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Grade Upto college levelTrigonometry

if the inequality sin²x +a cosx +a²> 1+cosx holds for any x belongs to R then the largest negative integral value of a is

A) -4 B) -3 C) -2 D) -1

ANAM V. RANGA REDDY

14 Years agoGrade Upto college level

1 Answer

Jitender Singh

12 Years ago

Ans: -2

Sol:

$\sin ^{2}x + a\cos x + a^{2} > 1 + \cos x$

$1 - \sin ^{2}x + (1-a)\cos x - a^{2} < 0$

$\cos ^{2}x + (1-a)\cos x - a^{2} < 0$

This is an simple quadratic equation in $\cos x$ ;

$f(x) = \cos ^{2}x + (1-a)\cos x - a^{2}$

$\Delta \geq 0$

$\Delta = (1-a)^{2} - 4(-a^{2})$

$\Delta = 5a^{2} - 2a + 1$

$So, \Delta \geq 0$

for all values of a.

Solution for $\cos x$

$\cos x = ((a-1) \pm (5a^{2} -2a + 1)^{1/2})/2$

$-1\leq \cos x\leq 1$

$-2\leq (a-1)\pm (5a^{2} -2a + 1)^{1/2}\leq 2$

$-1-a\leq \pm (5a^{2} - 2a + 1)^{1/2}\leq 3-a$

Square; we have

$a^{2} + 2a + 1\leq 5a^{2} - 2a + 1\leq a^{2} - 6a + 9$

We have two equation from here

$a(a-1)\geq 0$ …..............(1)

$(a+2)(a-1)\leq 0$ …...........(2)

Solution of (1) & (2) is

$a \epsilon [-2,0]$

So, max. negative integer value for a is -2.

Cheer!

Thanks & Regards

Jitender Singh

IIT Delhi

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