Mathematics NEB Class 11 2076 question paper

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1. 120766 marksNumericalHideAnswer

   (a) Find the truth value of the biconditional statement of p and q where p represents '2 is an even number' and q represents '4 is an even number'. (b) If f: R -> R be defined by f(x) = 2x - 3, find f^-1. (c) Test the even or odd nature and symmetricity of the function y = 8x^2.

   (a) Here $p$ (2 is even) is true and $q$ (4 is even) is true. The biconditional $p\Leftrightarrow q$ is true when both have the same truth value, so its truth value is T (true). (b) Let $y=2x-3\Rightarrow x=\dfrac{y+3}{2}$. Hence

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2. 220766 marksHideAnswer

   (a) Express tan^-1(x) in terms of inverse of sin function. (b) Prove by mathematical induction: 1 + 3 + 5 + ... + (2n - 1) = n^2. (c) Find the adjoint of the matrix (2 5; 3 -7).

   (a) Let $\tan^{-1}x=\theta\Rightarrow\tan\theta=x$. Then $\sin\theta=\dfrac{x}{\sqrt{1+x^2}}$, so $$\tan^{-1}x=\sin^{-1}!\left(\frac{x}{\sqrt{1+x^2}}\right).$$ (b) $P(n):\ 1+3+5+\cdots+(2n-1)=n^2$. $n=1$: $1=1^2$. True. Assume $P(k)$: s...

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3. 320766 marksNumericalHideAnswer

   (a) Using Cramer's rule, solve: 3x + 2y + 9 = 0, 2x - 3y + 6 = 0. (b) Express sqrt(3) + i in polar form. (c) If one root of the equation ax^2 + bx + c = 0 be twice the other, show that 2b^2 = 9ac.

   (a) Write $3x+2y=-9,\ 2x-3y=-6$. $$ \begin{aligned} D &= \begin{vmatrix}3&2\2&-3\end{vmatrix} \ &= -9-4 \ &= -13, \ \ Dx &= \begin{vmatrix}-9&2\-6&-3\end{vmatrix} \ &= 27+12 \ &= 39, \ \ Dy &= \begin{vmatrix}3&-9\2&-6\end{vmatri...

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4. 420766 marksNumericalHideAnswer

   (a) Find the equation of the straight line that has y-intercept '2' and is parallel to 4x - 8y + 3 = 0. (b) Find the equation of the circle which passes through origin and makes equal intercepts on both axes. (c) Evaluate: lim(x -> 0) (1 - cos mx)/(1 - cos nx).

   (a) $4x-8y+3=0\Rightarrow y=\tfrac12x+\tfrac38$, slope $\tfrac12$. A parallel line with $y$-intercept $2$ is $$y=\tfrac12x+2\Rightarrow\boxed{x-2y+4=0}.$$

   (b) A general circle through the origin is $x^2+y^2+2gx+2fy=0$ (constant term $0$). Its $x$-intercept is $-2g$ and $y$-intercept is $-2f$. Equal intercepts $\Rightarrow g=f$, so $$x^2+y^2+2g(x+y)=0,$$ i.e. $x^2+y^2=a(x+y)$ where $a=-2g$ is the common intercept.

   (c) Using $1-\cos\theta=2\sin^2\dfrac{\theta}{2}$: $$ \begin{aligned} \lim_{x\to0}\frac{1-\cos mx}{1-\cos nx} &= \lim_{x\to0}\frac{2\sin^2\frac{mx}{2}}{2\sin^2\frac{nx}{2}} \ &= \lim_{x\to0}\left(\frac{\sin\frac{mx}{2}}{\frac{mx}{2}}\right)^2\left(\frac{\frac{nx}{2}}{\sin\frac{nx}{2}}\right)^2\cdot\frac{m^2}{n^2} \ &= \frac{m^2}{n^2}. \end{aligned} $$

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5. 520766 marksNumericalHideAnswer

   (a) Find dy/dx when x = 2at and y = 2at^2. (b) Evaluate: integral of sin 2x . sin 3x dx. (c) Determine the interval in which the function f(x) = (1/2)x^2 - x is increasing or decreasing.

   (a) $\dfrac{dx}{dt}=2a,\ \dfrac{dy}{dt}=4at$, so $\dfrac{dy}{dx}=\dfrac{4at}{2a}=2t.$

   (b) Using $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$: $$ \begin{aligned} \sin2x\sin3x &= \tfrac12[\cos x-\cos5x]. \ \int\sin2x\sin3x,dx &= \frac12\left[\sin x-\frac{\sin5x}{5}\right]+C \ &= \frac{\sin x}{2}-\frac{\sin5x}{10}+C. \end{aligned} $$

   (c) $f'(x)=x-1$.

   • $f'(x)>0\Rightarrow x>1$: increasing on $(1,\infty)$.
   • $f'(x)<0\Rightarrow x<1$: decreasing on $(-\infty,1)$.

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6. 620768 marksHideAnswer

   (a) For any non-empty subsets A, B, C of universal set U, prove that: A - (B U C) = (A - B) n (A - C). OR Define absolute value of a real number. Also prove that |x| - |y| <= |x + y|, for all x, y in R. (b) Sketch the graph of f(x) = (x - 1)^3 indicating its characteristics.

   (a) $x\in A-(B\cup C)\iff x\in A\text{ and }x\notin(B\cup C)\iff x\in A\text{ and }(x\notin B\text{ and }x\notin C)$. That is $x\in A-B$ and $x\in A-C$, i.e. $x\in(A-B)\cap(A-C)$. Hence $A-(B\cup C)=(A-B)\cap(A-C)$.

   OR $|x|=\begin{cases}x,&x\ge0\-x,&x<0\end{cases}$. To prove $|x|-|y|\le|x+y|$: write $x=(x+y)+(-y)$, so by the triangle inequality $$ \begin{aligned} |x| &= |(x+y)+(-y)|\le|x+y|+|{-y}| \ &= |x+y|+|y|. \end{aligned} $$ Subtracting $|y|$ gives $|x|-|y|\le|x+y|$.

   (b) $f(x)=(x-1)^3$.

   • Only $x$-intercept at $x=1$ (a triple root, where the curve flattens); $y$-intercept $(-1)^3=-1$.
   • $f'(x)=3(x-1)^2\ge0$, so $f$ is increasing everywhere, with a horizontal tangent at $x=1$.
   • $f''(x)=6(x-1)=0$ at $x=1$: point of inflection at $(1,0)$; the curve is concave down for $x<1$ and concave up for $x>1$, and is point-symmetric about $(1,0)$.

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7. 720768 marksNumericalHideAnswer

   (a) Solve for general values of theta: sin(theta) + sin(2theta) + sin(3theta) = cos(theta) + cos(2theta) + cos(3theta). OR If a = 2, b = sqrt(6) and c = sqrt(3) - 1, find the angles of the triangle ABC. (b) Prove that determinant |a^2+1 ab ac; ab b^2+1 bc; ac bc c^2+1| = 1 + a^2 + b^2 + c^2.

   (a) Group first and third terms: $$ \begin{aligned} \sin\theta+\sin3\theta+\sin2\theta &= 2\sin2\theta\cos\theta+\sin2\theta \ &= \sin2\theta(2\cos\theta+1), \ \cos\theta+\cos3\theta+\cos2\theta &= 2\cos2\theta\cos\theta+\cos2\theta \ &= \cos2\theta(2\cos\theta+1). \end{aligned} $$ Equation: $(2\cos\theta+1)(\sin2\theta-\cos2\theta)=0$.

   • $2\cos\theta+1=0\Rightarrow\cos\theta=-\tfrac12\Rightarrow\theta=2n\pi\pm\dfrac{2\pi}{3}$.
   • $\sin2\theta=\cos2\theta\Rightarrow\tan2\theta=1\Rightarrow2\theta=n\pi+\dfrac{\pi}{4}\Rightarrow\theta=\dfrac{n\pi}{2}+\dfrac{\pi}{8}$.

   OR $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. With $b^2=6,\ c^2=(\sqrt3-1)^2=4-2\sqrt3,\ a^2=4$: numerator $=6+4-2\sqrt3-4=6-2\sqrt3$, denominator

   $$ \begin{aligned} 2bc &= 2\sqrt6(\sqrt3-1) \ &= \sqrt2(3-\sqrt3)\cdot2 \ &= 2(3\sqrt2-\sqrt6). \end{aligned} $$

   Then

   $$ \begin{aligned} \cos A &= \dfrac{6-2\sqrt3}{2\sqrt6(\sqrt3-1)} \ &= \dfrac{2\sqrt3(\sqrt3-1)}{2\sqrt6(\sqrt3-1)} \ &= \dfrac{\sqrt3}{\sqrt6} \ &= \dfrac{1}{\sqrt2} \ &\Rightarrow A=45^\circ. \ \cos B &= \dfrac{a^2+c^2-b^2}{2ac} \ &= \dfrac{4+4-2\sqrt3-6}{2\cdot2(\sqrt3-1)} \ &= \dfrac{2-2\sqrt3}{4(\sqrt3-1)} \ &= \dfrac{-2(\sqrt3-1)}{4(\sqrt3-1)} \ &= -\dfrac12 \ &\Rightarrow B=120^\circ. \end{aligned} $$

   $C=180^\circ-45^\circ-120^\circ=15^\circ.$ Hence $A=45^\circ,\ B=120^\circ,\ C=15^\circ.$

   (b) Let $\Delta=\begin{vmatrix}a^2+1&ab&ac\ab&b^2+1&bc\ac&bc&c^2+1\end{vmatrix}$. Multiply $R_1,R_2,R_3$ by $a,b,c$ and divide by $abc$: $$ \begin{aligned} \Delta &= \frac{1}{abc}\begin{vmatrix}a(a^2+1)&a^2b&a^2c\ab^2&b(b^2+1)&b^2c\ac^2&bc^2&c(c^2+1)\end{vmatrix} \ &= \frac{abc}{abc}\begin{vmatrix}a^2+1&a^2&a^2\b^2&b^2+1&b^2\c^2&c^2&c^2+1\end{vmatrix}. \end{aligned} $$ Apply $C_1\to C_1+C_2+C_3$ (common factor $1+a^2+b^2+c^2$ in column 1): $$=(1+a^2+b^2+c^2)\begin{vmatrix}1&a^2&a^2\1&b^2+1&b^2\1&c^2&c^2+1\end{vmatrix}.$$ Now $R_2\to R_2-R_1,\ R_3\to R_3-R_1$ gives $\begin{vmatrix}1&a^2&a^2\0&1&0\0&0&1\end{vmatrix}=1$. Hence $\Delta=1+a^2+b^2+c^2$.

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8. 820768 marksNumericalHideAnswer

   (a) Solve by inverse matrix or row equivalent method: x - y = 0, 2x - y + 4z = 18, -3x + z + 2 = 0. (b) Find the equation whose roots are reciprocal of the roots of x^2 - x + 1 = 0.

   (a) From $x-y=0$: $y=x$. From $-3x+z+2=0$: $z=3x-2$. Substitute in the second: $$ \begin{aligned} 2x-y+4z &= 18\Rightarrow2x-x+4(3x-2) \ &= 18\Rightarrow x+12x-8 \ &= 18\Rightarrow13x \ &= 26\Rightarrow x \ &= 2. \end{aligned} $$ Then $y=2$ and $z=3(2)-2=4$. $$\boxed{x=2,\ y=2,\ z=4}$$ Check: $2(2)-2+4(4)=4-2+16=18$. Correct.

   (b) If $\alpha,\beta$ are the roots of $x^2-x+1=0$, then $\alpha+\beta=1,\ \alpha\beta=1$. The reciprocal roots $\tfrac1\alpha,\tfrac1\beta$ have sum $\dfrac{\alpha+\beta}{\alpha\beta}=1$ and product $\dfrac1{\alpha\beta}=1$. Required equation: $$x^2-x+1=0\quad(\text{same as the original reciprocal equation}).$$

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9. 920768 marksNumericalHideAnswer

   (a) Prove that the circles x^2 + y^2 + 2ax + c^2 = 0 and x^2 + y^2 + 2by + c^2 = 0 touch each other if 1/a^2 + 1/b^2 = 1/c^2. (b) Evaluate: lim(x -> theta) (x cos theta - theta cos x)/(x - theta). OR Given f(x) = 3 - x for 0 <= x < 3/2, and -3 + x for x >= 3/2. Test f(x) for continuity at x = 3/2. If f(x) is not continuous at x = 3/2, how would you make it continuous?

   (a) Circle 1: centre $C_1(-a,0)$, radius $r_1=\sqrt{a^2-c^2}$. Circle 2: centre $C_2(0,-b)$, radius $r_2=\sqrt{b^2-c^2}$. Distance $C_1C_2=\sqrt{a^2+b^2}$. For the circles to touch, $C_1C_2=r_1\pm r_2$. Squaring: $$ \begin{aligned} a^2+b^2 &= (a^2-c^2)+(b^2-c^2)\pm2\sqrt{(a^2-c^2)(b^2-c^2)}\Rightarrow2c^2 \ &= \pm2\sqrt{(a^2-c^2)(b^2-c^2)}. \ c^4 &= (a^2-c^2)(b^2-c^2) \ &= a^2b^2-a^2c^2-b^2c^2+c^4\Rightarrow a^2b^2 \ &= a^2c^2+b^2c^2. \end{aligned} $$ Dividing by $a^2b^2c^2$: $\dfrac{1}{c^2}=\dfrac{1}{b^2}+\dfrac{1}{a^2}$, i.e. $\dfrac1{a^2}+\dfrac1{b^2}=\dfrac1{c^2}$.

   (b) $\dfrac00$ form. Split: $$ \begin{aligned} \frac{x\cos\theta-\theta\cos x}{x-\theta} &= \cos\theta-\theta,\frac{\cos x-\cos\theta}{x-\theta}\xrightarrow{x\to\theta}\cos\theta-\theta(-\sin\theta) \ &= \cos\theta+\theta\sin\theta. \end{aligned} $$

   OR At $x=\tfrac32$: LHL $=\lim_{x\to(3/2)^-}(3-x)=\tfrac32$; RHL $=\lim_{x\to(3/2)^+}(-3+x)=-\tfrac32$; $f!\left(\tfrac32\right)=-\tfrac32$. Since LHL $\neq$ RHL, $f$ has a jump discontinuity at $x=\tfrac32$. Because the left and right limits differ, the discontinuity is not removable, so $f$ cannot be made continuous there merely by redefining $f!\left(\tfrac32\right)$.

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10. 1020768 marksNumericalHideAnswer

    (a) Find from first principles, the derivative of (bx - a)^n. (b) Find the area of the circle x^2 + y^2 = 36, using method of integration.

    (a) Let $f(x)=(bx-a)^n$. Then $$ \begin{aligned} f'(x) &= \lim_{h\to0}\frac{(bx+bh-a)^n-(bx-a)^n}{h} \ &= \lim_{h\to0}\frac{(bx-a)^n\Big[\big(1+\tfrac{bh}{bx-a}\big)^n-1\Big]}{h}. \end{aligned} $$ Expanding by the binomial theorem, $\left(1+\tfrac{bh}{bx-a}\right)^n-1=\dfrac{nbh}{bx-a}+O(h^2)$, so $$ \begin{aligned} f'(x) &= (bx-a)^n\cdot\frac{nb}{bx-a} \ &= nb(bx-a)^{n-1}. \end{aligned} $$

    (b) By symmetry, area $=4\displaystyle\int_0^6\sqrt{36-x^2},dx$: $$ \begin{aligned} &= 4\left[\frac{x}{2}\sqrt{36-x^2}+18\sin^{-1}\frac{x}{6}\right]_0^6 \ &= 4\left[0+18\cdot\frac{\pi}{2}\right] \ &= 36\pi\ \text{square units}. \end{aligned} $$

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11. 1120766 marksNumericalHideAnswer

    Define function. Distinguish relation and function with example. Find the domain and range of f(x) = sqrt(x^2 - 2x - 8), x in R.

    Function. A relation $f$ from set $A$ to set $B$ is a function if every element of $A$ is associated with exactly one element of $B$.

    Relation vs function. A relation is any set of ordered pairs; a function is a special relation in which no first component is repeated with different second components. Example: $R={(1,2),(1,3)}$ is a relation but not a function (input $1$ has two outputs); $f={(1,2),(2,3)}$ is a function.

    Domain and range of $f(x)=\sqrt{x^2-2x-8}$. Require $x^2-2x-8\ge0\Rightarrow(x-4)(x+2)\ge0\Rightarrow x\le-2\text{ or }x\ge4$. $$\text{Domain}=(-\infty,-2]\cup[4,\infty).$$ Since a square root is non-negative and the radicand ranges over $[0,\infty)$ on the domain, $f(x)\ge0$ and takes all such values: $$\text{Range}=[0,\infty).$$

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12. 1220766 marksHideAnswer

    If A, G and H are A.M., G.M. and H.M. respectively between any two unequal positive numbers then prove that (i) A > G > H and (ii) G^2 = A.H.

    Let the two unequal positive numbers be $a$ and $b$. Then $$ \begin{aligned} A &= \frac{a+b}{2} \ \qquad G &= \sqrt{ab} \ \qquad H &= \frac{2ab}{a+b}. \end{aligned} $$

    (ii) $G^2=A\cdot H$: $A\cdot H=\dfrac{a+b}{2}\cdot\dfrac{2ab}{a+b}=ab=(\sqrt{ab})^2=G^2$. Hence $G^2=AH$, i.e. $G$ is the geometric mean of $A$ and $H$.

    (i) $A>G>H$: Since $a\neq b$, $(\sqrt a-\sqrt b)^2>0\Rightarrow a+b>2\sqrt{ab}\Rightarrow\dfrac{a+b}{2}>\sqrt{ab}$, so $A>G$. From $G^2=AH$ with $A>G>0$: $H=\dfrac{G^2}{A}<\dfrac{G^2}{G}=G$, so $G>H$. Therefore $A>G>H$.

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13. 1320766 marksNumericalHideAnswer

    Find the equations of the bisectors of the angles between the lines 4x - 3y + 1 = 0 and 12x - 5y + 7 = 0. Also prove that the bisectors are at right angle to each other, and identify the bisector which contains the origin. OR Prove that the bisectors of the angles between the pair of straight lines ax^2 + 2hxy + by^2 = 0 is (x^2 - y^2)/(a - b) = xy/h.

    Main. Bisectors: $\dfrac{4x-3y+1}{5}=\pm\dfrac{12x-5y+7}{13}\Rightarrow13(4x-3y+1)=\pm5(12x-5y+7)$. ($+$): $4x+7y+11=0$. ($-$): $7x-4y+3=0$. Slopes $-\tfrac47$ and $\tfrac74$; product $=-1$, so the bisectors are perpendicular. At the origin, $4(0)-3(0)+1=1>0$ and $12(0)-5(0)+7=7>0$ (both constants positive), so the origin-containing bisector is the $+$ one: $\boxed{4x+7y+11=0}$.

    OR Let $ax^2+2hxy+by^2=0$ represent lines $y=m_1x$ and $y=m_2x$ with $m_1+m_2=-\dfrac{2h}{b},\ m_1m_2=\dfrac ab$. The equation of the pair of angle bisectors is the locus of points equidistant from the two lines, which reduces to the standard result $$\frac{x^2-y^2}{a-b}=\frac{xy}{h}.$$ Derivation: the bisectors make equal angles with both lines; combining $\tan\alpha$ for each line and eliminating leads to $h(x^2-y^2)=(a-b)xy$, i.e. $\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$.

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14. 1420766 marksNumericalHideAnswer

    Define conjugate of a complex number. Using De Moivre's theorem, find the square root of -2 + 2 sqrt(3) i.

    Conjugate. For $z=x+iy$, the conjugate is $\bar z=x-iy$ (the sign of the imaginary part is reversed).

    Square roots of $-2+2\sqrt3,i$. Modulus $r=\sqrt{(-2)^2+(2\sqrt3)^2}=\sqrt{4+12}=4$. The point is in the second quadrant; reference angle $\tan^{-1}\dfrac{2\sqrt3}{2}=60^\circ$, so $\arg z=120^\circ=\dfrac{2\pi}{3}$. Thus $z=4\big(\cos120^\circ+i\sin120^\circ\big)$. By De Moivre's theorem the square roots ($k=0,1$) are $$z^{1/2}=2\left(\cos\frac{120^\circ+360^\circ k}{2}+i\sin\frac{120^\circ+360^\circ k}{2}\right).$$ $k=0$:

    $$ \begin{aligned} 2(\cos60^\circ+i\sin60^\circ) &= 2\left(\tfrac12+\tfrac{\sqrt3}{2}i\right) \ &= 1+\sqrt3,i. \end{aligned} $$ $k=1$: $-(1+\sqrt3,i)$. $$\boxed{\pm(1+\sqrt3,i)}.$$

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15. 1520766 marksNumericalHideAnswer

    Write the criteria for the function y = f(x) to have the local maxima and local minima at a point. Find the local maxima and local minima of f(x) = 2x^3 - 15x^2 + 36x + 10. Also find the point of inflection. OR A water flows into an inverted conical tank at the rate of 42 cm^3/sec. When the depth of the water is 8 cm, how fast is the level rising? Assume that the height of the tank is 12 cm and the radius of the top is 6 cm.

    Criteria. At $x=c$ with $f'(c)=0$: local maximum if $f''(c)<0$; local minimum if $f''(c)>0$.

    Main. $f(x)=2x^3-15x^2+36x+10$. $$ \begin{aligned} f'(x) &= 6x^2-30x+36 \ &= 6(x-2)(x-3) \ &= 0\Rightarrow x \ &= 2 \ \ 3.\qquad f''(x) &= 12x-30. \end{aligned} $$ At $x=2$: $f''=-6<0$ (maximum), $f(2)=16-60+72+10=38$. At $x=3$: $f''=6>0$ (minimum), $f(3)=54-135+108+10=37$. $$ \begin{aligned} & \text{Local maximum }=38\ (x=2) \ & \text{Local minimum }=37\ (x=3). \end{aligned} $$ Point of inflection: $f''=0\Rightarrow x=\tfrac52$,

    $$ \begin{aligned} f!\left(\tfrac52\right) &= 2\cdot\tfrac{125}{8}-15\cdot\tfrac{25}{4}+36\cdot\tfrac52+10 \ &= \tfrac{125}{4}-\tfrac{375}{4}+100 \ &= \tfrac{75}{2}. \end{aligned} $$

    Inflection $\left(\tfrac52,\tfrac{75}{2}\right)$.

    OR For the cone, radius and height satisfy $\dfrac rh=\dfrac{6}{12}=\dfrac12\Rightarrow r=\dfrac h2$. Volume $$ \begin{aligned} V &= \frac13\pi r^2h \ &= \frac13\pi\left(\frac h2\right)^2h \ &= \frac{\pi h^3}{12}\Rightarrow\frac{dV}{dt} \ &= \frac{\pi h^2}{4}\frac{dh}{dt}. \end{aligned} $$ With $\dfrac{dV}{dt}=42$ and $h=8$:

    $$ \begin{aligned} 42 &= \dfrac{\pi(64)}{4}\dfrac{dh}{dt} \ &= 16\pi\dfrac{dh}{dt}, \end{aligned} $$

    so $$ \begin{aligned} \frac{dh}{dt} &= \frac{42}{16\pi} \ &= \frac{21}{8\pi}\ \text{cm/sec}\approx0.835\ \text{cm/sec}. \end{aligned} $$

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