Question

f(x) = sinx - 4sinx - 1. a and b are it's least and greatest values respectively, then - (A) a = 4 (B) a = -5 (C) b = 4 (D) b =1​

Answer 1

Answer:

The correct answer is option (A),(C)

Step-by-step explanation:

Given

      $f(x)=\sin^2x-4\sin x -1$

Do completion of square

              $=(\sin x)^2-2(\sin x)(2)+(2)^2-2^2-1\\=(\sin x-2)^2-4-1\\=(\sin x-2)^2-5$

Now since we know that, $-1\le \sin x\le 1$

so maximum value $b=(-1-2)^2-5=9-5=4$

and minimum value $a=(1-2)^2-5=-4$

Answer 2

Answer:

(a=-4)(b=4)

Step-by-step explanation:

(i) We Have;

f(x)=sin²x-4sinx-1=0

let, (sinx=m), then [f(x)=m²-4m-1=0]

If x∈(-∞,∞) then t(sinx)∈[-1,1]

(ii) diff. the function w.r.t. to x;

d[f(x)]/dx=2m-4

d[f(x)]/dx=2(m-2)

for min/max d[f(x)]/dx=0

So; (m=2) (mini point)

(iii) but for t(sinx)∈[-1,1]-

f(-1)=(-1)²-4(-1)-1

f(-1)=1+4-1=(4)

f(1)=(1)²-4(1)-1

f(1)=1-4-1=(-4)

As (2) is not included in t(sinx)∈[-1,1],

no need to find range for f(2).

Therefore, f(x)∈[-4,4]. Comparing with [a,b] as least and greatest values, we get (a=-4)(b=4)

Hope It Helps!