5 Marks Questions

Question 15 Marks

If $\alpha, \beta$ are two different values of $x$ lying between $0$ and $2 \pi$ which satisfy the equation $6 \cos x +8 \sin x =9$, find the value of $\sin (\alpha+\beta)$

Answer

We have to find the value of $\sin (\alpha+\beta)$
It is given that
$6 \cos x + 8 \sin x = 9$
$\Rightarrow 6 \cos x=9-8 \sin x$
$\Rightarrow 36 \cos ^2 x=(9-8 \sin x)^2$
$\Rightarrow 36\left(1-\sin ^2 x\right)=81+64 \sin ^2 x-144 \sin x$
$\Rightarrow 100 \sin ^2 x-144 \sin x+45=0$
Now, $\alpha$ and $\beta$ are the roots of the given equation;
therefore, $\cos \alpha$ and $\cos \beta$ are the roots of the above equation.
$\Rightarrow \sin \alpha \sin \beta=\frac{45}{100}$
$($Product of roots of a quadratic equation $a x^2+b x+c=0$ is $\frac{c}{a}$
Again, $6 \cos x + 8 \sin x = 9$
$\Rightarrow 8 \sin x=9-6 \cos x$
$\Rightarrow 64 \sin ^2 x=(9-6 \cos x)^2$
$\Rightarrow 64\left(1-\cos ^2 x\right)=81+36 \cos ^2 x-108 \cos x$
$\Rightarrow 100 \cos ^2 x-108 \cos x+17=0$
Now, $\alpha$ and $\beta$ are the roots of the given equation;
therefore, $\sin \alpha$ and $\sin \beta$ are the roots of the above equation.
Therefore, $\cos \alpha \cos \beta=\frac{17}{100}$
Hence, $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$
$=\frac{17}{100}-\frac{45}{100}$
$=-\frac{28}{100}$
$=-\frac{7}{25}$
$\sin (\alpha+\beta)=\sqrt{1-\cos ^2(\alpha+\beta)}$
$=\sqrt{1-\left(\frac{-7}{25}\right)^2}$
$=\sqrt{\frac{576}{625}}$
$=\frac{24}{25}$

View full question & answer→

Question 25 Marks

Prove that $\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}$

Answer

$\text { LHS }=\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$
$=\cos \frac{2 \pi}{15} \cos 2\left(\frac{2 \pi}{15}\right) \cos 4\left(\frac{2 \pi}{15}\right) \cos 8\left(\frac{2 \pi}{15}\right)$
Put $\frac{2 \pi}{15}=\alpha$
$\Rightarrow \text { LHS }=\cos \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha$
$=\frac{2 \sin \alpha[\cos \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha]}{2 \sin \alpha}$
$[$ multiplying numerator and denominator by $2 \sin \alpha]$
$=\frac{(2 \sin \alpha \cdot \cos \alpha) \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha}{2 \sin \alpha}$
$=\frac{2(\sin 2 \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha)}{2(2 \sin \alpha)}$
$[\because 2 \sin \alpha \cos \alpha=\sin 2 \alpha$ and multiplying numerator and denominator by $2]$
$=\frac{(2 \sin 2 \alpha \cdot \cos 2 \alpha) \cdot \cos 4 \alpha \cdot \cos 8 \alpha}{4 \sin \alpha}$
$=\frac{2(\sin 4 \alpha \cdot \cos 4 \alpha) \cos 8 \alpha}{2(4 \sin \alpha)}$
$[\because 2 \sin \alpha \cos \alpha=\sin 2 \alpha$ and multiplying numerator and denominator by $2]$
$=\frac{2(\sin 8 \alpha \cdot \cos 8 \alpha)}{2(8 \sin \alpha)}$
$=\frac{\sin 16 \alpha}{16 \sin \alpha}=\frac{\sin (15 \alpha+\alpha)}{16 \sin \alpha}$
Now, $15 \alpha=2 \pi,$
$=\frac{\sin (2 \pi+\alpha)}{16 \sin \alpha}=\frac{\sin \alpha}{16 \sin \alpha}=\frac{1}{16}=\text { RHS }$
$\therefore \text { LHS }=\text { RHS }$
Hence proved.

View full question & answer→

Question 35 Marks

Find four numbers in $GP,$ whose sum is $85$ and product is $4096.$

Answer

Let the four numbers in $GP$ be
$\frac{a}{r^3}, \frac{a}{r}, a r, a r^3 \ldots (i)$
Product of four numbers $= 4096 [$ given $]$
$\Rightarrow\left(\frac{a}{r^3}\right)\left(\frac{a}{r}\right)( ar )\left( ar ^3\right)=4096$
$\Rightarrow a ^4=4096$
$\Rightarrow a ^4=8^4$
On comparing the base of the power $4,$ we get
$\Rightarrow \frac{a}{r^3}+\frac{a}{r}+ ar + ar ^3=85$
$\Rightarrow a \left[\frac{1}{r^3}+\frac{1}{r}+r+r^3\right]=85$
$\Rightarrow 8\left[ r ^3+\frac{1}{r^3}\right]+8\left[ r +\frac{1}{r}\right]=85[\because a =8]$
$\Rightarrow 8\left[\left(r+\frac{1}{r}\right)^3-3\left(r+\frac{1}{r}\right)\right]+8\left(r+\frac{1}{r}\right)=85\left[\because a ^3+ b ^3=( a + b )^3-3( a + b )\right]$
$\Rightarrow 8\left(r+\frac{1}{r}\right)^3-16\left(r+\frac{1}{r}\right)-85=0 \ldots \text { (ii) }$
On putiing $\left(r+\frac{1}{r}\right)= x$ in Eq. $(ii),$ we get
$8 x^3-16 x-85=0$
$\Rightarrow(2 x-5)\left(4 x^2+10 x+17\right)=0$
$\Rightarrow 2 x-5=0 [\because 4 x^2+10 x+177=0$ has imaginary roots $]$
$\Rightarrow x=\frac{5}{2}$
$\Rightarrow r+\frac{1}{r}=\frac{5}{2} [$ put $x=r+\frac{1}{r}]$
$\Rightarrow 2 r^2-5 r+2=0$
$\Rightarrow(r-2)(2 r-1)=0$
$\Rightarrow r=2$ or $r=\frac{1}{2}$
On putting $a =8$ and $r =2$ or $r =\frac{1}{2}$ in Eq. $(i),$
we obtain four numbers as $\frac{8}{2^3}, \frac{8}{2}, 8 \times 2,8 \times 2^3$
$\text { or } \frac{8}{(1 / 2)^3}, \frac{8}{(1 / 2)}, 8 \times \frac{1}{2}, 8 \times\left(\frac{1}{2}\right)^3$
$\Rightarrow 1,4,16,64$ or $64,16,4,1$

View full question & answer→

Question 45 Marks

Differentiate log sin $x$ from first principles.

Answer

Let $f (x) = \log \sin x.$ Then, $f (x + h) = \log \sin (x + h)$
$\therefore \frac{d}{d x}(f(x))=\underset{{h \rightarrow 0}}{\lim} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \sin (x+h)-\log \sin x}{h}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{\frac{\sin (x+h)}{\sin x}\right\}}{h}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)}{\sin x}-1\right\}}{h}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\lim _h \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{\sin x}\right\}}{h}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{\sin x}\right\}}{h\left\{\frac{\sin (x+h)-\sin x}{\sin x}\right\}} \times\left\{\frac{\sin (x+h)-\sin x}{\sin x}\right\}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \frac{\sin (x+h)-\sin x}{h} \times \frac{1}{\sin x}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \underset{{h \rightarrow 0}}{\lim} \frac{2 \sin \frac{h}{2} \cos \left(x+\frac{h}{2}\right)}{h} \times \frac{1}{\sin x}$
$\Rightarrow \frac{d}{d x}( f ( x ))=\underset{{h \rightarrow 0}}{\lim} \frac{\log \left\{1+\frac{\sin (x+h)-\sin x}{h}\right\}}{\left\{\frac{\sin (x+h)-\sin x}{h}\right\}} \times \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(\frac{h}{2}\right) \cos \left(x+\frac{h}{2}\right)}{\frac{h}{2}} \times \frac{1}{\sin x}$
$\Rightarrow \frac{d}{d x}( f ( x ))=1 \times \cos x \times \frac{1}{\sin x}$
$=\cot x .$

View full question & answer→

Question 55 Marks

Solve: $\lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}$

Answer

Dividing $x^4-3 x^3+2$ by $x^3-5 x^2+3 x+1$

$\Rightarrow \lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}=\lim _{x \rightarrow 1}(x+2)+\lim _{x \rightarrow 1} \frac{7 x^2-7 x}{x^3-5 x^3+3 x+1}$
$=\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{x^3-5 x^3+3 x+1}$
$=\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{(x-1)\left(x^2-4 x-1\right)}$
$=\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x}{\left(x^2-4 x-1\right)}$
$=1+2+\frac{7}{(1-4-1)}$
$=3-\frac{7}{4}$
$=\frac{12-7}{4}$
$=\frac{5}{4}$

View full question & answer→

Question 65 Marks

$20$ cards are numbered from $1$ to $20.$ One card is drawn at random. What is the probability that the number on the card drawn is,
$i.$ A prime number
$ii.$ An odd number
$iii.$ A multiple of $5$
$iv.$ Not divisible by $3.$

Answer

Let $S$ be the sample space
$S = \{1,2,3,4,5.....20\} $
Let $\ce{E_1, E_2}$ and $\ce{E_3, E_4}$ are the event of getting prime number, an odd number, multiple of $5$ and not divisible by $3$ respectively.
$P\left(E_1\right)=\frac{8}{20}=\frac{2}{5}, E _1=\{2,3,5,7,11,13,17,19\}$
$P\left(E_2\right)=\frac{10}{20}=\frac{1}{2}, E _2=\{1,3,5,7,9,11,13,15,17,19\}$
$P\left(E_3\right)=\frac{4}{20}=\frac{1}{5}, E _3=\{5,10,15,20\}$
$P\left(E_4\right)=\frac{14}{20}=\frac{7}{10}, E _4=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}$

View full question & answer→

Maths 5 Marks Questions

Take a timed test