Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Question 1. }\text{Write the number of solutions of the equation }\\ \tan x+\sec x=2\cos x\text{ in the interval }[0,2\pi].$
$\displaystyle \text{Answer:}$
$\displaystyle \tan x+\sec x=2\cos x$
$\displaystyle \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$
$\displaystyle \frac{\sin x+1}{\cos x}=2\cos x$
$\displaystyle \sin x+1=2\cos^2x$
$\displaystyle \sin x+1=2(1-\sin^2x)$
$\displaystyle 2\sin^2x+\sin x-1=0$
$\displaystyle (2\sin x-1)(\sin x+1)=0$
$\displaystyle \therefore \sin x=\frac{1}{2}\qquad \text{or}\qquad \sin x=-1$
$\displaystyle \sin x=\frac{1}{2}\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6}$
$\displaystyle \sin x=-1\Rightarrow x=\frac{3\pi}{2}$
$\displaystyle \text{But at }x=\frac{3\pi}{2},\ \cos x=0,\text{ so }\tan x\text{ and }\sec x\text{ are not defined}$
$\displaystyle \therefore \text{Number of solutions}=2$
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$\displaystyle \textbf{Question 2. }\text{Write the number of solutions of the equation } \\ 4\sin x-3\cos x=7.$
$\displaystyle \text{Answer:}$
$\displaystyle 4\sin x-3\cos x\leq\sqrt{4^2+(-3)^2}$
$\displaystyle \therefore 4\sin x-3\cos x\leq5$
$\displaystyle \text{But RHS }=7$
$\displaystyle \therefore \text{No solution exists.}$
$\displaystyle \therefore \text{Number of solutions }=0$
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$\displaystyle \textbf{Question 3. }\text{Write the general solution of }\tan^22x=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \tan^22x=1$
$\displaystyle \therefore \tan2x=\pm1$
$\displaystyle \therefore 2x=n\pi+\frac{\pi}{4}$
$\displaystyle \therefore x=\frac{n\pi}{2}+\frac{\pi}{8},\quad n\in Z$
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$\displaystyle \textbf{Question 4. }\text{Write the set of values of }a\text{ for which }\sqrt3\sin x-\cos x=a\text{ has no solution.}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt3\sin x-\cos x\in[-2,2]$
$\displaystyle \therefore \text{The equation has no solution if }a<-2\text{ or }a>2$
$\displaystyle \therefore a\in(-\infty,-2)\cup(2,\infty)$
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$\displaystyle \textbf{Question 5. }\text{If }\cos x=k\text{ has exactly one solution in }[0,2\pi]\text{ then write the value(s) of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos x=k\text{ has exactly one solution in }[0,2\pi]\text{ only when }k=-1$
$\displaystyle \therefore k=-1$
$\\$

$\displaystyle \textbf{Question 6. }\text{Write the number of points of intersection of } \\ 2y=1\text{ and }y=\cos x,\ 0\leq x\leq2\pi.$
$\displaystyle \text{Answer:}$
$\displaystyle 2y=1\Rightarrow y=\frac{1}{2}$
$\displaystyle \therefore \cos x=\frac{1}{2}$
$\displaystyle x=\frac{\pi}{3},\ \frac{5\pi}{3}$
$\displaystyle \therefore \text{Number of points of intersection }=2$
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$\displaystyle \textbf{Question 7. }\text{Write the values of }x\in[0,\pi]\text{ for which }\sin2x,\ \frac{1}{2}\text{ and }\cos2x\text{ are in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \sin2x,\ \frac{1}{2},\ \cos2x\text{ are in A.P.}$
$\displaystyle \therefore \sin2x+\cos2x=1$
$\displaystyle \therefore \sqrt2\sin\left(2x+\frac{\pi}{4}\right)=1$
$\displaystyle \therefore x=0,\ \frac{\pi}{4},\ \pi$
$\\$

$\displaystyle \textbf{Question 8. }\text{Write the number of points of intersection of }2y=-1\text{ and } \\ y=\mathrm{cosec}\,x.$
$\displaystyle \text{Answer:}$
$\displaystyle 2y=-1\Rightarrow y=-\frac{1}{2}$
$\displaystyle \mathrm{cosec}\,x=-\frac{1}{2}\Rightarrow \sin x=-2$
$\displaystyle \text{This is not possible.}$
$\displaystyle \therefore \text{Number of points of intersection }=0$
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$\displaystyle \textbf{Question 9. }\text{Write the solution set of }(2\cos x+1)(4\cos x+5)=0\text{ in }[0,2\pi].$
$\displaystyle \text{Answer:}$
$\displaystyle 2\cos x+1=0\Rightarrow \cos x=-\frac{1}{2}$
$\displaystyle \therefore x=\frac{2\pi}{3},\ \frac{4\pi}{3}$
$\displaystyle 4\cos x+5=0\Rightarrow \cos x=-\frac{5}{4}$
$\displaystyle \text{This is not possible.}$
$\displaystyle \therefore \text{Solution set }=\left\{\frac{2\pi}{3},\frac{4\pi}{3}\right\}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Write the number of values of }x\in[0,2\pi]\text{ satisfying } \\ \sin^2x-\cos x=\frac{1}{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle 1-\cos^2x-\cos x=\frac{1}{4}$
$\displaystyle \therefore 4\cos^2x+4\cos x-3=0$
$\displaystyle \therefore (2\cos x-1)(2\cos x+3)=0$
$\displaystyle \therefore \cos x=\frac{1}{2}$
$\displaystyle x=\frac{\pi}{3},\ \frac{5\pi}{3}$
$\displaystyle \therefore \text{Number of values }=2$
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$\displaystyle \textbf{Question 11. }\text{If }3\tan\left(x-\frac{\pi}{12}\right)=\tan\left(x+\frac{\pi}{12}\right),\ 0<x<\frac{\pi}{2},\text{ find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\tan x=t\text{ and }\tan\frac{\pi}{12}=2-\sqrt3$
$\displaystyle 3\cdot\frac{t-(2-\sqrt3)}{1+t(2-\sqrt3)}=\frac{t+(2-\sqrt3)}{1-t(2-\sqrt3)}$
$\displaystyle \therefore (t-1)^2=0$
$\displaystyle \therefore t=1$
$\displaystyle \therefore \tan x=1$
$\displaystyle \therefore x=\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 12. }\text{If }2\sin^2x=3\cos x,\text{ where }0\leq x\leq2\pi,\text{ then find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle 2(1-\cos^2x)=3\cos x$
$\displaystyle \therefore 2\cos^2x+3\cos x-2=0$
$\displaystyle \therefore (2\cos x-1)(\cos x+2)=0$
$\displaystyle \therefore \cos x=\frac{1}{2}$
$\displaystyle \therefore x=\frac{\pi}{3},\ \frac{5\pi}{3}$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }\sec x\cos5x+1=0,\text{ where }0<x\leq\frac{\pi}{2},\text{ find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \sec x\cos5x+1=0$
$\displaystyle \therefore \frac{\cos5x}{\cos x}=-1$
$\displaystyle \therefore \cos5x=-\cos x$
$\displaystyle \therefore \cos5x+\cos x=0$
$\displaystyle 2\cos\frac{5x+x}{2}\cos\frac{5x-x}{2}=0$
$\displaystyle 2\cos3x\cos2x=0$
$\displaystyle \therefore \cos3x=0\qquad \text{or}\qquad \cos2x=0$
$\displaystyle \cos3x=0\Rightarrow 3x=\frac{\pi}{2},\frac{3\pi}{2}$
$\displaystyle \therefore x=\frac{\pi}{6},\frac{\pi}{2}$
$\displaystyle \cos2x=0\Rightarrow 2x=\frac{\pi}{2}$
$\displaystyle \therefore x=\frac{\pi}{4}$
$\displaystyle \therefore x=\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{2}$
$\\$