If f(x)

Question: If $f(x)=\frac{\sin ^{4} x+\cos ^{2} x}{\sin ^{2} x+\cos ^{4} x}$ for $x \in \mathrm{R}$, then $f(2002)=$ (a) 1 (b) 2 (c) 3 (d) 4 Solution: (a) 1 Given: $f(x)=\frac{\sin ^{4} x+\cos ^{2} x}{\sin ^{2} x+\cos ^{4} x}$ On dividing the numerator and denominator by $\cos ^{4}$ $f(x)=\frac{\tan ^{4} x+\sec ^{2} x}{1+\tan ^{2} x \sec ^{2} x}=\frac{1+\tan ^{4} x+\tan ^{2} x}{1+\tan ^{2} x\left(1+\tan ^{2} x\right)}=\frac{1+\tan ^{4} x+\tan ^{2} x}{1+\tan ^{4} x+\tan ^{2} x}=1$ (For every $x \i...

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Let f(x) = x,

Question: Let $f(x)=x, g(x)=\frac{1}{x}$ and $h(x)=f(x) g(x)$. Then, $h(x)=1$ (a)x R (b)x Q (c)x R Q (d)x R,x 0 Solution: (d)x R,x 0 Given: $f(x)=x, g(x)=\frac{1}{x}$ and $h(x)=f(x) g(x)$ Now, $h(x)=x \times \frac{1}{x}=1$ We observe that the domain of $f$ is $\mathbb{R}$ and the domain of $g$ is $\mathbb{R}-\{0\}$. $\therefore$ Domain of $h=$ Domain of $f \cap$ Domain of $g=\mathbb{R} \cap[\mathbb{R}-\{0\}]=\mathbb{R}-\{0\}$ $\Rightarrow x \in \mathrm{R}, x \neq 0$...

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4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950.

Question: 4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950. Find the cost of 2 chairs and 1 table. Solution: Given: (i) Cost of 4 tables and 3 chairs = Rs 2250. (ii) Cost of 3 tables and 4 chairs = Rs 1950. To find: The cost of 2 chairs and 1 table. Suppose, the cost of 1 table = Rsx. The cost of 1 chair = Rsy. According to the given conditions, 4x+ 3y= 2250, 4x+ 3y 2200 = 0 (1) 3x+ 4y= 1950, 3x+ 4y 1950 = 0 (2) Solving eq. (1) and Eq. (2) by cross multiplica...

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If f(x) = cos

Question: If $f(x)=\cos \left(\log _{e} x\right)$, then $f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f(x y)+f\left(\frac{x}{y}\right)\right\}$ is equal to (a) cos (xy) (b) log (cos (xy)) (c) 1 (d) cos (x+y) Solution: Given: $f(x)=\cos \left(\log _{e} x\right)$ $\Rightarrow f\left(\frac{1}{x}\right)=\cos \left(\log _{e}\left(\frac{1}{x}\right)\right)$ $\Rightarrow f\left(\frac{1}{x}\right)=\cos \left(-\log _{e}(x)\right)$ $\Rightarrow f\left(\frac{1}{x}\right)=\cos \left...

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Show that

Question: $\int_{0}^{1} x e^{x^{2}} d x$ Solution: Let $I=\int_{0}^{1} x e^{x^{2}} d x$ Put $x^{2}=t \Rightarrow 2 x d x=d t$ As $x \rightarrow 0, t \rightarrow 0$ and as $x \rightarrow 1, t \rightarrow 1$ $\therefore I=\frac{1}{2} \int_{0}^{1} e^{t} d t$ $\frac{1}{2} \int e^{t} d t=\frac{1}{2} e^{t}=\mathrm{F}(t)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(1)-\mathrm{F}(0)$ $=\frac{1}{2} e-\frac{1}{2} e^{0}$ $=\frac{1}{2}(e-1)$...

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Show that

Question: $\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$ Solution: Let $I=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$ $\int \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \int \frac{5(2 x+3)}{5 x^{2}+1} d x$ $=\frac{1}{5} \int \frac{10 x+15}{5 x^{2}+1} d x$ $=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5 x^{2}+1} d x$ $=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5\left(x^{2}+\frac{1}{5}\right)} d x$ $=\frac{1}{5} \log \left(5 x^{2}+1\right)+\frac{3}{5} \cdot \frac{1}{\frac{1}...

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If x ≠ 1 and f(x)=

Question: If $x \neq 1$ and $f(x)=\frac{x+1}{x-1}$ is a real function, then $f(f(f(2)))$ is (a) 1 (b) 2 (c) 3 (d) 4 Solution: (c) 3 $f(x)=\frac{x+1}{x-1}$ $f(f(f(2)))$ $=f\left(f\left(\frac{2+1}{2-1}\right)\right)$ $=f(f(3))$ $=f\left(\frac{3+1}{3-1}\right)$ $=f(2)=3$...

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If x ≠ 1 and f(x)=

Question: If $x \neq 1$ and $f(x)=\frac{x+1}{x-1}$ is a real function, then $f(f(f(2)))$ is (a) 1 (b) 2 (c) 3 (d) 4 Solution: (c) 3 $f(x)=\frac{x+1}{x-1}$ $f(f(f(2)))$ $=f\left(f\left(\frac{2+1}{2-1}\right)\right)$ $=f(f(3))$ $=f\left(\frac{3+1}{3-1}\right)$ $=f(2)=3$...

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The range of the function

Question: The range of the function $f(x)=\frac{x^{2}-x}{x^{2}+2 x}$ is (a) R (b) R {1} (c) R {1/2, 1} (d) None of these Solution: (c) R {-1/2,1} $f(x)=\frac{x^{2}-x}{x^{2}+2 x}$ Let $y=\frac{x^{2}-x}{x^{2}+2 x} \quad[$ Also,$x \neq 0]$ $\Rightarrow y=\frac{x(x-1)}{x(x+2)}$ $\Rightarrow y=\frac{(x-1)}{(x+2)}$ $\Rightarrow x y+2 y=x-1$ $\Rightarrow x=\frac{2 y+1}{1-y}$ Here, $1-y \neq 0$ or, $y \neq 1$ Also, $x \neq 0$ $\Rightarrow \frac{2 y+1}{1-y} \neq 0$ $\Rightarrow y \neq-\frac{1}{2}$ Thus, ...

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The range of the function

Question: The range of the function $f(x)=\frac{x^{2}-x}{x^{2}+2 x}$ is (a) R (b) R {1} (c) R {1/2, 1} (d) None of these Solution: (c) R {-1/2,1} $f(x)=\frac{x^{2}-x}{x^{2}+2 x}$ Let $y=\frac{x^{2}-x}{x^{2}+2 x} \quad[$ Also,$x \neq 0]$ $\Rightarrow y=\frac{x(x-1)}{x(x+2)}$ $\Rightarrow y=\frac{(x-1)}{(x+2)}$ $\Rightarrow x y+2 y=x-1$ $\Rightarrow x=\frac{2 y+1}{1-y}$ Here, $1-y \neq 0$ or, $y \neq 1$ Also, $x \neq 0$ $\Rightarrow \frac{2 y+1}{1-y} \neq 0$ $\Rightarrow y \neq-\frac{1}{2}$ Thus, ...

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Let f :

Question: Letf: R R be defined byf(x) = 2x+ |x|. Thenf(2x) +f(x) f(x) = (a) 2x (b) 2|x| (c) 2x (d) 2|x| Solution: (b) 2|x|f(x) = 2x+ |x| Then,f(2x) +f(x) f(x) $=2(2 x)+2|x|+(-2 x)+|-x|-2 x+|x|$ $=4 x-2 x-2 x+2|x|+|-x|-|x|$ $=0+2|x|+|x|-|x|$ $=2|x|$...

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If 2f (x) − 3f

Question: If $2 f(x)-3 f\left(\frac{1}{x}\right)=x^{2}(x \neq 0)$, then $f(2)$ is equal to (a) $-\frac{7}{4}$ (b) $\frac{5}{2}$ (c) 1 (d) None of these Solution: (a) $-\frac{7}{4}$ $2 f(x)-3 f\left(\frac{1}{x}\right)=x^{2} \quad(x \neq 0)$ Replacing $x$ by $\frac{1}{x}$ : $2 f\left(\frac{1}{x}\right)-3 f(x)=\frac{1}{x^{2}}$ ....(2) Solving equations $(1) \(2)$ $-5 f(x)=\frac{3}{x^{2}}+2 x^{2}$ $\Rightarrow f(x)=\frac{-1}{5}\left(\frac{3}{x^{2}}+2 x^{2}\right)$ Thus, $f(2)=\frac{-1}{5}\left(\frac...

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If f(x)=

Question: If $f(x)=\frac{2^{x}+2^{-x}}{2}$, then $f(x+y) f(x-y)$ is equal to (a) $\frac{1}{2}[f(2 x)+f(2 y)]$ (b) $\frac{1}{2}[f(2 x)-f(2 y)]$ (c) $\frac{1}{4}[f(2 x)+f(2 y)]$ (d) $\frac{1}{4}[f(2 x)-f(2 y)]$ Solution: (a) $\frac{1}{2}[f(2 x)+f(2 y)]$ Given: $f(x)=\frac{2^{x}+2^{-x}}{2}$ Now, $f(x+y) f(x-y)=\left(\frac{2^{x+y}+2^{-x-y}}{2}\right)\left(\frac{2^{x-y}+2^{-x+y}}{2}\right)$ $\Rightarrow f(x+y) f(x-y)=\frac{1}{4}\left(2^{2 x}+2^{-2 y}+2^{2 y}+2^{-2 x}\right)$ $\Rightarrow f(x+y) f(x-y...

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If f(x) = cos (log x),

Question: If $1(x)=\cos (\log x)$, then value of $f(x) f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f(4 x)\right\}$ is (a) 1 (b) 1 (c) 0 (d) 1 Solution: (c) 0Given :f(x) = cos (logx) Then, $f(x) f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f(4 x)\right\}$ $=\cos (\log x) \cos (\log 4)-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos (\log 4 x)\right\}$ $=\frac{1}{2}[\cos (\log x+\log 4)+\cos (\log x-\log 4)]-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos (\log 4 x)\ri...

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If f(x)=log

Question: If $f(x)=\log \left(\frac{1+x}{1-x}\right)$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ is equal to (a) $\{f(x)\}^{2}$ (b) $\{f(x)\}^{3}$ (c) $2 f(x)$ (d) $3 f(x)$ Solution: (c) $2 f(x)$ $f(x)=\log \left(\frac{1+x}{1-x}\right)$ Then, $f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}\right)$ $=\log \left(\frac{\frac{1+x^{2}+2 x}{1+x^{2}}}{\frac{1+x^{2}-2 x}{1+x^{2}}}\right)$ $=\log \left(\frac{(1+x)^{2}}{(1-x)^{2}}\right)$ $=2 \log \left(\fr...

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Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm,

Question: Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and the diagonal BD = 20 cm. Solution: Given AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and the diagonal BD = 20 cm. Now, for the area of triangle ABD Perimeter of triangle ABD 2s = AB + BD + DA 2s = 34 cm + 42 cm + 20 cm s = 48 cm By using Herons Formula, Area of the triangle $\mathrm{ABD}=\sqrt{\mathrm{s} \times(\mathrm{s}-\mathrm{a}) \times(\mathrm{s}-\mathrm{b}) \times(\mathrm{s}-\mathr...

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Prove

Question: $\int_{2}^{3} \frac{x d x}{x^{2}+1}$ Solution: Let $I=\int_{2}^{3} \frac{x}{x^{2}+1} d x$ $\int \frac{x}{x^{2}+1} d x=\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x=\frac{1}{2} \log \left(1+x^{2}\right)=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(3)-\mathrm{F}(2)$ $=\frac{1}{2}\left[\log \left(1+(3)^{2}\right)-\log \left(1+(2)^{2}\right)\right]$ $=\frac{1}{2}[\log (10)-\log (5)]$ $=\frac{1}{2} \log \left(\frac{10}{5}\right)=\frac{1}{2} \log 2$...

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Reena has pend and pencils which together are 40 in number.

Question: Reena has pend and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils. Solution: Given: (i) Total numbers of pens and pencils = 40. (ii) If she has 5 more pencil and 5 less pens, the number of pencils would be 4 times the number of pen. To find: Original number of pens and pencils. Suppose original number of pencil =x And original number of pen =y...

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Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90° and BCD forms an

Question: Find the area of the quadrilateral $\mathrm{ABCD}$ in which $\mathrm{AD}=24 \mathrm{~cm}$, angle $\mathrm{BAD}=90^{\circ}$ and $\mathrm{BCD}$ forms an equilateral triangle whose each side is equal to $26 \mathrm{~cm}$. [Take $\sqrt{3}=1.73$ ] Solution: Given that, in a quadrilateral ABCD in which AD = 24 cm, Angle BAD =90 BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm In triangle BAD, by applying Pythagoras theorem, $B A^{2}=B D^{2}-A D^{2}$ $B A^{2}=26^{2}+242$ $B A...

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Prove

Question: $\int_{2}^{3} \frac{x d x}{x^{2}+1}$ Solution: Let $I=\int_{2}^{3} \frac{x}{x^{2}+1} d x$ $\int \frac{x}{x^{2}+1} d x=\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x=\frac{1}{2} \log \left(1+x^{2}\right)=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(3)-\mathrm{F}(2)$ $=\frac{1}{2}\left[\log \left(1+(3)^{2}\right)-\log \left(1+(2)^{2}\right)\right]$ $=\frac{1}{2}[\log (10)-\log (5)]$ $=\frac{1}{2} \log \left(\frac{10}{5}\right)=\frac{1}{2} \log 2$...

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7 audio cassettes and 3 video cassettes cost Rs 1110, while 5 audio cassettes and 4 video cassettes cost Rs 1350. Find the cost of an audio cassette and a video cassette.

Question: 7 audio cassettes and 3 video cassettes cost Rs 1110, while 5 audio cassettes and 4 video cassettes cost Rs 1350. Find the cost of an audio cassette and a video cassette. Solution: Given: (i) 7 Audio cassettes and 3 Video cassettes cost is 1110. (ii) 5 Audio cassettes and 4 Video cassettes cost Rs. 1350. To Find: Cost of 1 audio cassette and 1 video cassettes. Let (i) the cost of 1 audio cassette = Rs.x. (ii) the cost of 1 video cassette = Rs.y. According to the given conditions, we ha...

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Show that

Question: $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$ Solution: Let $I=\int_{1}^{\frac{\pi}{2}} \cos ^{2} x d x$ $\int \cos ^{2} x d x=\int\left(\frac{1+\cos 2 x}{2}\right) d x=\frac{x}{2}+\frac{\sin 2 x}{4}=\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\left[\mathrm{F}\left(\frac{\pi}{2}\right)-\mathrm{F}(0)\right]$ $=\frac{1}{2}\left[\left(\frac{\pi}{2}-\frac{\sin \pi}{2}\right)-\left(0+\frac{\sin 0}{2}\right)\right]$ $=\frac...

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A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per meter square. Find the cost of painting.

Question: A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per meter square. Find the cost of painting. Solution: Given that, Perimeter of a rhombus = 32 m We know that, Perimeter of a rhombus =4 side 4 side= 32 m 4 a= 32 m a = 8 m Let AC = 10 m OA =12 AC OA =12 10 OA = 5 m By using Pythagoras theorem $O B^{2}=A B^{2}-O A^{2}$ $O B^{2}=8^{2}-5^{2}$ $O B=\sqrt{39} m$ $B D=2 \times O B$ $B D=2 \sqrt{39} m$ Area of the sheet ...

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5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of 1 pen and 1 pencil.

Question: 5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of 1 pen and 1 pencil. Solution: Given: (i) 5 pens and 6 pencils together cost of Rs. 9. (ii) 3 pens and 2 pencils cost Rs. 5. To Find: Cost of 1 pen and 1 pencil. Let (i) The cost of 1 pen = Rsx. (ii) The cost of 1 pencil = Rsy. According to question $5 x+6 y=9$$\ldots \ldots(1)$ $3 x+2 y=5$$\ldots \ldots(2)$ Thus we get the following system of linear equation $5 x+6 y-9=0$...(3) from eq. 1 $3 x+...

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Show that

Question: $\int_{2}^{3} \frac{d x}{x^{2}-1}$ Solution: Let $I=\int_{2}^{3} \frac{d x}{x^{2}-1}$ $\int \frac{d x}{x^{2}-1}=\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|=\mathrm{F}(x)$ By second fundamental theorem of calculus, we obtain $I=\mathrm{F}(3)-\mathrm{F}(2)$ $=\frac{1}{2}\left[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\right]$ $=\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]$ $=\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{...

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