ISC Maths 2012 Class-12 Solved Previous Year Question Paper for practice. Step by step Solutions with section-A, B and C. Visit official website CISCE for detail information about ISC Board Class-12 Maths.

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## ISC Maths 2012 Class-12 Previous Year Question Papers Solved

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Section-A

Section-B

Section-C

Time Allowed: 3 Hours
Maximum Marks: 100

(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.)

• The Question Paper consists of three sections A, B and C.
• Candidates are required to attempt all questions from Section A and all questions either from Section B or Section C.
• Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.
• Section B: Internal choice has been provided in two questions of four marks each.
• Section C: Internal choice has been provided in two questions of four marks each.
• All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Mathematical tables and graph papers are provided.

### Section – A (80 Marks)

ISC Maths 2012 Class-12 Previous Year Question Papers Solved

Que 1:
(i) Solve for x if $\left(\begin{array}{c}{x^{2}} \\ {y^{2}}\end{array}\right)+2\left(\begin{array}{l}{2 x} \\ {3 y}\end{array}\right)=3\left(\frac{7}{-3}\right)$ [3] (ii) Prove that $\sec ^{2}\left(\tan ^{-1} 2\right)+\csc ^{2}\left(\cot ^{-1} 3\right)=15$ [3] (iii) Find the equation of the hyperbola whose Transverse and Conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13. [3] (iv) From the equations of the two regression lines, 4x + 3y + 7 = 0 and 3x + 4y + 8 = 0, find: [3] (a) Mean of x and y.
(b) Regression coefficients.
(c) Coefficient of correlation.
(v) Evalulate: $\int e^{x}(\tan x+\log \sec x) d x$ [3] (vi) Evaluate: [3]
(vii) Find the locus of the complex number, Z = x + iy given $\left|\frac{x+i y-2 i}{x+i y+2 i}\right|=\sqrt{2}$ [3] (viii) Evaluate: $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$ [3] (ix) Three persons A, B and C shoot to hit a target. If in trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. Find the probability that: [3]
(a) Exactly two persons hit the target.
(b) At least two persons hit the target.
(x) Solve the differential equation: [3] (xy2 + x)dx + (x2y + y) dy = 0
Solution 1:

Que 2:
(a) Using properties of determinants, prove that: [5]
$\left|\begin{array}{ccc}{a} & {a+b} & {a+b+c} \\ {2 a} & {3 a+2 b} & {4 a+3 b+2 c} \\ {3 a} & {6 a+3 b} & {10 a+6 b+3 c}\end{array}\right|=a^{3}$
(b) Find the product of the matrices A and B where: [5]
$A=\left(\begin{array}{ccc}{-5} & {1} & {3} \\ {7} & {1} & {-5} \\ {1} & {-1} & {1}\end{array}\right), B=\left(\begin{array}{lll}{1} & {1} & {2} \\ {3} & {2} & {1} \\ {2} & {1} & {3}\end{array}\right)$
Hence, solve the following equations by matrix method:
x + y + 2z = 1
3x + 2y + z = 7
2x + y + 3z = 2
Solution 2:

Que 3:
(a) Prove that: $\cos ^{-1} \frac{63}{65}+2 \tan ^{-1} \frac{1}{5}=\sin ^{-1} \frac{3}{5}$ [5] (b)

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(i) Write the Boolean expression corresponding to the circuit given below: [5]

(ii) Simplify the expression using laws of Boolean Algebra and construct the simplified circuit.
Solution 3:

(b)

(i) The statement using the given switching circuits is as:
CA + A(B + C) (C + A) (C + B) ….. (i)
using laws of Boolean Algebra, we have
CA + A(B + C) (C + A) (C + B) = (CA + AB + AC) (C + A) (C + B)
= (AC + AB + AC) (C + A) (C + B)
= ACC + ACA + ABC + ABA (C + B)
= AC + AC + ABC + AB (C + B)
= AC + ABC + ABC + AB
= AC + ABC + AB
= AC + AB (1 + C)
= AC + AB (1)
= AC + AB
= A(C + B)
Hence, the simplified switching network can be shown as in the figure.

Que 4:
(a) Verify Rolle’s theorem for the function: [5]
$f(x)=\log \left\{\frac{x^{2}+a b}{(a+b) x}\right\}$ in the interval [a, b] where, 0 ∉ [a, b].
(b) Find the equation of the ellipse with its centre at (4, -1) focus at (1, -1) and given that it passes through (8, 0). [5] Solution 4:
(a) Given
$f(x)=\log \left(\frac{x^{2}+a b}{x(a+b)}\right) \log \left(x^{2}+a b\right)-\log x-\log (a+b)$
Algorithmic function is differentiable and so continuous on .its domain. Therefore f(x) is continuous on [a, b] and differentiable on (a, b)
f(a) = f(b)

(b) Coordinate of the centre and focus are the same.
Therefore both lie on y = -1 & hence the major axis of the ellipse is parallel to the x-axis. & minor axis is parallel to the y-axis.
Let 2a and 2b be the length of major & minor axes respectively. Then the equation of the ellipse is

Que 5:
(a) If ey (x + 1) = 1, then show that: [5] $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$
(b) A printed page is to have a total area of 80 sq. cm with a margin of 1 cm at the top and on each side and a margin of 1.5 cm at the bottom. What should be the dimensions of the page so that the printed area will be maximum? [5] Solution 5:

Que 6:
(a) Evaluate $\int \frac{d x}{x\left\{6(\log x)^{2}+7 \log x+2\right\}}$ [5] (b) Find the area of the region bounded by the curve x = 4y – y2 and the y-axis. [5] Solution 6:

Que 7:
(a) Ten candidates received percentage marks in two subjects as follows: [5]

Calculate Spearman’s rank correlation coefficient and interpret your result.
(b) The following results were obtained with respect to two variables x and y: [5]
Σx = 30, Σy = 42, Σxy = 199, Σx2 = 184, Σy2 = 318, Σn = 6
Find the following:
(i) The regression coefficients.
(ii) Correlation coefficient between x and y.
(iii) Regression equation ofy on x.
(iv) The likely value ofy when x = 10.
Solution 7:
(a) In the case of Mathematics:
88 is scored by 1 student, so we assign rank 1 to it.
Again, 80 is scored by the two students So we assign common rank $\frac{2+3}{2}=2.5$ to each of them.
And 76 is scored by only one thus we assign rank 4 to him.
74 is scored by only one, so we assign rank 5 to him.
68 is scored by only one so, we assign rank 6 to him.
65 is scored by only one so, we assign rank 7 to him.
43 is scored by only one so, we assign rank 8 to him.
40 is scored by two persons so, we assign common rank $\frac{9+10}{2}=9.5$ to each of them.
In Statistics
90 is scored by only one thus we assign rank 1 to him.
84 is scored by only one thus we assign rank 2 to him.
72 is scored by only one thus we assign rank 3 to him.
66 is scored by only one thus we assign rank 4 to him.
54 is scored by two candidates thus we assign common rank $\frac{5+6}{2}=5.5$ to both of them.
50 is scored by only one thus we assign rank 7 to him.
43 is scored by only one thus we assign rank 8 to him.
38 is scored by only one thus we assign rank 9 to him.
30 is scored by only one thus we assign rank 10 to him.

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