JAC Class 12 Mathematics Past Paper 2013 PDF

## Mathematics - 2013 सामान्य निर्देश : General Instructions: इस प्रश्न-पत्र में 29 प्रश्न हैं, जो तीन खण्डों-अ, ब और स में घंटे हुए हैं। खण्ड-अ में 10 प्रश्न हैं, जिनमें प्रत्येक 1 अंक का है, खण्ड-य में 12 प्रश्न है जिनमें प्रत्येक 4 अंक का है तथा खण्ड-स में 7 प्रश्न हैं जिनमें प्रत्येक 6 अंक का है। कैलकुलेटर के उपयोग की अनुमति नहीं है। आवश्यकता हो अथवा परीक्षार्थी के माँग पर लघुगणकीय अथवा सांख्यिकीय सारणी उपलब्ध करायी जा सकती है। The question paper consists of 29 questions divided into three Sections-A, B and C. Section-A comprises of 10 questions of 1 mark each. Section -B comprises of 12 questions of 4 marks each and Section-C comprises of 7 questions of 6 marks each. Use of calculator is not permitted. However, you may ask for logarithmic and statistical tables, if required. ### Section-A (खण्ड-अ) (Objective questions) (वस्तुनिष्ठ प्रश्न) **Q.1.** Find x and y if (2x, x + y) = (6, 2). यदि (2x, x + y) = (6, 2) हो, तो x और y ज्ञात करें। **Ans.** If two matrices are equal then the corresponding elements would be equal. or, $2x = 6$ and $x + y = 2$ $x = 3$ $y = 2 - 3 = -1$ $x = 3, y = -1$ Ans. **Q.2.** A bag contains 3 white and 2 black balls. Find the probability of drawing a white ball at random. एक थैले में 3 उजली तथा 2 काली गोलियाँ हैं। यदृच्छया एक उजली गोली निकालने की प्रायिकता निकालें। **Ans.** Total nos. of events = 5 The probability of drawing a white ball at random $\frac{3c_1}{5c_1} = \frac{3}{5}$ Ans. **Q.3.** Evaluate : $\int \frac{1}{x^2} dx$ मान ज्ञात करें : $\int \frac{1}{x^2} dx$ **Ans.** $\int \frac{1}{x^2} dx = \int x^{-2}dx$ $=\frac{x^{-2+1}}{-2+1}$ $= -\frac{1}{x}$ Ans. **Q.4.** Prove that $sin^{-1}x + cos^{-1}x = \frac{\pi}{2}$ सिद्ध करें कि $sin^{-1}x + cos^{-1}x = \frac{\pi}{2}$ **Ans.** Prove that $sin^{-1}x + cos^{-1}x = \frac{\pi}{2}$ Let $sin^{-1}x = \theta$ $sin\theta = x$ $\therefore cos(\frac{\pi}{2} - \theta) = sin\theta = x$ $cos^{-1}x = (\frac{\pi}{2} - \theta)$ or $cos^{-1}x = \frac{\pi}{2} - sin^{-1}x$ $sin^{-1}x + cos^{-1}x = \frac{\pi}{2}$, Proved. **Q.5.** Evaluate: $i^{-71}$ मान निकालें : $i^{-71}$ **Ans.** $i^{-71} = \frac{1}{i^{71}} = \frac{1}{(i^2)^{35}i}$ $= \frac{1}{(-1)^{35}i}$ $= \frac{1}{-i} = \frac{i}{(-1)i\cdot i} = i$ Ans. **Q.6.** If $A = \begin{bmatrix} 2 & 7 \\ 9 & 8 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}$, then find $A - B$. यदि $A = \begin{bmatrix} 2 & 7 \\ 9 & 8 \end{bmatrix}$ और $B = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}$ हों, तो A - B ज्ञात करें। **Ans.** $A - B = \begin{bmatrix} 2-2 & 7-(-1) \\ 9-0 & 8-3 \end{bmatrix}$ $= \begin{bmatrix} 0 & 8 \\ 9 & 5 \end{bmatrix}$ Ans. **Q.7.** If $y = sin(cot x)$, then find $\frac{dy}{dx}$. यदि $y = sin(cot x)$ हो, तो $\frac{dy}{dx}$ निकालें। **Ans.** $\frac{dy}{dx} = \frac{d(sin(cotx))}{dcotx} \cdot \frac{dcot x}{dx}$ $= cos(cot x) \cdot -cosec^2 x$ $= - cos(cot x) \cdot cosec^2 x$ Ans. **Q.8.** Solve: $ydx - xdy = xydx$. हल करें : $ydx - xdy = xydx$. **Ans.** From the given equation, we have $xdy = (1-x)dx$ $\frac{dy}{y} = \frac{(1-x)}{x} dx = (\frac{1}{x} - 1)dx$ Integrating, we get $\int \frac{dy}{y} = \int (\frac{1}{x} - 1)dx$ $Log\ y = log\ x - x + c$ $Log\ y - Log\ x = c - x$ $Log\ (\frac{y}{x}) = c - x$ $\frac{y}{x} = e^{c-x}$ $y = xe^{c-x}$ This is required eqn. **Q.9.** Evaluate : $\begin{vmatrix} 4 & 9 &7 \\ 3 & 5 & 7 \\ 5 & 4 & 5 \end{vmatrix}$ मान निकालें : $\begin{vmatrix} 4 & 9 &7 \\ 3 & 5 & 7 \\ 5 & 4 & 5 \end{vmatrix}$ **Ans.** Let $A = \begin{vmatrix} 4 & 9 &7 \\ 3 & 5 & 7 \\ 5 & 4 & 5 \end{vmatrix}$ $= 4(5\cdot 5 - 7\cdot 4) - 9(3\cdot 5 - 7\cdot 5) + 7(3\cdot 4 - 5\cdot 5)$ $= 4(25 - 28) - 9(15 - 35) + 7(12 -25)$ $= 4(-3) - 9(-20) + 7(-13)$ $=-12 + 180 - 91 = 77$ Ans. **Q.10.** Prove that the vectors $i -2j +5k$ and $-2i +4j + 2k$ are mutually perpendicular. सिद्ध कीजिए कि सदिश $i -2j +5k$ तथा $-2i +4j + 2k$ परस्पर लम्ब हैं। **Ans.** Here $(i -2j +5k) \cdot (-2i +4j + 2k)$ $= 1(-2) + (-2)(4) + 5(2)$ $= -2 -8 +10 = 0$ $\therefore$ the given vectors are mutually perpendicular. ### Section - B (खण्ड- य) **Q.11.** If $f(x) = 4x^2 - 5x$, find the value of $\frac{f(x+h) - f(x)}{h}$. यदि $f(x) = 4x^2 - 5x$, तो $\frac{f(x+h) - f(x)}{h}$ का मान निकालें। **Ans.** $f(x) = 4x^2 - 5x$ $f(x + h) = 4(x+h)^2 - 5(x+h) $ $= 4(x^2 + h^2 + 2xh) - 5x - 5h$ $f(x + h) - f(x) = 4x^2 + 4h^2 + 8xh - 5x - 5h - 4x^2 + 5x$ $= 4h^2 + 8xh - 5h$ $\frac{f(x + h) - f(x)}{h} = \frac{(4h^2 + 8x - 5)h}{h}$ $= 4h + 8x -5$ Ans. **Q.12.** Prove that $2tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{4} = tan^{-1}\frac{32}{43}$. सिद्ध करें कि $2tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{4} = tan^{-1}\frac{32}{43}$. **Ans.** L.H.S. $= 2tan^{-1}\frac{1}{5} + tan^{-1}\frac{1}{4}$ $= tan^{-1}(\frac{2 \cdot \frac{1}{5}}{1 - (\frac{1}{5})^2})+ tan^{-1}\frac{1}{4}$ $= tan^{-1}(\frac{\frac{2}{5}}{1 - \frac{1}{25}})+ tan^{-1}\frac{1}{4}$ $= tan^{-1}(\frac{\frac{2}{5}}{\frac{25-1}{25}})+ tan^{-1}\frac{1}{4}$ $= tan^{-1}\frac{5}{12} + tan^{-1}\frac{1}{4}$ $= tan^{-1} \frac{\frac{5}{12} + \frac{1}{4}}{1- \frac{5}{12} \cdot \frac{1}{4}}$ $= tan^{-1}\frac{\frac{8}{12}}{\frac{48-5}{48}} = tan^{-1}\frac{\frac{8}{12}}{\frac{43}{48}}$ $= tan^{-1} \frac{8}{12} \cdot \frac{48}{43}$ $= tan^{-1}\frac{32}{43} = R.H.S.$ dy 1-cosx **Q.13.** Find $\frac{dy}{dx}$, if $y = \frac{1-cosx}{1+cosx}$. dy निकालें, यदि $y = \frac{1-cosx}{1+cosx}$. **Ans.** $\frac{d}{dx}(\frac{1-cosx}{1+cosx}) = \frac{(1-cosx)'(1+cosx) - (1+cosx)'(1-cosx)}{(1+cosx)^2}$ $=\frac{0-(-sinx)(1+cosx) - (+sinx)(1-cosx)}{(1+cosx)^2}$ $=\frac{sinx + sint.cosx + sinx - sint.cosx}{(1+cosx)^2}$ $=\frac{2sinx}{(1+cosx)^2}$ $=\frac{2sinx}{(cos^2 \frac{x}{2} + sin^2 \frac{x}{2} + cos^2 \frac{x}{2} - sin^2 \frac{x}{2})^2}$ $=\frac{2sin(\frac{x}{2})cos(\frac{x}{2})}{(2cos^2 \frac{x}{2})^2} = \frac{tan \frac{x}{2}}{2}$ **Q.14.** Factorize: $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$ गुणनखंड करें : $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$ **Ans.** By doing $c_1 \rightarrow c_1 - c_2$ and $c_3 \rightarrow c_3 - c_2$ $\Delta = \begin{vmatrix} a-b-c - 2a & 2a & 2a - 2a \\ 2b - (b-c-a) & b-c-a & 2b - (b-c-a) \\ 2c-2c & 2c & c-a-b - 2b \end{vmatrix}$ $= \begin{vmatrix} -a-b-c & 2a & 0 \\ b+c+a & b-c-a & a+c-b \\ 0 & 2c & c-a-b \end{vmatrix}$ taking the factor $(a+b+c)$, from $c_1 \& c_3$ $= (a+b+c) * \begin{vmatrix} -1 & 2a & 0 \\ 1 & b-c-a & -1 \\ 0 & 2c & c-a-b \end{vmatrix}$ on expanding, we get $A = (a + b + c)^2 [-1 {(-1) (c-a-b)-2b}+2a (1-0)] $ $= (a + b + c)^2 [-1 {(c + a + b)-2b}+2a] $ $= (a + b + c)^2 [-1(a - b - c) + 2a]$ $= (a + b + c)^2 [- a + b + c + 2a]$ $= (a + b + c)^2 [a + b + c]$ $= (a + b + c)^3 $ Ans. **Q.15.** Test the continuity of the function $f (x)$ at $x = 0$, where $f(x) = \frac{sin x}{x}$, when $x \ne 0$ $= 2$, when $x = 0$. फलन $f (x)$ का $x = 0$ पर संततता की जाँच करें जहाँ $f(x) = \frac{sin x}{x}$ जब $x \ne 0$ $= 2$, जब $x = 0$. **Ans.** $f(x)$ would be continuous at $x = 0$ when $\lim_{h \to 0} f(0 + h) = f(0) = \lim_{h \to 0} f(0 - h)$. Now, $\lim_{h \to 0} f(0 + h) = \lim_{h \to 0} \frac{sin(0 + h)}{h} = \lim_{h \to 0}\frac{sin h}{h}=1$ $\lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \frac{sin(0 - h)}{-h} = \lim_{h \to 0}\frac{sin(-h)}{-h} = \lim_{h \to 0}\frac{sinh}{h} = 1$ $f(0) = 2$ $\lim_{h \to 0} f(0 + h) = \lim_{h \to 0} f(0 - h) \ne f(0)$ so, $f(x)$ is not continuous at $x = 0$. **Q.16.** Find the acute angle between two lines that have the direction ratios (1, 1, 0) and (2, 1, 2). दो सरल रेखाओं के बीच का न्यूनकोण ज्ञात कीजिए जिनके दिक् अनुपात (1, 1, 0) और (2, 1, 2) हैं। *Diagram of two lines AB and CD intersecting at Q* let AB and CD be two lines whose d.c. are (1, 1, 0) & (2, 1, 2) respectively. We have to find out the acute angle between lines AB & CD. Let the acute angle be Q. The direction cosines of AB and CD are $l_1=\frac{1}{\sqrt{1^2 + 1^2 + 0^2}}, m_1= \frac{1}{\sqrt{1^2 + 1^2 + 0^2}}, n_1=\frac{0}{\sqrt{1^2 + 1^2 + 0^2}}$ (suppose) and $l_2=\frac{2}{\sqrt{2^2 + 1^2 + 2^2}}, m_2=\frac{1}{\sqrt{2^2 + 1^2 + 2^2}}, n_2=\frac{2}{\sqrt{2^2 + 1^2 + 2^2}}$ (suppose) $=(m,n)$ suppose :: $l_1=\frac{1}{\sqrt{2}}, m_1 = \frac{1}{\sqrt{2}}, n_1 = 0$ and $l_2= \frac{2}{3}, m_2=\frac{1}{3}, n_2=\frac{2}{3}$ cos Q =$l_1l_2+m_1m_2+n_1n_2$ $=\frac{1}{\sqrt{2}} \cdot \frac{2}{3}+ \frac{1}{\sqrt{2}}\cdot \frac{1}{3} + 0$ $=\frac{3}{\sqrt{2} \cdot 3}$ $=\frac{3}{3\sqrt{2}} = cos45°$ **Q.17.** Evaluate any one: किसी एक का घार हात करें: (i) $\int \frac{1}{1 + cosx} dx$ (ii) $\int tan^3x. dx$ **Ans.** (i) $\int \frac{1}{1 + cosx} dx$ $=\frac{1}{2}\int \frac{1}{cos^2x/2}dx$ = $\frac{1}{2}\int sec^2\frac{x}{2} dx$ =$\frac{1}{2} tan^2 \frac{x}{2} + c$ Ans. (ii) $\int tan^3x. dx$ I = $\int tan^2x. tanx. dx$ = $\int tan^2x (sec^2 - 1)d$ = $\int tan^2x sec^2 dx - $\int tan x dx) = 1,-I, (suppose) $I_1$= $\int tan^2x sec^2 dx$ pulling $tan^2 x$ = z sec² x dx = dz ..$I_1$= $\int z. dz = \frac{z^3}{3} = \frac{tan^2x}{3}$ $I₂$= $\int tan²x dx = \int (sec²x-1) dx$ = tan x-x I-= I,-I, =$\frac{tan^2x}{3} - tanx+x+c$ Ans. and when $tan \frac{x}{2}=1$, then cot 2$\theta$=1 or 2$\theta$=0 or $\theta=0$ $1 - cos2\theta = 2sin^2\theta$ **Q.18.** Evaluate any one: किसी एक का मान ज्ञात करें : (1) $\int \frac{\sqrt{1-x}}{1 + x} dx $ (ii) $\int_0^{\frac{\pi}{2}} \frac{1-sinx}{1-cosx}dx$ **Ans.** (1) $\int_0^{1}\frac{\sqrt{1-x}}{1 + x} dx$ let x = $\cos 2\theta dx = - 2 sin 2\theta.d\theta$ Also when x = 0 the $\cos 2\theta$ = 0 or 2$\theta$ = $\frac{\pi}{2}$ or $\theta$ = $\frac{\pi}{4}$ *Calculations involving trigonometric equalities* **(ii)** $\int \frac{1-sinx}{1-cosx}dx$ Evaluate $\int \frac{1-sinx}{1-cosx}dx$Let $\frac{1-sinx}{1-cosx}$ = $\frac{1-2sin^x/2 \cos x/2}{2sin^2 x/2}$ = $\frac{1}{2}\int \frac{(1-cos x)}{(1-cos x)} dx$ = [-cot^x/2 + log(sinx/2)]0 = -log(sinx/2) - (cot/2) -(cot/k -logic == 1 to 2 * **Q.19.** Evaluate any one *Equations with trigonometric functions* **Q.20.** If $a=2\hat{i} -3\hat{j} -5\hat{k} $ and $b= -7\hat{i} +6\hat{j} +8\hat{k}$. Then find $a \times b $ . यदि $a=2\hat{i} -3\hat{j} -5\hat{k} $ and $b= -7\hat{i} +6\hat{j} +8\hat{k}$. Then find $a \times b $. **Q.21.** Find the equation of the plane perpendicular to the line joining the points (2,-1,2) and (3,2,-1) and passing through and point (4,-3,1) बिन्दुओं (2,-1,2) और (3,2,-1) को मिलाने वाली रेखा पर लप्य तथा यिन्दु (4,-3,1) से गुजरनेवाले तल का समीकरण निकालें। *Visual representation of a plane.* **Ans.** The equation of a plane passing through the point (4,-3,1) a) + b(y+3) + c 2-1( = 0 ...... 1) where a bc are cosine of the perpendicular to the plane. Now, the direction cosines of the line perpendicular going the points2,-1, and 3 are 3. So the lines whose d.c. are a,b,c and 1,-3,3 are parallel abc =2 putting the values is (1) →80 4. or, x+ 3x + 4/20/ **Q.22.** 10 coins are tossed. What is the probability that exactly 5 heads appear? 10 सिक्कों को उछाला जाता है। ठीक 5 शीर्ष आने की क्रायिकता है? **Ans.** Let p = probability of coming head in one chance = 1\ q = probability of coming tail in one chance = 1\ 2 **Q.23.** If $A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}$, then show that $(AB)^{-1} = B^{-1}A^{-1}$ or/अथवा If $A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix}$ then find A-1. Hence solve the following equations: $x + 2y -3z = -4$ $2x + 3y + 2z = 2$ $3x -3y - 4z =11$ *Calculations* **Q.24.** Find the maximum and minimum values of function f(x) = 2x²-15x²+36x + 11. फलन f(x) = 2x-15x²+36x + 11 का सर्वोच्च तथा सर्वनिम्न मान ज्ञात करें। **Ans.** *(equations)* for x-4x9 axis of the pasbola y" = -axis or Ans WEknow that 3- *Visual representation of a parabola* **Q.26.** Solve: xdx + ydy = xdy - ydx. हल करें : xdx + ydy = xdy - ydx. Ans. *(equations)*

JAC Class 12 Mathematics Past Paper 2013 PDF

Document Details

Tags

Related

Summary

Full Transcript