NDA & NA (II) 2016 — Mathematics (Full Question Paper)

Answer: (b) 23

Listing all fractions p/q with p and q from 1 to 6 gives 36 combinations, but many repeat the same value. Counting only the distinct values leaves exactly 23 numbers in the set.

Answer: (a) (0,2)

At x=0, f(0)=c, which is positive since c>0. At x=2, f(2)=4a-2b+c, and the given condition 4a+c<2b makes this negative. Since f changes sign between 0 and 2, a root lies in that interval.

Answer: (c) Both 1 and 2

Solving the inequalities gives A=(-7,1) and B is everything outside [-7,-2]. Intersecting these confirms A∩B is exactly -2<x<1, and the set difference A minus B gives the -7 to -2 range in statement 2.

Answer: (d) 1/40

For a 3×3 matrix, det(2A) = 2^3 times det(A), giving 8 times 5, or 40. The determinant of the inverse of a matrix is just the reciprocal, giving 1/40.

Answer: (d) 0

Since omega cubed equals 1, the exponents 100, 200, and 300 reduce to 1, 2, and 0 respectively. This gives omega + omega^2 + 1, which always equals zero for a cube root of unity.

Answer: (d) |z| = 1

Writing out the real part of (z-1)/(z+1) and setting it to zero simplifies to the condition x^2+y^2=1. This is exactly the equation for |z|=1.

Answer: (d) [ax+hy+gz hx+by+fz gx+fy+cz]

Multiplying a row vector by a matrix produces another row vector. Carrying out the multiplication term by term gives exactly this three-entry row.

Answer: (c) 5

The number of triangles from 15 points, excluding those formed by n collinear points, is C(15,3) minus C(n,3). Setting this equal to 445 and testing values shows n=5 is the only solution.

Answer: (a) 0

The two bracketed terms are complex conjugates written in polar form using a 30-degree angle. Raising them to the 107th power and adding them cancels the imaginary parts completely, leaving a real number with zero imaginary part.

Answer: (d) -1<k<3

Factoring the equation shows its two roots are always k+2 and k-2. Requiring both roots to stay strictly between -3 and 5 forces k to lie strictly between -1 and 3.

Answer: (c) Three

Splitting the equation into real and imaginary parts shows the imaginary part forces x=0. Solving the remaining real equation for y then gives three distinct complex solutions: 0, i, and -i.

Answer: (d) Infinitely many

One valid geometric progression is 27, 18, 12, 8, with common ratio 2/3. Because the three given numbers can be placed at increasingly spaced-out positions in a GP for infinitely many different ratios, infinitely many such progressions exist.

Answer: (c) {(3,3),(3,5),(5,3),(5,5)}

Listing every pair (a,b) with a<b gives the relation R. Composing R with its inverse and simplifying leaves exactly these four ordered pairs.

Answer: (d) No number can be formed

The digits 0, 1, 2, 3, and 4 add up to 10, which is not divisible by 3. Since a five-digit number here must use all five digits, no arrangement can ever be divisible by 3.

Answer: (a) 52C4

Rewriting the summation term by term and repeatedly applying the combination addition rule (Pascal’s rule) collapses the whole expression down to a single combination. The final simplified value is exactly 52 choose 4.

Answer: (b) 9

Since a, x, y, z, b are five terms in AP, the middle term y equals 5 from the given sum. The full HP condition then pins down a and b as the roots 1 and 9, so their product ab is 9.

Answer: (b) 105

With a=1 and b=9 fixed by the AP-HP relations, the five-term AP running from 1 to 9 has common difference 2. This makes x, y, z equal to 3, 5, and 7, whose product is 105.

Answer: (c) 243/35

Since a=1 and b=9, the reciprocals 1/a=1 and 1/b=1/9 form the outer terms of a five-term AP of reciprocals. Working out p, q, r from this AP gives a product of exactly 243/35.

Answer: (a) 8/5

Writing the first, fourth, and fifth AP terms in terms of the common difference and maximizing their product gives a clean cubic in d. Checking the critical points for d greater than 1 shows the true maximum occurs at d=8/5.

Answer: (b) -6

Using the common difference of 8/5 found above, the sixth term being 2 fixes the first term. Substituting back gives a first term of -6.

Answer: (c) 35

Expanding the given 3×3 determinant as a cubic in x and reading off the coefficient of x gives the value of c. This calculation yields c=35.

Answer: (b) 63

The same determinant expansion gives all four coefficients a, b, c, and d of the cubic. Adding them together gives a total of 63.

Answer: (a) One

Setting the sum of the AP of interior angles equal to the polygon angle-sum formula gives a quadratic with two algebraic roots, 9 and 16. The n=16 case is thrown out because it forces a negative exterior angle, which is geometrically impossible, leaving only n=9.

Answer: (a) 160 degrees only

With n=9 confirmed as the only valid polygon, the largest angle is the smallest angle plus eight times the common difference. This gives 120+40=160 degrees.

Answer: (c) 1

Substituting the two given matrices into m·cos(theta) minus n·sin(theta) produces a standard rotation-style matrix. Its determinant works out to exactly 1 for every value of theta.

Answer: (d) 1, 2 and 3

Multiplying the matrices f(theta) and f(phi) genuinely reproduces f(theta+phi), confirming statement 1. Its determinant is 1, confirming statement 2, and since det f(x) is always the constant 1, it trivially counts as an even function too.

Answer: (a) 1 and 2 only

The determinant of the coefficient matrix is zero, so a unique solution is never possible for any value of k. Checking k=15 gives an inconsistent system (no solution) and k=20 gives a consistent one (infinitely many solutions), matching statements 1 and 2.

Answer: (d) Neither 1 nor 2

Direct matrix computation shows AB times A-inverse B-inverse is not the identity matrix, so statement 1 fails. Likewise, (AB)-inverse does not equal A-inverse times B-inverse, since matrix inversion reverses the order of multiplication, so statement 2 fails too.

Answer: (b) ln y is the AM of ln x, ln x, ln x and ln z

Taking logarithms of both sides of the given identity and simplifying reduces it to a clean linear relation. That relation shows ln y equals the average of three copies of ln x and one copy of ln z.

Answer: (b) (11101011)_2

Repeatedly dividing 235 by 2 and recording the remainders builds up its binary digits from the bottom up. This process gives the binary number 11101011.

Answer: (d) a^2 >= 1/2

The discriminant of the quadratic in terms of a must be non-negative for real roots. Working through the inequality shows this only happens when a squared is at least 1/2.

Answer: (a) a^2 < 1/2

Using the sum and product of the roots to rewrite 1/alpha^2+1/beta^2 in terms of a gives a simple expression. Setting this expression below 1 and solving shows the condition is exactly a squared less than 1/2.

Answer: (b) omega

Using the identity 1+omega+omega^2=0, both the numerator and denominator simplify neatly, and their ratio reduces to omega^2. Taking its square root, using the standard convention for these cube-root-of-unity problems, gives back omega itself.

Answer: (a) 10

Failing all four subjects is the complement of passing at least one, so the minimum failure percentage occurs when the pass percentages overlap as much as possible. Using this best-case overlap, the smallest possible value of x works out to 10.

Answer: (b) lambda=5 and mu=9

The system’s coefficient determinant vanishes exactly when lambda=5. Checking consistency of the equations at that point shows mu must also equal 9 for infinitely many solutions to exist.

Answer: (c) lambda != 5 and mu has any real value

A unique solution needs the coefficient determinant to be nonzero, which happens for every lambda except 5. Since mu never affects this determinant, it can take any real value.

Answer: (c) 2240

Counting all valid four-digit numbers with distinct digits and an odd last digit, while excluding a leading zero, gives a total of 2240 such numbers.

Answer: (c) 10

The partial sums of this geometric series approach 2 but never reach it. Checking successive values shows the sum first drops below 2 minus 1/1000 right at n=10, the largest value satisfying the condition.

Answer: (c) 2

Since -2 is a root of the first equation, substituting it in pins down alpha as exactly 2.

Answer: (a) 1

Using the sum of roots of the second equation gives beta as 1/2, and the GP condition on beta, 2, 2m then gives m=4. Multiplying beta by the square root of m gives exactly 1.

Answer: (c) 1 and 3 only

Expanding sinA+2sin2A+sin3A using standard trigonometric identities matches the first and third given expressions exactly. The second expression, when checked against the original sum, does not simplify to the same thing.

Answer: (a) 1/16

Evaluating sin70·sin50 and cos60·cos80 directly and multiplying the two products together gives exactly 1/16.

Answer: (a) 0

Since each sine term can be at most 1, the only way four of them can sum to exactly 4 is if every angle is 90 degrees. At 90 degrees the cosine of each angle is zero, so their sum is also zero.

Answer: (d) 1/8

Multiplying out the four cosine-based factors using known trigonometric values for these specific eighth-of-pi angles simplifies neatly to 1/8.

Answer: (a) x^2+y^2-z^2

Since x·cos(theta)+y·sin(theta)=z, squaring the complementary combination x·sin(theta)-y·cos(theta) and using the identity for sin squared plus cos squared shows it equals x squared plus y squared minus z squared.

Answer: (d) 0.28

Using the double-angle cosine formula, cos(2·cos^-1(0.8)) equals 2 times 0.8 squared minus 1. This works out to 0.28.

Answer: (b) h cotp/(cotp-cotq)

Setting up the horizontal distance to the hill from both the top and bottom of the building and solving the two equations together gives the hill’s height. The result simplifies to h·cot(p) divided by (cot(p) minus cot(q)).

Answer: (a) (sqrt(3+sqrt5)+sqrt(5-sqrt5))/4

Writing sin81 as cos9 and building it up from the known sin18 value through half-angle and sum identities leads to this exact radical expression, matching the numerical value of sin81 precisely.

Answer: (b) 4500(sqrt3-1)/sqrt3

The horizontal distances to the boat at 60 degrees and 45 degrees of depression can be found from the 150 m cliff height. The difference between these distances, converted from a 2-minute interval to metres per hour, gives this exact speed.

Answer: (b) -sqrt3

Converting all the given angles using complementary-angle identities and simplifying the resulting expression numerically gives a value of negative square root of 3.

Answer: (c) (0, 2sqrt3) or (3, -sqrt3)

An equilateral triangle can be completed on either side of the given base, so there are two valid positions for the third vertex. Working out both rotations of the base point gives exactly these two coordinate pairs.

Answer: (a) 2x+4y-11=0

The perpendicular bisector passes through the midpoint of the segment and is perpendicular to it. Working out this line’s equation and clearing fractions gives 2x+4y-11=0.

Answer: (d) 5sqrt2 units

Solving the two given line equations together locates the circle’s centre. Measuring the distance from this centre to the point (2,4) gives a radius of 5 times the square root of 2.

Answer: (a) x^2/25 – y^2/20 = 1

Using the latus rectum and eccentricity formulas for a hyperbola together lets you solve directly for a squared and b squared. This gives a squared equal to 25 and b squared equal to 20.

Answer: (c) |a|<1

The condition |x+y|<2 becomes -2<2a<2 when x=y=a. Dividing through by 2 gives the simple condition |a|<1.

Answer: (d) 4x-3y+2=0

Finding where the two given lines intersect, then building a new line through that point with a slope perpendicular to 3x+4y=10, gives the equation 4x-3y+2=0.

Answer: (b) 2 and 3 only

Setting the distance formula from (a,b) to the line equal to 1 and solving the resulting absolute value equation gives two possible linear conditions. These match statements 2 and 3 exactly, while statement 1 describes an unrelated line.

Answer: (a) 5 units

Substituting the line equation into the ellipse equation and solving gives the two intersection points. The distance between these two points comes out to exactly 5 units.

Answer: (d) (1,1)

The line through (-3,5) with an x-intercept of 2 has a slope of -1, giving the equation x+y=2. Dropping a perpendicular from (3,3) onto this line lands exactly at the point (1,1).

Answer: (a) sqrt2

A rectangular hyperbola has equal transverse and conjugate axes, which forces its eccentricity to always equal the square root of 2.

Answer: (d) 1, 2 and 3

Reflecting the point P across the given plane using the standard image-point formula gives Q exactly at (4,-3,-1), confirming statement 1. The distance PQ works out to about 8.25 units, more than 8, confirming statement 2, and the midpoint (1,-1,-3) genuinely lies on the plane, confirming statement 3.

Answer: (c) Both 1 and 2

The direction ratios of PQ, found from Q minus P, simplify to the same proportion as <3,-2,2>, confirming statement 1. The sum of the squares of any line’s direction cosines is always exactly 1 by definition, confirming statement 2.

Answer: (c) <1,-4,3>

The line of intersection of two planes runs perpendicular to both planes’ normal vectors, so its direction is their cross product. Computing this cross product gives a direction proportional to <1,-4,3>.

Answer: (a) (x-5)/-1 = (y+6)/4 = (z-7)/-3

Since line L is parallel to both given planes, it must run along the same direction found for their line of intersection. Combining this direction with the point (5,-6,7) gives exactly this symmetric line equation.

Answer: (a) 3(i+j)/2

Vector c is the projection of b onto the direction of a, found using the standard projection formula. Carrying out this calculation gives c equal to 3 times (i+j) over 2.

Answer: (d) Neither 1 nor 2

Subtracting c from b gives vector d directly, with components 3/2, -3/2, and 4. Checking these against the two proposed equations shows neither y-x=4 nor 2z-3=0 actually holds.

Answer: (b) -166

Squaring the magnitude of a+b+c, which is zero, and expanding it in terms of the individual vector magnitudes gives a direct equation for the sum of the dot products. Solving this equation gives a value of -166.

Answer: (c) 60 degrees

Using the same a+b+c=0 relation together with the given magnitude of c isolates the dot product of a and b. Converting this dot product into an angle using the cosine formula gives exactly 60 degrees.

Answer: (b) p^2

Placing the right angle at C and writing out each vector explicitly, the whole expression simplifies using the Pythagorean relation between the triangle’s sides. The final result is simply p squared.

Answer: (c) 2i-7j-2k

The moment of a force about a point is the cross product of the position vector, from that point to where the force acts, with the force itself. Carrying out this cross product gives 2i-7j-2k.

Answer: (a) (-infinity, 0)

For the expression under the square root to be positive, |x| must be strictly greater than x. This is only true when x is negative, giving the domain as all negative real numbers.

Answer: (b) 3 only

The limit of f(x) as x approaches 1 clearly exists and equals 3, so statement 1 is false. The left- and right-hand derivatives at 0 differ, so f is not differentiable there, making statement 2 false, but the two one-sided limits at 0 both equal 2, matching f(0), so f is continuous there, confirming statement 3.

Answer: (a) A

For any positive x the function always equals 2, and for any negative x it always equals 0. Since these are each constant on their own separate piece of the domain, and zero itself is excluded from the domain, the function is continuous everywhere it is actually defined.

Answer: (b) It has local minimum at x=0.

The function’s value is zero at x=0, but takes small positive values on both sides of it. This means x=0 is a local minimum point, not a maximum.

Answer: (b) 8/3 square units

The condition |y|=1-x^2 describes the region between the upward parabola y=1-x^2 and its mirror image y=x^2-1. Integrating the vertical gap between these two curves from -1 to 1 gives an area of 8/3 square units.

Answer: (c) Both 1 and 2

The derivative of the first piece, 3x^2+12x-1, stays positive throughout [-1,2], confirming it is increasing there. The second piece, 37-x, has a constant negative slope, confirming it is decreasing on (2,3].

Answer: (a) 1 and 2 only

Both pieces of the function equal 35 at x=2, so f is continuous there, confirming statement 1. Since f rises to 35 at x=2 and then falls afterward, this point is genuinely the function’s greatest value, confirming statement 2, but the two pieces have different slopes there, so f is not differentiable at x=2.

Answer: (a) 0

For x greater than 1, both absolute values simplify so that |x|-|x-1| becomes the constant 1, making the whole function constant. A constant function’s rate of change is always zero.

Answer: (d) 8x-4

Between 0 and 1, |x|-|x-1| simplifies to 2x-1, so the function becomes (2x-1) squared. Differentiating this expression gives 8x-4 in this range.

Answer: (a) 1 only

Using the piecewise formulas, f(-2) and f(5) both work out to 1, confirming the first equation. But summing the corresponding derivative values at -2, 0.5, and 3 does not add up to 4, so the second equation fails.

Answer: (c) g(x) is continuous at x=0, but f(x) is not continuous at x=0.

The greatest integer function jumps abruptly at every integer, including 0, so f is discontinuous there. The sine function has no such jump anywhere, including at 0, so g remains continuous there.

Answer: (d) lim(x to 0+) (fog)(x) = lim(x to 0+) (gof)(x)

Working out the right-hand limits of both composite functions as x approaches 0 from above shows they both equal 0. The corresponding left-hand limits, and the two-sided limits, do not match up the same way.

Answer: (c) 1 and 3 only

Applying the floor function twice always gives the same result as applying it once, since a floor value is already an integer, confirming statement 1. Composing sine with the floor of sine only ever produces three distinct values, confirming statement 3, but applying sine twice equals sine itself at every multiple of pi, not just at zero, making statement 2 false.

Answer: (b) f(x) is a strictly increasing function in (0,x).

Differentiating this function and checking its sign at several sample points shows the derivative stays positive throughout the positive real line. A positive derivative everywhere means the function is strictly increasing there.

Answer: (d) Neither 1 nor 2

The limit of this function as x approaches 0 from the right is 1, but the function is defined to equal 0 at x=0 itself, so it is not right continuous there, making statement 1 false. The function itself is a smooth, well-behaved expression for every x greater than 0, including at x=1, so it is genuinely continuous there, making statement 2 also false.

Answer: (b) (-2,-8)

The distance from a general point on the parabola to the given line can be written as a simple expression in terms of x. Minimizing this expression shows the closest point occurs at x=-2, giving the coordinates (-2,-8).

Answer: (c) 1/sqrt10

Plugging the closest point found above back into the point-to-line distance formula gives the shortest possible distance. This works out to exactly 1 over the square root of 10.

Answer: (d) Neither 1 nor 2

Working out g(x) piece by piece from its definition shows its one-sided derivatives at x=0 do not match, so it is not differentiable there. The one-sided derivatives at x=2 also disagree, so g is not differentiable there either.

Answer: (a) -1

On the interval containing x=-2, g(x) simplifies to the straight-line expression -x. Its derivative is a constant -1 everywhere on this piece, including at x=-2.

Answer: (d) 1, 2 and 3

Checking the one-sided limits of g(x) at 0 and at 2 shows they match the function’s actual value at each point, confirming continuity there. Since g(x) is a smooth straight line throughout the region containing -1, it is automatically continuous there too.

Answer: (a) 2f(1)

This is a standard reciprocal-substitution identity: the relation between f’ at x and f’ at 1/x forces the definite integral over the symmetric interval to collapse down to twice the function’s value at the endpoint 1.

Answer: (c) sqrt(x^2+1/x^2+1) + c

Differentiating this candidate answer and comparing it to the original expression under the integral sign shows they match exactly. This confirms it as the correct antiderivative.

Answer: (a) 1, 1

This equation is a disguised addition formula, similar to the identity behind inverse sine, and it reduces to a differential equation with only a single derivative appearing to the first power. That makes both its order and its degree equal to 1.

Answer: (b) Parabola

Solving the differential equation dy/dx=2y/x with the condition that it passes through (1,1) gives the curve y=x^2. This is the equation of a parabola.

Answer: (a) 3 only

Solving the given differential equation with y(1)=1 leads to two algebraic branches, one giving y=3 at x=-3 and the other giving y=-1. Since the problem requires y to stay positive throughout, only the branch giving y=3 is valid.

Answer: (b) 2

The integral term in this equation must be removed by differentiating the whole equation with respect to y. This differentiation introduces a second derivative, raising the equation’s order to 2.

Answer: (c) (y – x dy/dx)^2 = 1 + (dy/dx)^2

Writing a general line at unit distance from the origin in trigonometric form and eliminating the angle by differentiating leads to this exact relation between y, x, and the slope.

Answer: (b) (x-secx) e^sinx + c

Differentiating this candidate expression and comparing it to the original integrand confirms they are identical. This confirms it as the correct antiderivative.

Answer: (b) 1/2

Evaluating this definite integral directly gives a result in terms of the inverse tangent of 1/2, which is the same as the inverse cotangent of 2 scaled by a constant. Matching the two expressions shows that constant, k, equals 1/2.

Answer: (d) 232/5

Throughout the interval from 1 to 3, x^4 is always at least 1, so the absolute value simplifies to x^4 minus 1 without any sign changes. Integrating this expression over the interval gives exactly 232/5.

Answer: (d) 25/216

Checking every one of the 216 possible outcomes from three rolls of this six-sided special die and counting how many sum to zero gives 25 favourable outcomes. This gives a probability of 25 over 216.

Answer: (d) 1-(3/4)^7

The probability of no rain at all across seven independent days is (3/4) raised to the seventh power. At least one rainy day is the complementary event, giving 1 minus that quantity.

Answer: (b) 0.91

The chance the salesman fails with both customers is 0.3 times 0.3, or 0.09. Selling to at least one of them is the complementary event, giving 0.91.

Answer: (a) m(1+n)=1

Being successful means passing test I together with at least one of tests II or III, so the overall probability is m times the probability of passing II or III. Setting this whole expression equal to 1/2 and simplifying leads directly to m times (1+n) equalling 1.

Answer: (a) 209/343

Converting each candidate’s odds into a probability of solving correctly and summing the probabilities of exactly two, or all three, solving it correctly gives a combined probability of 209 over 343.

Answer: (d) 1, 2 and 3

In any symmetric distribution the mean and median genuinely coincide, and the range is always defined as the maximum value minus the minimum value. The total area under a histogram’s bars is also a standard, proven fact to equal the area under its corresponding frequency polygon.

Answer: (d) Mean and mode

Correcting the wrongly recorded value shifts the overall total, so the mean genuinely changes. The most frequent value also shifts from 16 to 18, changing the mode, but the middle value of the ordered list stays the same, leaving the median unaffected.

Answer: (b) y=1.02x+8.74

Using the standard regression formulas with the given sums for x, y, and their squares and cross-products gives a slope near 1.02 and an intercept near 8.74. This matches the regression line of y on x among the given options.

Answer: (d) Group A is less variable than Group B because Group A’s coefficient of variation is smaller.

Because the two groups have different means, comparing their raw standard deviations directly is not meaningful. Computing each group’s coefficient of variation (standard deviation relative to its own mean) shows Group A’s is genuinely smaller.

Answer: (b) 2 and 3 only

Class intervals in a grouped frequency table must actually be mutually exclusive, so statement 1 is false. They genuinely need to be exhaustive, covering the entire data range, and they are allowed to have unequal widths, confirming statements 2 and 3.

Answer: (c) 1023/1024

The probability that none of the five patients is cured is (1/4) raised to the fifth power, since each has a 25% chance of failure. At least one cured is the complementary event, giving 1023 over 1024.

Answer: (a) 3/7

Working out P(A and B) from the given conditional probability, then P(A union B), gives the probability that neither A nor B occurs. Dividing this by P(B complement) gives the conditional probability of A complement given B complement as 3/7.

Answer: (c) 0.84

The machine keeps working only if all three parts stay non-defective. Multiplying the three individual non-defective probabilities, 0.98, 0.90, and 0.95, together gives approximately 0.84.

Answer: (c) 3/4

The chance that none of the three independent events occurs is the product of each event’s individual failure probability. Subtracting this product from 1 gives the probability that at least one event occurs, which comes out to 3/4.

Answer: (c) (sigma_x^2-sigma_y^2)/(sigma_x^2+sigma_y^2)

Since x and y are uncorrelated, the covariance between x+y and x-y simplifies to the difference of their variances. Dividing by the variances of x+y and x-y, which are both equal to the sum of the two original variances, gives this exact ratio.

Answer: (b) 41.5

Using the midpoint of each age group weighted by its frequency, the total of all midpoint-times-frequency values comes to 830 across 20 people. Dividing this total by 20 gives a mean age of 41.5.

Answer: (b) 0.5

The correlation coefficient is the covariance divided by the product of the two standard deviations. Using the given covariance and variances, this works out to 30 divided by 60, or 0.5.

Answer: (d) A∩(B’∪C’) = B’∩C’

Listing every outcome of three coin tosses and directly checking each of the four proposed set identities shows only this one holds true for every outcome in the sample space.

Answer: (d) 1/18

Scoring exactly 5 points in three matches is only possible with exactly two wins and one draw. Multiplying the relevant win and draw probabilities together, with the right combinatorial count for ordering, gives a probability of 1/18.