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701.

Find general solution of `4cos^2 x=1`.A. `npipm(4pi)/(3), ninZ`B. `npipm(5pi)/(3), ninZ`C. `npipm(pi)/(3), ninZ`D. `npipm(2pi)/(3), ninZ`

Answer» Correct Answer - C
Discussion
702.

Let`(sin(theta-alpha))/("sin"(theta-beta))=a/b a n d(cos(theta-alpha))/("cos"(theta-beta))=c /d t h e n(a c+b d)/(a d+b c)=`(a)`cos(alpha-beta)`(b) `sin(alpha-beta)``sin(alpha+beta)`(d) none of these

Answer» `sin(theta-alpha)/(sin(theta-beta)) = a/b`
`=>a/sin(theta-alpha) = b/(sin(theta-beta)) `
Let `a/sin(theta-alpha) = b/(sin(theta-beta)) = k_1`
`=>a = k_1sin(theta-alpha), b = k_1sin(theta-beta)`
Similarly,
`=>c = k_2cos(theta-alpha), d = k_2cos(theta-beta)`
Now,
`(ac+bd)/(ad+bc) = (k_1sin(theta-alpha)k_2cos(theta-alpha)+k_1sin(theta-beta)k_2cos(theta-beta))/(k_1sin(theta-alpha)k_2cos(theta-beta)+k_1sin(theta-beta)k_2cos(theta-alpha))`
`=(k_1k_2(sin(theta-alpha)cos(theta-alpha)+sin(theta-beta)cos(theta-beta)))/(k_1k_2(sin(theta-alpha)cos(theta-beta)+sin(theta-beta)cos(theta-alpha)))`
`=(1/2(sin(2(theta-alpha))+sin(2(theta-beta))))/(sin(theta-alpha+theta-beta))`
`=(sin(2(theta-alpha))+sin(2(theta-beta)))/(2sin(2theta-(alpha+beta))`
`=(2sin((2(theta-alpha+theta-beta))/2)cos((2(theta-alpha-theta+beta))/2))/(2sin(2theta-(alpha+beta))`
`=(2sin((2theta-(alpha+beta))cos(beta-alpha)))/(2sin(2theta-(alpha+beta))`
`=cos(beta-alpha)`
`=cos(-(beta-alpha))...[As cos(-theta) = cos theta]`
`=cos(alpha-beta)`
So, option `(a)` is the correct option.
Discussion
703.

The general solution of the equation `4cos^2x+6sin^2x=5`A. `npipm(pi)/(6), ninZ`B. `npipm(pi)/(4), ninZ`C. `npipm(pi)/(3), ninZ`D. `npipm(pi)/(2), ninZ`

Answer» Correct Answer - B
Discussion
704.

Find the range of `y=sin^3x-6sin^2x+11sinx-6.`

Answer» `y=sin^3x-6sin^2x+11sinx-6`
Put `sinx=t`
`y=t^3-6t^2+11t-6`
diff. with respect to t
`dy/dt=3t^2-12t+11`
`3(t-2)^2-1>0`
Range[f(-1),f(1)]
Range[-24,0]
`f(-1)=-1-6-11-6=-24`
`f(1)=1-6+11-6=0`.
Discussion