Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Continuity and Differentiability

Short Answer Type

471.

Differentiate space tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight a squared space straight x squared end root minus 1 over denominator straight a space straight x end fraction close square brackets space straight w. straight r. straight t. straight x.

88 Views

472.

If space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses plus sec to the power of negative 1 end exponent open parentheses fraction numerator 1 plus straight x squared over denominator 1 minus straight x squared end fraction close parentheses comma space prove space that space dy over dx equals fraction numerator 4 over denominator 1 plus straight x squared end fraction

93 Views

473.

If space straight y equals sin open square brackets 2 space tan to the power of negative 1 end exponent square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root close square brackets comma space prove space that space dy over dx equals negative fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction

90 Views

474.

If space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x squared end root minus square root of 1 minus straight x squared end root over denominator square root of 1 plus straight x squared end root plus square root of 1 minus straight x squared end root end fraction close square brackets comma space find space dy over dx

78 Views

475.

If space space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x squared end root plus square root of 1 minus straight x squared end root over denominator square root of 1 plus straight x squared end root minus square root of 1 minus straight x squared end root end fraction close square brackets comma space prove space that space dy over dx equals negative fraction numerator straight x over denominator square root of 1 minus straight x to the power of 4 end root end fraction

75 Views

476.
Find the equation of the normal to the curve x² = 4 y which passes through the point (1, 2).

The equation of the curve is x² = 4y ...(1)
$therefore space space space space space space space space space space straight y space equals space straight x squared over 4 space space space space space space space rightwards double arrow space space space space space dy over dx space equals space fraction numerator 2 straight x over denominator 4 end fraction space equals space straight x over 2$
Let normal at (h, k) pass through (1, 2).
Since (h, k) lies on (1)
$therefore space space space space straight h squared space equals space 4 straight k$ ...(2)
Slope of tangent at (h, k) = $straight h over 2$
$therefore space space space space space space slope space of space normal space at space left parenthesis straight h comma space straight k right parenthesis space equals space minus 2 over straight h$
Equation of normal at (h, k) is $straight y minus straight k space equals space minus 2 over straight h left parenthesis straight x minus straight h right parenthesis$
$because space space space space space space space it space passes space through space left parenthesis 1 comma space 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space because space space space space space space space 2 minus straight k space equals space minus 2 over straight h left parenthesis 1 minus straight h right parenthesis$
$or space space space space space 2 straight h minus hk space equals space minus 2 plus 2 straight h space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight h space straight k space equals space 2 From space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get comma space straight h open parentheses straight h squared over 4 close parentheses space equals space 2. therefore space space space space space space space space straight h squared space equals space 8 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight h space equals space 2 space space space space space space space space space space space therefore space space space straight k space equals space 2 over straight h space equals space 2 over 2 space equals space 1 therefore space space space space space space equation space of space normal space is space straight y minus 1 space equals space minus 2 over 2 left parenthesis straight x minus 2 right parenthesis or space space space space space space straight y minus 1 space equals space minus straight x plus 2 space space space space space space space space space space space space space space or space space space space space space space straight x plus straight y minus 3 space equals space 0$