The value of limx→(3π)/4(4sin^2xcosx-cosx+sinx)/(sinx+cosx) is

Correct answer: (A) -1. L=limx→(3π)/4(4sin^2xcosx-cosx+sinx)/(sinx+cosx) L=limx→(3π)/4(2sinxsin^2x+cos^2x+2sinxcosx-1-cosx+sinx)/(sinx+cosx) L=limx→(3π)/4(2sinx(sinx+cosx)^2-1-cosx+sinx)/(sinx+cosx) L=limx→(3π)/4(2sinx(sinx+cosx)^2-2sinx-cosx+sinx)/(sinx+cosx) L=limx→(3π)/4(sinx+cosx2sinxsinx+cosx-1)/(sinx+cosx) L=limx→(3π)/42sinxsinx+cosx-1=-1

Q51 / 125 · BITSAT 2019 · Mathematics

The value of $\lim_{x \to \frac{3 π}{4}} \frac{4 \sin^{2} x \cos x - \cos x + \sin x}{\sin x + \cos x}$ is

Correct answer
A
$- 1$
B
$0$
C
$1$
D
None of these

Solution

$L = \lim_{x \to \frac{3 π}{4}} \frac{4 \sin^{2} x \cos x - \cos x + \sin x}{\sin x + \cos x}$

$L = \lim_{x \to \frac{3 π}{4}} \frac{2 \sin x (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1) - \cos x + \sin x}{\sin x + \cos x}$

$L = \lim_{x \to \frac{3 π}{4}} \frac{2 \sin x [(\sin x + \cos x)^{2} - 1] - \cos x + \sin x}{\sin x + \cos x}$

$L = \lim_{x \to \frac{3 π}{4}} \frac{2 \sin x (\sin x + \cos x)^{2} - 2 \sin x - \cos x + \sin x}{\sin x + \cos x}$

$L = \lim_{x \to \frac{3 π}{4}} \frac{(\sin x + \cos x) \{2 \sin x (\sin x + \cos x) - 1\}}{\sin x + \cos x}$

$L = \lim_{x \to \frac{3 π}{4}} 2 \sin x (\sin x + \cos x) - 1 = - 1$

Hence, the correct answer is option: 'A'.

The value of limx→3π44sin2xcosx-cosx+sinxsinx+cosx is

Solution

Relevant Questions

The value of $\lim_{x \to \frac{3 π}{4}} \frac{4 \sin^{2} x \cos x - \cos x + \sin x}{\sin x + \cos x}$ is