ExamSpark Class 12 Mathematics

Mock Test 06 Performance Solutions

Subject: CBSE Class 12 Mathematics (NCERT Full-Syllabus Practice)

Total Score

--/20

Correct

--

Incorrect

--

Unattempted

--

Detailed Question Analysis

Q1. Let A={1,2,3,4}. The total number of bijective functions from set A to itself is:

Correct Answer: Option C (24)

Concept Explanation: Bijective functions matlab function strictly one-one aur onto dono hona chahiye. A function from a set of n elements to itself is bijective if and only if it is a permutation of the set. So, total bijective functions =n!=4!=24.

Q2. The principal value of tan-1(tan67 π ) is:

Correct Answer: Option B (6π)

Concept Explanation: Yeh ek classic trap hai! tan-1x ki principal value branch (-2π ,2π ) hoti hai, aur 67π is range ke bahar hai. Hum isko likhenge tan( π+6π )= tan(6 π ). So, correct principal value is 6π .

Q3. If A and B are symmetric matrices of the same order, then AB-BA is always a:

Correct Answer: Option B (Skew-symmetric matrix)

Concept Explanation: Let's take the transpose: (AB-BA)T=(AB)T-(BA)T=BTAT-ATBT. Kyunki dono symmetric hain, BT=B aur AT=A. Toh yeh ban gaya BA-AB=-(AB-BA). Jab XT=-X hota hai, toh matrix skew-symmetric hoti hai.

Q4. If a is a non-zero vector of magnitude a, and λ is a non-zero scalar, then λ a represents a unit vector if:

Correct Answer: Option D (a= ∣ λ ∣ 1)

Concept Explanation: Unit vector ka magnitude 1 hota hai. Isliye, ∣ λ a ∣ =1 ⟹∣ λ ∣∣ a ∣ =1. Hamein pata hai ∣ a ∣ =a, toh equation bani ∣ λ ∣ a=1 ⟹ a= ∣ λ ∣ 1 .

Q5. If A is a square matrix such that A2=A, then (I+A)3-7A is equal to:

Correct Answer: Option C (I)

Concept Explanation: Binomial expansion use karenge: (I+A)3=I3+3I2A+3IA2+A3. Since A2=A, we also get A3=A. Yeh expand hoke banega I+3A+3A+A=I+7A. Ab expression mein daalo: (I+7A)-7A=I.

Q6. The value of k for which the function f(x)=8x21-cos4x for x=0, and f(0)=k, is continuous at x=0 is:

Correct Answer: Option B (1)

Concept Explanation: Continuous hone ke liye limx→0 f(x)=f(0) hona chahiye. 1-cos4x=2sin2(2x). Limit banegi 8x22sin2(2x) =4x2sin2(2x) =(2xsin2x )2. Standard limit lim θ→0 θ sin θ =1 se, answer 12=1 aayega. So, k=1.

Q7. The Integrating Factor (I.F.) for the differential equation xdxdy -y=2x2 is:

Correct Answer: Option C (x1)

Concept Explanation: Pehle standard form dxdy +Py=Q mein laane ke liye x se divide karo: dxdy -x1 y=2x. Yahan P=-x1 . I.F. ka formula hai e∫Pdx=e∫-1/xdx=e-logx=elog(x-1)=x-1=x1 .

Q8. The value of the integral ∫ex(x1 -x21 )dx is:

Correct Answer: Option B (xex +C)

Concept Explanation: CBSE directly tests the competency form ∫ex[f(x)+f′(x)]dx=exf(x)+C. Yahan observe karo, agar f(x)=x1 hai, toh uska exact derivative f′(x)=-x21 hota hai. So direct answer is ex(x1 )+C.

Q9. In a Linear Programming Problem, the objective function is Z=4x+3y. The corner points of the bounded feasible region are (0,0),(0,40),(20,40),(60,20),(60,0). The maximum value of Z occurs at:

Correct Answer: Option C ((60,20))

Concept Explanation: Har corner point par Z ki value check karenge. Z(20,40)=80+120=200. Z(60,0)=240+0=240. Z(60,20)=240+60=300. Maximum value 300 hai, jo (60,20) par aati hai.

Q10. If A and B are two independent events such that P(A)=0.3 and P(B)=0.6, then P(A and not B) is:

Correct Answer: Option A (0.12)

Concept Explanation: P(A and not B) ko P(A∩B′) likhte hain. Kyunki events independent hain, P(A∩B′)=P(A)×P(B′). Hamein pata hai P(B′)=1-P(B)=1-0.6=0.4. Ab multiply kar do: 0.3×0.4=0.12.

Q11. A spherical balloon is being inflated such that its volume is increasing at a constant rate of 8 cm3/s. The rate of increase of its radius when the radius is 2 cm is:

Correct Answer: Option B (2π1 cm/s)

Concept Explanation: Volume V=34 π r3. Differentiate with respect to time t: dtdV =4 π r2dtdr . Given dtdV =8 and r=2. Values put karne par: 8=4π(4) dtdr ⟹ 8=16 π dtdr ⟹ dtdr =16 π8 =2π1 .

Q12. The area of the region bounded by the curve x=y2 and the y-axis, between y=1 and y=2 is:

Correct Answer: Option A (37 sq units)

Concept Explanation: Real-life geometry application. Jab y-axis ke respect mein area nikalna ho, toh hum ∫xdy use karte hain. Limits 1 se 2 hain. Area =∫12 y2dy=[3y3 ]12 =38 -31 =37 .

Q13. Two aircraft flight paths are represented by the straight lines r =(i^-j^ )+ λ(2 i^+k^) and r =(2i^-j^ )+ μ( i^+j^ -k^). To guarantee that these aircraft will never collide in mid-air, their shortest distance must be:

Correct Answer: Option C (Non-zero)

Concept Explanation: 3D Geometry application! Do lines (flight paths) space mein sirf tabhi intersect karti hain (collide hoti hain) jab unke beech ki shortest distance perfectly zero ho. Collision avoid karne ke liye, shortest distance strictly non-zero honi chahiye.

Q14. Assertion (A): The degree of the differential equation x(dxdy )3+y(dx2d2y )=ex is 1. Reason (R): The highest power of the highest order derivative is called the degree of the differential equation, provided it is a polynomial equation in derivatives.

Correct Answer: Option A (Both A and R are true and R is the correct explanation of A)

Concept Explanation: Is equation mein highest order derivative dx2d2y hai, aur uski total power 1 hai. Assertion bilkul sahi hai. Reason bhi perfectly degree ki definition de raha hai aur explain kar raha hai ki humne power 3 wali term ko kyu ignore kiya.

Q15. Assertion (A): If A is a square matrix of order 3 and ∣ A ∣ =4, then ∣ 2A ∣ =8. Reason (R): For a square matrix A of order n, and any scalar k, ∣ kA ∣ =kn ∣ A ∣ .

Correct Answer: Option D (A is false but R is true)

Concept Explanation: Reason ka rule perfectly correct hai ( ∣ kA ∣ =kn ∣ A ∣ ). Agar yeh rule Assertion pe lagayein: order n=3, k=2, aur ∣ A ∣ =4. Toh ∣ 2A ∣ =23×4=8×4=32 hona chahiye. Assertion keh raha hai value 8 hai, jo ki galat hai. Passage for Q16 & Q17: A marketing agency is designing a promotional rectangular poster. The central printed area must exactly be 50 cm2. The design requires blank margins at the top and bottom of 2 cm each, and at the left and right sides of 1 cm each. Let the width of the inner printed area be x cm.

Q16. The total area of the poster paper A(x), expressed as a function of x, is:

Correct Answer: Option A ((x+2)(x50 +4))

Concept Explanation: Printed area dimensions: width = x, height = y. We know xy=50 ⟹ y=x50 . Total poster ki width (margins add karke) =x+1+1=x+2. Total poster ki height =y+2+2=y+4=x50 +4. Total Area =width×height.

Q17. To minimize the cost of paper, the total area A(x) must be minimized. What width x of the printed area minimizes the total poster area?

Correct Answer: Option B (5 cm)

Concept Explanation: A(x)=(x+2)(x50 +4)=50+4x+x100 +8. Differentiating w.r.t x: A′(x)=4-x2100 . For maxima/minima, A′(x)=0 ⟹ 4=x2100 ⟹ x2=25 ⟹ x=5 cm.

Q18. For any 3×3 non-singular matrix A, if ∣ A ∣ =3, what is the exact value of ∣ adj(adjA) ∣ ?

Correct Answer: Option C (81)

Concept Explanation: A brilliant HOTS question based on advanced determinant properties. The direct formula is ∣ adj(adjA) ∣ = ∣ A ∣ (n-1)2. Yahan order n=3 hai. So power will be (3-1)2=22=4. Evaluating it: 34=81.

Q19. The value of the definite integral ∫03 ∣ x-2 ∣ dx is:

Correct Answer: Option B (25)

Concept Explanation: Modulus function hai, toh interval break karna padega jahan x-2=0 (i.e., at x=2). ∫02 (-(x-2))dx+∫23 (x-2)dx. Isko geometrically do triangles ke area ki tarah calculate karo: 21 (2)(2)+21 (1)(1)=2+0.5=2.5 or 25 .

Q20. The general solution of the differential equation dxdy =ex+y is:

Correct Answer: Option A (ex+e-y=C)

Concept Explanation: Highly repeated exam pattern! Expand the exponent: dxdy =ex ⋅ ey. Separation of variables karo: eydy =exdx ⟹ e-ydy=exdx. Dono side integrate karne par: -e-y=ex+C1 , which rearranges beautifully to ex+e-y=C.

ExamSpark Quick Mathematics Revision Signals

Quick Links & ExamSpark Resources

← Back to Global Scorecard