f (x)$
$f (- \frac{11 π}{6}) = - [3 \sin \frac{11 π}{6} + 4 \cos \frac{11 π}{6}]$
$= - 3 \sin (2 π - \frac{π}{6}) - 4 \cos (2 π - \frac{π}{6})$
$= 3 \sin \frac{π}{6} - 4 \cos \frac{π}{6}$
$= \frac{3}{2} - 4 \frac{\sqrt{3}}{2} = \frac{3}{2} - 2 \sqrt{3}$