Maths Airforce X Group Navy SSR AA Quiz | 03 April 2020

mathematics questions for defence exams preparation

Maths Airforce is an important subject in various exams like Airforce X Group, Indian Navy AA & Navy SSR, NDA, CDS and other Defence Exams. Here at DefencePrep we provide you with the quizzes based on the syllabus and pattern of these exams.

Maths Airforce  Questions 

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  1. Question 1 of 10
    1. Question

    Evaluate $\log_x2 =4$

    गणना कीजिए $\log_x2 =4$

    Correct

    Given

    $\log_x2 =4$

    Taking antilog

    x4 = 2

    x = 2(1/4)

    दिया गया है

    $\log_x2 =4$

    प्रतिलघुगुणक लेने पर

    x4 = 2

    x = 2(1/4)

    Incorrect

    Given

    $\log_x2 =4$

    Taking antilog

    x4 = 2

    x = 2(1/4)

    दिया गया है

    $\log_x2 =4$

    प्रतिलघुगुणक लेने पर

    x4 = 2

    x = 2(1/4)

  2. Question 2 of 10
    2. Question

    Evaluate log10 (x2 – 6x + 45) = 2

    गणना कीजिए log10 (x2 – 6x + 45) = 2

    Correct

    Given

    log10 x2 – 6x + 45 = 2

    Taking antilog

    x2 – 6x + 45 = 100

    x2 – 6x – 55 = 0

    x2 – 11x + 5x – 55 = 0

    x(x – 11) + 5(x – 11) = 0

    (x – 11)(x + 5) = 0

    x = 11 or -5

    दिया गया है
    log10 x2 – 6x + 45 = 2

    प्रतिलघुगुणक लेने पर

    x2 – 6x + 45 = 100

    x2 – 6x – 55 = 0

    x2 – 11x + 5x – 55 = 0

    x(x – 11) + 5(x – 11) = 0

    (x – 11)(x + 5) = 0

    x = 11 या -5

    Incorrect

    Given

    log10 x2 – 6x + 45 = 2

    Taking antilog

    x2 – 6x + 45 = 100

    x2 – 6x – 55 = 0

    x2 – 11x + 5x – 55 = 0

    x(x – 11) + 5(x – 11) = 0

    (x – 11)(x + 5) = 0

    x = 11 or -5

    दिया गया है
    log10 x2 – 6x + 45 = 2

    प्रतिलघुगुणक लेने पर

    x2 – 6x + 45 = 100

    x2 – 6x – 55 = 0

    x2 – 11x + 5x – 55 = 0

    x(x – 11) + 5(x – 11) = 0

    (x – 11)(x + 5) = 0

    x = 11 या -5

  3. Question 3 of 10
    3. Question

    Evaluate $x$ for $\frac{{\log 256}}{{\log 16}} = \log x$

    $x$ के लिये गणना कीजिए $\frac{{\log 256}}{{\log 16}} = \log x$

    Correct

    Concept:

    ${\log _a}b = \frac{{\log b}}{{\log a}}$

    Calculation:

    Given: $\frac{{\log 256}}{{\log 16}} = \log x$

    $⇒ \frac{{\log {{\left( {16} \right)}^2}}}{{\log \left( {16} \right)}} = \log x$

    As we know that, ${\log _a}b = \frac{{\log b}}{{\log a}}$ and ${\log _a}{a^n} = n$ where a ≠ 1

    ⇒ 2 = log x

    Raising both sides to the power 10, we get

    ⇒ x = 102

    ⇒ x = 100

    Incorrect

    Concept:

    ${\log _a}b = \frac{{\log b}}{{\log a}}$

    Calculation:

    Given: $\frac{{\log 256}}{{\log 16}} = \log x$

    $⇒ \frac{{\log {{\left( {16} \right)}^2}}}{{\log \left( {16} \right)}} = \log x$

    As we know that, ${\log _a}b = \frac{{\log b}}{{\log a}}$ and ${\log _a}{a^n} = n$ where a ≠ 1

    ⇒ 2 = log x

    Raising both sides to the power 10, we get

    ⇒ x = 102

    ⇒ x = 100

  4. Question 4 of 10
    4. Question

    Find the value of $\frac{{x – 4}}{{x + 4}} + \frac{{x + 4}}{{x – 4}} = \frac{{10}}{3}$

    $\frac{{x – 4}}{{x + 4}} + \frac{{x + 4}}{{x – 4}} = \frac{{10}}{3}$ में x का मान ज्ञात कीजिए।

    Correct

    $\frac{{x – 4}}{{x + 4}} + \frac{{x + 4}}{{x – 4}} = \frac{{10}}{3}$

    $\frac{{{{\left( {x – 4} \right)}^2} + {{\left( {x + 4} \right)}^2}}}{{\left( {x + 4} \right)\left( {x – 4} \right)}} = \frac{{10}}{3}$

    $\frac{{{x^2} + 16 – 8x + {x^2} + 16 + 8x}}{{{x^2} – 16}} = \frac{{10}}{3}$

    $\frac{{\left( {{x^2} + 16} \right)}}{{{x^2} – 16}} = \frac{5}{3}$

    3x2 + 48 = 5x2 – 80

    2x2 = 128

    x2 = 64

    x = ± 8

    Incorrect

    $\frac{{x – 4}}{{x + 4}} + \frac{{x + 4}}{{x – 4}} = \frac{{10}}{3}$

    $\frac{{{{\left( {x – 4} \right)}^2} + {{\left( {x + 4} \right)}^2}}}{{\left( {x + 4} \right)\left( {x – 4} \right)}} = \frac{{10}}{3}$

    $\frac{{{x^2} + 16 – 8x + {x^2} + 16 + 8x}}{{{x^2} – 16}} = \frac{{10}}{3}$

    $\frac{{\left( {{x^2} + 16} \right)}}{{{x^2} – 16}} = \frac{5}{3}$

    3x2 + 48 = 5x2 – 80

    2x2 = 128

    x2 = 64

    x = ± 8

  5. Question 5 of 10
    5. Question

    4th term of a G. P is 8 and 10th term is 27. Then its 6th term is?

    एक G. P का चौथा पद 8 और दसवां पद 27 है। तो इसका छठा पद क्या है?

    Correct

    We know that

    हम जानते हैं कि

    For G. P.

    Tn = a rn-1

    ∴ T4 = a. r3 = 8      —-(1)

    T10 = a r9 = 27      —-(2)

    (2) ÷ (1)

    ${r^6} = \frac{{27}}{8}$

    ${r^2} = \frac{3}{2}$

    T6 = a r5

    = a r3.r2

    =8.32=8.32

    = 12

    Incorrect

    We know that

    हम जानते हैं कि

    For G. P.

    Tn = a rn-1

    ∴ T4 = a. r3 = 8      —-(1)

    T10 = a r9 = 27      —-(2)

    (2) ÷ (1)

    ${r^6} = \frac{{27}}{8}$

    ${r^2} = \frac{3}{2}$

    T6 = a r5

    = a r3.r2

    =8.32=8.32

    = 12

  6. Question 6 of 10
    6. Question

    $y = {\tan ^{ – 1}}\sqrt {\frac{{1 + \sin x}}{{1 – \sin x}}} {\rm{then}}\frac{{dy}}{{dx}} = ?{\rm{\;}}$

    Correct

    $\sqrt {\frac{{1 + \sin x}}{{1 – \sin x}}} = \sqrt {{{\tan }^2}(\frac{\pi }{4} + \frac{x}{2}} )$

    $y = {\tan ^{ – 1}}\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)$

    $y = \frac{\pi }{4} + \frac{x}{2}$

    $\frac{{dy}}{{dx}} = \frac{1}{2}$

    Incorrect

    $\sqrt {\frac{{1 + \sin x}}{{1 – \sin x}}} = \sqrt {{{\tan }^2}(\frac{\pi }{4} + \frac{x}{2}} )$

    $y = {\tan ^{ – 1}}\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)$

    $y = \frac{\pi }{4} + \frac{x}{2}$

    $\frac{{dy}}{{dx}} = \frac{1}{2}$

  7. Question 7 of 10
    7. Question

    Find the coefficient of x2 in  ${\left( {3x – \frac{1}{{{x^2}}}} \right)^6}$

    ${\left( {3x – \frac{1}{{{x^2}}}} \right)^6}$ में x2 का गुणांक ज्ञात कीजिए।

    Correct

    Tn = nCr (a)n-r (b)r

    $= {\left( {3x – \frac{1}{{{x^2}}}} \right)^6}{ = ^6}{C_0}{\left( {3x} \right)^6}.{\left( {\frac{-1}{{{x^2}}}} \right)^0}{ + ^6}{C_1}{\left( {3x} \right)^5}.{\left( {\frac{-1}{{{x^2}}}} \right)^1}{ + ^6}{C_2}{\left( {3x} \right)^4}{\left( {\frac{-1}{{{x^2}}}} \right)^2}$

    ${ + ^6}{C_3}{\left( {3x} \right)^3}{\left( {\frac{-1}{{{x^2}}}} \right)^3}{ + ^6}{C_4}{\left( {3x} \right)^2}.{\left( {\frac{-1}{{{x^2}}}} \right)^4}{ + ^6}{C_5}{\left( {3x} \right)^1}.{\left( {\frac{-1}{{{x^2}}}} \right)^5}{ + ^6}{C_6}{\left( {3{\rm{x}}} \right)^0}{\left( {\frac{-1}{{{{\rm{x}}^2}}}} \right)^6}$

    = 3x6 – 6.35.x3 + 15.34 – 20.33.x-3 + 15.32.x-6 – 6.3.x-9 + x-12

    We can see that there is not term of x2 so the coefficient is 0.

    हम देख सकते हैं कि यहाँ x2 का कोई पद नहीं है, इसलिए गुणांक 0 है।

    Incorrect

    Tn = nCr (a)n-r (b)r

    $= {\left( {3x – \frac{1}{{{x^2}}}} \right)^6}{ = ^6}{C_0}{\left( {3x} \right)^6}.{\left( {\frac{-1}{{{x^2}}}} \right)^0}{ + ^6}{C_1}{\left( {3x} \right)^5}.{\left( {\frac{-1}{{{x^2}}}} \right)^1}{ + ^6}{C_2}{\left( {3x} \right)^4}{\left( {\frac{-1}{{{x^2}}}} \right)^2}$

    ${ + ^6}{C_3}{\left( {3x} \right)^3}{\left( {\frac{-1}{{{x^2}}}} \right)^3}{ + ^6}{C_4}{\left( {3x} \right)^2}.{\left( {\frac{-1}{{{x^2}}}} \right)^4}{ + ^6}{C_5}{\left( {3x} \right)^1}.{\left( {\frac{-1}{{{x^2}}}} \right)^5}{ + ^6}{C_6}{\left( {3{\rm{x}}} \right)^0}{\left( {\frac{-1}{{{{\rm{x}}^2}}}} \right)^6}$

    = 3x6 – 6.35.x3 + 15.34 – 20.33.x-3 + 15.32.x-6 – 6.3.x-9 + x-12

    We can see that there is not term of x2 so the coefficient is 0.

    हम देख सकते हैं कि यहाँ x2 का कोई पद नहीं है, इसलिए गुणांक 0 है।

  8. Question 8 of 10
    8. Question

    Evaluate $\mathop \smallint \limits_{ – 1}^1 \log \left[ {x + \sqrt {1 + {x^2}} } \right]dx$

    गणना कीजिए $\mathop \smallint \limits_{ – 1}^1 \log \left[ {x + \sqrt {1 + {x^2}} } \right]dx$

    Correct

    $= \mathop \smallint \limits_a^b f\left( x \right)\;dx = \mathop \smallint \limits_a^b f\left( {a + b – x} \right)dx$

    $= \mathop \smallint \limits_{ – 1}^1 \log \left( {x + \sqrt {1 + {x^2}} } \right)dx$

    $I = \mathop \smallint \limits_{ – 1}^1 \log \left( {\sqrt {1 + {x^2}} – x} \right)dx$

    $2I = \mathop \smallint \limits_{ – 1}^1 \log \left( {\sqrt {1 + {x^2}} + x} \right)\left( {\sqrt {1 + {x^2}} – x} \right)dx$

    $2I = \mathop \smallint \limits_{ – 1}^1 \log \left( {1 + {x^2} – {x^2}} \right)dx$

    2I = 0

    I = 0

    Incorrect

    $= \mathop \smallint \limits_a^b f\left( x \right)\;dx = \mathop \smallint \limits_a^b f\left( {a + b – x} \right)dx$

    $= \mathop \smallint \limits_{ – 1}^1 \log \left( {x + \sqrt {1 + {x^2}} } \right)dx$

    $I = \mathop \smallint \limits_{ – 1}^1 \log \left( {\sqrt {1 + {x^2}} – x} \right)dx$

    $2I = \mathop \smallint \limits_{ – 1}^1 \log \left( {\sqrt {1 + {x^2}} + x} \right)\left( {\sqrt {1 + {x^2}} – x} \right)dx$

    $2I = \mathop \smallint \limits_{ – 1}^1 \log \left( {1 + {x^2} – {x^2}} \right)dx$

    2I = 0

    I = 0

  9. Question 9 of 10
    9. Question

    If Z1 = 2 + 3i and Z2 = 1 + 2i then value of $\frac{{{Z_1}}}{{{Z_2}}}$ is

    यदि Z1 = 2 + 3i और Z2 = 1 + 2i है, तो $\frac{{{Z_1}}}{{{Z_2}}}$ का मान क्या है?

    Correct

    $\frac{{{Z_1}}}{{{Z_2}}} = \frac{{2 + 3i}}{{1 + 2i}} \times \frac{{\left( {1 – 2i} \right)}}{{\left( {1 – 2i} \right)}}$

    $= \frac{{2 – 4i + 3i – 6{i^2}}}{{1 + 4}}$

    $= \frac{{8 – i}}{5}$

    Incorrect

    $\frac{{{Z_1}}}{{{Z_2}}} = \frac{{2 + 3i}}{{1 + 2i}} \times \frac{{\left( {1 – 2i} \right)}}{{\left( {1 – 2i} \right)}}$

    $= \frac{{2 – 4i + 3i – 6{i^2}}}{{1 + 4}}$

    $= \frac{{8 – i}}{5}$

  10. Question 10 of 10
    10. Question

    If |a| = 10 and |b| = 2 and a.b = 12 then |a × b| = ?

    यदि |a| = 10 और |b| = 2 और a.b = 12 है, तो |a × b| = ? क्या है?

    Correct

    |a| = 10, |b| = 2

    a.b = 12

    a.b = |a|.|b|. cos θ

    $\theta = \frac{{12}}{{10 \times 2}} = \frac{3}{5}$

    $\sin \theta = \frac{4}{5}$

    |a × b| = |a| |b| sin θ

    $= \left( {10} \right)\left( 2 \right)\left( {\frac{4}{5}} \right)$

    = 16

    Incorrect

    |a| = 10, |b| = 2

    a.b = 12

    a.b = |a|.|b|. cos θ

    $\theta = \frac{{12}}{{10 \times 2}} = \frac{3}{5}$

    $\sin \theta = \frac{4}{5}$

    |a × b| = |a| |b| sin θ

    $= \left( {10} \right)\left( 2 \right)\left( {\frac{4}{5}} \right)$

    = 16

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