Oswal 36 Sample Question Papers ISC Class 12 Maths Solutions
Section-A
Answer 1.
(i) (b) ± 1, 0
Explanation :
From the graph, it is clear that f(x) is not differentiable at x = –1, 0 and 1.
(ii) (b)
Explanation :
f : A → B is function or not it can be checked by a graph of the relation. If it is possible to draw a vertical line which cuts the given curve at more than one point then the given relation is not a function and when this vertical line means line parallel to Y-axis cuts the curve at only point then it is a function.
$$\textbf{(iii)\space\textbf{(a)}\space}\frac{\textbf{1}}{\textbf{2}}$$
Explanation :
$$\text{We have,}\\\text{sin}\begin{bmatrix}\frac{\pi}{2} -\text{sin}^{\normalsize-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)\end{bmatrix}\\\text{Let}\space\text{sin}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg) = x\\\Rarr\space \text{sin x = }\frac{-\sqrt{3}}{2}\\=-\text{sin}\frac{\pi}{3} =\text{sin}\bigg(-\frac{\pi}{3}\bigg)\\\Rarr\space x=-\frac{\pi}{3}\\\Rarr\space\text{sin}^{\normalsize-1}\bigg(\frac{-\sqrt{3}}{2}\bigg) =-\frac{\pi}{3}$$
$$\therefore\space\text{sin}\begin{bmatrix}\frac{\pi}{2}-\bigg(-\frac{\pi}{3}\bigg)\end{bmatrix}\\=\text{sin}\bigg(\frac{\pi}{2} + \frac{\pi}{3}\bigg)\\\text{cos}\frac{\pi}{3} =\frac{1}{2}$$
(iv) (c) Determinant is a number associated to a square matrix
$$\textbf{(v) (b)}\space\begin{bmatrix}\textbf{1} &\textbf{0}\\\textbf{0} &\textbf{1}\end{bmatrix}$$
Explanation :
$$\text{Given, \space A =}\begin{bmatrix}0 &1\\1 &0\end{bmatrix}\\\therefore\space\text{A}^{2} =\begin{bmatrix}0 &1\\1 &0\end{bmatrix}\begin{bmatrix}0 &1\\1 &0\end{bmatrix}\\=\begin{bmatrix}1 &0\\0 &1\end{bmatrix}$$
(vi) (b) 3
Explanation :
$$\text{Let\space A =}\begin{bmatrix}1 &2\\k &6\end{bmatrix}$$
∴ A– 1 does not exist if | A | = 0
$$\therefore\qquad\begin{vmatrix}1 &2\\k &6\end{vmatrix} = 0$$
$$\Rarr\space\text{6 – 2k = 0}\\\Rarr\space\text{2k = 6}\\\Rarr\space\text{k = 3.}$$
$$\textbf{(vii)\space(b)\space}\frac{\textbf{5}}{\textbf{2}}$$
Explanation :
Given curve x2 + 3y + y2 = 5
$$\Rarr\space 2x+3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\\\Rarr\space(3+2y)\frac{dy}{dx} = -2x\\\Rarr\space\frac{dy}{dx} =\frac{-2x}{3+2y}\\\therefore\space\text{Slope of the normal at (1, 1) =}\\-\frac{1}{\frac{dy}{dx}}\\= -\frac{1}{\frac{\normalsize-2x}{3+2y}} =\frac{3+2y}{2x}\\=\frac{3+2×1}{2×1} =\frac{5}{2}.$$
(viii) (c) x + y = 0
Explanation :
Given : Curve y = sin x
$$\therefore\space\frac{dy}{dx} =\text{cos x}$$
$$\Rarr\space\frac{dy}{dx}\biggm\vert_{(0,0)} = 1\\\Rarr\space\text{Slope of normal is – 1.}\\\therefore\space\text{Equation is,}\\\text{y – 0 = – 1(x – 0).}\\\Rarr\space\text{x + y = 0.}$$
(ix) (d) Statement (1) is false and statement (2) is true.
From the graph, it is clear that statement 1 is false, and statement 2 is true.
(x) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
Explanation :
Reason is true now
$$\text{P(E}_{1}) = \space^{4}\text{C}_{3}p^{3}q\\= 4×\bigg(\frac{1}{2}\bigg)^{3}×\frac{1}{2}\\\bigg(\because\space \text{P = q}=\frac{1}{2}\bigg)\\=\frac{4}{16}=\frac{1}{4}\\\Rarr\space\text{P(E}_{2}) = ^\space{8}\text{C}_{5}\bigg(\frac{1}{2}\bigg)^{5}×\bigg(\frac{1}{2}\bigg)^{3}\\=\frac{56}{2^{8}} =\frac{7}{32}\\\text{Therefore,}\qquad\frac{1}{4}\gt\frac{7}{32}$$
P(E1) > P(E2)
Assertion is true.
Reason is correct explanation of Assertion.
(xi) Consider,
$$\text{L.H.S = sin}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg) +\\2\text{tan}^{\normalsize-1}\bigg(\frac{1}{\sqrt{3}}\bigg)\\=\frac{\pi}{3} + 2×\frac{\pi}{6}\\=\frac{\pi}{6} + \frac{\pi}{3}\\=\frac{2\pi}{3}$$
= R.H.S. Hence Proved.
(xii) Given | A | = 4
$$\therefore\space\text{|-2A|} =(-2)^{3}|\text{A}|$$
= – 8 × 4 = – 32. Ans.
(xiii) (b) 1 and 2
Explanation :
x = 1 and x = 2 is non differentiable.
(xiv) Here,
$$\text{P(A) =}\frac{4}{5},\space\text{P(B)}=\frac{1}{3}\\\therefore\space\text{P(A')} =\frac{1}{5},\text{P(B')} =\frac{2}{3}\\\therefore\space\text{The required probability =}\\\text{P(A').P(B')} =\frac{1}{5}×\frac{2}{3}=\frac{2}{15}\space\textbf{Ans.}$$
(xv) Multiples of 5 are 5, 10, 15 and 20
Multiples of 7 are 7, 14
Total favourable events = 4 + 2 = 6
Total number of possible outcomes = 20
∴ Probability that the ball drawn is marked with a number multiple of 5 or 7
$$=\frac{6}{20}= \frac{3}{10}.\space\textbf{Ans.}$$
Answer 2.
(i) Here, R = {(a, b) : b = a + 1}
∴ R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6)}
⇒ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
(a) R is not reflexive as (a, a) ∉ R
(b) R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R
(c) R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R
OR
(ii) Given, f(x) = x2 + 4
Let f(x1) = f(x2)
⇒ x21 + 4 = x22 +4
⇒ x1 = x2
Thus, f(x) is one-one.
Since, x2 + 4 is a real number. Thus, for every y in the co-domain of f, there exists a number x in R+ such that f(x) = y = x2 + 4.
Thus, we can say that f(x) is onto.
Now, f(x) is one-one and onto. Hence, f(x) is invertible.
Let f(x) = y ⇒ x2 + 4 = y
$$\Rarr\space x^{2}= y-4\\\text{i.e\space} x =\sqrt{y-4}\\\text{Also,\space}x = f^{\normalsize-1}(y)\\\text{f}^{\normalsize-1}(y) =\sqrt{y-4}\space\\\textbf{Hence Proved.}$$
Answer 3.
Given : Equation of curve is,
y2 = px3 + q
Differentiating both sides w.r.t. x, we get
$$2y\frac{dy}{dx} =3px^{2}\\\Rarr\space\frac{dy}{dx} =\frac{3px^{2}}{2y}\\\therefore\space\bigg(\frac{dy}{dx}\bigg)_{(2,3)} =\frac{3p×2^{2}}{2×3}\\\text{or}\space m=\bigg(\frac{dy}{dx}\bigg)_{(2,3)} \\=\frac{12p}{6} = 2p$$
Since, y = 4x – 7 is the tangent to the curve at point (2, 3).
So, on comparing with y = mx + c, we get m = 4
Now, 2p = 4
$$\Rarr\space p =\frac{4}{2} = 2$$
Since, point (2, 3) lies on the curve,
∴ 32 = p × 23 + q
⇒ 9 = 2 × 8 + q
⇒ 9 – 16 = q
⇒ q = –7
Hence, p = 2 and q = – 7 Ans.
Answer 4.
We know that
sin–1(sin x) = x
$$\text{So,\space}\text{sin}^{\normalsize-1}\begin{bmatrix}\text{sin}\bigg(\frac{-17\pi}{8}\bigg)\end{bmatrix}\\=\text{sin}^{\normalsize-1}\begin{bmatrix}-\text{sin}\frac{17\pi}{8}\end{bmatrix}\\=\text{sin}^{\normalsize-1}\begin{bmatrix}-\text{sin}\bigg(2\pi + \frac{\pi}{8}\bigg)\end{bmatrix}\\=\text{sin}^{\normalsize-1}\bigg(-\text{sin}\frac{\pi}{8}\bigg)\\=\text{sin}^{\normalsize-1}\begin{bmatrix}\text{sin}\bigg(-\frac{\pi}{8}\bigg)\end{bmatrix}\\=-\frac{\pi}{8}\space\textbf{Ans.}$$
Answer 5.
(i) Given,
$$y =\frac{e^{2x} + e^{\normalsize-2x}}{e^{2x} - e^{-2x}}\\=\frac{e^{4x} +1}{e^{4x}-1}$$
On differentiating w.r.t. x, we get
$$\frac{dy}{dx} =\frac{4e^{4x}(e^{4x}-1)-4e^{4x}(e^{4x} +1)}{(e^{4x}-1)^{2}}\\=\frac{-8e^{4x}}{(e^{4x}-1)^{2}}\space\textbf{Ans.}$$
OR
(ii) Given,
x = a sin3t
$$\therefore\space\frac{dx}{dt} =\text{3a sin}^2 \text{t.cos t}$$
Also, y = a cos3t
$$\therefore\space\frac{dy}{dt} =\text{3a cos}^{2}t(-\text{sin t})\\\therefore\space\frac{dy}{dx} =\frac{-3a \text{cos}^{2}\text{t. sint}}{\text{3a sin}^{2}\text{t. cos t}}\\\begin{bmatrix}\because\space\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{bmatrix}\\\Rarr\frac{dy}{dx} =-\text{cot t.}\space\textbf{Ans.}$$
Answer 6.
Given, y = 2 cos (log x) + 3 sin (log x)
On differentiating both sides w.r.t. x, we get
$$\frac{dy}{dx} =-2\text{sin}\space(\text{log\space x}).\frac{1}{x} +\\\text{3 cos(log x)}.\frac{1}{x}\\\Rarr\space x\frac{dy}{dx} =\\-\text{2 sin (log x) + 3 cos (log x)}$$
Again, differentiating both sides w.r.t. x, we get
$$x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} =\\\text{-2 cos(log x)}.\frac{1}{x} -3\space\text{sin}(\text{log\space x}).\frac{1}{x}\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x.\frac{dy}{dx} =\\\text{-(2 cos(log x)) + 3 sin (log x)})\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} =-y\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + y = 0$$
Hence Proved.
Answer 7.
Given function is
$$f(x) =\frac{x^{4}}{4}-x^{3}-5x^{2} +24x +12\\\Rarr\space f'(x) =\frac{4x^{3}}{4}-3x^{2}-10x + 24$$
For critical points, put f ′(x) = 0
∴ x3 – 3x2 – 10x + 24 = 0
(x – 2) (x2 – x – 12) = 0
(x – 2) (x – 4) (x + 3) = 0
$$\Rarr\space x=2,4,\normalsize-3$$
Therefore, we have the intervals (– ∞, – 3), (– 3, 2), (2, 4) and (4, ∞).
Since f ’(x) > 0 in (– 3, 2) ∪ (4, ∞).
∴ f(x) is increasing in interval (– 3, 2) ∪ (4, ∞).
And f ′(x) < 0 in (– ∞, – 3) ∪ (2, 4)
∴ f(x) is decreasing in (– ∞, – 3) ∪ (2, 4). Ans.
Answer 8.
Sample space n(S) = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
(i) When A throws two heads
n(A) = (H, H)
n(A) = 1
$$\text{P(A)} =\frac{n(A)}{n(S)} =\frac{1}{4}\\\textbf{Ans.}$$
(ii) P(B does not throw two heads) =
$$1 -\frac{1}{4} =\frac{3}{4}.\space\textbf{Ans.}$$
(iii) If A starts the game, then A can win in either 1st or 3rd or 5th trial and so on.
$$\text{P(A wins the game)}=\\\frac{1}{4} + \bigg(\frac{3}{4}\bigg)^{2}×\frac{1}{4} +\bigg(\frac{3}{4}\bigg)^{4}×\frac{1}{4} +.....\\=\frac{1}{4}\begin{bmatrix}1 + \bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4} + ......\end{bmatrix}\\=\frac{1}{4}\begin{bmatrix}\frac{1}{1 -\frac{9}{16}}\end{bmatrix}\\\begin{bmatrix}\because\space\text{For infinite G.P.S.} =\frac{a}{1-r}\end{bmatrix}\\=\frac{1}{4}×\frac{16}{7} =\frac{4}{7}\qquad\textbf{Ans.}$$
(iv) If B starts the game and A wins, then A can win in either 2nd or 4th or 6th trial and so on.
∴ P(A wins the game) =
$$\frac{3}{4}×\frac{1}{4} +\bigg(\frac{3}{4}\bigg)^{3}×\frac{1}{4} +\\\bigg(\frac{3}{4}\bigg)^{5}×\frac{1}{4} + .......\\=\frac{3}{16} +\frac{3}{16}×\bigg(\frac{3}{4}\bigg)^{2}+\\\frac{3}{16}×\bigg(\frac{3}{4}\bigg)^{4} +........\\=\frac{3}{16}\begin{bmatrix}1 +\bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4} +.....\end{bmatrix}\\=\frac{3}{16}×\frac{1}{1-\frac{9}{16}} =\frac{3}{16}×\frac{16}{7}\\=\frac{3}{7}\qquad\textbf{Ans.}$$
Answer 9.
(i) Let
$$\text{I =}\frac{1}{2}\int(\text{2 cos 2x cos 4x})\\\space\text{cos 6x}\space dx\\\Rarr\space\text{I =}\frac{1}{2}\int\text{(cos 6x + cos 2x)}\\\text{cos 6x dx}$$
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
$$=\frac{1}{2}\int\text{(cos 6x cos6x dx)} +\\\frac{1}{2}\int\text{cos 2x cos6x dx}\\=\frac{1}{4}\int\text{2 cos 6x cos 6x dx} +\\\frac{1}{4}\int\text{2 cos 2x cos6x dx}\\=\frac{1}{4}\int(\text{cos 12 x + cos 0})dx +\\\frac{1}{4}\int\text{cos8x cos 4x dx}\\=\frac{1}{4}\\\begin{bmatrix}\int(\text{cos 12 x +1 +cos 8x +cos 4x})dx\end{bmatrix}$$
$$=\frac{1}{4}\begin{bmatrix}\frac{\text{sin 12 x}}{12} + \frac{\text{sin 8 x}}{8} + \frac{\text{sin 4 x}}{4}\end{bmatrix}\\+\space\text{C}\space\textbf{Ans.}$$
OR
$$\textbf{(ii)\space}\int\frac{\text{cos 2x - cos 2}\alpha}{\text{cos x - cos}\alpha}dx =\\\int\frac{2 \text{cos}^{2}x - 1-2\space\text{cos}^{2}\alpha +1}{\text{cos x - cos}\alpha}dx\\\lbrack\because\space\text{cos}2\theta = 2 \text{cos}^{2}\theta -1\rbrack\\=\int\frac{\text{2 cos}^{2}x - 2 \text{cos}^{2}\alpha}{\text{cos x - cos}\alpha}dx\\=\int\frac{\text{2(cos x - cos}\alpha)(\text{cos}x + cos\alpha)}{(\text{cos x - cos}\alpha)}dx\\= 2\int(\text{cos x + cos}\space\alpha)dx\\= 2\begin{bmatrix}\int\text{cos x dx +}\int\text{cos}\alpha dx\end{bmatrix}$$
= 2 sin x + 2x cosα + C Ans.
Answer 10.
(i) Given differential equation is
$$\frac{dy}{dx} = 1+x^{2}+ y^{2} +x^{2}y^{2}$$
= (1 + x2) + y2 (1 + x2)
$$\Rarr\space\frac{dy}{dx} =(1 +x^{2})(1 +y^{2})\\\Rarr\space\frac{dy}{1 +y^{2}}=(1 + x^{2})dx$$
On integrating both sides, we have
$$\int\frac{dy}{1 +y^{2}} =\int(1 + x^{2})dx\\\text{tan}^{-1}y = x + \frac{x^{3}}{3} + \text{C}\\\text{...(i)}$$
put y = 1 and x = 0 in equation (i),
tan–1 1 = 0 + 0 + C
$$\text{C} =\frac{\pi}{4}\\\text{Equation (i) becomes:}\\\text{tan}^{\normalsize-1}y = x + \frac{x^{3}}{3} + \frac{\pi}{4}\\\Rarr\space y =\text{tan}\bigg(x + \frac{x^{3}}{3} + \frac{\pi}{4}\bigg)$$
is the required particular solution of given equation. Ans.
OR
(ii) The given differential equation is
(1 – y2) (1 + log x) dx + 2xy dy = 0
$$\frac{(\text{1 + log x})}{x}dx =\frac{-2y}{(1 - y^{2})}dy$$
On integrating both sides, we have
$$\int\frac{1 + log x}{x}dx =\int\frac{-2y}{(1-y^{2})}dy$$
In first integral,
put 1 + log x = t
$$\Rarr\space\frac{1}{x}dx= dt$$
Also in second integral,
put 1 – y2 = u
$$\Rarr\space\text{-2ydy} = du\\\therefore\qquad\int t.dt =\int\frac{1}{u}du\\\Rarr\space\frac{t^{2}}{2} -\text{log}|u| = \text{C}\\\text{or}\space\frac{1}{2}(1 + log\space x)^{2} -\\\text{log}|1 - y^{2}| = \text{C}$$
It is given that y = 0 when x = 1
$$\text{So,\space}\frac{1}{2}(1 + \text{log 1})^{2} -\text{log}|1 - 0^{2}| \\=\text{C}\\\Rarr\space\text{C} =\frac{1}{2}\\\therefore\space\frac{(1 + log x)^{2}}{2} -\\\text{log}|1-y^{2}| =\frac{1}{2}$$
or (1 + log x)2 – 2 log|1 – y2| = 1
It is the required particular solution. Ans.
Answer 11.
Let the number of children be x and amount distributed by Ishan to each child be `y.
As per the given information
(x – 8) (y + 10) = xy
⇒ xy + 10x – 8y – 80 = xy
⇒ 5x – 4y = 40 ...(i)
⇒ (x + 16) (y – 10) = xy
⇒ –10x + 16y + xy – 160 = xy
⇒ –5x + 8y = 80 ...(ii)
Now, we can express equation (i) and equation
(ii) in matrix form as follows:
$$\text{A =}\begin{bmatrix}5 &\normalsize-4\\\normalsize-5 &8\end{bmatrix},\text{X} =\begin{bmatrix}x\\y\end{bmatrix}\\\text{and B =}\begin{bmatrix}40\\80\end{bmatrix}\\\Rarr\space \text{X = A}^{\normalsize-1}\text{B}\\\Rarr\space\text{X} =\frac{|\text{adj A}|}{|\text{A}|}\text{B}$$
Now for adj A
A11 = 8, A12 = 4
A21 = 5, A22 = 5
$$\therefore\space\text{adjA =}\begin{bmatrix}8 &4\\5 &5\end{bmatrix}$$
and | A | = 40 – 20 = 20
$$\Rarr\space X =\frac{1}{20}\begin{bmatrix}8 &4\\5 &5\end{bmatrix}\begin{bmatrix}40\\80\end{bmatrix}$$
$$\Rarr\space\text{X} =\frac{1}{20}\begin{bmatrix}640\\600\end{bmatrix}\\\Rarr\space\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}32\\30\end{bmatrix}$$
$$\Rarr\space\text{x = 32, y = 30}$$
Number of children = 32 Ans.
Amount donate by Ishan = ₹(32 × 30) = ₹960 Ans.
Answer 12.
(i) Let ex3 = t
∴ 3x2 ex3 dx = dt
$$\therefore\space\int x^{2}(e^{x^{3}})\text{cos}(2e^{x^{3}})dx=\\\int\frac{1}{3}\text{cos 2t dt}\\=\frac{1}{3}.\frac{1}{2}\text{sin 2t + C}\\=\frac{1}{6}\text{sin}(2e^{x^{3}}) + \text{C}\space\textbf{Ans.}$$
OR
$$\text{(ii)\space}\int x(\text{tan}^{\normalsize-1}x)^{2}dx\\\text{Let}\space\text{tan}^{\normalsize-1}x = t\\\Rarr\space x =\text{tan t}$$
dx = sec2 t dt
$$\int x(\text{tan}^{\normalsize-1}x)\space dx =\\\int\space t^{2}.\text{tan t sec}^{2}t\space dt\\\lbrack\text{Integrating by parts}\rbrack\\= t^{2}.\frac{\text{tan}^{2}t}{2}-\int\frac{\text{2t tan}^{2}t}{2}dt\\=\frac{1}{2} t^{2}\text{tan}^{2}t -\int t(\text{sec}^{2}t -1)dt\\=\frac{1}{2}t^{2}\text{tan}^{2}t -\int\text{t sec}^{2}t\space dt\\+\int t\space dt\\=\frac{1}{2} t^{2}\text{tan}^{2}t + \frac{t^{2}}{2} -\int t\space sec^{2}t dt $$
$$=\frac{1}{2}t^{2}(\text{tan}^{2}t +1) -\\\begin{Bmatrix}\text{t tan t} -\\\int\text{tan t} dt\end{Bmatrix}\\=\frac{1}{2}t^{2}(\text{tan}^{2} + 1) -\\\text{t tan t + log}|\text{sec t}| +\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} -\\x\text{tan}^{\normalsize-1}x + \text{log}(\sqrt{1 + x}^{2}) +\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} - x\space\text{tan}^{\normalsize-1}x+\\\text{log}(\sqrt{1 + x^{2}})+\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2}- $$
$$\text{x tan}^{\normalsize-1} x +\text{log}(\sqrt{1 +x^{2}})+ \text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2}-\\\text{x tan}^{\normalsize-1}x +\frac{1}{2}\text{log}(1 + x^{2}) + \text{C}\space\\\textbf{Ans.}$$
Answer 13.
(i) Given,
$$(1 + x^{2})\frac{dy}{dx} \\=(e^{m} tan^{\normalsize-1}x - y)\\\frac{dy}{dx} =\frac{e^{m tan^{\normalsize-1}x}}{1 +x^{2}}-\frac{y}{1 +x^{2}}\\\Rarr\space\frac{dy}{dx} + \frac{1}{1 + x^{2}}y\\=\frac{e^{m tan^{\normalsize-1}x}}{1 +x^{2}}$$
This equation is of the form
$$\frac{dy}{dx} +\text{Py = Q(x)}\\\text{where P}=\frac{1}{1 +x^{2}}\text{and}\space\\\text{Q(x)} =\frac{e^{m tan^{\normalsize-1}x}}{1 + x^{2}}\\\text{I.F. = e}^{\int\text{P dx}} =\int e^{\bigg(\frac{1}{1+x^{2}}\bigg)dx}\\= e^{\text{tan}\normalsize-1\space x}$$
Hence, solution of linear differential equation is given by
$$y×\text{I.F. =}\int\text{I.F × Q(x)dx}\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x} =\\\int e^{\text{tan}^{\normalsize-1}x}.\frac{e^{m\space tan^{-1}x}}{1 +x^{2}}dx\\\Rarr\space y× e^{tan^{\normalsize-1}x} \\=\int\frac{e^{(1 + m)tan^{\normalsize-1}x}}{1 +x^{2}}dx$$
Let e(1 + m) tan -1 x = t
$$\Rarr\space(1 + m)\frac{e^{(1+m)\text{tan}^{\normalsize-1}x}}{1 +x^{2}}dx\\= dt\\\therefore\space y×e^{tan^{\normalsize-1}x}\\=\int\frac{1}{1+m}dt\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x}\\=\frac{1}{1 + m}t + \text{C}\\=\frac{1}{1 +m} e^{(1+m)\text{tan}^{\normalsize-1}x} +\text{C}$$
Now, when x = 0, then y = 1,
$$\therefore\space\text{1× e}^{\text{tan}^{\normalsize-1}0}\\=\frac{1}{\text{1 + m}}e^{(1+m)\text{tan}^{\normalsize-1}0} +\text{C}\\\Rarr\space 1×e^{0} =\frac{1}{1+m}e^{0} +\text{C}\\\Rarr\space1 -\frac{1}{\text{1 -m}} =\text{C}\\\therefore\space\text{C} = \frac{m}{m+1}$$
Hence, particular solution of given differential equation is
$$y×e^{\text{tan}^{\normalsize-1}x} =\\\frac{1}{m+1}e^{(1+m)\text{tan}^{\normalsize-1}x} + \frac{m}{m+1}$$
[where m ≠ – 1] Ans.
If m = – 1 then particular solution of the given differential equation is
$$y×e^{\text{tan}^{\normalsize-1}x} =\\\int e^{tan^{\normalsize-1}x}.\frac{e^{-\text{tan}^{\normalsize-1}x}}{1 +x^{2}}dx\\\Rarr\space y×e^{tan^{\normalsize-1}x} =\\\int\frac{e^{0}}{1 + x^{2}}dx =\int\frac{dx}{1 +x^{2}}\\\Rarr\space y×e^{tan^{\normalsize-1}x}\\=\text{tan}^{-1}x +\text{C}$$
When x = 0, then y = 1,
∴ 1 × etan-1 0 = tan–1 0 + C
$$\Rarr\space 1×e^{0} = 0 +\text{C}\\\Rarr\space\text{C = 1}$$
Hence, particular solution of differential equation is
y × etan−1x = tan−1 x+1. Ans.
OR
(ii) Let ABC be an isosceles triangle with AB = AC and a circle of radius r unit with centre I is inscribed in ΔABC.
Now, AD ⊥ BC
∴ BD = DC
Let AE = AF = x units,
[Tangents drawn from an external point are equal in length]
CE = CD = BD = y units
∴ BD = BF = y units
Perimeter of triangle = AB + BC + AC
= x + y + 2y + x + y
P = 2x + 4y ...(i)
In ΔAIE, ∠AEI = 90°
$$\therefore\space\frac{\text{AE}}{\text{IE}} =\text{cot}\space\theta$$
AE = r cot θ = x ...(ii)
In ΔADC, ∠ADC = 90°
$$\frac{y}{x+y} =\text{sin}\space\theta\\\Rarr\space\text{x sin θ + y sin θ = y}\\\Rarr\space\text{r cot}\space\theta\space\text{sin}\space\theta = y(\text{1 - sin}\space\theta)\\\lbrack\text{Using (ii)}\rbrack\\\Rarr\space\frac{\text{r cos}\theta}{\text{1 - sin}\theta} = y\space\text{...(iii)}$$
From equations (i), (ii) and (iii),
P = 2x + 4y
$$\text{P = 2r cot}\theta + \frac{4r\space cos\space\theta}{\text{1 - sin\space}\theta}\\\=\frac{dP}{d\theta} =\frac{d}{d\theta}\text{2r cot}\theta +\frac{d}{d\theta}\bigg(\frac{4r cos\space\theta}{\text{1 - sin\space}\theta}\bigg)\\=-\text{2r\space cosec}^{2}\theta + 4r\\\begin{bmatrix}\frac{(1 - sin\space\theta)(-\text{sin}\theta) -\text{cos}\theta(-\text{cos}\theta)}{(\text{1 - sin}\theta)^{2}}\end{bmatrix}\\= -\text{2r cosec}^{2}\theta + 4r\\\begin{bmatrix}\frac{-\text{sin}\space\theta + \text{sin}^{2}\theta + \text{cos}^{2}\theta}{(\text{1 - sin}\theta)}\end{bmatrix}\\=\text{-2r cosec}^{2}\theta +4r\\\begin{bmatrix}\frac{\text{1 - sin}\theta}{(\text{1 - sin}\theta)^{2}}\end{bmatrix}\\=\text{-2r cosec}^{2}\theta + 4r\\\begin{bmatrix}\frac{\text{1 - sin}\theta}{(\text{1 - sin}\theta)^{2}}\end{bmatrix}$$
$$=\space\text{-2r cosec}^{2}\theta + \frac{4r}{\text{1 - sin}\theta}\\= 2r\begin{bmatrix}\frac{\normalsize-1}{\text{sin}^{2}\theta} + \frac{2}{\text{1 - sin}\theta}\end{bmatrix}\\= 2r\space\begin{bmatrix}\frac{-1 + \text{sin}\theta + 2\text{sin}^{2}\theta}{\text{sin}^{2}\theta(1 - sin)\theta}\end{bmatrix}\\\therefore\qquad\frac{\text{dP}}{\text{d}\theta} =\frac{\text{2r}(\text{2 sin}\theta-1)(\text{sin}\space\theta +1)}{\text{sin}^{2}\theta(\text{1 - sin}\space\theta)}$$
For maxima and minima,
$$\text{Put,\qquad}\frac{\text{dP}}{\text{d}\theta} = 0\\\therefore\qquad\frac{\text{2r(2 sin}\theta-1)(\text{sin}\space\theta +1)}{\text{sin}^{2}\theta(\text{1 - sin}\space\theta)}\\=0$$
2r(2sinθ − 1)(sinθ + 1) = 0
∴ 2 sin θ – 1 = 0
$$\Rarr\space\text{sin}\theta =\frac{1}{2}\\\Rarr\space\theta = 30\degree\space\text{or}\frac{\pi}{6},$$
or sin θ + 1 = 0
$$\Rarr\space\text{sin}\space\theta =-1\\\Rarr\space\theta =-\frac{\pi}{2}$$
$$\because\space\theta =-\frac{\pi}{2}\space\\\text{is not possible}\\\therefore\space\theta = 30\degree \text{or}\frac{\pi}{6}\\\text{Now,\qquad}\frac{d^{2}\text{P}}{d\theta^{2}}=\\\frac{d}{d\theta}\begin{bmatrix}\text{-2r cosec}^{2}\theta + \frac{4r}{\text{1 - sin}\theta}\end{bmatrix}\\= \text{-2r(2 cosec}\space\theta)(\text{- cosec}\theta\space cot\theta)-4r.\\\frac{1}{(\text{1 - sin}\theta)^{2}}(\text{ - cos}\space\theta)\\=\space 4r\begin{bmatrix}\text{cosec}^{2}\theta cot\theta + \frac{\text{cos}\theta}{(\text{1 - sin}\space\theta)^{2}}\end{bmatrix}\\= 4r\begin{bmatrix}\text{cosec}^{2}\frac{\pi}{6}\text{cot}\frac{\pi}{6} + \frac{\text{cos}\frac{\pi}{6}}{\bigg(1 - \text{sin}\frac{\pi}{6}\bigg)}\end{bmatrix}$$
$$= 4r(4.\sqrt{3} + 2\sqrt{3})\\\therefore\space\bigg(\frac{d^{2}\text{P}}{d\theta}\bigg)_{\theta =\frac{\pi}{6}}\gt0$$
Hence, perimeter is minimum at π/6.
Now, Perimeter of ΔABC = 2x + 4y
$$ = \text{2r cot}\theta + \frac{\text{4r cos}\theta}{\text{1 - sin}\theta}\\=\text{2r cot}\frac{\pi}{6} + \frac{\text{4r cos}\frac{\pi}{6}}{\text{1 - sin}\frac{\pi}{6}}\\= \text{2r.}\sqrt{3} +\frac{4r.\frac{\sqrt{3}}{2}}{1 -\frac{1}{2}}\\= r(2\sqrt{3} + 4\sqrt{3}) = 6\sqrt{3r}$$
$$\therefore\space\text{Least perimeter of}\\\Delta\text{ABC is}\space 6\sqrt{3r}\space\text{units.}$$
Hence Proved.
Answer 14.
(i) n(W|A) = 2
n(S) = 5
$$\text{P(W|A)} =\frac{2}{5}\\\text{Similarly}\\\text{P(W|B) =}\frac{1}{5},\space\text{P(W|C)} =\frac{4}{5}$$
$$\text{(ii)\space Given P(A) =}\frac{2}{5},\space\text{P(B) =}\frac{2}{5},\\\text{P(C) =}\frac{1}{5}$$
P(Probability that the drawn ball is white) = P(A).P(W|A) + P(B).P (W|B) + P(C).P(W|C)
(∵ Theorem of total probability)
$$= \frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} +\frac{1}{5}×\frac{4}{5}\\=\frac{4}{25} +\frac{2}{25} +\frac{4}{25}\\=\frac{10}{25} = \frac{2}{5}\qquad\textbf{Ans.}$$
(iii) Let E denote that drawn ball is black then
P(E) = P(A).P(E|A) + P(B).P(E|B) + P(C).P(E|C)
$$\text{P(E) =}\frac{2}{5}×\frac{3}{5} + \frac{2}{5}×\frac{4}{5} +\frac{1}{5}×\frac{1}{5}\\\text{P(E) =}\frac{6}{25} +\frac{8}{25} +\frac{1}{25}\\=\frac{15}{25} =\frac{3}{5}\qquad\textbf{Ans.}$$
(iv) By Baye’s theorem, we have
$$\text{P(C|W) =}\\\frac{\text{P(C)P(W|C)}}{\text{P(A)P(W|A)} + \text{P(B)P(W|B) + P(C)P(W|C)}}\\=\\\frac{\frac{1}{5}×\frac{4}{5}}{\frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} + \frac{1}{5}×\frac{4}{5}}\\=\frac{4}{10} =\frac{2}{5}\qquad\textbf{Ans.}$$
Section-B
Answer 15.
$$\text{(i)\space(c)\space}\vec{b},\vec{c}\space\text{and\space}\vec{d}\\\text{Vector\space}\vec{b},\vec{c}\space\text{and}\space\vec{d}\\\text{start from same point.}$$
(ii) (b) 60°
Angle between two planes a1 x + b1 y + c1 = 0 and a2x + b2y + c2 = 0 is given by,
$$\text{cos}\space\theta =\begin{vmatrix}\frac{a_{1}a_{2} +b_{1}b_{2} +c_{1}c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} +c_{1}^{2}}\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}\end{vmatrix}$$
Here a1= 2, b1 = – 1, c1 = 1
and a2 = 1, b2 = 1, c2 = 2
$$\therefore\space\text{cos}\space\theta = \begin{vmatrix}\frac{2(1) + (\normalsize-1)(1) + (1)(2)}{\sqrt{2^{2} + 1^{2} +1^{2}}\sqrt{1^{2} + 1^{2} + 2^{2}}}\end{vmatrix}\\\text{cos}\space\theta =\begin{vmatrix}\frac{2-1+2}{\sqrt{6}\sqrt{6}}\end{vmatrix}=\frac{1}{2}\\\therefore\theta =60\degree =\frac{\pi}{3}$$
$$\text{(iii)\space For}\space\bigg(\frac{1}{\sqrt{2}},\frac{1}{2},k\bigg)\space\text{to represent}$$
direction cosines, we should have
$$\bigg(\frac{1}{\sqrt{2}}\bigg)^{2} + \bigg(\frac{1}{2}\bigg)^{2}+ k^{2}=1\\\text{or}\space\frac{1}{2} + \frac{1}{4}+k^{2} =1\\\Rarr\space k =\pm\frac{1}{2}.\\\text{(iv)\space}\vec{a} =2\hat{i} -\hat{j} + \hat{k}, \vec{b} =\hat{i} + 2\hat{j} - 3\hat{k}\\\text{and}\space\vec{c} = 3\hat{i} + x\hat{j} +5\hat{k}\\\text{If\space}\vec{a},\vec{b}\space\text{and}\space\vec{c}\space\text{are coplanar, then}\\\begin{vmatrix}2 &\normalsize-1 &1\\1 &2 &\normalsize-3\\ 3 &x &5\end{vmatrix} = 0\\\Rarr\space 2\begin{vmatrix}2 &\normalsize-3\\x &5\end{vmatrix} + 1\begin{vmatrix}1 &\normalsize-3\\3 &5\end{vmatrix} +\\ 1\begin{vmatrix}1 &2\\ 3 &x\end{vmatrix} = 0$$
$$\Rarr\space\text{2(10 + 3x) + (5 + 9)}\\\text{+ (x - 6)} = 0\\\Rarr\space\text{20 + 6x + 14 + x – 6 = 0}\\\Rarr\space\text{7x = 28}\\\therefore\space\text{x = 4.}\qquad\textbf{Ans.}\\\text{(v)\space}\vec{\text{OP}×\text{OQ}} =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 &2 &1\\2 &1 &3\end{vmatrix}\\=\space 5\hat{i} -\hat{j} -3\hat{k}\\\vec{\text{PQ}}×\vec{\text{PR}} =\begin{vmatrix}\hat{i} &\hat{j}&\hat{k}\\1 &\normalsize-1 &2\\\normalsize-2 &\normalsize-1 &1\end{vmatrix}\\=\hat{i} -5\hat{j} - 3\hat{k}$$
Required angle cos θ =
$$\frac{5+5+9}{\sqrt{25 +1 +9}\sqrt{1 + 25 +9}}\\\text{cos}\space\theta =\frac{19}{35}\\\theta =\text{cos}^{\normalsize-1}\bigg(\frac{19}{35}\bigg)\space\textbf{Ans.}$$
Answer 16.
(i) The given lines are
$$\frac{x+1}{\normalsize-3} =\frac{y-3}{2} =\frac{z+2}{1}\space\text{and}\\\frac{x}{1} =\frac{y-7}{\normalsize-3} =\frac{z+7}{2}$$
Direction ratios of the given lines are (– 3, 2, 1) and (1, – 3, 2) and these are not proportional.
∴ Given lines are not parallel.
Hence, these lines are intersecting. Hence Proved.
If lines are coplanar then
$$\begin{vmatrix}x_{2} -x_{1} &y_{2} -y_{1} &z_{2}-z_{1}\\a_{1} &b_{1} &c_{1}\\a_{2} &b_{2} &c_{2}\end{vmatrix}\\ = 0\\\begin{vmatrix}0-(\normalsize-1) &7-3 &-7-(-2)\\\normalsize-3 &2 &1\\1 &\normalsize-3 &2\end{vmatrix}\\= 0\\\begin{vmatrix}1 &4 &\normalsize-5\\\normalsize-3 &2 &1\\ 1 &\normalsize-3 &2\end{vmatrix} = 0\\\Rarr\space 1\begin{vmatrix}2 &1\\\normalsize-3 &2\end{vmatrix} -4\begin{vmatrix}-3 &1\\1 &2\end{vmatrix} +\\$$
$$(\normalsize-5)\begin{vmatrix}\normalsize-3 &2\\1 &\normalsize-3\end{vmatrix} = 0\\\Rarr\space\text{(4+3) - 4(-6-1) -5(9-2)}\\= 0\\\Rarr\space\text{7 + 28 -35} = 0$$
Hence, the given lines are coplanar.
Hence Proved.
Equation of the plane containing these lines is
$$\begin{vmatrix}x -x_{1} &y-y_{1} &z-z_{1}\\x_{2} -x_{1} &y_{2}-y_{1} &z_{2}-z_{1}\\a_{1} &b_{1} &c_{1}\end{vmatrix}\\= 0\\\begin{vmatrix}x-(\normalsize-1) &y-3 &z-(\normalsize-2)\\0-(\normalsize-1) &7-3 &-7-(\normalsize-2)\\\normalsize-3 &2 &1\end{vmatrix}\\= 0\\\begin{vmatrix}x+1 &y-3 &z+2\\1 &4 &\normalsize-5\\\normalsize-3 &2 &1\end{vmatrix}\\= 0\\\Rarr\space (x+1)\begin{vmatrix}4 &\normalsize-5\\2 &1\end{vmatrix}-(y-3)\begin{bmatrix}1 &\normalsize-5\\\normalsize-3 &1\end{bmatrix} +$$
$$(z+2)\begin{vmatrix}1 &4\\\normalsize-3 &2\end{vmatrix} = 0$$
$$\Rarr\space (x+1)(4+10)-(y-3)(1- 15) +\\(z+2)(2+12) = 0\\\Rarr\space 14(x+1)-(\normalsize-14)(y-3)\\+ 14(z+2) = 0\\\Rarr\space\text{14x + 14 + 14y – 42 +}\\\text{14z + 28 = 0}\\\Rarr\space\text{14x + 14y +14z = 0}\\\Rarr\space\text{x + y +z} = 0$$
Hence, equation of the plane is x + y + z = 0. Ans.
OR
(ii) Given line is
5x – 25 = 14 – 7y = 35z
$$\Rarr\space\text{5(x - 5)} =-7(y-2) = 35z\\\Rarr\space\frac{x-5}{\frac{1}{5}} =\frac{y-2}{\frac{\normalsize-1}{7}} =\frac{z-0}{\frac{1}{35}}\\\Rarr\space \frac{x-5}{7} =\frac{y-2}{-5} =\frac{z-0}{1}$$
Direction ratios of this line are 7, – 5, 1.
∴ Vector equation of the line which passes through the point A(1, 2, – 1) and whose direction ratios are proportional to 7, – 5, 1 is
$$\vec{r} =\hat{i} + 2\hat{j} - \hat{k} +\\\lambda(7\hat{i} -5\hat{j} +\hat{k})\space\textbf{Ans.}$$
Answer 17.
(i) Given : Coordinates of vertices of ΔABC are A(1, 2, 3), B (2, – 1, 4) and C(4, 5, – 1)
$$\therefore\space\vec{\text{AB}} = \hat{i} -3\hat{j} +\hat{k}\\\text{and}\space\vec{\text{AC}} = 3\hat{i} + 3\hat{j}-4\hat{k}$$
We know that,
$$\text{ar}(\Delta\text{ABC}) =\frac{1}{2}\begin{vmatrix}\vec{\text{AB}}×\vec{\text{AC}}\end{vmatrix}\\\text{Now,\qquad}\vec{\text{AB}}×\vec{\text{AC}} =\\\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\ 1 &\normalsize-3 &1\\3 &3 &\normalsize-4\end{vmatrix}\\=\space\hat{i}(12-3) -\hat{j}(-4-3) +\hat{k}(3+9)\\ =9\hat{i} + 7\hat{j}+12\hat{k}\\\therefore\space|\vec{\text{AB}}×\vec{\text{AC}}| =\sqrt{9^{2} + 7^{2} + (12)^{2}}\\=\sqrt{274}\\\text{So,}\qquad ar(\Delta \text{ABC})=\\\frac{1}{2}\sqrt{274}\space\text{sq. units}\qquad\textbf{Ans.}$$
OR
(ii) Here,
$$\vec{a} =(\hat{i} +\hat{j} +\hat{k}).\hat{n}\\\text{is unit vector}\\\vec{b} = 2\hat{i} + 4\hat{j} -5\hat{k}\\\vec{c} =\lambda\hat{i} +2\hat{j} +3\hat{k}\\\vec{b} +\vec{c} =(2 +\lambda)\hat{i} + 6\hat{j}-2\hat{k}\\\text{Then,\space}\hat{n} =\frac{(2+\lambda)\hat{i} + 6\hat{j}-2\hat{k}}{\sqrt{(2 + \lambda)^{2} +36 +4}}\\\text{Given,\space}\vec{a}.\hat{n} = 1\\(\hat{i} + \hat{j} +\hat{k}).\begin{pmatrix}\frac{(2 +\lambda)\hat{i} +6\hat{j}-2\hat{k}}{\sqrt{(2 +\lambda)^{2} +40}}\end{pmatrix}\\= 1\\\Rarr\space(2 +\lambda) +6-2 =\\\sqrt{(2 + \lambda)^{2} +40}$$
$$\lbrack\because\space\hat{i}.\hat{j}=1, \hat{j}.\hat{j} =1, \hat{k}.\hat{k}=1\rbrack\\\Rarr\space (2 + \lambda) +4 =\\\sqrt{(2 + \lambda)^{2} +40}\\\Rarr\space\lambda +6 =\\\sqrt{(2 + \lambda)^{2} +40}$$
On squaring both sides, we get
(6 + λ)2 = (2 + λ)2 + 40
$$\Rarr\space 36 +\lambda^{2} + 12\lambda =\\ 4 + \lambda^{2} +4\lambda+40\\\Rarr\space 36 +12\lambda -4-4\lambda -40 = 0\\\Rarr\space 8\lambda -8 = 0\\\Rarr\space\lambda = 1\\\text{Then,\space}\vec{b} + \vec{c} =\\(2 +\lambda)\hat{i} + 6\hat{j} -2 \hat{k}\\ = 3\hat{i} +6\hat{j}-2\hat{k}\\\text{unit vector along}\space(\vec{b} +\vec{c})\\=\frac{3\hat{i} +6\hat{j}-2 \hat{k}}{\sqrt{49}}\\=\frac{3\hat{i} +6\hat{j}-2\hat{k}}{7}\qquad\textbf{Ans.}$$
Answer 18.
Given, {(x, y) : 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 2}
⇒ y ≤ x2 ...(i)
y ≤ x ...(ii)
x ≤ 2 ...(iii)
x ≥ 0 ...(iv)
y ≥ 0 ...(v)
Considering inequalities as equation :
y = x2, y = x, x = 2, x = 0, y = 0
Solving y = x2 and y = x
⇒ x2 = x
⇒ x(x – 1) = 0
⇒ x = 0, x = 1
∴ y = 0, y = 1
∴ Points of intersection of curve (i) and the (ii) are (0, 0) and (1, 1).
$$\therefore\space\text{Required area =}\int^{1}_{0}\text{(y of the parabola) dx} +\\\int^{2}_{1}\text{(y of the line)}dx\\\therefore\text{Required area =}\\\int^{1}_{0} x^{2}dx + \int^{2}_{1}\text{x dx}\\=\begin{bmatrix}\frac{x^{3}}{3}\end{bmatrix}^{1}_{0} + \begin{bmatrix}\frac{x^{2}}{2}\end{bmatrix}^{2}_{1}\\=\bigg(\frac{1}{3}-0\bigg) +\bigg(\frac{4}{2}-\frac{1}{2}\bigg)\\=\frac{1}{3} + \bigg(2 -\frac{1}{2}\bigg)$$
$$=\frac{1}{3} +\frac{3}{2} =\frac{2 +9}{6}\\=\frac{11}{6}\space\text{square units.}$$
Section-C
Answer 19.
(i) (b) ₹30.015
C(x) = 0.005x3 – 0.02x2 + 30x + 5000
$$\text{M.C. =}\frac{d}{dx}\lbrace\text{C(x)}\rbrace =\\\frac{d}{dx}(0.005x^{3} - 0.02x^{2} +30x + 5000)$$
= 0.015x2 – 0.04x + 30 + 0
∴ M.C. = 0.015x2 – 0.04x + 30
(M.C.)x = 3 = 0.015(3)2 – 0.04(3) + 30
= 0.135 – 0.120 + 30
= 30.015
∴ (M.C.)x = 3 = ₹30.015. Ans. (b)
(ii) (a)
(iii) C(x) = 0.007x3 – 0.003x2 + 15x + 4000
$$\text{M.C. =}\frac{d}{dx}\lbrace\text{C(x)}\rbrace =\\\frac{d}{dx}(0.007 x^{3} -0.003x^{2} + 15x + 4000)$$
MC at x = 17
M.C. = 0.021x2 – 0.006x + 15 + 0
M.C. = 0.021(17)2 – 0.006(17) + 15
M.C. = 0.021(289) – 0.006(17) + 15
M.C. = 6.069 – 0.102 + 15
M.C. = ₹20.967. Ans.
(iv) (c) Regresion
(v) Given, X = 0.85Y and Y = 0.89X
∴ bxy = 0.85 and byx = 0.89
Coefficient of correlation is given as,
$$r =\pm\sqrt{b_{xy} × b_{yx}}\\=\pm\sqrt{0.85×0.89}\\=\pm\sqrt{0.7565}\\=\pm 0.87\\\because\space b_{xy}.b_{yx}\gt 0\\\therefore\space \text{r = 0.87}\space\textbf{Ans.}$$
Answer 20.
(i) Given, Cost price of x items =
$$₹\bigg(\frac{x}{5} +500\bigg)\\\text{Selling price of x items =}\\₹\bigg(5 -\frac{x}{100}\bigg)x\\= ₹\bigg(5x -\frac{x^{2}}{100}\bigg)$$
∴ Profit = S.P. – C.P.
$$= ₹\bigg(5x -\frac{x^{2}}{100}-\frac{x}{5}-500\bigg)\\\text{P(x) =}\frac{24 x}{5} -\frac{x^{2}}{100}-500$$
Differentiating w.r.t. x, we get
$$\frac{dP}{dx} =\frac{d}{dx}\bigg(\frac{24x}{5}\bigg) -\frac{d}{dx}\bigg(\frac{x^{2}}{100}\bigg)-\\\frac{d}{dx}(500)\\\frac{\text{dP}}{\text{dx}} =\frac{24}{5} -\frac{x}{50}-0$$
For P to be maximum,
$$\frac{dP}{dx} =0\\\therefore\space\frac{24}{5}-\frac{x}{50} = 0\\\Rarr\space\frac{x}{50} = \frac{24}{5}\\\Rarr\space x =\frac{24×50}{5} = 240\\\Rarr\space\frac{d^{2}\text{P}}{dx^{2}} =\frac{d}{dx}\bigg(\frac{24}{5}\bigg) -\frac{d}{dx}\bigg(\frac{x}{50}\bigg)\\\Rarr\space\frac{d^{2}\text{P}}{dx^{2}} =\frac{\normalsize-1}{50}\\\because\space\frac{d^{2}\text{P}}{\text{dx}^{2}}\lt0$$
∴ Profit is maximum when 240 items are sold.
Maximum profit at x = 240, Ans.
$$\text{P}(240) =\frac{24}{5}×240 -\\\frac{240×240}{100}-500$$
= 24 × 48 – 576 – 500
= 1152 – 1076
= ₹ 76
Hence, maximum profit is ₹ 76. Ans.
OR
(ii) Let the annual subscription be increased by ₹ x.
∴ Charges per subscriber = ₹(300 + x)
and Number of subscribers = (500 – x)
Annual income (I) = ₹(500 – x)(300 + x)
= ₹(150000 + 200x – x2)
I = 150000 + 200x – x2
Differentiating w.r.t. x, we get
$$\frac{\text{dI}}{\text{dx}} =\frac{\text{d}}{\text{dx}}(1500000) +\frac{d}{dx}(200 x)-\\\frac{d}{dx}(x^{2})\\\Rarr\space\frac{\text{dI}}{\text{dx}} =0 +200-2x$$
For income to be maximum,
$$\frac{\text{dI}}{\text{dx}} = 0\\\Rarr\space\text{200 - 2x = 0}\\\Rarr\space\space\text{x = 100}\\\Rarr\space\frac{d^{2}\text{I}}{\text{dx}^{2}} =\frac{d}{dx}(200 - 2x)\\\Rarr\space\frac{d^{2}\text{I}}{\text{dx}^{2}} = 0-2= -2\\\Rarr\space\frac{d^{2}\text{I}}{dx^{2}}\lt 0$$
Hence, annual income is maximum at increment of ₹100.
New annual income = ₹400 × 400 = ₹160000
Old annual income = ₹500 × 300 = ₹150000
Increase in annual income = ₹10000. Ans.
Answer 21.
| Marks in Physics x | Marks in Chemistry y | dx = x – 40 | dy = y – 36 | (dx)2 | (dy)2 | dxdy |
| 46 | 40 | 6 | 4 | 36 | 16 | 24 |
| 42 | 38 | 2 | 2 | 4 | 4 | 4 |
| 44 | 36 | 4 | 0 | 16 | 0 | 0 |
| 40 | 35 | 0 | -1 | 0 | 1 | 0 |
| 43 | 39 | 3 | 3 | 9 | 9 | 9 |
| 41 | 37 | 1 | 1 | 1 | 1 | 1 |
| 45 | 41 | 5 | 5 | 25 | 25 | 25 |
| Σdx = 21 | Σdy = 14 | Σ(dx)2 = 91 | Σ(dy)2 = 56 | Σdxdy = 63 |
Here, n = 7
$$\text{b}_{xy} =\frac{n\Sigma dx.dy - \Sigma dx\Sigma dy}{n\Sigma(dy)^{2} -(\Sigma dy)^{2}}\\ b_{xy} =\frac{7×63-21×4}{7×56-(14)^{2}}\\b_{xy} =\frac{\text{441 - 294}}{\text{392 - 196}}\\\text{b}_{xy} =\frac{147}{196} = 0.75\\ b_{yx} =\frac{n\Sigma dx.dy -\Sigma dx\Sigma dy}{n\Sigma(dx)^{2} -(\Sigma dx)^{2}}\\b_{yx} =\frac{7× 63-21×14}{7×91-(21)^{2}}\\=\frac{441 -294}{637 - 441} =\frac{147}{196}\\= 0.75$$
$$\bar{x} = A +\frac{\Sigma dx}{n},\space\bar y = A +\frac{\Sigma dy}{n}\\\bar {x} = 40 +\frac{21}{7} = 43,\\\bar{y} = 36 +\frac{14}{7} = 38$$
Equation of x on y
$$x -\bar x = b_{xy}(y - \bar y)$$
x – 43 = 0.75(y – 38)
x = 0.75y – 28.5 + 43
x = 0.75y + 14.5
Equation of y on x
$$y -\bar y = b_{yx}(x -\bar x)$$
y – 38 = 0.75(x – 43)
y = 0.75x – 32.25 + 38
y = 0.75x + 5.75
Hence, the regression coefficients are 0.75 and 0.75 and the regression equations are x = 0.75y + 14.5 and
y = 0.75x + 5.75. Ans.
Answer 22.
(i) Let the number of type A cake made be x and the number of type B cake made be y.
To maximise the number of cakes.
∵ Z = x + y
Subject to constraint :
200x + 100y ≤ 5000
$$\Rarr\space\text{2x+y}\le 50\\\text{...(i)}\\\text{Also\space 25x + 50 y}\leq 1000\\\Rarr\space \text{x +2y}\leq 40\space\text{...(ii)}\\\text{and}\space \text{x}\geq0, y\geq0$$
Consider, 2x + y = 50
| x | 0 | 25 | 10 |
| y | 50 | 0 | 30 |
x + 2y = 40
| x | 0 | 40 | 10 |
| y | 20 | 0 | 15 |
The feasible region is the shaded region.
| Corner Points | Z = x + y |
| A (25, 0), | Z = 25 + 0 = 25 |
| B (20, 10), | Z = 20 + 10 = 30 |
| C (0, 20), | Z = 0 + 20 = 20 |
Hence, maximum number of cakes = 20 + 10 = 30.
Ans.
OR
(ii) We have,
Maximise Z = 34x + 45y
Subject to the constraints:
x + y ≤ 300
2x + 3y ≤ 70
x ≥ 0, y ≥ 0
Converting the given inequalities into equations, we obtain the following equations:
x + y = 300
2x + 3y = 70
Then, x + y = 300 and
| x | 0 | 300 |
| y | 300 | 0 |
2x + 3y = 70
| x | 0 | 35 |
| y | $$\frac{70}{3}$$ | 0 |
Plotting these points on the graph, we get the shaded feasible region i.e., OCDO.
| Corner point | Value of Z = 34x + 45y |
| O (0, 0) | 34(0) + 45(0) = 0 |
| C (35, 0) | 34(35) + 45(0) = 1190 |
| D (0, 70/3) | 34(0) + 45(70/3) = 1050 |
Clearly, the maximum value of Z is 1190 at (35, 0).
Ans.
ISC 36 Sample Question Papers
All Subjects Combined for Class 12 Exam 2024
ISC 36 Sample Question Papers
All Subjects Combined for Class 12 Exam 2024
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