JEE Mathematics Previous Year Questions MCQs Set 04

Answer: (c) HP

Explanation:
1. \( x=\frac1{1-a},\ y=\frac1{1-b},\ z=\frac1{1-c} \), so \( \frac1x=1-a,\ \frac1y=1-b,\ \frac1z=1-c \).
2. Since a, b, c are in AP, so are \( 1-a,1-b,1-c \), so the reciprocals of x, y, z are in AP, which means x, y, z are in HP.

Teacher's Note:
1. Sum each infinite geometric series first.
2. Reciprocals in AP means the numbers are in HP.

Answer: (c) \( I-A \)

Explanation:
1. \( A=A^2+I\Rightarrow A^2-A+I=0\Rightarrow A(I-A)=I \).
2. Hence \( A^{-1}=I-A \).

Teacher's Note:
1. Factor the given matrix equation to get A times something equals I.
2. The something is the inverse.

Answer: (d) 1 : 2 : 3

Explanation:
1. \( 1^2+(\sqrt3)^2=4=2^2 \), so the triangle is right angled, with angles \( 30^\circ,60^\circ,90^\circ \).
2. The ratio of the angles is \( 1:2:3 \).

Teacher's Note:
1. Test the Pythagoras relation on the sides.
2. The side ratio 1 : root 3 : 2 gives a 30-60-90 triangle.

Answer: (a) \( a(b+c)\lt 0 \)

Explanation:
1. Sum \( -\frac{b}{a}>2 \) and product \( \frac{c}{a}\lt 2 \).
2. If \( a>0 \): \( b\lt -2a \) and \( c\lt 2a \), so \( b+c\lt 0 \), giving \( a(b+c)\lt 0 \).
3. If \( a\lt 0 \): \( b>-2a \) and \( c>2a \), so \( b+c>0 \), again \( a(b+c)\lt 0 \).

Teacher's Note:
1. Split into the cases a positive and a negative.
2. Multiplying an inequality by a negative number reverses it.

Answer: (a) y - x = 3

Explanation:
1. The hyperbola is \( \frac{x^2}{18}-\frac{y^2}{9}=1 \). A tangent \( y=x+c \) needs \( c^2=a^2m^2-b^2=18-9=9 \), so \( c=\pm3 \).
2. \( y=x+3 \) is \( y-x=3 \), which is option (a). The other lines are not tangents.

Teacher's Note:
1. Use the condition c squared = a squared m squared - b squared.
2. Compare each option with the lines y = x + 3 and y = x - 3.

Answer: (a) 24

Explanation:
1. \( y=50-x \) must be even and lie in \( B \) (so \( 2\le y\le100 \)), hence x is even.
2. \( x=2,4,\dots,48 \) gives \( y=48,\dots,2 \). For \( x=50 \), \( y=0\notin B \).
3. So there are 24 pairs.

Teacher's Note:
1. List the allowed values of x after fixing y = 50 - x.
2. Check the boundary values in both sets.

Answer: (a) \( \left(2,\frac18\right) \)

Explanation:
1. P lies on AB with \( AP:PB=1:3 \), so \( x=\frac34x_A \), i.e. \( x_A=\frac43x \).
2. The tangent at P meets the x-axis where \( x-\frac{y}{y'}=\frac43x \), so \( -\frac{y}{y'}=\frac x3 \).
3. Hence \( x\,y'+3y=0 \), so \( \frac{dy}{y}=-3\frac{dx}{x} \) and \( y=\frac{c}{x^3} \).
4. \( f(1)=1\Rightarrow c=1 \), so \( y=\frac1{x^3} \), which passes through \( \left(2,\frac18\right) \).

Teacher's Note:
1. Use the section formula to relate the point to the intercepts.
2. Solve the resulting variable separable equation.

Answer: (b) \( \left[-\frac\pi4,\frac\pi4\right] \)

Explanation:
1. \( g\{f(x)\}=(\sin x+\cos x)^2-1=\sin2x \).
2. \( \sin2x \) is strictly increasing for \( 2x\in\left[-\frac\pi2,\frac\pi2\right] \), i.e. \( x\in\left[-\frac\pi4,\frac\pi4\right] \), so it is one-one there.
3. On the other intervals it is not monotonic.

Teacher's Note:
1. Simplify the composite function first.
2. A function is invertible where it is strictly monotonic.

Answer: (a) 3

Explanation:
1. Rearrange: \( \frac{y\,dx-x\,dy}{y^2}=-dy\Rightarrow d\left(\frac xy\right)=-dy \).
2. Integrating, \( \frac xy=-y+C \). With \( x=1,y=1 \): \( C=2 \), so \( x=2y-y^2 \).
3. For \( x=-3 \): \( y^2-2y-3=0\Rightarrow y=3 \) (since \( y>0 \)).

Teacher's Note:
1. Recognise the exact differential of x over y.
2. Use the condition to find the constant.

Answer: (c) \( \hat i \)

Explanation:
1. Check option (c): \( \vec b=\hat i \). \( \vec a\cdot\vec b=1 \) and \( \vec a\times\hat i=(\hat i+\hat j+\hat k)\times\hat i=-\hat k+\hat j=\hat j-\hat k \), both conditions hold.
2. The other options fail: (a) gives \( \vec a\times\vec b=2\hat i-2\hat k \); (b) gives \( \vec a\times\vec b=-3\hat i+\hat j+2\hat k \); (d) gives \( \vec a\cdot\vec b=2 \).

Teacher's Note:
1. Verify each option against both given conditions.
2. The dot and cross product together determine b uniquely.

Answer: (a) 364

Explanation:
1. Put \( y_i=-x_i\ge1 \). Then \( y_1+y_2+y_3+y_4=15 \) with each \( y_i\ge1 \).
2. Number of solutions \( ={}^{14}C_3=364 \).

Teacher's Note:
1. Convert to positive integer solutions.
2. Use the stars and bars formula with each part at least 1.

Answer: (a) \( n(2c)^{1/n} \)

Explanation:
1. Apply AM-GM to the n numbers \( a_1,\dots,a_{n-1},2a_n \).
2. Their product is \( 2c \), so the sum is at least \( n(2c)^{1/n} \).

Teacher's Note:
1. Choose the n terms so that their product is constant.
2. Equality needs all the n terms equal.

Answer: (b) f(x) is continuous everywhere

Explanation:
1. sin x
2. is a composition of continuous functions, so f is continuous everywhere.
3. It is differentiable except at multiples of pi, so (c) is false, and f(0) = 1 exists, so (d) is false.

Teacher's Note:
1. Modulus of a continuous function is continuous.
2. Corners occur only where sin x = 0.

Answer: (c) 4

Explanation:
1. \( y'=(3-x)(3-3x) \), so critical points are \( x=1 \) (maximum, \( y=4 \)) and \( x=3 \) (minimum, \( y=0 \)).
2. Area \( =\int_1^3x(3-x)^2dx=\left[\frac{x^4}4-2x^3+\frac{9x^2}2\right]_1^3=\frac{27}4-\frac{11}{4}=4 \).

Teacher's Note:
1. Find the abscissae of maximum and minimum by differentiating.
2. Integrate between them.

Answer: (c) 3

Explanation:
1. \( |u+1|-|u-1| \) (with \( u=\frac x2 \)) is at most the distance between the points \( -1 \) and \( 1 \), which is 2, attained for \( u\ge1 \).
2. So the greatest value of f is \( 2+1=3 \).

Teacher's Note:
1. The difference of two moduli is bounded by the distance between the shifts.
2. Check the case u at least 1.

Answer: (c) \( \left(\frac12,3\right) \)

Explanation:
1. The point lies in the angle if \( \frac x2\lt y\lt 3x \) with \( x>0 \).
2. \( \frac a2\lt a^2\lt 3a \) with \( a>0 \) gives \( \frac12\lt a\lt 3 \).

Teacher's Note:
1. Write the region as two inequalities in x and y.
2. The condition x greater than 0 fixes the sign of a.

Answer: (b) f(x) is bijective

Explanation:
1. A function has an inverse exactly when it is one-one (injective) and onto (surjective), i.e. bijective.
2. Monotonic functions are one-one but need not be onto, and an injective function need not be onto.

Teacher's Note:
1. Invertible means bijective.
2. A monotonic function is invertible only on its range.

Answer: (c) \( \frac34 \)

Explanation:
1. Each particle chooses one of two directions, so there are \( 2^3=8 \) equally likely outcomes.
2. They avoid collision only if all move clockwise or all move anticlockwise: 2 outcomes.
3. Probability of collision \( =1-\frac28=\frac34 \).

Teacher's Note:
1. Count the outcomes in which nobody collides and subtract from one.
2. All same direction is the only safe case.

Answer: (a) 40

Explanation:
1. Total arrangements \( =\frac{6!}{3!\,2!}=60 \).
2. With the two N's together (treated as one letter): \( \frac{5!}{3!}=20 \).
3. Required number \( =60-20=40 \).

Teacher's Note:
1. Count the total and subtract the arrangements with the N's together.
2. Treat NN as a single block for the unwanted case.

Answer: (d) reflection in the line through the origin with slope \( \tan\frac\alpha2 \)

Explanation:
1. P has polar angle \( \theta \) and Q has polar angle \( \alpha-\theta \), both on the unit circle.
2. Reflection in a line through the origin at angle \( \varphi \) takes angle \( \theta \) to \( 2\varphi-\theta \).
3. Setting \( 2\varphi=\alpha \) gives \( \varphi=\frac\alpha2 \), slope \( \tan\frac\alpha2 \).

Teacher's Note:
1. Reflection in a line at angle phi maps polar angle theta to 2 phi minus theta.
2. Compare with alpha minus theta.

Answer: (b) 2

Explanation:
1. Normal from \( P(h,s) \): \( m^3+(2-h)m+s=0 \). So \( m_1+m_2+m_3=0 \), \( \sum m_1m_2=2-h \), \( m_1m_2m_3=-s \).
2. With \( m_1m_2=k \): \( m_3=-\frac sk \), \( m_1+m_2=\frac sk \), and \( k-\frac{s^2}{k^2}=2-h \), so \( s^2=k^2(h+k-2) \).
3. For this to be \( s^2=4h \) we need \( k^2=4 \) and \( k-2=0 \), so \( k=2 \).

Teacher's Note:
1. Use the cubic satisfied by the slopes of normals from a point.
2. Compare the locus with y squared equals 4x.

Answer: (c) \( \frac3{32} \)

Explanation:
1. Each person has 4 choices, so there are \( 4^4=256 \) equally likely outcomes.
2. All different choices: \( 4!=24 \) outcomes.
3. Probability \( =\frac{24}{256}=\frac3{32} \).

Teacher's Note:
1. Count the favourable arrangements as permutations.
2. Total outcomes are four choices for each of the four persons.

Answer: (a) one-to-one and onto

Explanation:
1. \( f'(x)=2+\cos x\ge1>0 \), so f is strictly increasing, hence one-to-one.
2. As \( x\to\pm\infty \), \( f(x)\to\pm\infty \) and f is continuous, so the range is all of R, so f is onto.

Teacher's Note:
1. Strictly increasing implies one-one.
2. A continuous function that tends to both infinities is onto R.

Answer: (c) 2

Explanation:
1. Expanding gives \( f(x)=(x-1)^2\left[1+(a^2+b^2+c^2+2)x\right] \).
2. Since \( a^2+b^2+c^2+2=0 \), \( f(x)=(x-1)^2 \), a polynomial of degree 2.

Teacher's Note:
1. Use row operations: add all rows and factor.
2. Substitute the given condition at the end.

Answer: (c) A is true, but R is false.

Explanation:
1. Since \( \tan x \) is increasing on \( \left(0,\frac\pi2\right) \), \( \tan\beta>\tan\alpha>0 \), so \( \frac{\tan\beta}{\tan\alpha}>1 \).
2. Also \( \frac\alpha\beta\lt 1 \), so the assertion is true.
3. \( f''>0 \) only says f is convex; f can still be decreasing (for example \( x^2 \) for negative x), so the reason is false.

Teacher's Note:
1. Convexity does not imply that a function is increasing.
2. Compare each ratio with 1.