Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

The value of the limit

$\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{4\sqrt{2}(sin3x+sinx)}{\left(2sin2xsin\frac{3x}{2}+cos\frac{5x}{2}\right)-\left(\sqrt{2}+\sqrt{2}cos2x+cos\frac{3x}{2}\right)}$

is ....

solution : It is multi - conceptual question. let's solve it step by step.

sin3x + sinx = 2sin(3x + x)/2 cos(3x - x)/2 = 2sin2x cosx = 4sinx cos²x [ using formula, sinA + sinB = 2sin(A + B)/2 cos(A - B)/2 and 2sinA cosA = sin2A ]

denominator is ... (2sin2x sin3x/2 + cos5x/2) - (√2 + √2cos2x + cos3x/2)

= (2sin2x sin3x/2) + (cos5x/2 - cos3x/2) - √2(1 + cos2x)

= (2sin2x sin3x/2) + 2sin(5x/2 + 3x/2)/2 sin(3x/2 - 5x/2)/2 - √2(2cos²x)

[ using formula, cosC - cosD = 2sin(C + D)/2 sin(D -C)/2 and 1 + 2cos2A = 2cos²A ]

= (2sin2x sin3x/2) - 2sin2x sinx/2 - 2√2cos²x

= 2sin2x (sin3x/2 - sinx/2) - 2√2 cos²x

= 4sinx cosx (2cosx sinx/2 ) - 2√2 cos²x

= 8sinx sinx/2 cos²x - 2√2 cos²x

= cos²x (8 sinx sinx/2 - 2√2)

now limit converts into

$\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{16\sqrt{2}sinx.cos^2x}{cos^2x\left(8sinxsin\frac{x}{2}-2\sqrt{2}\right)}$

$\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{16\sqrt{2}sinx}{8sinxsin\frac{x}{2}-2\sqrt{2}}$

now apply x = π/2

we get, 16√2 × sin(π/2)/[8 sinπ/2 sinπ/4 - 2√2]

= 16√2 × 1/[8 × 1 × 1/√2 - 2√2 ]

= 16√2/[2√2]

= 8

Therefore the value of limit is 8