JEE Mathematics IIT Questions MCQs Set 01

Answer: (a) equal to 1

Explanation:
1. Since \( |z_i| = 1 \), \( \frac{1}{z_i} = \overline{z_i} \).
2. So \( \left|\overline{z_1} + \overline{z_2} + \overline{z_3}\right| = |z_1 + z_2 + z_3| = 1 \).

Teacher's Note:
1. For a unit complex number the reciprocal is the conjugate.
2. A number and its conjugate have the same modulus.

Answer: (b) \( c^2 + a^2 - b^2 \)

Explanation:
1. \( A + C = \pi - B \), so \( \frac{A - B + C}{2} = \frac\pi2 - B \).
2. \( 2ac\sin\left(\frac\pi2 - B\right) = 2ac\cos B = a^2 + c^2 - b^2 \) by the cosine rule.

Teacher's Note:
1. Replace A + C using the angle sum.
2. Recognise the cosine rule.

Answer: (c) \( f(\theta) \geq 0 \) for all real \( \theta \)

Explanation:
1. \( \sin\theta + \sin3\theta = 2\sin2\theta\cos\theta \).
2. So \( f(\theta) = 2\sin\theta\cos\theta\sin2\theta = \sin^2 2\theta \geq 0 \) for all real \( \theta \).

Teacher's Note:
1. Use the sum-to-product formula.
2. 2 sinθ cosθ is sin 2θ.

Answer: (a) (i) and (iii)

Explanation:
1. The ellipse has \( e = \frac35 \) and foci \( (\pm3, 0) \). The hyperbola has \( e = \frac53 \) and passes through \( (3, 0) \), so \( a = 3 \).
2. \( c = ae = 5 \) and \( b^2 = c^2 - a^2 = 16 \): the hyperbola is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) with foci \( (\pm5, 0) \).
3. So (i) and (iii) are correct.

Teacher's Note:
1. The eccentricities multiply to 1, so e = 5/3.
2. A vertex at the ellipse focus gives a = 3.

Answer: (c) continuous if f is continuous

Explanation:
1. The modulus of a continuous function is continuous.
2. Counterexamples: \( f(x) = x \) gives \( g = |x| \), which is not one-one and not differentiable at 0; and \( f(x) = x \) is onto but \( |x| \) is not.

Teacher's Note:
1. Composition with
2. ·
3. preserves continuity only.
4. Test with f(x) = x.

Answer: (c) 3

Explanation:
1. Let the roots be \( \alpha \) and \( \alpha^2 \). The product is \( \alpha^3 = 1 \). A real root gives \( \alpha = 1 \) and \( p = -6 \), which is negative.
2. So \( \alpha \) is a complex cube root of unity: the roots are \( \omega, \omega^2 \), with sum \( -1 = -\frac p3 \), giving \( p = 3 \).

Teacher's Note:
1. Use the product of roots first.
2. Reject the solution with p negative.

Answer: (a) 0

Explanation:
1. \( U = A^{-1}\begin{bmatrix}1&2&2\\0&3&3\\0&0&1\end{bmatrix} = \begin{bmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{bmatrix} \)
2. \( U^{-1} = \frac13\begin{bmatrix}-1&-2&0\\-7&-5&-3\\9&6&3\end{bmatrix} \) (verify: \( UU^{-1} = I \)).
3. Sum of entries \( = \frac{-3 - 15 + 18}{3} = 0 \).

Teacher's Note:
1. Find U first, then invert.
2. Sum of all entries is the product of the row of ones with U^-1 and the column of ones.

Answer: (a) 1

Explanation:
1. Both curves pass through \( (1, 0) \).
2. Slope of the first: \( 3^{x-1}\left(\frac1x + \ln3\ln x\right) = 1 \) at \( x=1 \). Slope of the second: \( x^x(\ln x + 1) = 1 \) at \( x=1 \).
3. Equal slopes mean the curves touch, so the angle is 0 and the cosine is 1.

Teacher's Note:
1. Find the common point first.
2. Equal slopes give an angle of zero.

Answer: (b) 1210

Explanation:
1. \( a + b = 10c \), \( ab = -11d \), \( c + d = 10a \), \( cd = -11b \).
2. Since \( a \) is a root of the first equation, \( a^2 - 10ca - 11d = 0 \); similarly \( c^2 - 10ac - 11b = 0 \). Subtracting: \( a^2 - c^2 = 11(d - b) \), and from the sums \( a + b - c - d = 10(c - a) \).
3. Solving gives \( a + c = 121 \), so \( a + b + c + d = 10c + 10a = 1210 \).

Teacher's Note:
1. Use sums and products of roots in both equations.
2. The solution comes from a + c = 121.

Answer: (a) \( \frac34 \)

Explanation:
1. Take the centre at the origin, vertices \( (\pm1, \pm1) \). For a point at distance \( \rho \) from the centre, \( \sum PA^2 = 4\rho^2 + 4\cdot 2 = 4\rho^2 + 8 \).
2. \( C_1 \) has \( \rho = 1 \): sum \( = 12 \). \( C_2 \) has \( \rho = \sqrt2 \): sum \( = 16 \).
3. Ratio \( = \frac{12}{16} = \frac34 \).

Teacher's Note:
1. Place the square on coordinate axes.
2. The sum of squared distances to the vertices depends only on the distance from the centre.

Answer: (b) \( \frac52 \)

Explanation:
1. Put \( u = \ln x \), \( du = \frac{dx}{x} \); the limits become \( -1 \) to \( 2 \).
2. \( \int_{-1}^{2}|u|\,du = \frac12 + 2 = \frac52 \).

Teacher's Note:
1. Substitute u = ln x.
2. Split the modulus at u = 0.

Answer: (d) \( \frac43 \)

Explanation:
1. Intersection: \( -4y^2 = 1 - 5y^2 \), so \( y^2 = 1 \) and \( y = \pm1 \).
2. Area \( = \int_{-1}^{1}\left[(1-5y^2) - (-4y^2)\right]dy = \int_{-1}^{1}(1 - y^2)\,dy = \frac43 \).

Teacher's Note:
1. Integrate with respect to y.
2. The curve with larger x is on the right.

Answer: (b) 5

Explanation:
1. \( a_n = \frac37\left[1 - \left(-\frac34\right)^n\right] \), so \( b_n > a_n \) means \( a_n \lt \frac12 \), i.e. \( \left(-\frac34\right)^n > -\frac16 \).
2. For even \( n \) this always holds. For odd \( n \) we need \( \left(\frac34\right)^n \lt \frac16 \), true for \( n \geq 7 \) (\( n = 5 \) gives \( 0.237 \), fails).
3. So the inequality holds for all \( n > 5 \) but fails at \( n = 5 \): \( n_0 = 5 \).

Teacher's Note:
1. Sum the GP with ratio -3/4.
2. Test odd and even n separately.

Answer: (a) a local maximum

Explanation:
1. Near 0 (but not at 0), \( f(x) = |x| \), which is close to 0 and less than 1.
2. \( f(0) = 1 \) is larger than all nearby values, so \( f \) has a local maximum at 0.

Teacher's Note:
1. Compare f(0) with values nearby.
2. A removable-type jump can create a local extremum.

Answer: (b) \( \alpha \lt 0 \lt \beta \lt |\alpha| \)

Explanation:
1. The product \( c \lt 0 \), so the roots have opposite signs: \( \alpha \lt 0 \lt \beta \).
2. The sum \( -b \lt 0 \), so the negative root is larger in size: \( \beta \lt |\alpha| \).

Teacher's Note:
1. Product tells the signs, sum tells which is bigger.
2. Draw the parabola if unsure.

Answer: (b) \( 12 + 7\sqrt3 \)

Explanation:
1. The 120° angle is the apex. Let the equal sides be \( s \); then the base is \( s\sqrt3 \), the area is \( \frac{\sqrt3}{4}s^2 \) and the semi-perimeter is \( \frac{s(2+\sqrt3)}{2} \).
2. \( r = \frac{\text{Area}}{\text{semi-perimeter}} = \frac{\sqrt3\,s}{2(2+\sqrt3)} = \sqrt3 \), so \( s = 4 + 2\sqrt3 \).
3. Area \( = \frac{\sqrt3}{4}(4 + 2\sqrt3)^2 = 12 + 7\sqrt3 \).

Teacher's Note:
1. Use r = area / semi-perimeter.
2. A 120° angle in an isosceles triangle must be the apex.

Answer: (a) 0

Explanation:
1. \( \vec a \times \vec b \) and \( \vec c \times \vec d \) are the normals of the two planes.
2. Their cross product is zero, so the normals are parallel, hence the planes are parallel and the angle is 0.

Teacher's Note:
1. Zero cross product means parallel vectors.
2. The angle between planes equals the angle between their normals.

Answer: (c) 2

Explanation:
1. On \( [-2, 2] \), \( f \) is odd, because \( e^{\cos x} \) is even and \( \sin x \) is odd, so that integral is 0.
2. On \( (2, 3] \), \( f = 2 \), giving \( 2 \times 1 = 2 \).

Teacher's Note:
1. Odd functions integrate to zero over symmetric limits.
2. Split the integral at the point where the definition changes.

Answer: (a) \( 2 \) or \( -\frac32 \)

Explanation:
1. Orthogonality: \( 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \). Here \( g_1 = 1, f_1 = k, c_1 = 6 \) and \( g_2 = 0, f_2 = k, c_2 = k \).
2. \( 2k^2 = 6 + k \), so \( 2k^2 - k - 6 = 0 \) and \( k = 2 \) or \( -\frac32 \).

Teacher's Note:
1. Use the orthogonality condition 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
2. Read g and f from the x and y coefficients.

Answer: (a) a + b

Explanation:
1. For a right angle at C, \( r = \frac{a + b - c}{2} \) and \( R = \frac c2 \) (the hypotenuse is a diameter).
2. \( 2(r + R) = a + b - c + c = a + b \).

Teacher's Note:
1. Use the right-triangle inradius formula.
2. The circumcentre is the midpoint of the hypotenuse.

Answer: (c) \( (1, 2) \)

Explanation:
1. \( f \) decreases where \( f'(x) \lt 0 \). Since \( e^x > 0 \), this needs \( (x-1)(x-2) \lt 0 \).
2. That holds for \( 1 \lt x \lt 2 \).

Teacher's Note:
1. The sign of f' decides monotonicity.
2. e^x never changes sign.

Answer: (d) \( (-\infty, 1) \)

Explanation:
1. \( 2^y = 2 - 2^x \) must be positive, so \( 2^x \lt 2 \).
2. Hence \( x \lt 1 \), and the domain is \( (-\infty, 1) \).

Teacher's Note:
1. An exponential is always positive.
2. This gives a strict inequality.

Answer: (d) 1

Explanation:
1. The normal has slope \( \tan\frac{3\pi}{4} = -1 \).
2. The tangent is perpendicular, so \( f'(3) = 1 \).

Teacher's Note:
1. Slope of tangent times slope of normal = -1.
2. tan(3π/4) = -1.

Answer: (a) \( \pi \)

Explanation:
1. For \( \arg z \in (-\pi, 0) \), \( -z \) is \( z \) rotated by \( \pi \), and the result \( \arg z + \pi \) lies in \( (0, \pi) \), which is the principal range.
2. So \( \arg(-z) - \arg z = \pi \).

Teacher's Note:
1. Principal arguments lie in (-π, π].
2. Add π when the argument is negative, subtract π when positive.

Answer: (d) 5

Explanation:
1. \( U = A^{-1}\begin{bmatrix}1&2&2\\0&3&3\\0&0&1\end{bmatrix} = \begin{bmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{bmatrix} \)
2. \( [3\ 2\ 0]U = [-1\ \ 4\ \ 4] \).
3. \( [-1\ \ 4\ \ 4]\begin{bmatrix}3\\2\\0\end{bmatrix} = -3 + 8 = 5 \).

Teacher's Note:
1. Multiply the row vector by U first.
2. The final product is a 1 × 1 matrix.

Answer: (b) \( 0 \leq g(2) \lt 2 \)

Explanation:
1. \( \int_0^1 f \in \left[\frac12, 1\right] \) and \( \int_1^2 f \in \left[0, \frac12\right] \).
2. So \( g(2) \in \left[\frac12, \frac32\right] \), which lies inside \( [0, 2) \) and in none of the other ranges.

Teacher's Note:
1. Bound each piece of the integral separately.
2. Pick the option that contains the whole interval.

Answer: (b) circumcentre

Explanation:
1. Equal angles of elevation with a common height mean equal horizontal distances from the foot to A, B and C.
2. The point equidistant from all three vertices is the circumcentre.

Teacher's Note:
1. Equal elevation angles give equal horizontal distances.
2. The circumcentre is equidistant from the vertices.

Answer: (a) \( 0 \lt M \leq 1 \)

Explanation:
1. \( (a+b) + (c+d) = 2 \), and for positive numbers with a fixed sum the product is at most the square of half the sum.
2. \( M \leq \left(\frac{2}{2}\right)^2 = 1 \), and \( M > 0 \).

Teacher's Note:
1. Use AM-GM on the two brackets.
2. Their sum is fixed at 2.

Answer: (c) 60

Explanation:
1. Odd digits 3, 3, 5, 5 go into the 4 even places in \( \frac{4!}{2!\,2!} = 6 \) ways.
2. Even digits 2, 2, 8, 8, 8 go into the 5 odd places in \( \frac{5!}{2!\,3!} = 10 \) ways.
3. Total \( 6 \times 10 = 60 \).

Teacher's Note:
1. Place the odd digits first, then the rest.
2. Divide by factorials of repeated digits.

Answer: (d) \( 2\sqrt2 \)

Explanation:
1. The normal is \( (2, -2, 1) \times (1, -1, 2) = (-3, -3, 0) \), i.e. direction \( (1, 1, 0) \).
2. The plane: \( x + y = 1 + (-2) = -1 \), i.e. \( x + y + 1 = 0 \).
3. Distance from \( (1, 2, 2) \): \( \frac{|1 + 2 + 1|}{\sqrt2} = 2\sqrt2 \).

Teacher's Note:
1. The normal of the required plane is the cross product of the two given normals.
2. Use the point-to-plane distance formula.