JEE Mathematics Mixed Practice Questions MCQs Set 01

Answer: (c) \( f(x) \) is continuous at \( x = 1 \) but not differentiable there

Explanation:
1. \( f(x) = 2-x \) for \( x \le 1 \) and \( f(x) = x \) for \( x \ge 1 \).
2. At \( x = 1 \) both pieces give 1, so f is continuous.
3. Left slope = −1 and right slope = +1, which differ, so f is not differentiable at \( x = 1 \).

Teacher's Note:
1. The max of two lines has a corner where they cross.
2. A corner is continuous but not differentiable.

Answer: (d) None of these

Explanation:
1. \( x^2 - x + 2 \) has discriminant \( 1 - 8 < 0 \), so it is always positive.
2. So \( f(x) > 0 \) iff \( x^2 > 4 \), i.e. \( x \in (-\infty,-2)\cup(2,\infty) \).
3. This union is not any of the first three options.

Teacher's Note:
1. A quadratic with negative discriminant and positive leading coefficient is always positive.
2. The solution is a union of two intervals.

Answer: (b) 0

Explanation:
1. Only the third row depends on r, so the sum is the same determinant with the third row replaced by the row sums.
2. \( \sum {}^nC_r = 2^n,\ \sum 2^r = 2^{n+1}-1,\ \sum(2r+1) = (n+1)^2 \).
3. Row 1 minus the new row 3 gives \( (1,\,1,\,-1) \) which equals row 2, so the determinant is 0.

Teacher's Note:
1. Use the linearity of the determinant in one row.
2. Rows that differ by another row give zero.

Answer: (c) \( \left(-3, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, 3\right) \)

Explanation:
1. Inside the square: \( -3 < a < 3 \).
2. Outside the circle: \( a^2 + a^2 > 1 \Rightarrow |a| > \frac{1}{\sqrt{2}} \).
3. So \( a \in \left(-3, -\frac{1}{\sqrt{2}}\right)\cup\left(\frac{1}{\sqrt{2}}, 3\right) \).

Teacher's Note:
1. Points (a, a) lie on the line y = x.
2. |a| > 1/√2 gives two intervals.

Answer: (b) \( (\sqrt{2}+1) : 1 \)

Explanation:
1. Take legs of length 1. The hypotenuse is \( \sqrt{2} \).
2. Circumradius \( R = \frac{\sqrt{2}}{2} \) (half the hypotenuse); inradius \( r = \frac{1+1-\sqrt{2}}{2} \).
3. \( R/r = \frac{\sqrt{2}}{2-\sqrt{2}} = \sqrt{2}+1 \).

Teacher's Note:
1. For a right triangle, \( r = \frac{a+b-c}{2} \).
2. Rationalise the denominator.

Answer: (d) \( -\vec{b} \)

Explanation:
1. \( \vec{a}\times(\vec{a}\times\vec{b}) = \vec{a}(\vec{a}\cdot\vec{b}) - \vec{b}(\vec{a}\cdot\vec{a}) \).
2. Here \( \vec{a}\cdot\vec{b} = 0 \) and \( \vec{a}\cdot\vec{a} = 1 \), so the result is \( -\vec{b} \).

Teacher's Note:
1. Use the vector triple product expansion (BAC-CAB).
2. Geometrically the result is perpendicular to a, opposite to b.

Answer: (b) 7

Explanation:
1. \( {}^nP_3 - {}^nC_3 = 5\,{}^nC_3 > 100 \), so \( {}^nC_3 > 20 \).
2. \( {}^6C_3 = 20 \) (not greater than 20), while \( {}^7C_3 = 35 \).
3. The least value is n = 7.

Teacher's Note:
1. nP3 = 6 × nC3.
2. Test consecutive values of n.

Answer: (a) 2, 1

Explanation:
1. The family is \( x^2 + (y-k)^2 = r^2 \) with two arbitrary constants k and r, so we differentiate twice.
2. Differentiating once gives \( x + (y-k)y_1 = 0 \); differentiating again gives \( 1 + y_1^2 + (y-k)y_2 = 0 \). Eliminating k (using \( y-k = -x/y_1 \)) gives \( y_1(1+y_1^2) = x\,y_2 \)
3. Order = 2 and degree = 1.

Teacher's Note:
1. Order = number of independent arbitrary constants.
2. The highest derivative appears linearly.

Answer: (c) \( \frac{\pi}{2\sqrt{2}}\log_e(\sqrt{2}+1) \)

Explanation:
1. \( \sqrt{1+\sin 2x} = \sin x + \cos x \) on \( [0, \pi/2] \).
2. Using \( I = \int_0^a f(x)dx = \int_0^a f(a-x)dx \): \( 2I = \frac{\pi}{2}\int_0^{\pi/2}\frac{dx}{\sin x+\cos x} \).
3. \( \int_0^{\pi/2}\frac{dx}{\sin x+\cos x} = \sqrt{2}\ln(\sqrt{2}+1) \), so \( I = \frac{\pi}{2\sqrt{2}}\ln(\sqrt{2}+1)\approx0.979 \).

Teacher's Note:
1. Use the property of definite integrals f(x) to f(a - x).
2. The result was checked numerically.

Answer: (d) \( -1 \le \lambda \le -\frac{3}{4} \)

Explanation:
1. Let \( s = \sin^2\theta \in [0,1] \). Then \( \cos^4\theta + \sin^2\theta = (1-s)^2 + s = s^2 - s + 1 \).
2. This has minimum \( \frac{3}{4} \) (at s = 1/2) and maximum 1 (at s = 0 or 1).
3. So \( -\lambda \in [\frac{3}{4}, 1] \), i.e. \( -1 \le \lambda \le -\frac{3}{4} \).

Teacher's Note:
1. Convert to a quadratic in sin²θ and use its range on [0, 1].
2. Find the vertex and the endpoints.

Answer: (d) \( 1 - \frac{\pi}{4} \)

Explanation:
1. \( 1-\sin 2 = (\cos 1-\sin 1)^2 \) and \( \cos 2 = (\cos 1-\sin 1)(\cos 1+\sin 1) \).
2. So \( \frac{\sin 2-1}{\cos 2} = \frac{\sin 1-\cos 1}{\sin 1+\cos 1} = \frac{\tan 1-1}{\tan 1+1} = \tan\left(1-\frac{\pi}{4}\right) \).
3. Since \( 1-\frac{\pi}{4}\approx0.2146 \) lies in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), the principal value is \( 1-\frac{\pi}{4} \).

Teacher's Note:
1. Check: the numerical value is 0.2146, which equals 1 − π/4.
2. The principal value of tan⁻¹ lies between −π/2 and π/2.

Answer: (b) odd

Explanation:
1. The integrand \( g(t) = \frac{t^2}{e^t+e^{-t}} \) is an even function.
2. The integral of an even function from 0 to x is odd: \( f(-x) = -f(x) \).

Teacher's Note:
1. Integral from 0 of an even function is odd; of an odd function is even.
2. The function is not periodic because it increases.

Answer: (a) 2

Explanation:
1. At (0, 1) the curve has a corner: the right-hand slope is +1 and the left-hand slope is −1.
2. The lines \( y = 1+x \) and \( y = 1-x \) are the two one-sided tangents, so there are 2 tangents.

Teacher's Note:
1. At a corner the curve has two one-sided tangents.
2. A smooth point has just one.

Answer: (b) \( \sqrt{2\sqrt{2}-2} : 1 \)

Explanation:
1. Centres \( (\lambda, 0) \) and \( (0, -\lambda) \); radii \( \sqrt{\lambda^2-c^2} \) and \( |\lambda| \).
2. Three common tangents means external contact: \( \sqrt{2}|\lambda| = \sqrt{\lambda^2-c^2} + |\lambda| \).
3. Squaring \( (\sqrt{2}-1)|\lambda| = \sqrt{\lambda^2-c^2} \) gives \( c^2 = (2\sqrt{2}-2)\lambda^2 \).

Teacher's Note:
1. Three common tangents occur when the circles touch externally.
2. Distance between centres = sum of radii.

Answer: (a) f(x) is continuous at x = 2

Explanation:
1. At x = 2: f(2) = 2, and the right limit is [x] = 2 for x slightly above 2, so f is continuous at 2.
2. Left slope is 1 but right slope is 0, so f is not differentiable at 2.
3. At x = 3, [x] jumps from 2 to 3, so f is not continuous in [−2, 4].

Teacher's Note:
1. Check one-sided limits and slopes at the joining point.
2. Greatest integer functions jump at integers.

Answer: (a) f(x) is odd and \( f(x_1) + f(x_2) = f\left(\frac{x_1+x_2}{1+x_1x_2}\right) \)

Explanation:
1. \( f(-x) = \log\frac{1-x}{1+x} = -f(x) \), so f is odd.
2. \( f(x_1)+f(x_2) = \log\frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \), and \( \frac{1+u}{1-u} \) with \( u = \frac{x_1+x_2}{1+x_1x_2} \) gives the same value.
3. So the first option is correct.

Teacher's Note:
1. The log of a reciprocal flips the sign.
2. This is the tanh-addition identity.

Answer: (c) \( x^2 - 3y^2 - 2x + 10 = 0 \)

Explanation:
1. Complete the square: \( (x-1)^2 - 3y^2 = 9 \).
2. The conjugate hyperbola is \( (x-1)^2 - 3y^2 = -9 \), i.e. \( x^2 - 3y^2 - 2x + 10 = 0 \).

Teacher's Note:
1. The conjugate is obtained by changing the sign of the constant on the right side.
2. Write the hyperbola in standard form first.

Answer: (c) 6

Explanation:
1. By symmetry/convexity the minimum is at the equilateral triangle, A = B = C = 60 degrees.
2. Each secant is 2, so the sum is 6.

Teacher's Note:
1. sec is convex on (0, π/2), so Jensen's inequality applies.
2. Equality holds for the equilateral triangle.

Answer: (d) 2

Explanation:
1. Put \( y = x^2 \): \( \frac{x^2 + \sin x^2}{x^4 + \sin^2 x} \).
2. Divide by \( x^2 \): numerator \( \to 1 + 1 = 2 \); denominator \( \to x^2 + 1 \to 1 \).
3. The limit is 2.

Teacher's Note:
1. Use sin(u)/u → 1.
2. Divide numerator and denominator by x².

Answer: (b) 1

Explanation:
1. The normal of the plane is the direction (3, 5, 7).
2. The plane contains the second line, so its direction (1, −2, k) is perpendicular to the normal.
3. \( 3 - 10 + 7k = 0 \Rightarrow k = 1 \).

Teacher's Note:
1. A line lies in a plane only if its direction is perpendicular to the normal.
2. Dot product equals zero.

Answer: (c) \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \)

Explanation:
1. \( g(f(x)) = (\sin x+\cos x)^2 - 1 = \sin 2x \).
2. \( \sin 2x \) is one-one only on intervals where it is monotonic, such as \( 2x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), i.e. \( x \in [-\frac{\pi}{4}, \frac{\pi}{4}] \).

Teacher's Note:
1. A function is invertible where it is strictly monotonic.
2. sin 2x increases on [-π/4, π/4].

Answer: (b) (−5, 2)

Explanation:
1. The quadratic \( x^2 + 2ax + 10 - 3a \) must be positive for all x, so its discriminant is negative.
2. \( 4a^2 - 4(10-3a) < 0 \Rightarrow a^2 + 3a - 10 < 0 \Rightarrow (a+5)(a-2) < 0 \).
3. So \( a \in (-5, 2) \).

Teacher's Note:
1. Positive for all x means D < 0 with a positive leading coefficient.
2. Solve the resulting quadratic inequality.

Answer: (d) (−2, 2)

Explanation:
1. \( y^2 - 4y = 2x \Rightarrow (y-2)^2 = 2x + 4 = 2(x+2) \).
2. This is a parabola \( Y^2 = 2X \) with \( X = x+2,\ Y = y-2 \), so the vertex is (−2, 2).

Teacher's Note:
1. Complete the square in y.
2. Vertex is where X = Y = 0.

Answer: (a) 1001

Explanation:
1. Let each person get \( 2y_i + 1 \) things. Then \( \sum(2y_i+1) = 25 \Rightarrow \sum y_i = 10 \).
2. The number of non-negative solutions is \( {}^{14}C_4 = 1001 \).

Teacher's Note:
1. Substitution for odd numbers: 2y + 1.
2. Stars and bars gives C(n+r-1, r-1).

Answer: (a) 11

Explanation:
1. By Bayes' theorem: \( p = \frac{\frac{13}{52}\cdot\frac{{}^{12}C_2}{{}^{51}C_2}}{\frac{13}{52}\cdot\frac{{}^{12}C_2}{{}^{51}C_2}+\frac{39}{52}\cdot\frac{{}^{13}C_2}{{}^{51}C_2}} \).
2. \( p = \frac{13\times66}{13\times66 + 39\times78} = \frac{858}{3900} = \frac{11}{50} \).
3. So 50p = 11.

Teacher's Note:
1. Condition on the type of the lost card.
2. Both spades drawn is the observed event.