JEE Main Practice — Mathematics Set 6

Explanation: Multiply and divide by $8\sin\frac\pi7$ and use $2\sin\theta\cos\theta=\sin2\theta$ three times: the product $=\frac{\sin\frac{8\pi}{7}}{8\sin\frac\pi7}$. Since $\sin\frac{8\pi}{7}=\sin\left(\pi+\frac\pi7\right)=-\sin\frac\pi7$, the value is $-\frac18$.

Explanation: Multiply by $\cos x$ (which must be non-zero): $\sin x+1=2\cos^2x=2-2\sin^2x\Rightarrow2\sin^2x+\sin x-1=0\Rightarrow(2\sin x-1)(\sin x+1)=0$. $\sin x=-1$ forces $\cos x=0$, which is not allowed. $\sin x=\frac12$ gives $x=\frac\pi6,\frac{5\pi}6$ (both valid). Hence 2 solutions.

Explanation: Squaring: $1+2\sin\theta\cos\theta=\frac{49}{25}\Rightarrow\sin\theta\cos\theta=\frac{12}{25}$. So $25\sin\theta\cos\theta=12$.

Explanation: $\vec a\cdot\vec b=3-2+2=3$, $|\vec b|=\sqrt{11}$. Projection $=\frac{\vec a\cdot\vec b}{|\vec b|}=\frac{3}{\sqrt{11}}$. (1 would be the projection of $\vec b$ on $\vec a$.)

Explanation: $(\hat i+\hat j)\times(\hat j+\hat k)=\hat i-\hat j+\hat k$, magnitude $\sqrt3$.

Explanation: Dot product $=2-2\lambda+3=0\Rightarrow\lambda=\frac52$.

Explanation: $\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1\end{vmatrix}=1(1-2)-1(-1-1)+1(2+1)=-1+2+3=4$. Volume $=4$ cubic units.

Explanation: $\begin{vmatrix}1&2&3\\2&1&1\\3&3&\lambda\end{vmatrix}=(\lambda-3)-2(2\lambda-3)+3(6-3)=12-3\lambda=0\Rightarrow\lambda=4$ (indeed the third vector is the sum of the first two).

Explanation: $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$. $\vec a\cdot\vec c=2$, $\vec a\cdot\vec b=0$, so the result is $2\vec b=2\hat i-2\hat j$. (Option B is $(\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a$.)

Explanation: $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cos\frac\pi3=4+9+6=19$.

Explanation: $\vec{AB}=(1,2,2)$, $\vec{AC}=(2,-2,1)$. $\vec{AB}\times\vec{AC}=(2+4,\,4-1,\,-2-4)=(6,3,-6)$, magnitude $\sqrt{36+9+36}=9$. Area $=\frac92$, so twice the area $=9$.