JEE Advanced 2022 · Paper 1, Mathematics question 14

First determinant. It equals $2(1+\sin^2\theta)$, so half of it is $1+\sin^2\theta$.

Second determinant. Its diagonal is $\sin\pi=-\cos\frac\pi2=\tan\pi=0$. With $d=\log_e\frac4\pi$ and $\log_e\frac\pi4=-d$, it reduces to
$$d\left[\cos(\theta+\tfrac\pi4)\cot(\theta+\tfrac\pi4)-\tan(\theta-\tfrac\pi4)\sin(\theta-\tfrac\pi4)\right].$$
Using $\cos(\theta+\frac\pi4)=-\sin(\theta-\frac\pi4)$ and $\sin(\theta+\frac\pi4)=\cos(\theta-\frac\pi4)$, both products equal $\frac{\sin^2(\theta-\pi/4)}{\cos(\theta-\pi/4)}$. So the determinant is 0.

The function g. $f(\theta)=1+\sin^2\theta$, so
$$g(\theta)=|\sin\theta|+|\cos\theta|=\sin\theta+\cos\theta\in[1,\sqrt2]\ \text{on }[0,\tfrac\pi2].$$

The quadratic. $p(x)=k(x-1)(x-\sqrt2)$, and $p(2)=k(2-\sqrt2)=2-\sqrt2$ gives $k=1$. So $p<0$ exactly on $(1,\sqrt2)$.

Options.

• $\frac{3+\sqrt2}4\approx1.10$: inside $(1,\sqrt2)$, so $p<0$. (A) true.
• $\frac{1+3\sqrt2}4\approx1.31$: inside, so $p<0$. (B) false.
• $\frac{5\sqrt2-1}4\approx1.52$: above $\sqrt2$, so $p>0$. (C) true.
• $\frac{5-\sqrt2}4\approx0.90$: below 1, so $p>0$. (D) false.