JEE Mathematics Miscellaneous Questions MCQs Set 01

Answer: (a) equilateral

Explanation:
1. For a circulant determinant, \( \Delta=-\tfrac12(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] \) with x, y, z = sin A, sin B, sin C.
2. The sum of the sines is positive, so \( \sin A=\sin B=\sin C \); hence \( a=b=c \) and the triangle is equilateral.

Teacher's Note:
1. Use the circulant determinant identity.
2. The sum of sines of a triangle is never zero.

Answer: (b) b

Explanation:
1. Sum of the three roots: \( a=p+\gamma \). Sum of products in pairs: \( b=\alpha\beta+\gamma(\alpha+\beta)=q+p\gamma \).

Teacher's Note:
1. Use Vieta's relations for the cubic.
2. Group \( \alpha+\beta=p \) and \( \alpha\beta=q \).

Answer: (a) c

Explanation:
1. The product of the three roots is \( \alpha\beta\gamma=q\gamma \).
2. By Vieta this equals c.

Teacher's Note:
1. Product of all roots of a cubic is the constant term with sign.
2. Here the sign gives +c.

Answer: (a) (4, -2)

Explanation:
1. B = (1 + t, 1 - t) lies on the line. \( BC^2=(t-1)^2+(2-t)^2 \).
2. \( AC^2=5 \). Setting \( BC^2=5 \): \( 2t^2-6t=0 \), t = 0 (that is A) or t = 3.
3. t = 3 gives B = (4, -2).

Teacher's Note:
1. Parametrise B on the line.
2. Equate BC² with AC².

Answer: (a) 36

Explanation:
1. Taking a, b, c out of the rows and columns gives \( a^2b^2c^2\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}=4a^2b^2c^2 \).
2. Three consecutive integers have a product divisible by 6, so \( (abc)^2 \) is divisible by 36, and the determinant is divisible by 144, hence by 36.

Teacher's Note:
1. Factor a from row 1 and column 1, and similarly for b and c.
2. Use the divisibility of consecutive products.

Answer: (a) -2

Explanation:
1. \( f(x)=-(x-2)^2+5a^2 \), so the maximum is \( 5a^2 \).
2. \( 5a^2=20 \) gives \( a=\pm2 \); the option available is -2.

Teacher's Note:
1. Complete the square.
2. Maximum of a downward parabola is at its vertex.

Answer: (c) 90\(^\circ\)

Explanation:
1. Sine rule: \( \sin C=\dfrac{c\sin A}{a}=\dfrac{3\cdot\frac12}{\sqrt3}=\dfrac{\sqrt3}2 \), so C = 60\(^\circ\) or 120\(^\circ\).
2. Then B = 90\(^\circ\) or 30\(^\circ\); 90\(^\circ\) is the option listed.

Teacher's Note:
1. Use \( a/\sin A=c/\sin C \).
2. The case is ambiguous; both C are valid.

Answer: (a) r \gt 0

Explanation:
1. Let \( f(x)=px^2+qx+r \). With no real roots f has constant sign, and f(2) = 4p + 2q + r \gt 0, so f \gt 0 everywhere.
2. Therefore f(0) = r \gt 0 (and also p \gt 0 and p + q + r \gt 0).

Teacher's Note:
1. Evaluate f at 2 and at 0.
2. No real roots means constant sign.

Answer: (d) f'(2) does not exist

Explanation:
1. f(2) = 35 from both pieces, so f is continuous at 2.
2. Left derivative \( 6x+12=24 \); right derivative = -1. They differ, so f'(2) does not exist.

Teacher's Note:
1. Compare the one-sided derivatives.
2. Check continuity first.

Answer: (a) p and q are both integers

Explanation:
1. At x = 0: q is an integer.
2. At x = 1: 1 + p + q is an integer, so p is an integer too.

Teacher's Note:
1. Substitute x = 0 and x = 1.
2. Subtract integers.